# ATAR Notes: Forum

## HSC Stuff => HSC Science Stuff => HSC Subjects + Help => HSC Chemistry => Topic started by: jakesilove on January 28, 2016, 08:00:25 pm

Post by: jakesilove on January 28, 2016, 08:00:25 pm

If you have general questions about the HSC Chemistry course or how to improve in certain areas, this is the place to ask! 👌

Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding!

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you. So you may even get multiple answers from different people offering their insights - very cool.

To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!

OTHER CHEMISTRY RESOURCES

Original post.
Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)

Hey everyone!

A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!

This is a community, so we want you to feel like you can post any type of Chemistry, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Maths is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.

Remember that Chemistry is a bloody difficult course. There will be lots of answers to the same questions, and I'll try give you the best or easiest to remember ones.

I got an ATAR of 99.80, and a mark of 93 in the Chemistry course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!
Post by: Happy Physics Land on January 28, 2016, 09:47:35 pm
In 2007 HSC exam for chemistry, there was a question regarding "explain one benefit of car batteries lasting several years" and the notes from marking centre recommended that "Better responses clearly linked the benefit to a property of lead or sulfuric acid. Weaker responses did not give a specific benefit, but rather general statements that could have many meanings, such as ‘convenient and reliable’. The better responses explained such words in the context of their answers." I have some trouble trying to relate the property of lead or sulfuric to the question. Please Help Jake!!!
Post by: jakesilove on January 28, 2016, 10:19:34 pm
In 2007 HSC exam for chemistry, there was a question regarding "explain one benefit of car batteries lasting several years" and the notes from marking centre recommended that "Better responses clearly linked the benefit to a property of lead or sulfuric acid. Weaker responses did not give a specific benefit, but rather general statements that could have many meanings, such as ‘convenient and reliable’. The better responses explained such words in the context of their answers." I have some trouble trying to relate the property of lead or sulfuric to the question. Please Help Jake!!!

Hey Happy Physics Land (love the name by the way).

Great question, because it's the perfect example of the fact that sometimes teachers mark question in a way that is arbitrary and, to be frank, just wrong. The question, word for word, is "Explain one benefit of car batteries lasting several years." (from the 2007 paper, question 20b). This is framed in relation to an image of a Lead-Acid battery, however really the question doesn't ask specifically about a car battery with components that are Sulfuric acid or Lead. So really, an answer such as durability, convenience etc. should answer the question.

For a two mark question, make sure you talk about the positive attribute AND the reason its beneficial (ie. convenience of not having to replace battery regularly, resulting in greater costs to consumer, and safety, so that the battery doesn't die mid-drive).

The way the answer that you quote is phrased makes it seem like the question was "Explain how the components of the above battery make them suitable for car batteries". I really think the markers got that wrong, and in more recent exam papers you won't find such obvious misrepresentation of what the question was asking for.

The answer to the question "Explain how the components of the above battery make them suitable for car batteries" has many answers, including

1. The battery does not discharge much voltage whilst not in use, unlike many other forms of re-chargeable batteries
2. Results in a high voltage-discharge when required
3. Not easily subject to internal damage through corrosion etc.

However, like I've said I don't think these answers should have been required given the question. That being said, including as much information as possible is always a good idea when answering Chemistry questions, and so throwing one of these in could be a good idea if you want to play it safe.

I hope this helps: remember that you're only supposed to answer the question and, in this case, I think the marker got it wrong.

Jake :)
Post by: Happy Physics Land on January 28, 2016, 11:23:30 pm
Thank you very much jake that was very helpful and it certainly assisted me in answering the question. Before you explained that to me I was also confused by the necessity to include "the properties of lead and sulfuric acid" in my answer. But that was really helpful a response, thank you very much matey!!!!

Best Regards

Happy Physics Land
Post by: thushan on January 29, 2016, 12:39:23 pm
I'm the corresponding ATARNotes Chemistry lecturer for VCE down in Victoria - I got a 50 (in VCE the maximum score you can get in a subject is 50).

Will lurk here and help answer questions too! But I will be coming from a Victorian slant.
Post by: ghey04 on January 31, 2016, 10:35:52 am
Hey! I was wondering for the q "Analyse a recently developed biopolymer in terms of progress and method of production. Provide an evaluation of its potential of use" (5 marks) in Production of Materials (basically the dotpoint I guess), how would you write up the answer? I'm getting thrown off which part I should focus and write more on.
Post by: Happy Physics Land on January 31, 2016, 11:19:34 am
Hey! I was wondering for the q "Analyse a recently developed biopolymer in terms of progress and method of production. Provide an evaluation of its potential of use" (5 marks) in Production of Materials (basically the dotpoint I guess), how would you write up the answer? I'm getting thrown off which part I should focus and write more on.

Hello ghey04:

I am a year 12 student currently undertaking chemistry, just like yourself, and I am happy to help you out :) ! Ok so let's break down this question first since it consists of so many parts. Essentially the question asks us to do 5 things:
1. Identify this biopolymer (and perhaps even the year it was first developed)
2. How this biopolymer has developed in the society (i.e. is it widely used? Use statistics? state a use with slight references to its properties as an example. Is it becoming increasingly economical to produce?)
3. Identify and describe the method of production (Identify the bacteria/biomass required to produce this biopolymer (VERY IMPORTANT), describe the every single step of the procedure that is taken to produce this biopolymer)
4. Describe its current usage (list about 2 uses with slight reference to its properties)
5. Evaluate (i.e. make a judgement) the potential of the biopolymer in the future, and state your reason to back up your argument (e.g. this biopolymer has a significant potential for the society because as current technologies improve, this biopolymer can be produced more efficiently and economically. In conjunction with its ability to bio-degrade and the fact that it is produced from renewable resources, the production process of this biopolymer will be very sustainable and hence this biopolymer has a large potential to replace petrochemical polymers in both the contemporary and future society)

In regards to which part you should focus on, I think that both parts should be the same amount of writing. However, because you will be describing the method of production, and because the first question consists of essentially 3 parts, then I would put more concentration upon the first question. The ratio of writing between the two questions I would recommend is about 4 : 3. But please dont ignore the importance of the second question because it assesses how we can theoretically apply what we have learnt to predict the continual progress of the biopolymer into the future.

I hope this helps ghey04 and I wish you the best of luck throughout your HSC year!

Best Regards
Happy Physics Land
Post by: jakesilove on January 31, 2016, 11:23:45 am
Hey! I was wondering for the q "Analyse a recently developed biopolymer in terms of progress and method of production. Provide an evaluation of its potential of use" (5 marks) in Production of Materials (basically the dotpoint I guess), how would you write up the answer? I'm getting thrown off which part I should focus and write more on.

Hey ghey04!

Fantastic question, because really what a lot of Chemistry questions come down to is knowing what you are actually supposed to include in your answer.

I would always start off by underlining the important words. In this case, you need to analyse... progress and methods of production, and evaluate potential use. So that's exactly how I would structure your answer.

Start off by discussing the development of the biopolymer. Talk about its history/discovery, and any progress that has been made. You can summarise this in maybe two or three sentences. Then discuss how you produce the polymer, again in two or three sentences. Finally, discuss potential uses. I would definitely use subheadings for a 5 mark questions: teachers absolutely love it!

For instance, the structure would be:

History/Progress
Two/three sentences summarising important points. Try to include some specific facts!

Method of production
Two/three sentences explaining, fairly specifically, how we CURRENTLY develop the polymer. Try to include specific names of compounds etc.

Potential use
Two/three current uses, including WHY the biopolymer is useful in that scenario. For instance, polystyrene is a good material for a cup because it is an insulator, and will not heat up when boiling liquid is poured in.

I hope that helps! If you want to write up a response and post it here based on the above I am happy to comment on it :)

Jake
Post by: ghey04 on January 31, 2016, 11:28:18 pm
Thanks HappyPhysicsLand and Jake! Really good stuff, I'll see how I go now :)
Post by: wesadora on February 07, 2016, 12:40:34 am
Consider the chemical equation below:
Br2(l) + 2Cl-(aq) --> 2Br-(aq) + Cl2(g)
Will this reaction occur spontaneously? Justify your response using relevant half-equations (3 marks)

Firstly, what defines a 'spontaneous reaction' in regards to galvanic cells, and how would you go about this question :/

Thanks!

Post by: jakesilove on February 07, 2016, 09:44:08 am
Consider the chemical equation below:
Br2(l) + 2Cl-(aq) --> 2Br-(aq) + Cl2(g)
Will this reaction occur spontaneously? Justify your response using relevant half-equations (3 marks)

Firstly, what defines a 'spontaneous reaction' in regards to galvanic cells, and how would you go about this question :/

Thanks!

Below is my solution to this (very difficult) question. Usually, you could get similar questions regarding metals (Sodium and Potassium Nitrate etc.) but I've never seen one like this. Hope it helps!

(http://i.imgur.com/wZuKnNZ.png?1)

Jake
Post by: Johny1234567 on February 07, 2016, 10:41:15 am
Hello!
I need help with the following question:

35mL of 0.86 M sulfuric acid was placed into a 25mL solution of 1.00 M sodium hydroxide.

b) Calculate the concentration hydrogen or hydrogen ions after this reaction. You may assume the reaction is complete. (3 marks)

c) Calculate the pH of the final solution. (1 mark)

d) If 5 drops of methyl orange was placed into the NaOH solution before the reaction took place, explain the colour changes that occur over the course of the experiment. (3 marks)

Any help would be greatly appreciated!
Post by: jakesilove on February 07, 2016, 11:43:20 am
Hello!
I need help with the following question:

35mL of 0.86 M sulfuric acid was placed into a 25mL solution of 1.00 M sodium hydroxide.

b) Calculate the concentration hydrogen or hydrogen ions after this reaction. You may assume the reaction is complete. (3 marks)

c) Calculate the pH of the final solution. (1 mark)

d) If 5 drops of methyl orange was placed into the NaOH solution before the reaction took place, explain the colour changes that occur over the course of the experiment. (3 marks)

Any help would be greatly appreciated!

Hey Johny!

These sort of "Concentration" questions are very typical, however this is one of the most difficult ones I've seen (mainly because of the difficulty of the chemical equation). However, if you are able to follow this working out and do lots of practice questions, you'll find this to be easy marks in the HSC! Hope that this helps!

If you don't quite understand a step, maybe submit another question that is similar, but easier. Once you have the general working out style, you'll be totally fine :)

(http://i.imgur.com/XSdZ1HL.png?1)
(http://i.imgur.com/fXnbvvS.png?1)

Thanks for the great question!

Jake
Post by: Happy Physics Land on February 07, 2016, 12:30:06 pm
Hey Johny!

These sort of "Concentration" questions are very typical, however this is one of the most difficult ones I've seen (mainly because of the difficulty of the chemical equation). However, if you are able to follow this working out and do lots of practice questions, you'll find this to be easy marks in the HSC! Hope that this helps!

If you don't quite understand a step, maybe submit another question that is similar, but easier. Once you have the general working out style, you'll be totally fine :)

(http://i.imgur.com/XSdZ1HL.png?1)
(http://i.imgur.com/fXnbvvS.png?1)

Thanks for the great question!

Jake

Farrrrrr Jake you are on Fire..... Loving those explanations!
Post by: Johny1234567 on February 07, 2016, 01:14:00 pm
Hey Johny!

These sort of "Concentration" questions are very typical, however this is one of the most difficult ones I've seen (mainly because of the difficulty of the chemical equation). However, if you are able to follow this working out and do lots of practice questions, you'll find this to be easy marks in the HSC! Hope that this helps!

If you don't quite understand a step, maybe submit another question that is similar, but easier. Once you have the general working out style, you'll be totally fine :)

(http://i.imgur.com/XSdZ1HL.png?1)
(http://i.imgur.com/fXnbvvS.png?1)

Thanks for the great question!

Jake

Hello Jake,
Your explanation was clear and concise. Thank you !
Post by: Sharms on February 07, 2016, 03:52:24 pm
Just completed the first term of HSC and got 45% in my chem exam, how hard would it be to to change into senior science having missed the whole first term and quickly approaching the half yearlies?
I am dedicated towards learning its just I never really go on with chemistry and all the concepts
Post by: jakesilove on February 07, 2016, 04:02:56 pm
Just completed the first term of HSC and got 45% in my chem exam, how hard would it be to to change into senior science having missed the whole first term and quickly approaching the half yearlies?
I am dedicated towards learning its just I never really go on with chemistry and all the concepts

Hey Sharms!

I'm sorry to say that I actually don't know anything about Senior Science, in terms of difficulty or anything else. Really the only thing I can recommend is talking to your respective teacher about changing subjects.

Sorry that I can't be of more help! Maybe someone else on the forum has more information :)

Jake
Post by: wesadora on February 07, 2016, 04:19:07 pm
thanks jake!! :D
Post by: kathywong on February 09, 2016, 12:52:19 am
Hi, what are some chemical properties of radioisotopes? The syllabus dot point says "explain their use in terms of their properties".
My teacher said that "half life and the type of radiation they emit are not chemical properties so im kinda lost. Please help thanks
Post by: jakesilove on February 09, 2016, 09:33:34 am
Hi, what are some chemical properties of radioisotopes? The syllabus dot point says "explain their use in terms of their properties".
My teacher said that "half life and the type of radiation they emit are not chemical properties so im kinda lost. Please help thanks

Hey Kathy!

In this case I would respectfully disagree with your teacher. Half life and type of radiation they emit is certainly a chemical property, two of which you should ALWAYS include in any band 6 response. To repeat; they are chemical properties!.

However, there are definitely more chemical properties you can discuss if you want to achieve top marks. Things like state (for Medicine, useful if the compound is a liquid etc.) and ability to latch onto certain compounds in the body (such as Glucose) so that you can track its movement and thereby detect various diseases and illnesses.

The main this to get out of this is that your teacher is wrong. If they want you to include even more, then definitely do, but in the HSC half life and type of radiation is definitely a chemical property.

I hope this helps! Absolutely great question, because often your teachers can be slightly misguided in their approach to the syllabus.

Jake :)
Post by: g98 on February 09, 2016, 06:54:34 pm
Hey Jake!
I have a quick question about studying in general:
What is your opinion on music while you study?

And your lectures in Sydney were great by the way!
Thank you!
Post by: jakesilove on February 09, 2016, 06:58:48 pm
Hey Jake!
I have a quick question about studying in general:
What is your opinion on music while you study?

And your lectures in Sydney were great by the way!
Thank you!

Hey g98!

Great question! Actually, the second time I've been asked that today oddly enough.

Throughout my HSC year, I literally listened to Music non-stop. Like seriously, if there wasn't music coming out of my room, I was probably asleep. I didn't feel distracted by it, and it helped time go more,,, manageable?

Once, I tried listening to Classical music and to be completely honest, it actually helped a lot! Still, I found it heaps boring so decided to revert back to Alt-J.

My overall recommendation is just to do whatever you feel comfortable with. If music distracts you, keep it off. If it helps you pass the time, blast it out the windows. There's no "set rule" or particular method that will help you study better.

I know this isn't a very specific answer, but I hope it helps! Great question by the way.

Jake
Post by: Happy Physics Land on February 09, 2016, 07:10:53 pm
Hey Jake!
I have a quick question about studying in general:
What is your opinion on music while you study?

And your lectures in Sydney were great by the way!
Thank you!

Actually if you just type up "study concentration music" on youtube, there's a whole list of those and pretty much all of them have a brain stimulation effect that would help you to concentrate. My recommendation is to turn it to a volume where you can still hear someone calling you but loud enough to cover all the unnecessary noises in the surrounding environment.
Post by: zoe.mckeon on February 10, 2016, 07:07:11 pm
Hey Jake,

I'm just wondering what you found to be an effective study method for Chemistry? I found during Year 11 that I did study but I could not remember much when in an exam, and what I did know was hard to apply to a question....

Post by: jakesilove on February 10, 2016, 09:24:01 pm
Hey Jake,

I'm just wondering what you found to be an effective study method for Chemistry? I found during Year 11 that I did study but I could not remember much when in an exam, and what I did know was hard to apply to a question....

Hey Zoe!

Fantastic question, and one which I will soon be devoting an entire resource too. Studying is the hardest part of any subject, and making sure that you tailor your study to each subject is absolutely vital.

I totally agree with you though: the hardest part of studying for Chemistry in particular isn't learning the content; rather, it is applying that content to a standard exam question!

My best advice is to DO PAST PAPERS and COMPLETE PAST QUESTIONS. When you finish a topic, don't just take notes, learn the content etc. Find past questions relating to the topic, answer them, mark them yourself (using marking guidelines), re answer the question so that it's perfect, and then include that answer in your notes.

I would say that, until you've done the above, you haven't actually finished a topic. Don't wait until you finish an entire block of the syllabus (eg. The Acidic Environment etc.); as soon as you finish a dot point or two, try answering questions.

Once you done this enough, you'll know exactly what each question is asking. It's all about interpreting what the markers want, and the only way to do that is to do enough past papers to know exactly what's going on.

To summarise

-     Write notes
-     Learn the content
-     Do past paper questions relating to the topic
-     Mark question

I really hope that this helps! Great question, and great to see the activity on this forum! Looking forward to more great questions, and answers from the community!

Jake :)
Post by: thushan on February 11, 2016, 09:00:16 am
You'll find you remember content much more easily if you use what you know - whether it be solving stoichiometric problems or synthesising your understanding of the material in more worded and extended response questions.
Post by: kathywong on February 11, 2016, 08:13:28 pm
My questions:
-are transuranic elements and radioisotopes produced in exactly same way ie.using nuclear reactor and particle accelerator?
-Is it not necessary for nuclear fission to occur in nuclear reactor and nuclear fusion to occur in particle accelerator ?
-Will knowing the location discovered and how it is produced for recent discoveries of elements be enough?

I have an assessment coming up so would you mind checking if my notes are correct and if the main important details are mentioned? thank you

Describe how transuranic elements are produced

Neutron bombardment:
- Natural elements are immersed in a nuclear reactor and bombarded with neutrons (or other particles) to produce higher elements as nuclear fission occurs

Particle accelerator
-transuranic element atomic number>96 are made by bombarding heavy nucleus with high speed small positive nucleus (e.g. He, C or Boron) in accelerators, synchrotrons, and cyclotrons.

Describe how commercial radioisotopes are produced

Nuclear reactors
-Bombard target nuclei with slow neutrons produced by fission of uranium-235. The nucleus absorbs neutrons hence producing a neutron-rich radioisotope.

Particle accelerators
-A machine uses electric and magnetic fields to accelerate charged particles such as protons, or nuclei (eg.Helium) or ion of other atoms (eg. Hydrogen-2 ions, Helium-3 ions) to collide with target nucleus. Nuclear fusion occurs and neutron-deficient radioisotope is produced.
eg. (nuclear fusion) Fluorine-18 is prepared in a cyclotron by bombarding nitrogen-14 with helium nuclei.

Identify instruments and processes that can be used to detect radiation
-Photographic film- Darkening of photographic film indicated the presence of radioactivity. The darkening is due to formation of silver halide crystals as radiation is absorbed by silver salts in the film.

-Cloud Chamber - device that consists a container of supersaturated vapour of water or alcohol. As radiation travels through the device, it ionises surrounding air molecules which causes vapour molecules to condense onto these ions, to produce visible cloud tracks/ trails.

-Scintillation counter:
- When certain substances like ZnS absorbed energy from alpha, beta and gamma rays, they emit flashes of light which can be collected and amplified in a photo multiplier. Thus, an electrical pulse can be generated and recorded by a  counter
-Used to measure the amount of radiation exposed to a person who has taken a body scan

Geiger-Muller counter
- Best for detection of alpha and beta radiation but cannot tell difference between alpha and beta particles
- Consists of a sealed glass/metal tube with inside filled with argon gas and with a thin mica window at one end to allow particles to enter
- Particles enter ionise gas molecules (e.g. argon), emitting electrons which accelerate towards the positive electrode, and ionising more gas as they proceed in their path and create an electrical pulse
- GM counter records the electrical pulse and convert it to audible clicks

Medical and radioisotopes uses and chemical properties

Cobalt-60:  (industrial/medical)
Chemical properties:
Chemically inert
Gamma emitter
Long half-life of 5.3 years

Relating uses to chemical properties:
- it is an emitter of gamma rays which can penetrate the materials such as metals, paper, rolled steel and detected on the other side using radioactive film,  therefore can measure thickness of these materials or detect faults in metal casting and cracks in aircraft wings.
-The gamma radiation it emits used to sterilise food eg. Strawberries to minimise spoilage and extend shelf life, and medical supplies such as disposable syringes and bandages. Food on a conveyor belt is gamma irradiated by passing it through a chamber containing a safely shielded Co-60 source.

Benefits
-it is chemically inert therefore has relatively low emission of radiation also limits the potential damage to anyone working with the radioisotope.
-potent gamma emitter  - Gamma rays can easily penetrate hermetically sealed packaging and the contents, killing harmful microbes such as bacteria, viruses and fungi. They have sufficient energy to destroy bacteria but not enough to make food radioactive.
-has a reasonably long half-life of 5.3 years since it is chemically inert hence do not require frequent replacement/do not need to be produced on site and can be readily used to treat cancer patients
-Provides effective treatment of many cancerous tumours in sensitive organs because it releases beta and gamma radiation which can easily penetrate and kills abnormal cells
-Cheaper methods of diagnosis and treatment of medical illnesses and cancer
-Provides non-invasive, convenient diagnostic techniques to trace medical illnesses and problems
-medical sterilisation which is more effective than the use of antiseptics and heat treatment

Concerns
-There are concerns that gamma radiation can destroy vitamin content in food and may lead to the formation of harmful compounds in the food. In addition, workers must be protected from irradiation.
-Continued exposure to radiation can lead to diseases such as tissue damage, tumours, cancer and genetic damage
-Without further research, the long-term effects in irradiation of food, are unknown

Technetium-99m (medical):
Chem properties:
-short half-life of 6 hours
-emits low energy gamma
-can be attached to a range of biological carriers
-relatively unstable and reactive

Relating chemical prop to uses:
-used as a biological tracer for diagnostic imaging.  It can be attached to biological molecule which will concentrate in the targeted organ such as brain, kidney, bones liver and spleen and injected into the body. Then, it will emit gamma radiation which can easily penetrate and detected in various part of body using a gamma camera (scintigraphy) and pinpoint blood clots, constrictions, heart defects and size and location of cancer growth, etc.
eg. Tc-99m phosphonates is used to diagnose skeletal bone problems

Benefits:
-short half-life of 6 hours which is short enough to minimise the exposure of radiation to patient and long enough to examine metabolic processes occurring in the body
-emits low energy gamma radiation which minimises damage to healthy cells/tissues, but can still be detected in the body by a gamma ray sensitive camera.
-Can be attached to a range of biological carriers which will concentrate in a targeted organ
-relatively reactive, so it can be reacted to form a compound with chemical properties that leads to concentration in the organ of interest such as the heart, liver, lungs or thyroid.
-Can be made on site from molybdenum-99 in a transportable generator – Mo-99 is a product of nuclear fission in a nuclear reactor
-Easy to detect using small quantities
-Provides a range of non-invasive, convenient diagnostic techniques  to trace medical illnesses and problems and patient experiences little or no discomfort during the test
-Cheaper methods of diagnosis and treatment of medical illnesses and cancer
-Scintigraphy can be used to produce real time images of body sections which enables observation from all angles instead of Xrays which are static images

Problems:
-It is produced in a nuclear reactor from Molybdenum which is not as dangerous as Plutonium-241 but usual safeguards are necessary (lead shielding) as long term exposure to any radiation can lead to diseases such as tissue damage, tumours, cancer and genetic damage (which leads to deformities in offspring)

Evaluation:
-Tc-99m may cause damage to tissues, or cancer when it is exposed in long term. However, it can be minimised with proper safeguards and in addition its short half-life and low frequency gamma radiation. More importantly, it can give value information concerning a wide variety of metabolic disorders or cancers without the need for invasive surgery. Therefore, there are more benefits associated with the use of Tc-99m and this makes it the most widely used medical radioisotope.

Americium-241 (industrial)
Chemical properties:
-long half-life of 432.7 years

Relating uses to chem prop: (medical and industrial):
-Used in smoke detectors because it is emits alpha and gamma radiation.
When no smoke is present, the alpha particles ionise nitrogen and oxygen in the air in the detector.
When smoke is present, the smoke absorbs the alpha particles emitted, so the rate of ionisation drops and this sets off the alarm.
-Gamma radiation released from Am-241 can be used for indirect analysis of materials radiography and for quality control in manufacturing fixed gauges eg. It has been used to measure glass thickness to help create flat glass
-Gamma rays also provide diagnosis of thyroid function
-It is recently suggested for use as a denaturing agent in plutonium reactor fuel rods to render the fuel unusable for conversion to nuclear weapons

Benefits:
-Long half-life of 432.7 years so do not need frequent replacement
-Alpha particles do not themselves pose a health hazard – as they are absorbed in a few cm of air or by the detector itself

Problems:
-emits alpha and gamma radiation  which can be a serious health hazard if ingested or inhaled
-Dangerous if swallowed as it would concentrate in the skeleton and continue to emit radiation which could cause some cellular damage
-Production is dangerous as it is made through the decay of plutonium-241 in a nuclear reactor which is a highly radioactive element and emits high energy gamma radiation as it decays. Hence, special precautions must be taken when handling plutonium, otherwise it can cause cancer and death

Evaluation:
-It is necessary as it can save lives by warning inhabitants of fires

Sodium-24 (industrial)
Chemical properties:
-Soluble in water
-Half life of 15 hours

Relating uses to chem prop:
-it is soluble in water and and used as a leak detector in damaged underground water, gas and oil pipes as it emits beta and gamma radiation. Minimal radiation is detected if it is from inside a pipe, but if the tracer leaks into the soil large radiation are detected

Benefits:
-emits beta radiation  - it does not pose a major health hazard to living things because the radiation will be absorbed by the pipe itself or by its surrounding/ pipe leaks can be located without having to dig up entire pipeline.
-half-life of 15 hours – lasts long enough to find leaks and short enough so that it does not cause any serious permanent pollution to water bodies
-Non-toxic to humans or animals- although it may form organic compounds and be absorbed into the blood stream, this will not strongly affect the health of humans or animals due factors such as its life span and low intensity emission of gamma radiation

Problems:
-Not a naturally occurring isotope hence relies upon nuclear reactors for production which may cause disastrous consequences if large amount of harmful radiation is accidently released
-Radiation cause undesirable reactions in living tissue and so can cause tissue damage, cancers and/or genetic damage leading to deformities in offsprings.

Evaluation:
-It is certainly valuable in detecting leakages in industry despite some minor disadvantages which can be minimised/eliminated with strict safety precautions

Strontium-90 (industry)
Chemical prop:
-Half-life of 28 years

Uses:
-used in thickness gauges to monitor the thickness of sheet materials. Radiation from the Sr-90 is passed through the material to a detector. The intensity of the radiation detected is an indicator of the thickness of the material.

Benefits:
-Long half-life of 28 years so it can be used for extended periods of time so no need frequent replacement

Problem:
-It is similar to Calcium therefore has tendency to replace calcium and this can lead of leukaemia and bone cancer

Iodine-131 (medical)
Chemical prop:
-Half-life of 8 days
-Can be attached to biological molecule

Relating uses to chem prop:
-used in medicine to diagnose and treat thyroid cancers.
-When it is tagged to NaI and injected into bloodstream, it will accumulate in the thyroid gland where it undergoes beta and Gamma decay to kill cancer cells or identify thyroid diseases by detection

Benefits:
-Relatively short half-life of 8 days which is long enough for treatment to be done but not too long such that the patient is exposed to excess radiation.

Describe recent discoveries of elements
Copernicium-227
-half-life of 0.24 milliseconds.
-It was discovered in 9th of February in 1996 at the GSI in Darmstadt Germany
-element was created by firing accelerated zinc-70 nuclei at a target made of lead-208 nuclei in a heavy ion accelerator
Post by: jakesilove on February 11, 2016, 08:50:47 pm
My questions:
-are transuranic elements and radioisotopes produced in exactly same way ie.using nuclear reactor and particle accelerator?
-Is it not necessary for nuclear fission to occur in nuclear reactor and nuclear fusion to occur in particle accelerator ?
-Will knowing the location discovered and how it is produced for recent discoveries of elements be enough?

I have an assessment coming up so would you mind checking if my notes are correct and if the main important details are mentioned? thank you

Describe how transuranic elements are produced

Neutron bombardment:
- Natural elements are immersed in a nuclear reactor and bombarded with neutrons (or other particles) to produce higher elements as nuclear fission occurs

Particle accelerator
-transuranic element atomic number>96 are made by bombarding heavy nucleus with high speed small positive nucleus (e.g. He, C or Boron) in accelerators, synchrotrons, and cyclotrons.

Describe how commercial radioisotopes are produced

Nuclear reactors
-Bombard target nuclei with slow neutrons produced by fission of uranium-235. The nucleus absorbs neutrons hence producing a neutron-rich radioisotope.

Particle accelerators
-A machine uses electric and magnetic fields to accelerate charged particles such as protons, or nuclei (eg.Helium) or ion of other atoms (eg. Hydrogen-2 ions, Helium-3 ions) to collide with target nucleus. Nuclear fusion occurs and neutron-deficient radioisotope is produced.
eg. (nuclear fusion) Fluorine-18 is prepared in a cyclotron by bombarding nitrogen-14 with helium nuclei.

Identify instruments and processes that can be used to detect radiation
-Photographic film- Darkening of photographic film indicated the presence of radioactivity. The darkening is due to formation of silver halide crystals as radiation is absorbed by silver salts in the film.

-Cloud Chamber - device that consists a container of supersaturated vapour of water or alcohol. As radiation travels through the device, it ionises surrounding air molecules which causes vapour molecules to condense onto these ions, to produce visible cloud tracks/ trails.

-Scintillation counter:
- When certain substances like ZnS absorbed energy from alpha, beta and gamma rays, they emit flashes of light which can be collected and amplified in a photo multiplier. Thus, an electrical pulse can be generated and recenient diagnostic techniques to trace medical
Relating uses to chem prop:
-used in medicine to diagnose and treat thyroid cancers.
-When it is tagged to NaI and injected into bloodstream, it will accumulate in the thyroid gland where it undergoes beta and Gamma decay to kill cancer cells or identify thyroid diseases by detection

Benefits:
-Relatively short half-life of 8 days which is long enough for treatment to be done but not too long such that the patient is exposed to excess radiation.

Describe recent discoveries of elements
Copernicium-227
-half-life of 0.24 milliseconds.
-It was discovered in 9th of February in 1996 at the GSI in Darmstadt Germany
-element was created by firing accelerated zinc-70 nuclei at a target made of lead-208 nuclei in a heavy ion accelerator

Hey Kathy!

They look like great notes. I will tell you for sure though that you have way too much content. There is no way you need to memorise that many radioisotopes, and you can definitely ease up on the specific dates and examples. Still, the content itself is great! Just figure out how many isotopes you actually need (ie. Two for med/Two for industrial) and limit your notes to that so you don't need to do unnecessary study.

The answer to all of your questions at the start of your post is that you really don't need to know the answer to that. Nowhere in the curriculum does it require you to have an in depth understanding of the production of transuranic elements (having a basic understanding of particle accelerators etc. is absolutely enough). Remember you are not supposed to be learning all of Chemistry, just the Chemistry in the curriculum dotpoints.

So my overall tip is that it looks like you're doing a little too much study/note taking. It's great to see that you're enthusiastic, but there's not need to go overboard.

Jake :)
Post by: kathywong on February 12, 2016, 09:33:26 pm
equilibrium  --->>>  Iron (III) ions (yellow colour) reacts with salicylic acid  C7H6O3(s)  producing  Fe(c7h5o3)(aq)(violet colour)    +   H+ (hydrogen ion)

why does increasing in pressure have no effect ? is it because they're not gas molecules

Why does adding Fe(NO3)3(s)/Na(OH)(s) to solution result more intense violet colour  because it thought adding solid/liquid won't affect concentration unless they're aqueous or gaseous

Post by: Happy Physics Land on February 12, 2016, 10:15:10 pm
equilibrium  --->>>  Iron (III) ions (yellow colour) reacts with salicylic acid  C7H6O3(s)  producing  Fe(c7h5o3)(aq)(violet colour)    +   H+ (hydrogen ion)

why does increasing in pressure have no effect ? is it because they're not gas molecules

Why does adding Fe(NO3)3(s)/Na(OH)(s) to solution result more intense violet colour  because it thought adding solid/liquid won't affect concentration unless they're aqueous or gaseous

Hey Kathy:

What a great question, honestly it just covers so many aspects of Le Chatelier's principle and I would indeed see this as a hard question. I really appreciate that you are actually asking this interesting question!

First I will have to congratulate you on correctly stating that the increase in pressure has no effect on the position of the equilibrium because there are no gas molecules involved in this system. Pressure would only affect gas particles because doesnt matter whether we increase or decrease the volume, the amount of the solid and liquids would always stay constant (after all, if you imagine having a rock in a coke can, the rock is not gonna get larger when you pull open the liquid to decrease the pressure). So you have correctly stated that the increase in pressure has no impact due to a lack of gas molecules involved.

Ok the second question is actually a lot more interesting. when you add Fe(NO3)3 into the solution, it is EXTREMELY CRUCIAL to recall something we've learnt back in year 11 chemistry. YES, DISSOCIATION, THATS WHAT IM TALKING ABOUT. Because Fe(NO3)3 is an ionic substance, it is going to undergo dissociation when added to a solution (because of water), forming Iron (III) ions and NO3- ions. This increase in concentration of iron ions then disturbs the system. According to Le Chatelier's Principle, the equilibrium will shift to right, favouring the product side to minimise this disturbance. Hence more Fe(C7H5O3)(aq) are produced and consequently the solution becomes more intensely violet since the product has a violet colour.

Such an awesome question, thanks for posting this. If you still have any concerns or confusions please dont hesitate to ask! :)

Best Regards
Happy Physics Land
Post by: Happy Physics Land on February 13, 2016, 12:32:10 am
is it a good idea to use other people s notes to study (but type own notes for questions like discuss, evaluate) because it takes too much time for me to make my own notes and it might not be as good as others

one ques about nuclear chem: will fusion always occur in reactor and fission occur in particle accelerator?

Hey Amanda:

Yep, nice question. Actually this question has been posted for quite a few times now on this forum, and Jake and I kinda have our own individual ways of responding to your concerns. Jake recommends to effectively use other people's notes from a variety of sources and choose and pick the best information for the particular syllabus dot point. For me, I usually type up my own notes. I would update my notes for every single subject (especially content based subjects such as chemistry and physics) every week. In this way I only have to spend 3 hours a week to basically make notes and revise through everything. As your notes accumulate you will be much better prepared for your exams. But of course this is only my approach and when Jake comes around he can give you more insights.

Regarding the nuclear chemistry question, I may like to inform you that this is definitely not always the case. In fact in nuclear reactors, fission would often be involved. I do understand where your perspective came from though, since our final desired radioisotope would only be produced through fusion with neutrons, which you have correctly stated. But if you consider how neutron bombardment is achieved, you can see the slight inaccuracy in your statement (sorry if lm being a bit too critical!)

Neutron bombardment which takes place in nuclear reactors actually begins with the FISSION of Uranium. When we add a neutron to Uranium-235, an extremely unstable state Uranium-236 is formed, which will only exist for 10^-12 seconds before splitting into fission fragments such as Barium-146 and Krypton-92. During this process, excess neutrons are also produced which then becomes the source for neutron bombardment. So as you can see, fission DO indeed take place inside nuclear reactors and fission of Uranium-236 is essential for fusion between neutron and other nuclei to produce the desired radioisotope.

In particle accelerators or cyclotrons you would have positive particles such as an alpha-particle or neutrons which are a result of spallation (fragmentation of other elements) being accelerated into a target nuclei to form a new radioisotope. Similar to what happens in nuclear reactors, we dont just get those neutrons or alpha particles from nowhere. We need another element to undergo fission to produce there particles that would allow us to form new isotopes. So your second statement about particle accelerators would be quite right, particle accelerators involve predominantly the nuclear process of fission.

So anyways I hope my explanation is clear to you and I really liked the question you asked because it is quite confusing since they are quite complicated to understand (those fancy gadgets these days). If you have any further concerns, please dont hesitate to ask!

Best Regards
Happy Physics Land
Post by: jakesilove on February 13, 2016, 03:08:56 pm
Hey Amanda:

Yep, nice question. Actually this question has been posted for quite a few times now on this forum, and Jake and I kinda have our own individual ways of responding to your concerns. Jake recommends to effectively use other people's notes from a variety of sources and choose and pick the best information for the particular syllabus dot point. For me, I usually type up my own notes. I would update my notes for every single subject (especially content based subjects such as chemistry and physics) every week. In this way I only have to spend 3 hours a week to basically make notes and revise through everything. As your notes accumulate you will be much better prepared for your exams. But of course this is only my approach and when Jake comes around he can give you more insights.

Regarding the nuclear chemistry question, I may like to inform you that this is definitely not always the case. In fact in nuclear reactors, fission would often be involved. I do understand where your perspective came from though, since our final desired radioisotope would only be produced through fusion with neutrons, which you have correctly stated. But if you consider how neutron bombardment is achieved, you can see the slight inaccuracy in your statement (sorry if lm being a bit too critical!)

Neutron bombardment which takes place in nuclear reactors actually begins with the FISSION of Uranium. When we add a neutron to Uranium-235, an extremely unstable state Uranium-236 is formed, which will only exist for 10^-12 seconds before splitting into fission fragments such as Barium-146 and Krypton-92. During this process, excess neutrons are also produced which then becomes the source for neutron bombardment. So as you can see, fission DO indeed take place inside nuclear reactors and fission of Uranium-236 is essential for fusion between neutron and other nuclei to produce the desired radioisotope.

In particle accelerators or cyclotrons you would have positive particles such as an alpha-particle or neutrons which are a result of spallation (fragmentation of other elements) being accelerated into a target nuclei to form a new radioisotope. Similar to what happens in nuclear reactors, we dont just get those neutrons or alpha particles from nowhere. We need another element to undergo fission to produce there particles that would allow us to form new isotopes. So your second statement about particle accelerators would be quite right, particle accelerators involve predominantly the nuclear process of fission.

So anyways I hope my explanation is clear to you and I really liked the question you asked because it is quite confusing since they are quite complicated to understand (those fancy gadgets these days). If you have any further concerns, please dont hesitate to ask!

Best Regards
Happy Physics Land

Love you HPL.

Just remember everyone that you don't really need to know all this stuff in as much depth as HPL has described above; but that shouldn't stop you from extending yourself! If there isn't an explicit dotpoint on the topic, you can't really be assessed on it. Look at past papers to figure out how much information you need to remember!

Jake
Post by: curious.egg on February 13, 2016, 09:35:49 pm
Helloo!

Question:
" 2NO(g) ⇌ N2(g) + O2(g)               0.800 moles of NO was placed into a 2.00L vessel at 2000 degrees Celsius and the equilibrium concentration of N2 was found to be 0.198 mol/L. Calculate the equilibrium constant for this reaction and use this value to describe the position of the equilibrium. "

(industrial chemistry)
Post by: amandali on February 13, 2016, 09:48:46 pm
"given a geiger counter, outline a procedure by which you could determine the proportions of alpha and beta emissions from a sample of ac-277"  3 marks    im not sure what to write about    is it just describing the radiation entering the tube, ionising air molecules and generating audible clicks
Post by: jakesilove on February 13, 2016, 09:51:56 pm
Helloo!

Question:
" 2NO(g) ⇌ N2(g) + O2(g)               0.800 moles of NO was placed into a 2.00L vessel at 2000 degrees Celsius and the equilibrium concentration of N2 was found to be 0.198 mol/L. Calculate the equilibrium constant for this reaction and use this value to describe the position of the equilibrium. "

(industrial chemistry)

Hey Mr. Egg!

I'll premise my answer with the fact that I didn't actually do Industrial chemistry, and despite having done Equilibrium constants at University would have no idea how to actually calculate that equilibrium constant.

However, your question isn't to calculate the constant but how we would then use this value to describe the position of the equilibrium. My university level Chemistry can definitely help with that part!

An equilibrium constant is used to describe whether there will be more of the products, more of the reactants, or an equal amount of both in a reaction.

The rule is as follows.

If
K > 1

then the equilibrium favors the products.

If
K < 1

then the equilibrium favors the reactants.

And that's it! So if you got an equilibrium constant of 1.1 in your calculations, you would just say that the position of the equilibrium lies slightly on the right (the side of the products). If you got an equilibrium constant of 0.1, you would say that the position of the equilibrium lies largely on the left (the side of the reactants).

If anyone that has actually DONE Industrial Chemistry would like to contribute, that would be supremely helpful!

Jake
Post by: jakesilove on February 13, 2016, 10:00:48 pm
"given a geiger counter, outline a procedure by which you could determine the proportions of alpha and beta emissions from a sample of ac-277"  3 marks    im not sure what to write about    is it just describing the radiation entering the tube, ionising air molecules and generating audible clicks

Hey Amandali!

This is a really tricky question, and to be honest one that I think falls largely outside the syllabus. Still, you could use your working knowledge of Geiger counters to come to some sort of answer!

I definitely agree that, first, you would explain how Geiger counters actually work (radiation entering the tube, ionising air molecules and generating audible clicks etc.). However, since it says "outline a procedure" and it is worth three marks, I would recommend actually pretending it was some sort of experiment. Knowing that ac-277 is radioactive, I would just spout something vague like "1. Place ac-277 in a location such that its radiation does not negatively effect human life. 2. Taking adequate precautions, direct a Geiger counter towards the ac-277. 3. As the (radiation entering the tube, ionising air molecules and generating audible clicks etc.), a certain value can be read and recorded from the scale placed on the Geiger counter."

Basically, just make sure that you identify that a specific amount of radiation can be detected (as per the scale on the counter) and that this is some sort of procedure, to make sure you get full marks!

It is really important that you read the question thoroughly and address each individual word. Hope this helps!

Jake
Post by: FD121 on February 13, 2016, 10:14:59 pm
Any advice for Gravimatric Analysis SAC? It is a prac write up.
Post by: jakesilove on February 13, 2016, 11:13:55 pm
Any advice for Gravimatric Analysis SAC? It is a prac write up.

Hey FD121!

I think you're looking for the VCE Chemistry thread: This is the HSC Chemistry thread! As I don't know what is in your curriculum, I can't really answer your question.

Jake
Post by: amandali on February 14, 2016, 04:27:17 pm
The question is "Describe how commercial radioisotopes are produced, and how transuranic elements are produced" - 4 marks

Isn't this to much for a 4 marker ques ?

Commercial isotopes are used in medicine, industry and research. It can be produced in nuclear reactors where the target nuclei is bombarded with slow-moving neutrons which are then absorbed by the nucleus, thus forming a neutron-rich isotope.
eg.the medical isotope Cobalt-60 is produced in nuclear reactor (equation)

Neutron-deficient isotopes can be made in particle accelerators where the target nuclei collides with accelerated positively charged particles, causing fusion to occur.
eg.fluorine-18 is produced in particle accelerator
(equation)

Transuranic elements which have atomic masses greater than 92, can also be produced in nuclear reactor and particle accelerator, however, the target nuclei  is usually already large
eg. Neptunium-93 is made in reactor
(equation)

eg.Curium-242 is produced in particle accelerator (equation)
Post by: jakesilove on February 14, 2016, 04:57:51 pm
The question is "Describe how commercial radioisotopes are produced, and how transuranic elements are produced" - 4 marks

Isn't this to much for a 4 marker ques ?

Commercial isotopes are used in medicine, industry and research. It can be produced in nuclear reactors where the target nuclei is bombarded with slow-moving neutrons which are then absorbed by the nucleus, thus forming a neutron-rich isotope.
eg.the medical isotope Cobalt-60 is produced in nuclear reactor (equation)

Neutron-deficient isotopes can be made in particle accelerators where the target nuclei collides with accelerated positively charged particles, causing fusion to occur.
eg.fluorine-18 is produced in particle accelerator
(equation)

Transuranic elements which have atomic masses greater than 92, can also be produced in nuclear reactor and particle accelerator, however, the target nuclei  is usually already large
eg. Neptunium-93 is made in reactor
(equation)

eg.Curium-242 is produced in particle accelerator (equation)

Hey Amandali!

I definitely don't think that's too much for a four marker. I think it's always best to be SURE you get every single mark. By including loads of scientific jargon (which you did), and including great examples (which you did), you will absolutely get full marks. I think that is probably slightly longer than absolutely necessary, but in this case there really isn't much harm spending an extra 30 seconds to be sure you don't lose a mark.

Jake
Post by: amandali on February 16, 2016, 07:57:53 am
hi  how do u do this question
"a small amount of pure sodium metal was dropped into 1.5 L of water. the collected gas occupied a volume of 18.6L at 25 degrees and 100 kpa. calculate concentration of hydroxide ions(aq) after reaction" ans:1mol/L
Post by: RuiAce on February 16, 2016, 08:48:39 am
hi  how do u do this question
"a small amount of pure sodium metal was dropped into 1.5 L of water. the collected gas occupied a volume of 18.6L at 25 degrees and 100 kpa. calculate concentration of hydroxide ions(aq) after reaction" ans:1mol/L

The reaction between sodium and water is given as thus:

2 Na + 2 H2O -> 2 NaOH + H2

The gas collected is simply hydrogen gas. As we are at 25deg 100kPa, we have the molar volume Vm = 24.79
Hence, to determine the moles of H2:
n = V/Vm = 18.6/24.79 = 0.7503025413... mol

For each mole of H2 yielded, 2 moles of NaOH are yielded. Hence, moles of NaOH:
n = 2 * n(H2) = 1.500605083... mol

Note that one mole of {OH-} is present in one mole of NaOH, so the above answer gives us the moles of hydroxide ions present.

Hence, to determine the concentration, use C = n/V where V is the volume of water here
C = 1.500605083.../1.5 = 1.000403388 mol L^-1
= 1.0 mol L^-1 (correct to 2 sig. fig.)
Post by: jakesilove on February 16, 2016, 10:23:14 am
The reaction between sodium and water is given as thus:

2 Na + 2 H2O -> 2 NaOH + H2

The gas collected is simply hydrogen gas. As we are at 25deg 100kPa, we have the molar volume Vm = 24.79
Hence, to determine the moles of H2:
n = V/Vm = 18.6/24.79 = 0.7503025413... mol

For each mole of H2 yielded, 2 moles of NaOH are yielded. Hence, moles of NaOH:
n = 2 * n(H2) = 1.500605083... mol

Note that one mole of {OH-} is present in one mole of NaOH, so the above answer gives us the moles of hydroxide ions present.

Hence, to determine the concentration, use C = n/V where V is the volume of water here
C = 1.500605083.../1.5 = 1.000403388 mol L^-1
= 1.0 mol L^-1 (correct to 2 sig. fig.)

RuiAce, looks like you'll be a big name on this community! Love your "Maths Challenge" forum, and this is a great answer, perfectly set out, that will really benefit everyone!

The only thing I would add, which I'm sure you would have done in an actual answer but didn't because of the formatting, is to make sure to always include states when writing out a chemical reaction. It's just an easy way to lose marks unnecessarily, so keep an eye out for that!

Again, thanks for posting the response! Keep at it!

Jake
Post by: RuiAce on February 16, 2016, 12:04:36 pm
RuiAce, looks like you'll be a big name on this community! Love your "Maths Challenge" forum, and this is a great answer, perfectly set out, that will really benefit everyone!

The only thing I would add, which I'm sure you would have done in an actual answer but didn't because of the formatting, is to make sure to always include states when writing out a chemical reaction. It's just an easy way to lose marks unnecessarily, so keep an eye out for that!

Again, thanks for posting the response! Keep at it!

Jake

Thank you! I'll stay around here more and do my best :)

As for the comment, yes I agree most definitely. I think if I were writing it out I would've just subconsciously written down the states.
Post by: Maz on February 16, 2016, 09:53:45 pm
Hi humans  :)
i did an experiment in class the other day- and i have a validation on it in the next couple of days...i was wandering if someone could please help me out? please
The experiment was on equilibrium and Le CHatelier's principle
first we got K2CrO7 and added HCl solution to it...then to the same solution we added the base NaOH. originally the K2CrO7 was yellow, with added HCl it went to a darker colour (around orange), then when we added NaOH to the solution it went back to yellow..
i did the first part of the experiment ok...but i need help with this bit
the question is: By referring to the collision theory, account for the observed colour change that occurred when NaOH solution and the HCl solution were added to the K2CrO7 solution
I'd appreciate any help   :)
Post by: wesadora on February 16, 2016, 10:34:05 pm
Got a really short question: (yet surprisingly complicated, I find  :-\ )

"Using the tabulated reduction potentials, calculate a cell emf for the disproportionation of H2O2 and into O2 and H2O.

thanks!
-Wes
Post by: RuiAce on February 16, 2016, 10:43:31 pm
Hi humans  :)
i did an experiment in class the other day- and i have a validation on it in the next couple of days...i was wandering if someone could please help me out? please
The experiment was on equilibrium and Le CHatelier's principle
first we got K2CrO7 and added HCl solution to it...then to the same solution we added the base NaOH. originally the K2CrO7 was yellow, with added HCl it went to a darker colour (around orange), then when we added NaOH to the solution it went back to yellow..
i did the first part of the experiment ok...but i need help with this bit
the question is: By referring to the collision theory, account for the observed colour change that occurred when NaOH solution and the HCl solution were added to the K2CrO7 solution
I'd appreciate any help   :)

Before I proceed with your question, do you mean K2CrO4 or K2Cr2O7? Because CrO7 does not exist whereas CrO42- and Cr2O72- are the chromate and dichromate ions respectively.

Either way, the only disadvantage here is that I cannot formulate an equation. I will just call it potassium compound for now.

It would appear, as though the HCl reacted with the potassium compound in an equilibrium reaction. The addition of HCl shifts this equilibrium to the right according to LCP, favouring the forward reaction which produces this new orange colour. But when we introduced NaOH, because NaOH will react with the HCl in a non-equilibrium reaction (we assert that the reaction goes to completion: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)), we remove the amount of HCl present in the mixture with the potassium compound. As the HCl got reduced, the equilibrium shifted to the left according to LCP, favouring the reverse reaction and reintroducing new potassium compound (yellow colour remade).
Post by: RuiAce on February 16, 2016, 10:47:43 pm
Got a really short question: (yet surprisingly complicated, I find  :-\ )

"Using the tabulated reduction potentials, calculate a cell emf for the disproportionation of H2O2 and into O2 and H2O.

thanks!
-Wes

Your question seems rather peculiarly worded and I can't make sense out of it. So these are the only tips I can provide at this point:
a) Hydrogen peroxide is a weak acid
b) This reaction in the standard reduction potential might be useful:
1/2 O2(g) + 2 H+ + 2 e- ⇌ H2O(l)
c) I found this equation on Wikipedia just now:
2 H2O2(l) -> 2 H2O(l) + 2 H2(g)
Note that H+ is an acid
Post by: Maz on February 19, 2016, 12:04:53 am
hey
im a little confused on how to do concentration vs time graphs...could you please help me out?
if a reaction is 2NO2
N2O4 (g) + 57kJ
how do you sketch the concentration vs time graphs for the effect of changing volume on the system and
the effect of changing temperature

how would these be different for rate vs time graphs for the same things
thankyou so much  :)
Post by: RuiAce on February 19, 2016, 09:01:48 am
hey
im a little confused on how to do concentration vs time graphs...could you please help me out?
if a reaction is 2NO2
N2O4 (g) + 57kJ
how do you sketch the concentration vs time graphs for the effect of changing volume on the system and
the effect of changing temperature

how would these be different for rate vs time graphs for the same things
thankyou so much  :)

So at this point, because you don't have an exact amount of info on the concentrations of NO2 OR N2O4, we are unable to draw very precise concentration-time graphs. We can only make our graphs relative.

We will assume that the mixture is already at equilibrium. Consider the reaction: N2O4 (g) ⇌ 2 NO2 (g). Note that technically I have flipped your equation around.

According to secondary sources, the equilibrium naturally lies WELL to the right at lower temperatures. This means, initially the concentration of NO2 is far superior to that of N2O4.

Because heat is released when dinitrogen tetroxide is produced, we realise that this reaction above is endothermic. Higher temperatures favour the production of NO2, and lower temperatures favour the production of N2O2. Diagram 1 will therefore show what happens when heat is added to the system.

According to Le Chatelier's Principle, the equilibrium will shift to the right to minimise the changes, thereby increasing the concentration of NO2.

Firstly, all of the reactants given are gaseous. Hence, changes in pressure (and by consequence, volume) will affect ALL the substances in the equation. Therefore, we immediately proceed to the equation.

Note that on the left of the reaction, we only have 1 mol of gas. Whereas, on the right, we have 2. This means that if we increase the pressure, the system wants to eliminate the amount of gas present and thus the equilibrium will shift to the left, producing more N2O4. On the contrary, if we decrease the pressure, there is more room for gas to exist so the equilibrium will shift to the right and produce more NO2.

By convention, when we talk about changing the pressure we mean adding more NO2 AND N2O4 in, or taking some of BOTH out. But another way to change the pressure is to change the volume of the vessel. Note that if we DECREASE the volume of the vessel, we give LESS room for the gas to occupy. Hence, DECREASING the volume of the vessel is essentially the same as INCREASING the pressure. On the contrary, this means when we INCREASE the volume of the vessel, we DECREASE the pressure.

Thus, by decreasing the volume of the vessel we promote yield of N2O4. And vice versa for increase.

Now, let's consider the formula C=n/V. If you look at this formula, you will realise that if we decrease the volume of the vessel, we effectively INCREASE the concentration of EVERYTHING momentarily! This happens because obviously the amount mof moles present already can't be changed.

Analogy: Say at equilibrium we had 1 mol of NO2 in a 1L container. The concentration of NO2 is 1mol L^-1.
But then we decrease the volume of the vessel to 500mL (1/2 of a litre). The concentration, at the time we decreased it, effectively becomes 2mol L^-1 now, because we still have 1 mol of nitrogen dioxide present!

So if we decrease the volume of the vessel, momentarily we will have a SPIKE in the concentration graph. THEN, the system will try to adjust its equilibrium according to LCP.

This can be seen in diagram 2.

Now, the rate of reaction is dependent on two factors. We will now assume that a system has NOT YET achieved equilibrium, and is trying to.

As you know, the catalyst speeds up the rate of reaction by providing an alternate pathway requiring a lower activation energy. Hence, if we add a catalyst, we are going to force the equilibrium to be achieved more quickly. Note that both the forward and reverse reaction use the same catalyst.

Say we pumped some argon gas into the reaction chamber. You could say that this is increasing the pressure as well. BUT, there is a difference.
Argon will not react, so it will not cause havoc (oxygen being introduced can cause explosions in all sorts of reactions e.g. Haber process for production of ammonia). But argon has a counter effect. When we increase the pressure by pumping in actual equilibrium mixture (NO2 and N2O4) we are promoting reactions on one side. But pumping in argon favours NEITHER side. Hence, if a system is not at equilibrium, the presence of argon is only going to SLOW DOWN the rate of reaction.

HOWEVER... In both situations, where the equilibrium is actually AT is unaffected. This only changes the rate, not the final result.
See diagram 3.

- Diagrams are coming. I don't have pen and paper with me beside me but what I do have is an ortho appointment so I'm running short of time right now
Post by: RuiAce on February 19, 2016, 03:29:40 pm
Post by: Maz on February 19, 2016, 07:18:37 pm
So at this point, because you don't have an exact amount of info on the concentrations of NO2 OR N2O4, we are unable to draw very precise concentration-time graphs. We can only make our graphs relative.

We will assume that the mixture is already at equilibrium. Consider the reaction: N2O4 (g) ⇌ 2 NO2 (g). Note that technically I have flipped your equation around.

According to secondary sources, the equilibrium naturally lies WELL to the right at lower temperatures. This means, initially the concentration of NO2 is far superior to that of N2O4.

Because heat is released when dinitrogen tetroxide is produced, we realise that this reaction above is endothermic. Higher temperatures favour the production of NO2, and lower temperatures favour the production of N2O2. Diagram 1 will therefore show what happens when heat is added to the system.

According to Le Chatelier's Principle, the equilibrium will shift to the right to minimise the changes, thereby increasing the concentration of NO2.

Firstly, all of the reactants given are gaseous. Hence, changes in pressure (and by consequence, volume) will affect ALL the substances in the equation. Therefore, we immediately proceed to the equation.

Note that on the left of the reaction, we only have 1 mol of gas. Whereas, on the right, we have 2. This means that if we increase the pressure, the system wants to eliminate the amount of gas present and thus the equilibrium will shift to the left, producing more N2O4. On the contrary, if we decrease the pressure, there is more room for gas to exist so the equilibrium will shift to the right and produce more NO2.

By convention, when we talk about changing the pressure we mean adding more NO2 AND N2O4 in, or taking some of BOTH out. But another way to change the pressure is to change the volume of the vessel. Note that if we DECREASE the volume of the vessel, we give LESS room for the gas to occupy. Hence, DECREASING the volume of the vessel is essentially the same as INCREASING the pressure. On the contrary, this means when we INCREASE the volume of the vessel, we DECREASE the pressure.

Thus, by decreasing the volume of the vessel we promote yield of N2O4. And vice versa for increase.

Now, let's consider the formula C=n/V. If you look at this formula, you will realise that if we decrease the volume of the vessel, we effectively INCREASE the concentration of EVERYTHING momentarily! This happens because obviously the amount mof moles present already can't be changed.

Analogy: Say at equilibrium we had 1 mol of NO2 in a 1L container. The concentration of NO2 is 1mol L^-1.
But then we decrease the volume of the vessel to 500mL (1/2 of a litre). The concentration, at the time we decreased it, effectively becomes 2mol L^-1 now, because we still have 1 mol of nitrogen dioxide present!

So if we decrease the volume of the vessel, momentarily we will have a SPIKE in the concentration graph. THEN, the system will try to adjust its equilibrium according to LCP.

This can be seen in diagram 2.

Now, the rate of reaction is dependent on two factors. We will now assume that a system has NOT YET achieved equilibrium, and is trying to.

As you know, the catalyst speeds up the rate of reaction by providing an alternate pathway requiring a lower activation energy. Hence, if we add a catalyst, we are going to force the equilibrium to be achieved more quickly. Note that both the forward and reverse reaction use the same catalyst.

Say we pumped some argon gas into the reaction chamber. You could say that this is increasing the pressure as well. BUT, there is a difference.
Argon will not react, so it will not cause havoc (oxygen being introduced can cause explosions in all sorts of reactions e.g. Haber process for production of ammonia). But argon has a counter effect. When we increase the pressure by pumping in actual equilibrium mixture (NO2 and N2O4) we are promoting reactions on one side. But pumping in argon favours NEITHER side. Hence, if a system is not at equilibrium, the presence of argon is only going to SLOW DOWN the rate of reaction.

HOWEVER... In both situations, where the equilibrium is actually AT is unaffected. This only changes the rate, not the final result.
See diagram 3.

- Diagrams are coming. I don't have pen and paper with me beside me but what I do have is an ortho appointment so I'm running short of time right now

Thankyou so so so much  :)
Post by: katherine123 on February 20, 2016, 08:54:34 pm
why does acetic acid produce the least concentration of H+ ? doesnt it produce the same amount as HCl since the formula is
CH3COOH --> CH3COO- + H+
HCl --> H+ + Cl-
Post by: RuiAce on February 20, 2016, 09:03:21 pm
why does acetic acid produce the least concentration of H+ ? doesnt it produce the same amount as HCl since the formula is
CH3COOH --> CH3COO- + H+
HCl --> H+ + Cl-

No. Because acetic acid is weak and hydrochloric acid is strong.
HCl → H+ + Cl-
CH3COOH ⇌ H+ + CH3COO-

Hydrochloric acid fully ionises in solution. (The reality is that it's about 90-99%, but we assume it goes to 100%).
However, with acetic acid, only about 1.3% or so ionises.
Post by: jakesilove on February 21, 2016, 12:52:10 pm
why does acetic acid produce the least concentration of H+ ? doesnt it produce the same amount as HCl since the formula is
CH3COOH --> CH3COO- + H+
HCl --> H+ + Cl-

Just to build on RuiAce's answer, basically the way you assess the answer to this multiple choice question is with an understanding of "Weak" and "Strong" Acids.

Strong acids ionise completely in solution. Like RuiAce says, that means that approximately 90-99% of the HCl converts into the products (ie. Hydrogen ions and Chloride ions).

Weak acids do not ionise completely in solution. Like RuiAce says, that means that approximately 1.3% of the Acetic acid converts into the products.

In the HSC, you sort of just need to know which acids are weak, and which are strong. From memory, the only strong acids you work with are HCl and Sulfuric acid. The rest, you can assume, are weak! Therefore, as HCl is the only Strong acid in the list, it will produce the highest concentration of Hydrogen ions, and therefore have the lowest pH.

Jake
Post by: RuiAce on February 21, 2016, 01:34:00 pm
Just to build on RuiAce's answer, basically the way you assess the answer to this multiple choice question is with an understanding of "Weak" and "Strong" Acids.

Strong acids ionise completely in solution. Like RuiAce says, that means that approximately 90-99% of the HCl converts into the products (ie. Hydrogen ions and Chloride ions).

Weak acids do not ionise completely in solution. Like RuiAce says, that means that approximately 1.3% of the Acetic acid converts into the products.

In the HSC, you sort of just need to know which acids are weak, and which are strong. From memory, the only strong acids you work with are HCl and Sulfuric acid. The rest, you can assume, are weak! Therefore, as HCl is the only Strong acid in the list, it will produce the highest concentration of Hydrogen ions, and therefore have the lowest pH.

Jake

A tiny bit to add on again: Nitric acid is strong.

But yep, the list does not go beyond HCl, H2SO4 and HNO3 with regards to the HSC.

(However, you can assume that HBr and HI are also strong. Beause Cl, Br and I are all halogens (group A VII elements) There is a trap though - HF is weak!)

To fully answer the question though: To justify why citric acid is stronger than acetic obviously depends on the fact that citric acid (2-hydroxypropane-1,2,3-tricarboxylic acid) is triprotic and has three protons to donate, whereas acetic acid (ethanoic acid) only has one. But I think we all already knew that as the answer was disregarded immediately.
Post by: amandali on February 22, 2016, 11:47:39 pm
do you round of answer according to the given data with lowest sigfig
eg.Student mixed 20ml of 0.02mol/L of H2SO4 with 500ml of 0.001mol/L of Ca(OH)2 find the ph
so you will round the answer to 1 sig fig because of "0.02ml/0.001ml"
Will i lose marks if i don't round them enough  eg. 4.1231x10^-2 when its meant to be 4.12x10^-2
Post by: RuiAce on February 23, 2016, 08:36:48 am
do you round of answer according to the given data with lowest sigfig
eg.Student mixed 20ml of 0.02mol/L of H2SO4 with 500ml of 0.001mol/L of Ca(OH)2 find the ph
so you will round the answer to 1 sig fig because of "0.02ml/0.001ml"
Will i lose marks if i don't round them enough  eg. 4.1231x10^-2 when its meant to be 4.12x10^-2
Correct. In the question you provided, because 1 sig. fig. is the lowest amount of figures given, you must round it to that.

In terms of losing marks? Some examiners care, and others don't. Hence, not rounding correctly is a risk that you can't take.
Post by: katherine123 on February 24, 2016, 04:29:05 pm
How do you solve this? thanks ^^
The solubility of calcium hydroxide is 0.12 g per 100ml of water at 25 degrees. Calculate maximum pH of a solution of Calcium hydroxide assuming that the addition of the solid changes the volume only negligibly.   ans: 11.5
Post by: jakesilove on February 24, 2016, 06:40:22 pm
How do you solve this? thanks ^^
The solubility of calcium hydroxide is 0.12 g per 100ml of water at 25 degrees. Calculate maximum pH of a solution of Calcium hydroxide assuming that the addition of the solid changes the volume only negligibly.   ans: 11.5

Hey Katherine!

I actually get a different answer (12.5) and have looked through my working a few times and am not sure where I've gone wrong. Perhaps your answers are wrong? Otherwise, hopefully someone in the community can help me out!

(http://i.imgur.com/gRiFDzh.png?1)

Jake
Post by: IkeaandOfficeworks on February 24, 2016, 08:40:37 pm
I just wanted to ask about how to answer an ASSESS question. Like the structure (i.e how you start the response, what to include etc.) in answering these questions. Thank you!
Post by: jakesilove on February 24, 2016, 08:46:54 pm
I just wanted to ask about how to answer an ASSESS question. Like the structure (i.e how you start the response, what to include etc.) in answering these questions. Thank you!

Hey!

To answer an assess question, you essentially need to include an extra step with is a "judgement" or "reasoning" step.

You won't be required to conclude with some sort of actual assessment. For instance, at the end of an "assess the use of Ethanol as an alternate fuel source" you don't need to say "Ethanol is definitely good as an alternative fuel source". Rather, at the end you should say something like "Ethanol has many benefits as an alternate fuel source, however further research is required to increase the efficiency" etc.

The difference between "Explain" and "Assess" is the "reasoning" step. For instance, with ethanol, if you need to assess advantages and disadvantages it is not enough to just state that Ethanol has a lower heat of combustion. Rather, you need to say "As Ethanol has a lower heat of combustion, it is less productive than regular fuel when used. This is a negative result, as it means an individual needs to purchase more fuel to travel the same distance, as compared to Octane". Notice I've explained WHY it is a disadvantage. That is the main extra component of an "assess" question!

Hope this helps!

Jake
Post by: IkeaandOfficeworks on February 24, 2016, 10:27:53 pm
Thanks Jake! One more question, how different is an Evaluate question from an Assess question? Thanks again!
Post by: Maz on February 25, 2016, 07:47:01 pm
Hey
i have a test on equilibrium tomorrow and i was wandering if someone could please help me?
The general idea is that if a system is in equilibrium the foreword and reverse reactions are occurring at the same rate...
if a reactant is added the system will move to reduce and counter-react this added reactant by producing more products.
but what happens if the newly added reactant is so much that the other reactant doesn't even have enough particles available to
react with it? does the excess just exist in a non-reacted state in the system? and how do you determine that, just by being given a
equilibrium reaction?
Would really appreciate and help with this...
thankyou so so much in advance  :)
Post by: jakesilove on February 25, 2016, 08:08:24 pm
Hey
i have a test on equilibrium tomorrow and i was wandering if someone could please help me?
The general idea is that if a system is in equilibrium the foreword and reverse reactions are occurring at the same rate...
if a reactant is added the system will move to reduce and counter-react this added reactant by producing more products.
but what happens if the newly added reactant is so much that the other reactant doesn't even have enough particles available to
react with it? does the excess just exist in a non-reacted state in the system? and how do you determine that, just by being given a
equilibrium reaction?
Would really appreciate and help with this...
thankyou so so much in advance  :)

You're absolutely right! Any "excess" particles just sit there in an unreacted state. You can only figure this out if you are given molar values. For instance, in the reaction between Hydrochloric acid and Sodium Hydroxide, the ratio of the two reactants is 1:1. Therefore, if there are more moles of one than the other, the "excess" will just sit there gathering dust (not literally, obviously!).

In a generic "Extra reactant x is added, what will happen", you can assume that this "excess" principle is totally irrelevant. It is only when you are expected to do molar calculations that you should even think about principles like excess reagents etc.

Hope this helps!

Jake
Post by: Maz on February 25, 2016, 09:41:21 pm
You're absolutely right! Any "excess" particles just sit there in an unreacted state. You can only figure this out if you are given molar values. For instance, in the reaction between Hydrochloric acid and Sodium Hydroxide, the ratio of the two reactants is 1:1. Therefore, if there are more moles of one than the other, the "excess" will just sit there gathering dust (not literally, obviously!).

In a generic "Extra reactant x is added, what will happen", you can assume that this "excess" principle is totally irrelevant. It is only when you are expected to do molar calculations that you should even think about principles like excess reagents etc.

Hope this helps!

Jake
it helps a lot
thankyou!  :)
Post by: amandali on February 26, 2016, 03:53:46 pm
why does redox reaction occur when the cathode and anode are not separated?
Post by: jakesilove on February 26, 2016, 03:59:26 pm
why does redox reaction occur when the cathode and anode are not separated?

Hey Amandali!

Redox reactions like the one in the picture you've attached actually don't need to be separated! As long as there are two dissimilar metals and a way for ions to travel back and forth (in this case, the solution connecting the two is that method) then a reaction will occur. The separation of the typical Galvanic cell is not actually required. Electrons still travel from one substance to the other, and ions travel in the opposite direction in the solution.

Jake
Post by: RuiAce on February 26, 2016, 07:25:43 pm
Hey Katherine!

I actually get a different answer (12.5) and have looked through my working a few times and am not sure where I've gone wrong. Perhaps your answers are wrong? Otherwise, hopefully someone in the community can help me out!

(http://i.imgur.com/gRiFDzh.png?1)

Jake

Hey Jake, with this question I found out your mistake.

When I typed 0.12/Ans into the calculator, where Ans=Molar Mass of Ca(OH)2, I found the answer to be 1.619...x10-3
Post by: RuiAce on February 26, 2016, 07:27:05 pm
Thanks Jake! One more question, how different is an Evaluate question from an Assess question? Thanks again!
To answer this question, it doesn't.
Post by: RuiAce on February 26, 2016, 07:33:30 pm
why does redox reaction occur when the cathode and anode are not separated?
Hey Amandali!

Redox reactions like the one in the picture you've attached actually don't need to be separated! As long as there are two dissimilar metals and a way for ions to travel back and forth (in this case, the solution connecting the two is that method) then a reaction will occur. The separation of the typical Galvanic cell is not actually required. Electrons still travel from one substance to the other, and ions travel in the opposite direction in the solution.

Jake
To reinforce what Jake said here, note that for the seperated galvanic cell, the salt bridge is there only for a migration of ions. Note that in the single unit, the migration of ions is equally possible!
Post by: amandali on February 26, 2016, 10:41:50 pm
0.300g of solid NaOH was added to 1.00L of 5.00*10^-3 mol/L HNO3
Assuming no volume change, what is the pH of final solution
ans: 11.4

what i did was:
-show that HNO3 is limiting reagent with moles (5*10^-3)
-n(H20) = (5*10^-3)  since 1 mol of HNO3 produce 1 mole of H20
-find n(H+)= n(H20)
-[H+]= n/v
-  pH=-log(H+)

Post by: Happy Physics Land on February 26, 2016, 11:08:42 pm
0.300g of solid NaOH was added to 1.00L of 5.00*10^-3 mol/L HNO3
Assuming no volume change, what is the pH of final solution
ans: 11.4

what i did was:
-show that HNO3 is limiting reagent with moles (5*10^-3)
-n(H20) = (5*10^-3)  since 1 mol of HNO3 produce 1 mole of H20
-find n(H+)= n(H20)
-[H+]= n/v
-  pH=-log(H+)

Hey Amanda:

Good question. Ok so for me firstly what I would have done is to establish a balanced neutralisation equation.

NaOH (s) + HNO3 (aq) --> NaNO3 (aq) + H2O (l)

So clearly with this equation no further balancing is required. Now we need to work out the moles of HNO3 and moles of NaOH to determine the limiting reagent.

n of NaOH = 0.3/(22.9 + 16 + 1.008) = 0.0075173 moles
n of HNO3 = CV = 5 x 10^-3 x 1 = 0.005 moles

Ok so quite clearly since the molar ratio in the balanced neutralisation equation is 1:1:1:1, HNO3 is the limiting reagent, as you have correctly identified, well done!

Now theres an extra step here we need to be careful of, because the H+ ions and OH- ions that are dissociated from HNO3 and NaOH would neutralise one another and become H2O, we need to calculate the amount of moles of NaOH that is in excess of:

Excess n(NaOH) = 0.0075173 - 0.005 = 0.0025173 moles
Assuming theres no volume change, the concentration of NaOH would be C = 0.0025173/1 = 0.0025173 mol/L

So now we know that there are 0.0025173 moles per litre of OH- ions in the solution. In order to figure out the concentration of H+ ions and the pH values, we need to use the water constant (Kw = 1.0 x 10^-14).

Kw = [H+] [OH-]
[H+] = 1.0x10^-14/0.0025173 = 3.97251 x 10^-12

Now we can calculate the pH:

pH = -log(3.97251 x 10^-12)
Hence pH = 11.4

Hope you understood my reasoning there amanda, a very good question indeed! If you have any further queries please dont hesitate to ask! :)

Best Regards
Happy Physics Land
Post by: katherine123 on February 27, 2016, 08:43:01 am
A student is given 25ml of 0.50mol/L of HCl(aq).  He was asked to prepare a solution with pH=1.0 . What will be the final volume of the solution? Ans: 125ml

I dont know what went wrong with my working

I found:
-  final [H+]=10^-1
since HCl --> H+ + Cl-
-  [HCl] = 10^-1
- initial v(HCl) = 25*10^-3 L
- initial [HCl] = 25*10^-3*0.5 = 12.5*10^-3

using CiVi=CfVf

Vf=0.003L
Post by: RuiAce on February 27, 2016, 08:54:57 am
A student is given 25ml of 0.50mol/L of HCl(aq).  He was asked to prepare a solution with pH=1.0 . What will be the final volume of the solution? Ans: 125ml

I dont know what went wrong with my working

I found:
-  final [H+]=10^-1
since HCl --> H+ + Cl-
-  [HCl] = 10^-1
- initial v(HCl) = 25*10^-3 L
- initial [HCl] = 25*10^-3*0.5 = 12.5*10^-3

using CiVi=CfVf

Vf=0.003L

[H+]=10-pH=10-1mol L-1
Strong acid - Assume the acid fully ionises (Degree of ionisation = 100%)
Note: Degree of ionisation is why this happens. NOT because HCl → H+ + Cl- because what if you had acetic acid: CH3COOH ⇄ H+ + CH3COO-
Therefore [HCl] = [H+] = 10-1mol L-1
In the dilutions formula, we now have Vf=10-1mol L-1 ----> This you already have

Vi=2.5*10-2 L   ----> This you already have
Ci=0.50 mol L-1   ---->This is what you did that confused me. I am unsure as to why you chose to do V * C when the concentration was given already. The formula n=CV does allow us to find the moles of acid present, but the concentration was already given to us.

0.50 * 2.5*10-2 = 0.1 * Vf
Vf = 0.125 L
Post by: RuiAce on February 27, 2016, 09:00:17 am
Hey Amanda:

Good question. Ok so for me firstly what I would have done is to establish a balanced neutralisation equation.

NaOH (s) + HNO3 (aq) --> NaNO3 (aq) + H2O (l)

So clearly with this equation no further balancing is required. Now we need to work out the moles of HNO3 and moles of NaOH to determine the limiting reagent.

n of NaOH = 0.3/(22.9 + 16 + 1.008) = 0.0075173 moles
n of HNO3 = CV = 5 x 10^-3 x 1 = 0.005 moles

Ok so quite clearly since the molar ratio in the balanced neutralisation equation is 1:1:1:1, HNO3 is the limiting reagent, as you have correctly identified, well done!

Now theres an extra step here we need to be careful of, because the H+ ions and OH- ions that are dissociated from HNO3 and NaOH would neutralise one another and become H2O, we need to calculate the amount of moles of NaOH that is in excess of:

Excess n(NaOH) = 0.0075173 - 0.005 = 0.0025173 moles
Assuming theres no volume change, the concentration of NaOH would be C = 0.0025173/1 = 0.0025173 mol/L

So now we know that there are 0.0025173 moles per litre of OH- ions in the solution. In order to figure out the concentration of H+ ions and the pH values, we need to use the water constant (Kw = 1.0 x 10^-14).

Kw = [H+] [OH-]
[H+] = 1.0x10^-14/0.0025173 = 3.97251 x 10^-12

Now we can calculate the pH:

pH = -log(3.97251 x 10^-12)
Hence pH = 11.4

Hope you understood my reasoning there amanda, a very good question indeed! If you have any further queries please dont hesitate to ask! :)

Best Regards
Happy Physics Land

I will suggest an alternate route at the final steps. However, take note that this requires the same amount of working so it's preference as to what you use.

---> Substitute directly into pOH = -log10[OH+]
---> Use pH + pOH = 14

Note that this is essentially the process of using Kw just in a different way. (And you use subtraction instead of division)
Post by: katherine123 on February 27, 2016, 10:11:37 am
[H+]=10-pH=10-1mol L-1
Strong acid - Assume the acid fully ionises (Degree of ionisation = 100%)
Note: Degree of ionisation is why this happens. NOT because HCl → H+ + Cl- because what if you had acetic acid: CH3COOH ⇄ H+ + CH3COO-
Therefore [HCl] = [H+] = 10-1mol L-1
In the dilutions formula, we now have Vf=10-1mol L-1 ----> This you already have

Vi=2.5*10-2 L   ----> This you already have
Ci=0.50 mol L-1   ---->This is what you did that confused me. I am unsure as to why you chose to do V * C when the concentration was given already. The formula n=CV does allow us to find the moles of acid present, but the concentration was already given to us.

0.50 * 2.5*10-2 = 0.1 * Vf
Vf = 0.125 L

so is it okay if i write "assuming full ionisation and since HCl is monoprotic"   (without HCl-->H+ + Cl-)
and will i lose marks if i omit the equilibrium sign for weak acids in exam?
Post by: RuiAce on February 27, 2016, 02:01:49 pm

so is it okay if i write "assuming full ionisation and since HCl is monoprotic"   (without HCl-->H+ + Cl-)
and will i lose marks if i omit the equilibrium sign for weak acids in exam?

Assuming full ionisation due to HCl being strong but yes also include monoprotic if you want full accuracy.

And yes.
Post by: amandali on February 28, 2016, 07:10:19 am
Name ONE type of cell, other than dry-cell or lead-acid cell, you have studied. Evaluate it in comparison with either the dry-cell or lead-acid cell, in terms of chemistry and the impact on society. Include relevant chemical equations in your answer.

-  For chemistry of batteries: is it okay if i make a table containing oxidation, reduction equation and voltage without describing in words
-  "Evaluate it in comparison.." - so does that mean i talk about both problems and benefits and then judge which one is better
-  For assess/evaluate ques in general - is it okay to state advantages only or do i have to include both advantages and disadvantages

This is my plan:
•   Dry cell
-  first small and easily-portable cell
-  cheap
Dis
-  does not give constant voltage and has a short-life therefore used for infrequent used electrical devices eg. Torches, radios
Impacts on society
-  increase development of electrical devices
•   Button cell
-  Smaller than dry cell and has long lifespan thus used in smaller, frequently used devices eg. Hearing aids and cameras
-  provides the same initial voltage (1.6V) as dry cell but maintains constant voltage over longer period of time since [OH-] remains constant so more useful for above mentioned devices
Dis
-  made of silver so more expensive than dry cell
Impacts on society
-  small size, long-lifespan, stable voltage allows it to be used in small, sensitive devices
•   Evaluate – neither is problematic environmentally
-  KOH in button cell is caustic
-  Zn in dry cell is toxic to plants
BUT they are contained in small quantities hence do not pose serious concern
Large scale (industrial) will be problematic
Post by: katherine123 on February 28, 2016, 08:34:24 am
Equal volumes of  four 0.1mol/L acids were titrated with the same sodium hydroxide solution.
Which one requires the greatest volume of base to change the colour of indicator
a)citric acid
b)acetic acid
c)sulfuric acid
d)HCl

i dont understand why it's A  :/
Post by: RuiAce on February 28, 2016, 08:56:51 am
Equal volumes of  four 0.1mol/L acids were titrated with the same sodium hydroxide solution.
Which one requires the greatest volume of base to change the colour of indicator
a)citric acid
b)acetic acid
c)sulfuric acid
d)HCl

i dont understand why it's A  :/

Because the reaction is with sodium hydroxide (strong base), the reaction ALWAYS goes to completion as a strong base will fully react with both a strong acid, AND a weak acid.
I.e. (Assume aqueous states for all but water)
HCl + NaOH → NaCl + H2O
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
CH3COOH + NaOH → CH3COONa + H2O
C3H4OH(COOH)3 + 3 NaOH → C3H4OH(COO)3Na3 + 3 H2O

Because all reactions go to completion, we are fully dependent on the amount of protons they can donate. It can be seen that citric acid is triprotic, as opposed to sulfuric being diprotic and the rest being monoprotic. This is why the answer is citric: You have more protons to neutralise. (Note that the volumes and concentrations are constant, so the moles of acid are constant. It's the moles of H+ that differs.)
Post by: RuiAce on February 28, 2016, 09:02:19 am
Name ONE type of cell, other than dry-cell or lead-acid cell, you have studied. Evaluate it in comparison with either the dry-cell or lead-acid cell, in terms of chemistry and the impact on society. Include relevant chemical equations in your answer.

-  For chemistry of batteries: is it okay if i make a table containing oxidation, reduction equation and voltage without describing in words
-  "Evaluate it in comparison.." - so does that mean i talk about both problems and benefits and then judge which one is better
-  For assess/evaluate ques in general - is it okay to state advantages only or do i have to include both advantages and disadvantages

This is my plan:
•   Dry cell
-  first small and easily-portable cell
-  cheap
Dis
-  does not give constant voltage and has a short-life therefore used for infrequent used electrical devices eg. Torches, radios
Impacts on society
-  increase development of electrical devices
•   Button cell
-  Smaller than dry cell and has long lifespan thus used in smaller, frequently used devices eg. Hearing aids and cameras
-  provides the same initial voltage (1.6V) as dry cell but maintains constant voltage over longer period of time since [OH-] remains constant so more useful for above mentioned devices
Dis
-  made of silver so more expensive than dry cell
Impacts on society
-  small size, long-lifespan, stable voltage allows it to be used in small, sensitive devices
•   Evaluate – neither is problematic environmentally
-  KOH in button cell is caustic
-  Zn in dry cell is toxic to plants
BUT they are contained in small quantities hence do not pose serious concern
Large scale (industrial) will be problematic

Yes. Many candidates draw up a table in the exam room to answer these types of questions. Although this is not my preference as I liked writing lengthy responses, many band 6s have been obtained using tables and "dot points" everywhere.

You must always discuss advantages and disadvantages to be able to make an informed assessment of what you are talking about. You cannot only speak about one side.

And then, you should always make some comparison. Because when you are evaluating, you need to communicate exactly why as to one cell is better/worse than the other.

The plan is mostly good, however you could have more impacts on society for the dry cell - the question specifies this so this should be more important. One may argue that for the dry cell, the impact on society is more prominent amongst devices that do require low voltages for operation and it's mainly aided in just their manufacture, however the impact is limited when it comes to larger scale powering.

More importantly, observe that your question asked for the chemistry of the cells, to which you seemingly have forgotten. The chemistry of the cell refers to the exact equations of the cell that occur at the anode and cathode (and if there is an equation #3 that occurs elsewhere, that too).
Post by: jakesilove on February 28, 2016, 09:22:46 am
Name ONE type of cell, other than dry-cell or lead-acid cell, you have studied. Evaluate it in comparison with either the dry-cell or lead-acid cell, in terms of chemistry and the impact on society. Include relevant chemical equations in your answer.

-  For chemistry of batteries: is it okay if i make a table containing oxidation, reduction equation and voltage without describing in words
-  "Evaluate it in comparison.." - so does that mean i talk about both problems and benefits and then judge which one is better
-  For assess/evaluate ques in general - is it okay to state advantages only or do i have to include both advantages and disadvantages

This is my plan:
•   Dry cell
-  first small and easily-portable cell
-  cheap
Dis
-  does not give constant voltage and has a short-life therefore used for infrequent used electrical devices eg. Torches, radios
Impacts on society
-  increase development of electrical devices
•   Button cell
-  Smaller than dry cell and has long lifespan thus used in smaller, frequently used devices eg. Hearing aids and cameras
-  provides the same initial voltage (1.6V) as dry cell but maintains constant voltage over longer period of time since [OH-] remains constant so more useful for above mentioned devices
Dis
-  made of silver so more expensive than dry cell
Impacts on society
-  small size, long-lifespan, stable voltage allows it to be used in small, sensitive devices
•   Evaluate – neither is problematic environmentally
-  KOH in button cell is caustic
-  Zn in dry cell is toxic to plants
BUT they are contained in small quantities hence do not pose serious concern
Large scale (industrial) will be problematic

Hey!

The reason for this is just the way the dotpoint is worded. You'll likely need to talk about just one, or both, so splitting them up makes it far clearer that you've actually addressed the question!

That being said, your general approach and structure is very good. Make sure to include the Chemistry if asked for the chemistry!

Good luck! Memorisation sections like this were the bane of my existence in Year 12...

Jake
Post by: Chemystery on February 28, 2016, 05:19:32 pm
Hey Jake!

I've left my studying to last minute and have a chem test tomorrow on half the prelim course (Chemical Earth and Metals) and was wondering what would be the best way to smash through some last-minute revision? After seeing your other post i know I won't leave things last minute again in the future, thank gosh this is only preliminary haha!
Post by: Happy Physics Land on February 28, 2016, 06:59:37 pm
Hey Jake!

I've left my studying to last minute and have a chem test tomorrow on half the prelim course (Chemical Earth and Metals) and was wondering what would be the best way to smash through some last-minute revision? After seeing your other post i know I won't leave things last minute again in the future, thank gosh this is only preliminary haha!

Hey Chemystery:

Yeah its certainly not a very good idea to be doing last minute revisions. But like you said luckily its preliminaries and this perhaps is a lesson that can be learnt for year 12. In year 12 its definitely not a good idea to study the night before. Even studying the week before would still be a little short. Personally I recommend to start revising bit by bit a month beforehand and more work during the 2 weeks before the exam. This way you wont feel too pressurised the days before your exam.

But I would give you some ideas of what you should probably focus on studying right now:

Chemical Earth:
- Trends in periodic tables
- Bohr's atomic model
- Valence Electrons and energy shells
- Lewis Dot Diagram for ionic and covalent bonds as well as for individual atoms
- Ionic Bonds and Covalent Bonds
- Atomic radius and Effective charge
- Physical and electrical properties of ionic compounds
- Physical, electrical and heat properties of metals
- Properties of Covalent network (e.g. diamond) substance
- Properties of Covalent molecular substances
- The multiple spheres of Earth (e.g. atmosphere, lithosphere, hydrosphere etc.) and the dominate molecules and elements that exist within each sphere
- What is a physical change? List some physical properties of a substance
- What is a chemical Change? List some chemical properties of a substance
- How do you know when a chemical change occurs?
- Distinguish between compound, solution, mixture and solute
- Boiling points and melting points
- All the separation techniques and their industrial/home uses

Metals:
- The historical development of metals
- The smelting of iron from Haematite / copper from Chalcopyrite
- Metals discovered throughout the centuries
- What are alloys?
- Provide some examples of alloys, and state their industrial/home uses
- Balancing chemical equations
- Reaction of metal + water
- Displacement reaction
- Reaction of metal + oxygen
- Reaction of metal with dilute acid
- Activity series of metals
- Full ionic, net ionic equations and half ionic equations (oxidation equation and reduction equation)
- Redox reactions
- Be familiar with how to calculate moles using molar mass and mass and how to calculate mass using molar mass and moles
- Avogadro and Gay Lussac's law memorised
- Calculate Volume using moles and molar volume, calculate moles using volume and molar volume
- Empirical formula and molecular formula, and how to find the molecular formula when the question provides you with molar mass and the empirical formula
- Gravimetric analysis
- Electronegativity
- Ionisation energy (1st, second and third)
- Electrical conductivity of substances

Thats pretty much all you will have to study for, and it will be an immense amount of content if you only have one day. Just try to skim through textbook content for every dotpoint above and you would at least have some knowledge of what should be going on. Anyways, good luck for tomorrow!

Best Regards
Happy Physics Land
Post by: jakesilove on February 28, 2016, 07:16:36 pm
Hey Jake!

I've left my studying to last minute and have a chem test tomorrow on half the prelim course (Chemical Earth and Metals) and was wondering what would be the best way to smash through some last-minute revision? After seeing your other post i know I won't leave things last minute again in the future, thank gosh this is only preliminary haha!

Hey!

So firstly, obviously I'm going to suggest you don't leave it to the night before again.

However,  moving on, I recommend using your set of notes (or finding someone else's set of notes), going through and handwriting out (under subheadings, in extremely summarised form, with colours) all of the information that you don't know. Then, just read over that booklet that you've written out. The act of writing out information will make it far more likely that you remember the content tomorrow!

Really, best of luck. You definitely need it, but like you say it is only preliminary!

Jake
Post by: RuiAce on February 28, 2016, 07:17:42 pm
Hey Chemystery:

Yeah its certainly not a very good idea to be doing last minute revisions. But like you said luckily its preliminaries and this perhaps is a lesson that can be learnt for year 12. In year 12 its definitely not a good idea to study the night before. Even studying the week before would still be a little short. Personally I recommend to start revising bit by bit a month beforehand and more work during the 2 weeks before the exam. This way you wont feel too pressurised the days before your exam.

But I would give you some ideas of what you should probably focus on studying right now:

Chemical Earth:
- Trends in periodic tables
- Bohr's atomic model
- Valence Electrons and energy shells
- Lewis Dot Diagram for ionic and covalent bonds as well as for individual atoms
- Ionic Bonds and Covalent Bonds
- Atomic radius and Effective charge
- Physical and electrical properties of ionic compounds
- Physical, electrical and heat properties of metals
- Properties of Covalent network (e.g. diamond) substance
- Properties of Covalent molecular substances
- The multiple spheres of Earth (e.g. atmosphere, lithosphere, hydrosphere etc.) and the dominate molecules and elements that exist within each sphere
- What is a physical change? List some physical properties of a substance
- What is a chemical Change? List some chemical properties of a substance
- How do you know when a chemical change occurs?
- Distinguish between compound, solution, mixture and solute
- Boiling points and melting points
- All the separation techniques and their industrial/home uses

Metals:
- The historical development of metals
- The smelting of iron from Haematite / copper from Chalcopyrite
- Metals discovered throughout the centuries
- What are alloys?
- Provide some examples of alloys, and state their industrial/home uses
- Balancing chemical equations
- Reaction of metal + water
- Displacement reaction
- Reaction of metal + oxygen
- Reaction of metal with dilute acid
- Activity series of metals
- Full ionic, net ionic equations and half ionic equations (oxidation equation and reduction equation)
- Redox reactions
- Be familiar with how to calculate moles using molar mass and mass and how to calculate mass using molar mass and moles
- Avogadro and Gay Lussac's law memorised
- Calculate Volume using moles and molar volume, calculate moles using volume and molar volume
- Empirical formula and molecular formula, and how to find the molecular formula when the question provides you with molar mass and the empirical formula
- Gravimetric analysis
- Electronegativity
- Ionisation energy (1st, second and third)
- Electrical conductivity of substances

Thats pretty much all you will have to study for, and it will be an immense amount of content if you only have one day. Just try to skim through textbook content for every dotpoint above and you would at least have some knowledge of what should be going on. Anyways, good luck for tomorrow!

Best Regards
Happy Physics Land

Wow this was a nice summary of the prelim chem course... now how do you still even remember all of this...

I have to be given a question to know what to do...
Post by: Happy Physics Land on February 28, 2016, 07:30:10 pm
Wow this was a nice summary of the prelim chem course... now how do you still even remember all of this...

I have to be given a question to know what to do...

Thats why you make your own notes in prelims mate
Post by: RuiAce on February 28, 2016, 08:10:07 pm
Thats why you make your own notes in prelims mate

BACK IN MY DAYS...

Ok,
Back when I was in prelim... I either winged it... or read other people's notes (biology)... and went into the exam... and did well in everything except English.
Post by: ProfLayton2000 on February 29, 2016, 04:56:05 pm
Hi there (and you have no idea how glad I am this website exists)

Regarding 5-7 mark 'assess' or 'evaluate' questions, is there a set structure you would recommend?
Post by: amandali on February 29, 2016, 05:45:12 pm
Name one type of cell, other than dry cell or lead acid cell. Evaluate it in comparison with either the dry cell or lead-acid cell, in terms of chemistry and the impact on society. Include relevant chemical equations in your answer.   7 marks

Dry cell is a small, easily-portable and cheap cell. It has a short-life and provides a non-constant voltage hence it is used for low-energy drained devices like torches and radios. Whereas, silver oxide cell provides constant voltage over longer period of time than dry cell since [OH-] remains constant hence it is used to operate sensitive devices like hearing aids and watches rather than dry cell. Silver oxide cell is also smaller and has a long-lifespan than dry cell which makes it more useful for above mentioned devices. However, it is made of silver so it is more expensive than dry cell.
The electrolyte KOH in silver oxide cell is caustic and the outer zinc casing of dry cell is toxic to plants but there are present in small quantities hence neither cell is problematic to the environment.

Including a table:
Dry cell: oxidation, reduction equations
SIlver oxide cell : oxidation, reduction equations

-how much am i suppose to write for 7 marks ques
-I feel like it s a bit brief but i dont know what else to add
Post by: jakesilove on February 29, 2016, 05:53:53 pm
Hi there (and you have no idea how glad I am this website exists)

Regarding 5-7 mark 'assess' or 'evaluate' questions, is there a set structure you would recommend?

I think it really depends on the question itself. For an ethanol question, a table with advantages and disadvantages always works best. You will always tailor it for what sort of question it is (are there two clear sides to the argument, and so can you use a table? Are there multiple important components (eg. for your Battery) and so can you use subheadings?).

The important part is that you actually explain WHY an aspect is good/bad etc. It is not enough to say Ethanol has a lower molar heat of combustion: Explain how this has a negative impact on consumers.

Also, make sure you address EACH COMPONENT of the question. It is really easy to skip a word in a question that ends up being 3 marks worth of content. Underline the important part of the question, and address everything.

I don't think that there is a set structure. For an assess/evaluate, you will almost always have advantages/disadvantages, so I think a table/subheadings is always appropriate. However I think the best way to make sure you are prepared for extended response question is to do one standard question from each topic, re write it until it is perfect (marked by us or your teachers) and then put it in your notes.

Good luck!

Jake
Post by: jakesilove on February 29, 2016, 05:57:44 pm
Name one type of cell, other than dry cell or lead acid cell. Evaluate it in comparison with either the dry cell or lead-acid cell, in terms of chemistry and the impact on society. Include relevant chemical equations in your answer.   7 marks

Dry cell is a small, easily-portable and cheap cell. It has a short-life and provides a non-constant voltage hence it is used for low-energy drained devices like torches and radios. Whereas, silver oxide cell provides constant voltage over longer period of time than dry cell since [OH-] remains constant hence it is used to operate sensitive devices like hearing aids and watches rather than dry cell. Silver oxide cell is also smaller and has a long-lifespan than dry cell which makes it more useful for above mentioned devices. However, it is made of silver so it is more expensive than dry cell.
The electrolyte KOH in silver oxide cell is caustic and the outer zinc casing of dry cell is toxic to plants but there are present in small quantities hence neither cell is problematic to the environment.

Including a table:
Dry cell: oxidation, reduction equations
SIlver oxide cell : oxidation, reduction equations

-how much am i suppose to write for 7 marks ques
-I feel like it s a bit brief but i dont know what else to add

Hey Amandali!

I think that this is an almost perfect response. Whilst it may seem brief, remember that when you handwrite all of that it will be nearly two thirds of a page! The only think that I would absolutely make clearer is the general impact on society. Have one sentence saying "Dry cell had a huge/small impact on society because of (insert property here) allowing for (insert affect here). Similarly/conversely, the Silver cell".... etc. Just make it utterly clear that you're answering the specific question.

Really, though, I think that this is a very good answer overall. Decide whether subheadings/table could work, write it out a few times and maybe send through your final answer, and we can assess its final mark!

Jake
Post by: ProfLayton2000 on February 29, 2016, 06:31:34 pm

I think it really depends on the question itself. For an ethanol question, a table with advantages and disadvantages always works best. You will always tailor it for what sort of question it is (are there two clear sides to the argument, and so can you use a table? Are there multiple important components (eg. for your Battery) and so can you use subheadings?).

The important part is that you actually explain WHY an aspect is good/bad etc. It is not enough to say Ethanol has a lower molar heat of combustion: Explain how this has a negative impact on consumers.

Also, make sure you address EACH COMPONENT of the question. It is really easy to skip a word in a question that ends up being 3 marks worth of content. Underline the important part of the question, and address everything.

I don't think that there is a set structure. For an assess/evaluate, you will almost always have advantages/disadvantages, so I think a table/subheadings is always appropriate. However I think the best way to make sure you are prepared for extended response question is to do one standard question from each topic, re write it until it is perfect (marked by us or your teachers) and then put it in your notes.

Good luck!

Jake

Post by: jakesilove on February 29, 2016, 10:47:02 pm

I'm really glad! Keep on posting great questions like this; as the community grows, we'll be able to answer more and more questions, and release more and more resources!

Jake
Post by: katherine123 on March 02, 2016, 07:16:09 am
0.1 mol/L solution of HCl has pH of 1.0, whereas, a 0.1mol/L solution of citric acid has pH of 1.6
Explain why the two solutions have different pH values
3 marks

HCl has lower pH since it fully ionises while citric partially ionises but HCl is monoprotic and citric is triprotic though :/
Post by: RuiAce on March 02, 2016, 08:10:58 am
0.1 mol/L solution of HCl has pH of 1.0, whereas, a 0.1mol/L solution of citric acid has pH of 1.6
Explain why the two solutions have different pH values
3 marks

HCl has lower pH since it fully ionises while citric partially ionises but HCl is monoprotic and citric is triprotic though :/

Citric acid's degree of ionisation is only about 8.6% whereas hydrochloric acid's is virtually 100%. This difference is huge. You'd need to concentrate citric acid by a factor of about 12 to make it's pH similar to HCl.
Post by: jakesilove on March 02, 2016, 08:20:54 am
0.1 mol/L solution of HCl has pH of 1.0, whereas, a 0.1mol/L solution of citric acid has pH of 1.6
Explain why the two solutions have different pH values
3 marks

HCl has lower pH since it fully ionises while citric partially ionises but HCl is monoprotic and citric is triprotic though :/

Hey all!

Recalling that this is a three mark question, I would definitely recommend including more information. Here is what I would write, including where I think the marks lie.

HCl is a strong acid, which means it ionises completely in solution (in that the equilibrium between HCl and Water lies completely to the right) [1]. As the concentration of the solution is 0.1 mol/L, and the ratio between HCl and Hydrogen ions is 1:1, the pH will be $pH=-log[0.1]$
$pH=1.0$ [1.5]

Conversely, Citric acid is a weak acid, which means it does not ionise completely in solution (in that the equilibrium between Citric acid and Water lies largely to the left) [2.5]. Whilst the concentration of the solution is the same, as it does not ionise completely the ratio between moles of acid and moles of Hydrogen ions will not be 1:1 (it will be a:b, a>>b). Therefore, the pH of this solution will be higher than the pH of an equally concentrated solution of HCl [3].

I think this question would also be easier if you write out the ionisation of each acid, but honestly I can't figure out how to make that look nice on the forums. Just make sure to fully explain the concepts in order to get the marks!

Hope this helps.

Jake
Post by: William3558 on March 02, 2016, 02:18:32 pm
Hi, I need some help with explaining an answer.

A student titrated an aliquot of standard sodium carbonate solution with HCl acid in a burette.
Would the concentration determined for HCl be higher than, lower than or unchanged from the actual value if the student had previously washed with water, but not dried:

a) The pipette used to deliver the aliquot of sodium carbonate solution

The answer was higher. But, wouldn't the water reduce the concentration and amount of mol in the sodium carbonate? Therefore, the amount of mol coming out of the HCl in the burette be less than expected, thus the concentration is less than expected?

b)

c) The burette

The answer was lower. As this would reduce the concentration of the HCl from its actual value

I think I might be overthinking it...
Post by: katherine123 on March 02, 2016, 02:51:34 pm
i dont get why it is B
Post by: jakesilove on March 02, 2016, 03:33:20 pm
Hi, I need some help with explaining an answer.

A student titrated an aliquot of standard sodium carbonate solution with HCl acid in a burette.
Would the concentration determined for HCl be higher than, lower than or unchanged from the actual value if the student had previously washed with water, but not dried:

a) The pipette used to deliver the aliquot of sodium carbonate solution

The answer was higher. But, wouldn't the water reduce the concentration and amount of mol in the sodium carbonate? Therefore, the amount of mol coming out of the HCl in the burette be less than expected, thus the concentration is less than expected?

b)

c) The burette

The answer was lower. As this would reduce the concentration of the HCl from its actual value

I think I might be overthinking it...

Hey William!

So, let's start with the pipette. Like you say, if there is excess water in the pipette, the sodium carbonate will become more dilute. If it is more dilute, then you will require LESS HCl to neutralise the solution (and thus end the titration). If the number of moles of acid used in LESS than it should be, that means it is going to be calculated as MORE concentrated. Think of it this way: If you figure out (through calculation) that 1 mol is required to neutralise the base, but all of a sudden you release LESS than a mole of acid (because the base is more diluted than expected), you will still assume that, for instance, there is one mole of acid in the 15mL of liquid you release. When you do your
$C=\frac{n}{V}$
calculation, your value for concentration will be HIGHER than it should be.

If there is extra liquid in the Burette, then the Acid becomes more dilute. Therefore, it requires MORE acid to neutralise the base. This means, again, that you will increase V but keep n the same, resulting in a lower concentration (as per the formula)

Hope this helps! If I haven't explained it enough, let me know :)

Jake
Post by: jakesilove on March 02, 2016, 03:37:55 pm
i dont get why it is B

Hey Katherine!

This is a really tough question, but hopefully my explanation makes sense.

The list is ordered in DECREASING strength. HCl is the strongest acid, and HCN is the weakest acid. You can tell because, whilst they have the same concentration, they ionise to different extents, with HCl ionising completely (as it is strong) and HCN hardly ionising at all (as it is very weak).

An acid and a base will react when, first, it breaks up into its components (ie. HCl ionises to become a Hydrogen ion and a Chlorine ion and NaOH breaks up into a sodium ion and a Hydroxide ions) and then these ions react with each other. However, HCl will ionise completely, straight away, without the need for a huge amount of solvent. This is because it is strong. Conversely, weak bases will not ionise completely without a lot of solution. This is why the weakest acid will take the most amount of base to react completely! It just takes a long time to ionise, as opposed to stronger acids.

Hope that this makes sense!

Jake
Post by: William3558 on March 02, 2016, 04:17:21 pm
Hey William!

So, let's start with the pipette. Like you say, if there is excess water in the pipette, the sodium carbonate will become more dilute. If it is more dilute, then you will require LESS HCl to neutralise the solution (and thus end the titration). If the number of moles of acid used in LESS than it should be, that means it is going to be calculated as MORE concentrated. Think of it this way: If you figure out (through calculation) that 1 mol is required to neutralise the base, but all of a sudden you release LESS than a mole of acid (because the base is more diluted than expected), you will still assume that, for instance, there is one mole of acid in the 15mL of liquid you release. When you do your
$C=\frac{n}{V}$
calculation, your value for concentration will be HIGHER than it should be.

If there is extra liquid in the Burette, then the Acid becomes more dilute. Therefore, it requires MORE acid to neutralise the base. This means, again, that you will increase V but keep n the same, resulting in a lower concentration (as per the formula)

Hope this helps! If I haven't explained it enough, let me know :)

Jake

Just what I needed to hear! Thanks! Very appreciated.  ;D
Post by: grace_joseph on March 03, 2016, 05:51:11 pm
Hi! I have an upcoming practical exam on titrations with a written component on the Acidic Environment Syllabus 1 (indicators) and Syllabus 4. I was just wondering if you had any tips for studying for these sort of exams, as well as any helpful information on those syllabus's? Sorry, this is really vague and open-ended, I'm just a bit uncertain as to how I prepare!

Also, are there any rules to determining whether a substance is a strong/weak acid/base when just given it's formula?

Thanks so much! :)
Post by: jakesilove on March 03, 2016, 06:27:30 pm
Hi! I have an upcoming practical exam on titrations with a written component on the Acidic Environment Syllabus 1 (indicators) and Syllabus 4. I was just wondering if you had any tips for studying for these sort of exams, as well as any helpful information on those syllabus's? Sorry, this is really vague and open-ended, I'm just a bit uncertain as to how I prepare!

Also, are there any rules to determining whether a substance is a strong/weak acid/base when just given it's formula?

Thanks so much! :)

Hey Grace!

I'll answer your last question first, as honestly it's just easier; no, there is no way to know whether something is a strong/weak substance based entirely on its formula (as far as I know, and definitely as far as your Curriculum goes). The only way you could be required to 'know' is if it is a substance you regularly use in the Syllabus (ie. HCl and H2SO4 are strong, Citric is weak etc.) or, experimentally, you could tell based on the pH of a substance.

What I mean by that is you could create a dilute/concentrated solution, test its pH, and determine its strength. If you put 0.1 Mol of monoprotic acid into 1L of water, you would expect the concentration of $H^+$ to be 0.1mol/L for a strong acid (and therefore the pH to be 1). If you test this experimentally, and find the pH to be higher, you know its a weak acid!

As for the general "How do I study for a practical exam!" question, I think that's definitely harder to answer. Have a working understanding of the dot points, to which I can only suggest writing your own notes, talking to friends about the dotpoints, and answering past HSC questions on those specific dotpoints. Then, make sure you are very, very comfortable with the terms Reliability, Validity and Accuracy in terms of first-hand experiments. If you're not sure, Reliability refers to repetition, Validity refers to experimental design testing the required variable (ie. controlling variables etc.) and Accuracy refers to the precision of instruments. Then, make sure you know which the independent and dependent variables are. Look through potential practical tasks that can be drawn from the dotpoints in those two sections, in case you can guess the question.

Other than that, I don't really have specific tips! If you have a good idea of the chemistry in those sections, understand experimental design, and bring a Ruler for your graph, you'll be totally fine!

An extra note: make sure to write out procedures in past tense, in case you have to!

Let me know if there is anything (more specific, preferably aha) else I can help you with!

Jake :)
Post by: grace_joseph on March 03, 2016, 06:42:46 pm
Thank you so much Jake! Sorry about being vague, but you have definitely put me on the right track.
Massive help!
Grace :)
Post by: William3558 on March 03, 2016, 09:15:13 pm
Quick Question, In a back titration with standard sodium hydroxide added in excess to the solution with ammonium... When you boil the solution to evaporate all of the ammonia. Why do you need to keep adding water and keep a constant volume? Is it so the solution doesn't go dry?
Post by: jakesilove on March 03, 2016, 09:32:45 pm
Quick Question, In a back titration with standard sodium hydroxide added in excess to the solution with ammonium... When you boil the solution to evaporate all of the ammonia. Why do you need to keep adding water and keep a constant volume? Is it so the solution doesn't go dry?

Hey William!

I think you are looking for the VCE Question Thread (which is here). We hardly do any of that in the HSC!

Jake
Post by: amandali on March 06, 2016, 12:56:48 am
Am i allowed to wrote this condensed ionisation equation for citric acid in hsc exam:
H3X <--> X- + 3H+

and can you check my response thanks:

Compare the environmental sustainability of producing ethanol from petroleum and sugar cane. Support you answer with relevant chemical equations

Petroleum is not environmentally sustainable as it is a non-renewable resource hence will eventually deplete in the future. Moreover, the production of ethanol from petroleum is unsustainable as it involves the process of fractional distillation C6H12O6(aq)--> C6H12O6(l)  and catalytic cracking C4H8(l)-->C2H2(g) + C2H6 (g)which requires enormous amount of energy that are commonly produced by the combustion of fossil fuels which releases CO2 that contributes to global warming.

In contrast, sugar cane may be sustainable because it can be regrown once harvested thus a renewable resource. However, its production is not completely carbon neutral as petroleum may be combusted to produce energy to operate crop harvesters and trucks which releases more CO2 than it is absorbed to make ethanol.

Post by: Happy Physics Land on March 06, 2016, 01:22:50 am
Am i allowed to wrote this condensed ionisation equation for citric acid in hsc exam:
H3X <--> X- + 3H+

and can you check my response thanks:

Compare the environmental sustainability of producing ethanol from petroleum and sugar cane. Support you answer with relevant chemical equations

Petroleum is not environmentally sustainable as it is a non-renewable resource hence will eventually deplete in the future. Moreover, the production of ethanol from petroleum is unsustainable as it involves the process of fractional distillation C6H12O6(aq)--> C6H12O6(l)  and catalytic cracking C4H8(l)-->C2H2(g) + C2H6 (g)which requires enormous amount of energy that are commonly produced by the combustion of fossil fuels which releases CO2 that contributes to global warming.

In contrast, sugar cane may be sustainable because it can be regrown once harvested thus a renewable resource. However, its production is not completely carbon neutral as petroleum may be combusted to produce energy to operate crop harvesters and trucks which releases more CO2 than it is absorbed to make ethanol.

Hey Amanda:

Great questions! In regards to your first question, l understand that you would be doing that in order to reduce time consumption. But it is crucial that we dont risk anything in our hsc exams, because you never know who will be marking your essay. Hence it is always good to write the full formula out despite it will cost you several seconds more.

Still referring to your equation, there are some errors that has been made using the condensed formula:
1) After ionisation/donating all the protons, the charge of X wouldnt be X-, but X3-, because citric acid has lost the three protons in its carboxylic bond.
2) The process of ionisation of citric acid is gradual, i.e. it cant be represented in 1 step, because in the chemistry that we learn, the three protons cannot be donated all at once. I.e. it would follow a three-step sequence of: C6H8O7(aq) <---> C6H7O7-(aq) + H(aq), C6H7O7-(aq) <---> C6H6O7(2-)(aq) + H+(aq), C6H6O7(2-)(aq) <---> C6H5O7(3-)(aq) + H+(aq). So yeah this is why I dont recommend the condensed method (unless your teacher reinforce it) because it can yield careless mistakes and it doesnt show the full chemical process.

Referring to your response now, I quite like it because you definitely compared sugar and petroleum to some depth and you've also provided the correct equation of catalytic cracking. If this is a 4 mark question I would definitely give it a 3 and if lm a strict marker l would give it 2. What l recommend you can improve on is to maintain a balance between the amount of pros and cons you give to petroleum and those you give to sugar cane. What you have done there is good, but if we write more on petroleum than sugar cane it would sound a bit biased in terms of your understanding. What you can add to enhance your statement of sugar cane being a environmentally sustainable source is to say "during the cultivation of sugar canes, photosynthesis takes place (provide an equation of photosynthesis) which consumes the carbon dioxide from atmosphere, reducing the amount of greenhouse gas and alleviating global warming". This would make your reasoning stronger for sugar cane. With the equation for fractional distillation, you can include it but that is not the point. A better equation to include is the combustion of ethanol which clearly demonstrates that carbon dioxide is released and it is released a large quantity (2CO2, meaning that for every mole of ethanol combusted, two moles are released).

But Amanda you have been posting a lot of questions on AN which really shows how dedicated you are to your studying. If I sounded a little harsh please dont be discouraged because right now you are doing a super awesome job. If you have any further questions please dont hesitate to ask!

Best Regards
Happy Physics Land
Post by: jakesilove on March 06, 2016, 09:07:43 am
Am i allowed to wrote this condensed ionisation equation for citric acid in hsc exam:
H3X <--> X- + 3H+

and can you check my response thanks:

Compare the environmental sustainability of producing ethanol from petroleum and sugar cane. Support you answer with relevant chemical equations

Petroleum is not environmentally sustainable as it is a non-renewable resource hence will eventually deplete in the future. Moreover, the production of ethanol from petroleum is unsustainable as it involves the process of fractional distillation C6H12O6(aq)--> C6H12O6(l)  and catalytic cracking C4H8(l)-->C2H2(g) + C2H6 (g)which requires enormous amount of energy that are commonly produced by the combustion of fossil fuels which releases CO2 that contributes to global warming.

In contrast, sugar cane may be sustainable because it can be regrown once harvested thus a renewable resource. However, its production is not completely carbon neutral as petroleum may be combusted to produce energy to operate crop harvesters and trucks which releases more CO2 than it is absorbed to make ethanol.

I completely agree with HPL's assessment of your answer, and I just have one thing to add. Definitely make sure that you write roughly the same amount for each "half" of the question, and part of that is chemical equations. Make sure to include some sort of chemical equation for the second half. Also, I think it is important to identify that it is biomass (ie. Cellulose) that is important in the production of ethanol from Sugar Cane; it is then easy to claim that this is renewable, as biomass makes up 50% of all plant matter!

Good answer overall though; however I would definitely expand on your second half.

Jake
Post by: lazydreamer on March 06, 2016, 12:42:34 pm

I think it really depends on the question itself. For an ethanol question, a table with advantages and disadvantages always works best. You will always tailor it for what sort of question it is (are there two clear sides to the argument, and so can you use a table? Are there multiple important components (eg. for your Battery) and so can you use subheadings?).

The important part is that you actually explain WHY an aspect is good/bad etc. It is not enough to say Ethanol has a lower molar heat of combustion: Explain how this has a negative impact on consumers.

Also, make sure you address EACH COMPONENT of the question. It is really easy to skip a word in a question that ends up being 3 marks worth of content. Underline the important part of the question, and address everything.

I don't think that there is a set structure. For an assess/evaluate, you will almost always have advantages/disadvantages, so I think a table/subheadings is always appropriate. However I think the best way to make sure you are prepared for extended response question is to do one standard question from each topic, re write it until it is perfect (marked by us or your teachers) and then put it in your notes.

Good luck!

Jake

hello, sorry it was an oldish post but a question about using tables...
Could i get away with incorporating tables for questions like assess, compare/contrast, discuss etc? it's so much easier to organise information that way and looks better than a huge chunk of words, would markers appreciate that or...nah? haha
aaand because i don't have time to write a detailed paragraph

thanks
Post by: Happy Physics Land on March 06, 2016, 01:00:52 pm
hello, sorry it was an oldish post but a question about using tables...
Could i get away with incorporating tables for questions like assess, compare/contrast, discuss etc? it's so much easier to organise information that way and looks better than a huge chunk of words, would markers appreciate that or...nah? haha
aaand because i don't have time to write a detailed paragraph

thanks

Hey LazyDreamer:

Definitely yes for compare and contrast. If you use a table for compare and contrast and include the suitable information and suitable titles, then its almost a guaranteed full mark. Cautions needs to be taken with questions that involve evaluate, discuss or assess. These questions require a logical flowing response, i.e. you need to use linking words. Even if you are using a table for these questions, you would still need linking words and words that show cause and effect. If you are using a table for evaluate a assess, you must also make two clear judgments in your response in order to secure full marks. But yes, generally saying, a table is a great way to reduce time consumption whilst still effective presenting your response to the question.

Best Regards
Happy Physics Land
Post by: lazydreamer on March 06, 2016, 01:29:55 pm
Hey LazyDreamer:

Definitely yes for compare and contrast. If you use a table for compare and contrast and include the suitable information and suitable titles, then its almost a guaranteed full mark. Cautions needs to be taken with questions that involve evaluate, discuss or assess. These questions require a logical flowing response, i.e. you need to use linking words. Even if you are using a table for these questions, you would still need linking words and words that show cause and effect. If you are using a table for evaluate a assess, you must also make two clear judgments in your response in order to secure full marks. But yes, generally saying, a table is a great way to reduce time consumption whilst still effective presenting your response to the question.

Best Regards
Happy Physics Land

thanks for answering, i'll keep that in mind for pre-trials and see how i go :)
and i agree that for evaluate/discuss/assess qns i would need to include a couple of sentences at the end that makes a judgement. As for linking cause and effect, couldn't that be included in the table, eg |Cause|Effect|Evaluation|

btw
Quote
If you are using a table for evaluate a assess, you must also make two clear judgments in your response in order to secure full marks.
what do you mean by 2 clear judgments?
Post by: Happy Physics Land on March 06, 2016, 01:38:47 pm
thanks for answering, i'll keep that in mind for pre-trials and see how i go :)
and i agree that for evaluate/discuss/assess qns i would need to include a couple of sentences at the end that makes a judgement. As for linking cause and effect, couldn't that be included in the table, eg |Cause|Effect|Evaluation|

btw what do you mean by 2 clear judgments?

Ok so this might sound a little dumb but for example if you are asked to assess the impact of polyethylene on the society, in your first dotpoint/establishing sentence, you must state that polyethylene has had a significant/mediocre/insignificant impact upon the society (making a judge). At the end of your answer, l would literally write "Final Judgement: Polyethylene has had a tremendous impact upon the society". You can just include one judgment if you want, but for me, having a judgment at the start and having another one in the end helps to make your statements sound stronger.
Post by: lazydreamer on March 06, 2016, 02:14:21 pm
Ok so this might sound a little dumb but for example if you are asked to assess the impact of polyethylene on the society, in your first dotpoint/establishing sentence, you must state that polyethylene has had a significant/mediocre/insignificant impact upon the society (making a judge). At the end of your answer, l would literally write "Final Judgement: Polyethylene has had a tremendous impact upon the society". You can just include one judgment if you want, but for me, having a judgment at the start and having another one in the end helps to make your statements sound stronger.

not at all  ;D reads like an essay paragraph to me:  thesis statement that answers the qn, then supporting points, then a link back to the question. Makes sense i guess, thanks!
Post by: jakesilove on March 06, 2016, 03:11:42 pm
not at all  ;D reads like an essay paragraph to me:  thesis statement that answers the qn, then supporting points, then a link back to the question. Makes sense i guess, thanks!

Just wanted to back up HPL here: Totally agree with what he's said! If you have any further questions, about this or anything else, please don't hesitate to post :)

Jake
Post by: lazydreamer on March 06, 2016, 03:32:26 pm
Just wanted to back up HPL here: Totally agree with what he's said! If you have any further questions, about this or anything else, please don't hesitate to post :)

Jake

this site is great, much easier to get help than on BoS (which is blocked on the school laptops lol) so when i found a forum which was actually unblocked with a helpful community, it felt like i died and went to hsc heaven. Thanks, i think i'll be using this consistently throughout the year :)
Post by: Happy Physics Land on March 06, 2016, 07:18:14 pm
this site is great, much easier to get help than on BoS (which is blocked on the school laptops lol) so when i found a forum which was actually unblocked with a helpful community, it felt like i died and went to hsc heaven. Thanks, i think i'll be using this consistently throughout the year :)

Hahaha its good to know that we are helping you lazydreamer!!! Your involvement will make our community an even greater place than it was! :D

Happy Posting!

Post by: RuiAce on March 07, 2016, 10:27:28 am
Ok so this might sound a little dumb but for example if you are asked to assess the impact of polyethylene on the society, in your first dotpoint/establishing sentence, you must state that polyethylene has had a significant/mediocre/insignificant impact upon the society (making a judge). At the end of your answer, l would literally write "Final Judgement: Polyethylene has had a tremendous impact upon the society". You can just include one judgment if you want, but for me, having a judgment at the start and having another one in the end helps to make your statements sound stronger.
Will add my 2 cents to this.

To be honest, I reckon when giving the evaluation a good evaluation could be something such as

Polyethylene has benefited society appreciably through it's wide ranges of applications (give reasoning) however further improvements could still be made in shifting away from these resources and using biopolymer alternatives.

(Usually evaluate questions at this time of the year seem to be ethanol though so I would say overall, it is the renewability of the resource that ultimately makes ethanol a viable alternate fuel source.)

(Do forgive me though - I'm sleep-deprived right now so I don't feel like reading too much of what I missed.)
Post by: Happy Physics Land on March 07, 2016, 05:06:10 pm
Hi Jake!

We did an experiment in class where we titrated sodium carbonate and hydrochloric acid, creating a neutral salt (I think... it's a strong acid and strong base right?). Our teacher told us that methyl orange was the best indicator, but this doesn't make sense to me. Shouldn't we use bromothymol blue or phenolphthalein as the indicator as the equivalence point would be around pH 7?
I thought maybe it was to do with the presence of carbon dioxide as that's an acidic oxide, but not sure if that's right or not. Unless I've gotten something completely wrong!

Thanks!

Hey Grace:

That was a great experiment you did, and this actually is a pretty interesting question. One thing I can assure you is that the existence of CO2(g) as an outcome of this reaction definitely wouldnt affect which indicator to use, because CO2(aq) is the one that  would react with water to form H2CO3(aq) which is acidic. Of course you may wanna argue that theres an equilibrium equation of CO2(g) --> CO2(aq) and some CO2(g) may have converted into CO2(aq) due to slight changes in temperature or pressure. Even if this equilibrium reaction does take place, its usually almost negligible and it definite wouldnt influence your choice of indicators.

In regards to which indicator to use, l think that your teacher hasnt been very accurate with her words. When we test the pH of things, we normally use two indicators at the same time to ascertain the pH range of this substance that we are testing. (If you need further elaboration on that point just ask again). So we dont usually only use one indicator, say for example we do use methyl orange and the NaCl we are testing in this case is a neutral substance, then the methyl orange would turn yellow. But hey, methyl orange's pH is defined over 3.1-4.4, so the yellow colour can possibly mean its a neutral substance, but at the same time it can also mean that NaCl is weakly acidic or even basic! Thats why we needs a second indicator for the purpose of determining the salt's pH range. So personally, if you have an option over which indicators use, I would use methyl red which ranges from 4.4 -6.3 pH and phenolphthalein which ranges from 8.3 -11.0 pH. So to test this salt is neutral, you would see that methyl red shows a yellow colour, and phenolphthalein showing no colour (i.e. colourless). This means that the salt has a pH that ranges from 6.3 to 8.3, and hence you can justify that the salt is neutral or just a little bit acidic or basic.

Thats my response to your question, if Jake comes around soon he can probably give you a better response than I can. But if you have any questions, dont hesitate to ask! :)

Best Regards
Happy Physics Land
Post by: neeshy900 on March 07, 2016, 06:57:33 pm
Any tips on titration prac exams?? Stressing out :( :(
Post by: Happy Physics Land on March 07, 2016, 08:19:18 pm
Any tips on titration prac exams?? Stressing out :( :(

Hello Neeshy:

Yes I can definitely understand how those titration prac exams can be extremely concerning, but dont worry, we are all here to help you! I will provide you with some specific tips below and hope they help.

Preparation of a Primary Standard Solution:
- A standard solution is a solution of known composition and concentration used for quantitative analytical procedures (in this case, titration)
- Standard solution made by dissolving a primary standard volume of water
- Solids substances are desirable for preparing standard solution because they are easier to weigh
- It is also desirable for the substances to be easily obtained in its pure form, which reduces the cost and makes it easier to perform calculations with minimal errors involving the existence of impurities
- Substances with large molar mass is desirable to allow for reduced weighing errors
- And finally, the substance must also have a high water solubility - obviously, because the substance is required to be dissolved in water for standard solution preparation

*Procedure for preparing primary standard solutions may be obtained from your textbook

Dilution Calculations

- Since we are adding only solvent, moles of solute remains unchanged
- Before we dilute (solution 1, concentrated) the equation that applies is n1 = C1V1
- After we dilute, V and C have both changed, hence the equation that applies is n2=C2V2
- But since the moles of solute is the same before and after, hence the equation that applies is C1V1 = C2V2

General Overview of Neutralisation Reaction

- HEAT is released in neutralisation reactions between strong acids and strong bases (e.g. NaOH + HCl, the most common example)
- ACID + BASE ----> SALT + WATER
- Neutralisation reaction between strong acid and strong base is essentially the reaction of a proton and OH ion to form water

Titration

- An analytical process which involves finding the volume of a solution of accurately known conc. that is required to react completely with a known volume of a solution of unknown concentration. Hence it is a volumetric analysis.
- Summary of Procedure:
1. A known volume of a solution of unknown conc. placed in a conical flask
2. A solution of known conc. added until all of the analyte has reacted, volume is recorded
3. Using the mole ratio of the reaction the unknown conc. is calculated

Calculation of Concentration of an Unknown Solution:

1. Write a balanced equation
2. Calculate number of moles of reactant A (n=CV)
3. Using the moles of reactant A and the stoichiometric molar ratio in the equation, calculate the number of moles of Reactant B used
4. Use C=n/V to calculate the conc. of Reactant B

These are basically all I have to say, a bit content heavy, if Jake happens to come around he can give you more advices. Anyways good luck my friend, if you have any questions, dont hesitate to ask!

Best Regards
Happy Physics Land
Post by: RuiAce on March 07, 2016, 08:25:28 pm
Hi Jake!

We did an experiment in class where we titrated sodium carbonate and hydrochloric acid, creating a neutral salt (I think... it's a strong acid and strong base right?). Our teacher told us that methyl orange was the best indicator, but this doesn't make sense to me. Shouldn't we use bromothymol blue or phenolphthalein as the indicator as the equivalence point would be around pH 7?
I thought maybe it was to do with the presence of carbon dioxide as that's an acidic oxide, but not sure if that's right or not. Unless I've gotten something completely wrong!

Thanks!

Sodium carbonate is not a strong base...

Methyl orange was used because you're titrating a WEAK base with a strong acid!
_________________________

HPL mentioned to use something in it's pure form to prepare a standard solution. An example of this would be to use ANHYDROUS sodium carbonate, not hydrated sodium carbonate.
Post by: RuiAce on March 07, 2016, 08:30:33 pm
Hey Grace:

That was a great experiment you did, and this actually is a pretty interesting question. One thing I can assure you is that the existence of CO2(g) as an outcome of this reaction definitely wouldnt affect which indicator to use, because CO2(aq) is the one that  would react with water to form H2CO3(aq) which is acidic. Of course you may wanna argue that theres an equilibrium equation of CO2(g) --> CO2(aq) and some CO2(g) may have converted into CO2(aq) due to slight changes in temperature or pressure. Even if this equilibrium reaction does take place, its usually almost negligible and it definite wouldnt influence your choice of indicators.

In regards to which indicator to use, l think that your teacher hasnt been very accurate with her words. When we test the pH of things, we normally use two indicators at the same time to ascertain the pH range of this substance that we are testing. (If you need further elaboration on that point just ask again). So we dont usually only use one indicator, say for example we do use methyl orange and the NaCl we are testing in this case is a neutral substance, then the methyl orange would turn yellow. But hey, methyl orange's pH is defined over 3.1-4.4, so the yellow colour can possibly mean its a neutral substance, but at the same time it can also mean that NaCl is weakly acidic or even basic! Thats why we needs a second indicator for the purpose of determining the salt's pH range. So personally, if you have an option over which indicators use, I would use methyl red which ranges from 4.4 -6.3 pH and phenolphthalein which ranges from 8.3 -11.0 pH. So to test this salt is neutral, you would see that methyl red shows a yellow colour, and phenolphthalein showing no colour (i.e. colourless). This means that the salt has a pH that ranges from 6.3 to 8.3, and hence you can justify that the salt is neutral or just a little bit acidic or basic.

Thats my response to your question, if Jake comes around soon he can probably give you a better response than I can. But if you have any questions, dont hesitate to ask! :)

Best Regards
Happy Physics Land

Unnecessary. Bromothymol blue will be GREEN at around pH7, which is a very clearly distinguishable colour from blue and yellow. As with litmus with purple and blue/red.

The rule of thumb is:

Strong B Weak A -> Phenolphthalein
Strong B Strong A -> Litmus/Bromothymol blue
Weak B Strong A -> Methyl orange
Post by: Happy Physics Land on March 07, 2016, 09:03:30 pm
Unnecessary. Bromothymol blue will be GREEN at around pH7, which is a very clearly distinguishable colour from blue and yellow. As with litmus with purple and blue/red.

The rule of thumb is:

Strong B Weak A -> Phenolphthalein
Strong B Strong A -> Litmus/Bromothymol blue
Weak B Strong A -> Methyl orange

Umm, you reckon l should just delete my post?.....
Post by: lazydreamer on March 07, 2016, 09:06:24 pm
Any tips on titration prac exams?? Stressing out :( :(

Great run-down from HPL! You know it's a titration, so make sure you know exactly how to carry it out, step-by-step. Play it out in your mind if you want. That will give you some confidence and reassurance going into it. But don't get complacent, 'cause anything can happen!
A few tips, not sure if it'll help but...:
- There are different ways to rinse your burette, pipette and standard flask, make sure you know them!
- If you can, label your glassware, good lab. practice and prevents confusion, which you don't need in an exam yea?
- Don't forget to close the stopcock when rinsing the burette, lol
- Rinse down the edges of the flask as you titrate, esp. nearing the end-point, in case some analyte/titrant gets stuck on the sides, which will result in error

ummm that's all i can think of right now, best of luck in your exam, i'm sure you'll do great :D
Post by: RuiAce on March 07, 2016, 09:06:58 pm
Umm, you reckon l should just delete my post?.....

No real point. Once it's posted it's there. It's still a learning experience and who knows if someone else will make a similar mistake.
Post by: amandali on March 09, 2016, 05:24:16 pm
why is this unacceptable
HSO3- + H20 <--> SO3(2-) + H3O+
HSO3- + H2O <--> H2SO3 + OH-

to show that hydrogen sulfite ion is amphiprotic
Post by: RuiAce on March 09, 2016, 06:03:19 pm
why is this unacceptable
HSO3- + H20 <--> SO3(2-) + H3O+
HSO3- + H2O <--> H2SO3 + OH-

to show that hydrogen sulfite ion is amphiprotic

Technically it should be. Because you are still showing how it can act as a base or an acid according to B-L theory.

Easiest thing to do, however, is to demonstrate it reacting with a different acid and a different base just to clear up ambiguity.
HSO3- + HCl(aq) -> H2SO3(aq) + Cl-
HSO3- + OH- -> SO32- + H2O(l)

If you got marked incorrect by a teacher, you will have to dispute it with them
Post by: Happy Physics Land on March 09, 2016, 11:06:36 pm
Hey guys, need a hand here if anyone dont mind!

The question is: with the aid of appropriate equations, explain why the dihydrogen phosphate ion H2PO4- is amphiprotic, yet an aqueous solution of KH2PO4 has a greater pH than 7?

Thank you very much in advance :D !!!

Post by: jakesilove on March 10, 2016, 07:12:35 am
Hey guys, need a hand here if anyone dont mind!

The question is: with the aid of appropriate equations, explain why the dihydrogen phosphate ion H2PO4- is amphiprotic, yet an aqueous solution of KH2PO4 has a greater pH than 7?

Thank you very much in advance :D !!!

Hey HPL!

In a nutshell, here is the way you need to answer the question in order to get full marks. There is some guess work here, but I only make fair assumptions based (lol) on the question.

First, you need to write out equations showing Dihydrogen Phosphate acting as both an acid (proton donor) and base (proton acceptor). To do this, write an equation with the Dihydrogen Phosphate and any acid, and then Dihydrogen Phosphate and any base. You need to explain what an amphiprotic substance is (ability to donate AND ability to accept protons), and then relate it to your equations.
Then, all you really need to say is that Dihydrogen Phosphate is a STRONGER base than water. As such, the water will act as an acid (as it is also amphiprotic), donating protons to the Dihydrogen Phosphate (which acts as a base). This produces hydroxide ions (due to the water losing a proton) and therefore the pH is greater than 7.

Hope this helps!

Jake
Post by: Happy Physics Land on March 10, 2016, 06:40:23 pm
Hey HPL!

In a nutshell, here is the way you need to answer the question in order to get full marks. There is some guess work here, but I only make fair assumptions based (lol) on the question.

First, you need to write out equations showing Dihydrogen Phosphate acting as both an acid (proton donor) and base (proton acceptor). To do this, write an equation with the Dihydrogen Phosphate and any acid, and then Dihydrogen Phosphate and any base. You need to explain what an amphiprotic substance is (ability to donate AND ability to accept protons), and then relate it to your equations.
Then, all you really need to say is that Dihydrogen Phosphate is a STRONGER base than water. As such, the water will act as an acid (as it is also amphiprotic), donating protons to the Dihydrogen Phosphate (which acts as a base). This produces hydroxide ions (due to the water losing a proton) and therefore the pH is greater than 7.

Hope this helps!

Jake

Ahhhh ok thank you Jake! We actually had a pretty intense discussion on the most suitable equations to include and my teacher agreed with your reasoning as welll mmmm. Thank you so much Jake, greatly appreciated! :D

P.S. nice pun!
Post by: amandali on March 13, 2016, 03:47:20 pm
isnt this reaction endothermic

Since Z increases as temperature decreases so :
X (g) + Y (g) + heat <--> Z(g)

Post by: Happy Physics Land on March 13, 2016, 07:03:21 pm
isnt this reaction endothermic

Since Z increases as temperature decreases so :
X (g) + Y (g) + heat <--> Z(g)

Hey Amanda!

Quite a tricky question actually, took me a while to figure out.
So we know the equilibrium equation is here is a certain moles of X + a certain moles of Y =  a certain moles of Z. For convenience's sake, lets just assume first that the equation is X + Y <--> Z.
So as we can see, as temperature increases in all three trends, the yield of Z decreases. Le Chatelier's principle tells us that as temperature of system increases, the system is disturbed and hence the equilibrium shifts to the endothermic reaction in order to consume the heat in the system to minimise this disturbance. Since yield of Z decreases as temperature increases, the yield of X and Y are favoured, and according to Le Chatelier we know that the endothermic process is favoured. Therefore we can conclude that the reverse process of Z---> X + Y is endothermic. Hence the forward process of X + Y ---> Z becomes exothermic.

We know now that the answer definitely cant be A or C because they both display an endothermic reaction. So this leaves us with B and D. Great! Now lets have a look at the pressure. As pressure increases, the yield of Z actually increases (as you can see, the graph representing 200 atm pressure is way above the two other graphs). According to La Chatelier's principle, as pressure increases, the reaction would favour the side with the least moles of gas in order to alleviate the pressure by having less gas particles occupying a reduced volume. Since in the graph the production of Z is favoured and according to Le Chatelier's principle the side with least moles of gas should be favoured, then it can be concluded that Z must have less moles of gas than the reactants X and Y.D is the only answer that suits both criteria of being exothermic and Z being the side with least moles of gas.

Hence the correct answer is D.

If you have any further questions dont hesitate to ask! :)

Best Regards
Happy Physics Land
Post by: amandali on March 13, 2016, 08:03:00 pm
can anyone explain how to determine whether an acid/base can act as a buffer pair? thanks

Post by: Neutron on March 13, 2016, 08:07:33 pm
can anyone explain how to determine whether an acid/base can act as a buffer pair? thanks

Hey Amandali! I believe a buffer pair is commonly made up of a weak acid and its conjugate base (so for example, CH3COOH and CH3COO- are buffer pairs! Hope that makes sense :D

Neutron
Post by: Neutron on March 13, 2016, 08:09:32 pm
Hey guys!

Sorry I have quite a long question, I was wondering whether someone could explain how a dry cell battery works or a lead-acid battery cell, I've read some textbooks about it but I'm still a bit confused! And also, what's a good way of knowing what citric acid and sodium hydroxide forms? Like what the products are.. Thank you! :D

Neutron
Post by: Happy Physics Land on March 13, 2016, 08:33:01 pm
can anyone explain how to determine whether an acid/base can act as a buffer pair? thanks

Hey Amanda:

Good question! Actually a few weeks before this "buffer" concept was a little confusing for me too. But yes, like what neutron said, a buffer pair would be a weak acid and the conjugate base of that weak acid. You may ask, why not a strong acid? Because strong acids dont produce reversible reactions! Buffers are required to be weak acids or bases because we need the equilibrium reaction for excess acids/bases to be converted into neutral substances in order to minimise the added acid/base's impact upon the normal pH level of the system.

In our blood, H2CO3 and HCO3- are prevalent examples of buffers. Other buffer pairs can include CH3COOH/CH3COO- or H2PO42-/HPO4-.

Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 13, 2016, 09:00:25 pm
Hey guys!

Sorry I have quite a long question, I was wondering whether someone could explain how a dry cell battery works or a lead-acid battery cell, I've read some textbooks about it but I'm still a bit confused! And also, what's a good way of knowing what citric acid and sodium hydroxide forms? Like what the products are.. Thank you! :D

Neutron

Hey Neutron!

First of all, thank you for answering amanda's question! It's really great to see everyone helping each other out! I certainly give you credit for that!

Ok, so for me, my chem teacher didnt really explain that dotpoint, so I really had to chew my jacaranda textbook and collect online informations. Im suspecting that you are perhaps thrown into the same situation too? (correct me if lm wrong haha) But dont worry, Im here to help!

If you understood galvanic cells, you should be able to reason with the chemical functionality of a dry cell battery. All it is, is just the electrode in the anode half-cell being oxidised and the electrode in the cathode half-cell being reduced, and hence producing a voltage. It works exactly like your galvanic cells, except with more complicated substances. In a dry-cell battery, the positive cathode at the centre consists of an inert graphite rod surrounded by MnO2(s) powder. The negative anode consists of the zinc casing of the cell. And there is a thick layer of moist electrolyte paste NH4Cl(aq) between the graphite rod and MnO2(s). What basically happens is essentially the reaction between manganese dioxide powder and Ammonium chloride and zinc casing and ammonium chloride . (2MnO2(s) + 2NH4+(aq) + 2e- --> Mn2O3(s) + H2O(l) + 2NH3(aq)) In this equation, the original oxidation state of Mn is +4 but in the product, the oxidation state of Mn becomes +3. This means that Mn has been reduced. In the zinc casing and ammonium chloride reaction (Zn(s) + 2NH4Cl-(aq) --> ZnCl2(s) + 2NH3(aq) + H2(g)), the original oxidation state of Zinc was 0, and after the reaction, its oxidation state becomes 2+. This means that Zinc has been oxidised. This redox reaction then produces a voltage, just like what happens in a galvanic cell.

In regards to citric acid + sodium hydroxide, you can very easily deduce what happens either by using the rule "acid + base --> salt + water" or just simply memorise what I got below! :)

C6H8O7(aq) + NaOH(s) --> NaC6H5O7(aq) + 3H2O(l)

Hope my answer has helped with your understanding of these areas of chemistry, and they are indeed hard concepts to master. Once again, thanks for being helpful in the community and if you have any further questions, please dont hesitate to ask! :)

Best Regards
Happy Physics Land
Post by: katherine123 on March 13, 2016, 10:05:30 pm
Am i right in saying that:
-If all products are weak eg. H2PO4- + H3O+ --> H3PO4 + H20 then the reaction goes to completion
-some weak some strong eg. H2PO4- +H20 <--> OH- (strong) + H3PO4 (weak)
then reaction undergoes equilibrium
-weak acid eg. HF will always produce a strong conjugate base ( F-  )
Post by: Happy Physics Land on March 13, 2016, 10:12:55 pm
Am i right in saying that:
-If all products are weak eg. H2PO4- + H3O+ --> H3PO4 + H20 then the reaction goes to completion
-some weak some strong eg. H2PO4- +H20 <--> OH- (strong) + H3PO4 (weak)
then reaction undergoes equilibrium
-weak acid eg. HF will always produce a strong conjugate base ( F-  )

Hey Katherine:

In regards to your first two points, I find it a bit difficult to memorise it that way. All you need to know is whenever there is a strong acid or a strong base involved in the reactants, then the reaction is always going to go to completion because strong bases/acids completely ionise in water. Whenever both reactants are weak acids/bases, the reaction is always going to become reversible (i.e. equilibrium reaction) because they dont fully ionise in water.

I do agree with your third point though! Very well stated! Weak acids will for strong conjugate bases because they only partially ionise in water.

Hope this give you a clearer insight into acid-base reactions! :)

Best Regards
Happy Physics Land
Post by: qwerty222 on March 13, 2016, 11:04:34 pm
hello,
if i wanted to do some self-study at home wat textbook would u recommend?
Post by: Happy Physics Land on March 14, 2016, 12:01:34 am
hello,
if i wanted to do some self-study at home wat textbook would u recommend?

Hey qwerty222!

Jacaranda, definitely jacaranda.
It has the most detailed information on each dotpoint, a whole load of content, if you make notes out of the textbook and be familiar with them, you will be very familiarised with all your modules. The two disadvantages are its cost (60-70 dollars l remember) and the overwhelming amount of content which may bore you. If you want to obtain a fast grasp onto the chemistry knowledge I will recommend Excel chemistry textbook because it has all the information you would need for each dotpoint and present it in very succint manner. But of course it is not as extensive as jacaranda. I havent really used any other textbooks as just yet, but l do recommend these two textbooks, especially jacaranda.

Best Regards
Happy Physics Land
Post by: katherine123 on March 14, 2016, 10:45:29 pm
why is HNO3 and NO2- not  a conjugate acid/base pair   and conjugate acid-base buffer pair
Post by: Happy Physics Land on March 14, 2016, 11:25:43 pm
why is HNO3 and NO2- not  a conjugate acid/base pair   and conjugate acid-base buffer pair

Hey Katherine!

A conjugate acid/base pair will have exactly the same chemical composition except it will either lose or gain a proton (i.e. a H+ atom). So in this case, when HNO3(aq) loses a H+ atom, it becomes NO3-, not NO2-. So HNO3 and NO3- in this case will be conjugate pairs.

Post by: katherine123 on March 15, 2016, 06:50:46 am
Is this info wrong in saying that CH3COO-   since weak acids always produce strong conjugate bases
Post by: Happy Physics Land on March 15, 2016, 08:52:00 am
Is this info wrong in saying that CH3COO-   since weak acids always produce strong conjugate bases

I would say that the information is wrong. The chemistry that we have been taught in year 12 tells us that a weak acid would yield a strong conjugate base. Hence CH3COO- is likely to be a strong conjugate base.

Sorry, update. The information is actually correct. We got taught that a weak acid would yield a strong conjugate base. But just like what rui said, only an extremely weak acid would yield a strong conjugate base. However, the conjugate base of the weak acetic acid is still stronger in comparison to water (i.e. it will react with water) despite compared to other bases it is still a weak base. Sorry for my misunderstanding of the knowledge!
Post by: WLalex on March 15, 2016, 09:06:35 am
Hi, i was wondering if you have any resources/information of the lead-acid battery. The chemistry of it was quite confusing to me and the way the teacher taught it to us was in a hurry and she didn't really explain much. We also did the silver button cell

Thanks
Post by: RuiAce on March 15, 2016, 10:10:57 am
Is this info wrong in saying that CH3COO-   since weak acids always produce strong conjugate bases

The information is very much correct.

A weak acid (CH3COOH) yields a weak base conjugate CH3COO-

A strong acid (HCl), specifically, however has an extremely weak conjugate base (Cl-).

I would say that the information is wrong. The chemistry that we have been taught in year 12 tells us that a weak acid would yield a strong conjugate base. Hence CH3COO- is likely to be a strong conjugate base.

It actually isn't, basically.

No pun intended.
Post by: Happy Physics Land on March 15, 2016, 10:13:22 am
The information is very much correct.

A weak acid (CH3COOH) yields a weak base conjugate CH3COO-

A strong acid (HCl), specifically, however has an extremely weak conjugate base (Cl-).

It actually isn't, basically.

No pun intended.

But isnt CH3COOH a weak acid?
Post by: RuiAce on March 15, 2016, 10:14:35 am
Hi, i was wondering if you have any resources/information of the lead-acid battery. The chemistry of it was quite confusing to me and the way the teacher taught it to us was in a hurry and she didn't really explain much. We also did the silver button cell

Thanks

Note that with such cells, you don't have to listen to your teacher. You can choose yourself to do the dry cell over lead-acid, and pick another cell over the button cell. (However, admittedly the button cell, whilst not the one I did, is probably the easiest to do.)

EasyChem offers some decent notes on the cell. http://www.easychem.com.au/production-of-materials/electrochemical-methods/comparison-of-battery-cells. I didn't study this one (and my HSC was last year) so I can't offer something comprehensive - I will leave that to another user.

Take strong note that the 'chemistry' of a cell specifically refers to the reactions at the anode and cathode (and any electrolyte if applicable) - thus it's mostly the EQUATIONS.
Post by: RuiAce on March 15, 2016, 10:14:58 am
But isnt CH3COOH a weak acid?

Acetic acid is very much a weak acid, yes.

And the acetate ion is ALSO a weak base.

You need to be aware of the fact that the conjugate of only an EXTREMELY weak acid/base is a strong base/acid.

This is because the degree of ionisation of a strong acid/base is virtually 100% (at least 90% but we assume 100% in our course), thus the presence of it's conjugate is NOT going to reform into the original thing. I.e. this reaction can't happen: Cl- + H3O+ → HCl(aq) + H2O(l). Only the reverse reaction can happen.

As opposed to the acetate ion, where we do have the equilibrium happening and both substances can form due to a low degree of ionisation.
CH3COO- + H3O+ ⇌ CH3COOH(aq) + H2O(l)
Post by: katherine123 on March 15, 2016, 06:02:10 pm
Acetic acid is very much a weak acid, yes.

And the acetate ion is ALSO a weak base.

You need to be aware of the fact that the conjugate of only an EXTREMELY weak acid/base is a strong base/acid.

This is because the degree of ionisation of a strong acid/base is virtually 100% (at least 90% but we assume 100% in our course), thus the presence of it's conjugate is NOT going to reform into the original thing. I.e. this reaction can't happen: Cl- + H3O+ → HCl(aq) + H2O(l). Only the reverse reaction can happen.

As opposed to the acetate ion, where we do have the equilibrium happening and both substances can form due to a low degree of ionisation.
CH3COO- + H3O+ ⇌ CH3COOH(aq) + H2O(l)

so basically
Very strong/weak acid/base produces very weak/strong conjugate base/acid
Moderately strong/weak acid/base produce moderately weak/strong base/acid   ?

Post by: Happy Physics Land on March 15, 2016, 06:41:21 pm

so basically
Very strong/weak acid/base produces very weak/strong conjugate base/acid
Moderately strong/weak acid/base produce moderately weak/strong base/acid   ?

Hey Katherine!

Very strong acid/base would produce very weak conjugate base/acid. Very weak acid/base would produce very strong base/acid. With weak acids and weak bases such as CH3COOH and NH3, their conjugate bases and acids will be strong, but it is only strong when compared to water. When compared to other strong bases or acids they might still be weak. To say "Moderately strong/weak acid/base produce moderately weak/strong base/acid" is justifiable but I think its better to say that these bases and acids are weak/strong compared to water. So yes I would largely agree with your statement here. Sorry for my initial incorrect response on this by the way!

Best Regards
Happy Physics Land
Post by: RuiAce on March 15, 2016, 07:24:42 pm

so basically
Very strong/weak acid/base produces very weak/strong conjugate base/acid
Moderately strong/weak acid/base produce moderately weak/strong base/acid   ?

Not exactly the best way to memorise it. Because there's no such thing as very strong or moderately strong.

There's only strong, weak and extremely weak.

Conjugate of a strong acid/base is an extremely weak base/acid.
Conjugate of a weak acid/base is a weak base/acid.
Post by: katherine123 on March 15, 2016, 10:20:06 pm
Is this okay?

Discuss factors that must be considered when using neutralization reactions to safely minimise damage in chemical spills 4 marks

Acid can be neutralised by adding solid amphiprotic substance eg. NaHCO3. Since it is a weak acid/base it will not constitute a hazard when it is in excess.
as an acid neutralise base: HCO3- (aq)+ OH- (aq)< -> CO3(2-)(aq) + H2O (l)
as a base neutralise acid: HCO3(-)(aq) + H30+(aq) <-> H2CO3 (aq) + H2O(l)
As a powder, it provides a great surface area for quick absorption of liquid acid. Moreover, it produces effervescence in acid therefore complete neutralisation of concentrated acid can be known when fizzing stops.   HCO3- + H3O + --> CO2 + 2H2O
Its disadvantages has minimal adverse impact as CO2 produced has negligible contribution to atmospheric CO2 and calcification of water pipes that is causes can be easily managed
Post by: RuiAce on March 16, 2016, 10:00:11 am
Is this okay?

Discuss factors that must be considered when using neutralization reactions to safely minimise damage in chemical spills 4 marks

Acid can be neutralised by adding solid amphiprotic substance eg. NaHCO3. Since it is a weak acid/base it will not constitute a hazard when it is in excess.
as an acid neutralise base: HCO3- (aq)+ OH- (aq)< -> CO3(2-)(aq) + H2O (l)
as a base neutralise acid: HCO3(-)(aq) + H30+(aq) <-> H2CO3 (aq) + H2O(l)
As a powder, it provides a great surface area for quick absorption of liquid acid. Moreover, it produces effervescence in acid therefore complete neutralisation of concentrated acid can be known when fizzing stops.   HCO3- + H3O + --> CO2 + 2H2O
Its disadvantages has minimal adverse impact as CO2 produced has negligible contribution to atmospheric CO2 and calcification of water pipes that is causes can be easily managed

On the note of NaHCO3 being a weak acid/base, you should contrast it to the disadvantage of using a strong acid/base. One of them is yes, causing a hazard of it's own when used in excess, but the other is that the process of neutralisation is highly exothermic and it's only worse when a strong acid/base is used. This occurs due to the reaction H+ + OH- -> H2O(l) having ∆H = approx -57 kJ mol-1, which is reduced to about -42 kJ mol-1 when a weak acid/base is present.

Another suggestion when neutralising the acid, however this being more or so on larger scale spills, is to soak the acid up into sand first. Have the sand absorb the acid, and then still using NaHCO3, neutralise it at a much safer location.
Post by: katherine123 on March 16, 2016, 03:14:02 pm
how do you answer this ques

explain why a solution of CH3COONa is basic, while a sodium sulfate solution of the same concentration has a pH of 7. Write ionic equations to describe any reactions
Post by: RuiAce on March 16, 2016, 06:04:53 pm
how do you answer this ques

explain why a solution of CH3COONa is basic, while a sodium sulfate solution of the same concentration has a pH of 7. Write ionic equations to describe any reactions

Preface. We know that sulfuric acid is a strong acid. It's first proton will always ionise no matter what.
H2SO4 + H2O(l) → HSO4- + H3O+

But we obviously know that sulfuric acid is diprotic. It has two protons that it's willing to donate. However, the second hydrogen atom does not always get fully ionised - it goes into equilibrium
HSO4- + H2O(l) ⇌ SO42- + H3O+

Except, under normal circumstances this equilibrium lies well to the right. The amount of sulfate ions present will always be appreciably greater than those of hydrogen sulfate. In fact, it almost ionises to completion as well.

The consequence is that whilst H2SO4 is obviously strongly acidic, HSO4- is reasonably acidic and SO42- is only a tiny bit basic.

In Na2SO4, obviously the sodium ion Na+ is neutral. But because the sulfate ion is only somewhat basic, it doesn't adjust the pH of the solution much at all. It is essentially neutral, just not exactly (which the question mistakenly implied).

By contrast, for sodium acetate, the acetate ion is definitely basic. (Obviously, it's a weak base. But it's not as weak as the sulfate ion.)
CH3COO- + H3O+ ⇌ H2Ol + CH3COOH(aq)
A simple analysis of that equation clearly shows why a solution of sodium acetate will be basic.
Post by: king_sanj on March 16, 2016, 06:49:29 pm
Hey Jake! For HSC chem, in the Production of Materials module, would we need to know anything about Hydrobromous acid in terms of the Bromine water test used to distinguish between alkenes and alkanes?
Post by: Happy Physics Land on March 16, 2016, 07:09:24 pm
Hey Jake! For HSC chem, in the Production of Materials module, would we need to know anything about Hydrobromous acid in terms of the Bromine water test used to distinguish between alkenes and alkanes?

Hey King_sanj:

Jake is temporarily away for a period of time, so I can answer the question here if you wouldnt mind. In Production of Materials alone, we wouldnt need to know anything about Hydrobromous acids (HBr), because when we are doing the Bromine water test we are using only Br2(aq). Because the reaction between bromine water and alkenes is an addition process, it doesnt form hydrobromous acid and hence it is not necessary to know about HBr for the module. However, what you do need to know is the final product between the alkene and the Br2(aq), because oftentimes you would be asked to provide with an equation of the reaction between the alkene and bromine water. For module 2 Acidic Environment you WOULD need to know about HBr and how it is a weak acid compared to other ones such as H2SO4 and HCl.

Best Regards
Happy Physics Land
Post by: RuiAce on March 16, 2016, 07:43:44 pm
Hey King_sanj:

Jake is temporarily away for a period of time, so I can answer the question here if you wouldnt mind. In Production of Materials alone, we wouldnt need to know anything about Hydrobromous acids (HBr), because when we are doing the Bromine water test we are using only Br2(aq). Because the reaction between bromine water and alkenes is an addition process, it doesnt form hydrobromous acid and hence it is not necessary to know about HBr for the module. However, what you do need to know is the final product between the alkene and the Br2(aq), because oftentimes you would be asked to provide with an equation of the reaction between the alkene and bromine water. For module 2 Acidic Environment you WOULD need to know about HBr and how it is a weak acid compared to other ones such as H2SO4 and HCl.

Best Regards
Happy Physics Land

I know what he's referring to.

In truth, whilst we prepare it using bromine itself, Br2, what we actually get and use for the practical is what is called bromine water - HOBr.
Observe the following equation:
Br2(l) + H2O(l) -> HOBr(aq) + HBr(aq)

Now, hydrobromic acid will not affect the experiment in any way whatsoever. It may be worth noting that it's a strong acid, but it just sits there.
We are only interested in bromine water - HOBr, which actually has the same molecular formula as hyprobromous acid (HBrO) - not hydrobromous acid. Hydrobromous acid isn't actually a thing. Technically, the experiment used to distinguish between the alkane and alkene series actually uses bromine water. This is where the colloquial name "bromine water experiment" is derived from.

For the sake of having an alkene, I will use hex-1-ene, with formula C6H12(l).

The HSC (generally) accepts your usage of bromine itself in the equation you use:
Alkene: C6H12 + Br2 -> C6H12Br2

But the ACTUAL equation of what goes on, is believe it or not, that with the bromine water
Alkene:  C6H12 + HOBr -> C6H12BrOH

Realistically speaking, the second equation would be more accurate.

However, in terms of what bromine water HOBr actually is? That is beyond the scope of the syllabus. It is merely a tool for us to use in our Production of Materials topic.

Aside:
Naturally, we would use hexane as our contrast for the experiment:
Alkane: C6H14 + Br2 -(UV)-> C6H12Br2 + H2
Alkane: C6H14 + HOBr -(UV)-> C6H12BrOH + H2
Post by: qwerty222 on March 18, 2016, 04:08:44 pm
Hey qwerty222!

Jacaranda, definitely jacaranda.
It has the most detailed information on each dotpoint, a whole load of content, if you make notes out of the textbook and be familiar with them, you will be very familiarised with all your modules. The two disadvantages are its cost (60-70 dollars l remember) and the overwhelming amount of content which may bore you. If you want to obtain a fast grasp onto the chemistry knowledge I will recommend Excel chemistry textbook because it has all the information you would need for each dotpoint and present it in very succint manner. But of course it is not as extensive as jacaranda. I havent really used any other textbooks as just yet, but l do recommend these two textbooks, especially jacaranda.

Best Regards
Happy Physics Land

Thank you! :D
Post by: katherine123 on March 18, 2016, 05:46:38 pm
I have asked my teacher about moderately weak/strong acids/base producing moderately conjugates and she told me that I would not come across those except the strong and weak acids/base. Is this true? She also told me to assume CH3COOH is a weak acid and produces strong conjugate in the exam.

Post by: Happy Physics Land on March 18, 2016, 06:13:46 pm
I have asked my teacher about moderately weak/strong acids/base producing moderately conjugates and she told me that I would not come across those except the strong and weak acids/base. Is this true? She also told me to assume CH3COOH is a weak acid and produces strong conjugate in the exam.

Hey Katherine:

If you have a look at past hsc problems, they are usually pretty straight forward. In 2009 multiple choice question 7, there was a question on the conjugate base of HSO4-. In 2014 question 10 there was a question on identifying the conjugate pair. So generally there wont be a lot of questions on solely conjugate pairs, but they usually come with buffers in our systems. For me I would just prepare for everything and yes you should write that CH3COOH produces strong conjugate in the exam but say that its strength is relative to water
Post by: DrShellgon on March 18, 2016, 08:10:29 pm
Hi, how does shaking a can of soda release CO2 in relation to the equilibrium concept?
Post by: sire123 on March 18, 2016, 10:00:41 pm
Hey could anyone explain q23 of 2013 chem hsc paper?
23a) is what Im after, cheers.
Post by: RuiAce on March 18, 2016, 10:05:03 pm
I have asked my teacher about moderately weak/strong acids/base producing moderately conjugates and she told me that I would not come across those except the strong and weak acids/base. Is this true? She also told me to assume CH3COOH is a weak acid and produces strong conjugate in the exam.
If your teacher said to assume that then even if it's incorrect go with what she said. Because that is what will get you marks in the exam
Post by: RuiAce on March 18, 2016, 10:11:58 pm
Hi, how does shaking a can of soda release CO2 in relation to the equilibrium concept?

Consider the reaction CO2(g) + H2O(l) ⇌ H2CO3(aq)   ∆H = -'ve

As we know (from being taught), the dissolution of carbon dioxide is indeed exothermic. That is, the more carbon dioxide dissolves, the more heat gets released.

When we are shaking a can of soda, we are providing the entire system with kinetic energy. As you would expect, since this is a closed environment some of this kinetic energy is transformed into heat energy. What is heat energy? Heat.

As we introduce heat into the system, according to Le Chatelier's principle the system will readjust it's equilibrium to counteract the disturbance in the system. In this case, as heat is released, when more heat is present the reverse reaction will be favoured. Effectively, this produces a greater concentration of CO2
Post by: RuiAce on March 18, 2016, 10:25:26 pm
Hey could anyone explain q23 of 2013 chem hsc paper?
23a) is what Im after, cheers.

Please copy and paste the actual question accompanying the reference to the HSC next time. :)

Our data sheet positions the reduction of silver at a much lower point than the reduction of lead. From this, we can infer that lead is indeed a more reactive metal than silver. Thus, we can anticipate that a metal displacement reaction will happen, where lead takes the place of silver. The equation can be determined easily (keeping in mind that the typical valence for lead is 2+) as follows:

Pb(s) + 2 AgNO3(aq) -> Pb(NO3)2(aq) + 2 Ag(s)

n(Pb) = m/M = 20.72/207.2 = 0.1 mol

Initial moles of silver nitrate:
n(AgNO3) = C*V = 0.1 * 1 = 0.1 mol

Observe the mole ratio:
1 mol of Pb reacts with 2 mol of AgNO3.
Obviously we don't have 2 mol of AgNO3 here, we only have 0.1. Let's fix up our mole ratio a bit then.

0.05 mol of Pb reacts with 0.1 mol of AgNO3.

This looks better.

Now, firstly observe that after 0.05 mol of Pb reacts, we still have a remaining 0.05 mol. Hence, n(Pb) = 0.0500 mol. That's another part of the question done.
(Take note that because the minimum amount of significant figures the question gives is 4, we must answer with 4 s.f.)
Then, observe immediately after that all of our moles of AgNO3 must've reacted (this is obviously not an equilibrium reaction, so everything reacts). Hence, n(AgNO3) = 0 mol
Obviously, 1 mol of Ag+ is present in 1 mol of AgNO3 as silver nitrate is only composed of one silver ion!
Thus, n(Ag+) = 0 mol

Now, the yield for 1 mol of Pb and 2 mol of AgNO3 is
1 mol of Pb(NO3)2, and 2 mol of Ag

This just means:
The yield for 0.05 mol of Pb and 0.1 mol of AgNO3 is
0.05 mol of Pb(NO3)2, and 0.1 mol of Ag

This directly tells us that the moles of Ag we have is:
n(Ag) = 0.100 mol

But observe that 1 mol of Pb(NO3)2 yields:
1 mol of Pb2+, and
2 mol of NO3-

Hence, n(Pb2+) = 0.0500 mol
and, n(NO3-) = 0.100 mol
Post by: sire123 on March 18, 2016, 10:39:40 pm
Orrite ill upload the question next time as well! Thanks Rui, but are NO3- ions the spectator ions?
And would this be called a neutral species equation?
Post by: liiz on March 19, 2016, 01:08:01 am
Hello  :)
I was just wondering whether I could please get some advice on how to structure and approach this question: "Assess current developments in the use of biopolymers (5 marks)". I'm not sure whether I'm meant to just refer to one biopolymer and its use or kind of just a general idea about it all and provide examples? Thanks!!
Post by: RuiAce on March 19, 2016, 06:31:27 am
Orrite ill upload the question next time as well! Thanks Rui, but are NO3- ions the spectator ions?
And would this be called a neutral species equation?

Certainly. To verify that, take note that the initial moles and final moles of the nitrate ions are both 0.100 mol.

And no. A neutral species equation is one which does not involve any changes in the charge. Note that this would actually be a redox reaction, where lead is being oxidised and silver ions are being reduced.
Post by: RuiAce on March 19, 2016, 06:33:16 am
Hello  :)
I was just wondering whether I could please get some advice on how to structure and approach this question: "Assess current developments in the use of biopolymers (5 marks)". I'm not sure whether I'm meant to just refer to one biopolymer and its use or kind of just a general idea about it all and provide examples? Thanks!!

There are factors about biopolymers (e.g. renewability, possible feasible properties) that hold true for all biopolymers. However, in the scope of the chemistry course you will only be asked to make specific reference to one. It may also help to contrast to traditional petroleum-derived polymers such as the obvious polyethylene.

The two most common ones I've found in the HSC are polyhydroxybutyrate and polylactic acid
Post by: sire123 on March 19, 2016, 01:03:15 pm
And also, I have another q:
A chemist analysed aspirin tablets for quality control. The initial step of the analysis was the standardisation of a NaOH solution. Three 25.00 mL samples of a 0.1034 mol L–1 solution of standardised HCl were titrated with the NaOH solution.The average volume required for neutralisation was 25.75 mL.
(a) Calculate the molarity of the NaOH solution.
...
...
...
...
Three flasks were prepared each containing a mixture of 25 mL of water and 10 mL
of ethanol. An aspirin tablet was dissolved in each flask. The aspirin in each solution
was titrated with the standardised NaOH solution according to the following equation:
C9H8O4(aq) + NaOH(aq) → C9H7O4Na(aq) + H O(l) 2
The following titration results were obtained.
Tablet Volume (mL)
1 16.60
2 16.50
3 16.55
(b) (i) Calculate the average mass (mg) of aspirin per tablet.
...
...
...
...
...
...
...
(ii) Why was it necessary to include the ethanol in the mixture?
...
...
Post by: missrissole on March 19, 2016, 06:20:51 pm
Heyyyaaa there!! I have a Question that has been bugging me.. would love to get your advice / help with it! :))

Q: Why would the neutralisation of HCl and Na2CO3 require the addition of methyl orange indicator? I thought the best suited indicator would be bromothymol blue because this titration involved a strong base and a strong acid - hence a neutral salt? (or is Na2CO3 a weak base)?

Got me reeaaallll caught up. Thanks in advance!  8)
Post by: RuiAce on March 19, 2016, 06:43:24 pm
And also, I have another q:
A chemist analysed aspirin tablets for quality control. The initial step of the analysis was the standardisation of a NaOH solution. Three 25.00 mL samples of a 0.1034 mol L–1 solution of standardised HCl were titrated with the NaOH solution.The average volume required for neutralisation was 25.75 mL.
(a) Calculate the molarity of the NaOH solution.
...
...
...
...
Three flasks were prepared each containing a mixture of 25 mL of water and 10 mL
of ethanol. An aspirin tablet was dissolved in each flask. The aspirin in each solution
was titrated with the standardised NaOH solution according to the following equation:
C9H8O4(aq) + NaOH(aq) → C9H7O4Na(aq) + H O(l) 2
The following titration results were obtained.
Tablet Volume (mL)
1 16.60
2 16.50
3 16.55
(b) (i) Calculate the average mass (mg) of aspirin per tablet.
...
...
...
...
...
...
...
(ii) Why was it necessary to include the ethanol in the mixture?
...
...

a)
NaOH(aq) + HCl(aq) -> H2O(l) + NaCl(aq)
The reaction immediately shows that the mole ratio between sodium hydroxide and hydrochloric acid is one to one.
n(NaOH) = n(HCl)
Using n=CV
[NaOH] * V(NaOH) = [HCl] * V(HCl)
[NaOH] * 25.75mL = 25.00mL * 0.1034 mol L-1
[NaOH] = 0.100388... mol L-1
Note that for this part of the question you would round to 4 significant figures and give your answer as 1.004*10-1 mol L-1 but for the remainder of the question you use the UNROUNDED answer in your calculator

b) (i) This can obviously be done by inspection, but as the HSC requires proper calculations the average is obviously
V(NaOH) = (16.60+16.50+16.55)mL/3 = 0.01655L
Important aside: It's not that easy to copy and paste a table. In the actual HSC question, we assume that the tablet is the same, but rather human error caused the differences in the calculated volumes for NaOH that it was titrated against. So the average volume is the volume of NaOH used for titration.

Observe that the mole ratio is once again one-to-one. This means
n(C9H8O4)
= n(NaOH)
= [NaOH] * V(NaOH)
= answer from part a) * 0.01655
= 1.6614278...*10-3 mol

Using the fact that m = n*MM
m = 1.6614278...*10-3 * 180.154
= 0.2993g (correct to 4 s. f. here)

b) (ii) The question was generous here to make this a one mark question. And so they should, because they didn't give us a diagram of what aspirin looks like on the molecular level.

So the only logical answer to using ethanol in the mixture is to promote the solubility of aspirin (that's the answer). Just by looking at C9H8O4 we can make a few assumptions:
a) There's quite a fair few carbons. Expect some kind of a chain, or benzene ring (tbh, yes benzene ring in the actual molecule) which ethanol will easily dissolve.
b) The 4 oxygens could all form -OH groups for all we know but obviously we don't know. Even then, both water AND ethanol interact through hydrogen bonding and dipole-dipole interactions when hydroxyl groups are present, so you can still safely assume that ethanol increases solubility.

But basically, it's cause there's obviously going to be carbon chains, which essentially makes ethanol a more effective solvent.
Heyyyaaa there!! I have a Question that has been bugging me.. would love to get your advice / help with it! :))

Q: Why would the neutralisation of HCl and Na2CO3 require the addition of methyl orange indicator? I thought the best suited indicator would be bromothymol blue because this titration involved a strong base and a strong acid - hence a neutral salt? (or is Na2CO3 a weak base)?

Got me reeaaallll caught up. Thanks in advance!  8)

Na2CO3 is of course, a weak base. The only strong bases in the course are essentially NaOH, KOH and all the other group 1 hydroxides.
Post by: Happy Physics Land on March 19, 2016, 07:16:15 pm
Hello  :)
I was just wondering whether I could please get some advice on how to structure and approach this question: "Assess current developments in the use of biopolymers (5 marks)". I'm not sure whether I'm meant to just refer to one biopolymer and its use or kind of just a general idea about it all and provide examples? Thanks!!

Hey liiz:

The way I would first approach this question is to highlight key words, its really useful because it tells you what to focus on. Assess tells us that we are making a judgment, current developments tells us we need to perhaps provide some statistics or relate the emergence of biopolymers to what is happening in the world right now. Use of biopolymers should tell us that we should provide two examples of uses of biopolymers (relate it to the polymer's properties). A lot of people would start to talk about how a specific biopolymer is made, and explain the complicated chemistry behind the production process and its properties. These descriptions however wouldnt reward you any marks because our focus is "current development" and "uses of biopolymer".

Personally, this is how I would structure this response, but of course there are many approaches that you can adopt.

1. Evaluate the current development (i.e. something like "use of biopolymers in commodities have undergone significant growth over the recent years (insert a statistic if possible))
2. Provide examples of uses of biopolymers (2 examples, include specific examples if possible, e.g. because of its rigid property, PLA has been extensively used in Danone yoghurt cups)
3. Describe the reasons in the current world which have encouraged the development of biopolymers (e.g. the depletion of the non-renewable petroleum have resulted in the rising in price of fossil fuels, making the production of biopolymers increasingly economical)
4. List 3 advantages of biopolymers in general
5. Assess the limitations of biopolymers (e.g. despite these positive aspects about biopolymers, companies are currently experiencing some difficulties with manufacturing them in large amounts and the process of production biopolymers is still not as efficient as using petrochemicals.)
6. Restate your assessment of the development of biopolymers (Judgment: There has a been a rapid and significant current development of biopolymers. I know this sounds a bit clumsy, but by writing this, the teacher knows you are achieving the "assess" part of the question)

Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 19, 2016, 07:25:30 pm
Heyyyaaa there!! I have a Question that has been bugging me.. would love to get your advice / help with it! :))

Q: Why would the neutralisation of HCl and Na2CO3 require the addition of methyl orange indicator? I thought the best suited indicator would be bromothymol blue because this titration involved a strong base and a strong acid - hence a neutral salt? (or is Na2CO3 a weak base)?

Got me reeaaallll caught up. Thanks in advance!  8)

Hey missrissole!

Completely agree with what Rui said, just want to drop in my two pennies here. Na2CO3 definitely is a weak base, and like what Rui said, majority of the strong bases come from hydroxides of group 1 alkali metals and group 2 alkaline earth metals. Since HCl is a strong acid, the pH of reaction mixture at equivalence point would be below 7 (i.e. acidic). Hence methyl orange would be the best indicator here because it is suitable for use in the lower acidic range. Bromothymol blue is useful indicator for strong acid/strong base titration because the equivalence point of the reaction has a pH of 7. Bromothymol blue displays a green colour for pH range 6.0 - 7.6, hence it would be useful for evaluating neutral equivalence points.

Best Regards
Happy Physics Land
Post by: DrShellgon on March 21, 2016, 09:31:04 pm
Hi, I was wondering, why do pure liquids not affect equilibrium? Isn't there a possibility that the pure liquid, say it's like a pure acid, could react with the reactants or the products in an equilibrium? How about pure water? Won't this potentially dilute either the reactants or the products, causing a change in concentration and causing the equilibrium to shift?

How about pure solids? Why do pure solids not affect equilibrium? E.g. what if we put in a block of pure sodium, or lithium, won't it react with a particular side of the equilibrium and cause some sort of shift?

Post by: RuiAce on March 21, 2016, 09:35:15 pm
Hi, I was wondering, why do pure liquids not affect equilibrium? Isn't there a possibility that the pure liquid, say it's like a pure acid, could react with the reactants or the products in an equilibrium? How about pure water? Won't this potentially dilute either the reactants or the products, causing a change in concentration and causing the equilibrium to shift?

How about pure solids? Why do pure solids not affect equilibrium? E.g. what if we put in a block of pure sodium, or lithium, won't it react with a particular side of the equilibrium and cause some sort of shift?

Solids don't carry much kinetic energy for them to start reacting with other things by themselves. All that you do by adding a solid is somewhat alter the surface area for the solid to react; solid molecules aren't going to bump into other liquid/gaseous/aqueous substances just like that.

Now, what exactly do you mean by pure liquids? Do you mean something such as 18M sulfuric acid (H2SO4(l))?
__________________

In terms of dilutions.

Aqueous solution substances are called that because, they are, dissolved water. All water does is reduce the concentration of the substance (out of the definition of dilution).
Post by: liiz on March 22, 2016, 06:26:30 pm
hello :) wondering if someone could please help explain this question to me "Explain why ethanol will dissolve in water and in pentane?" - I know it's a simple question but I'm still struggling to understand the whole polar and non-polar concept! Thanks so much
Post by: Johny1234567 on March 22, 2016, 07:01:36 pm
hello :) wondering if someone could please help explain this question to me "Explain why ethanol will dissolve in water and in pentane?" - I know it's a simple question but I'm still struggling to understand the whole polar and non-polar concept! Thanks so much
Firstly, hydrocarbons are non-polar substances. Water are polar substances. Remember the theory behind 'like dissolves like', i.e polar substances dissolve other polar substances and vice versa.

Ethanol, if you draw out the structure, has a non-polar carbon end, as well as a polar hydoxyl functional group. Because ethanol has both non-polar and polar-ends, it will dissolve in water (polar) and pentane (non-polar).
Post by: Johny1234567 on March 22, 2016, 07:05:14 pm
Attached is a question in chemical monitoring and management. I've done up to e) but am stuck on f). Could someone please help me c:

P.S Not entirely necessary but would prefer if someone completed the entire question so I can confirm my answers.

Thanks!
Post by: Happy Physics Land on March 22, 2016, 07:26:49 pm
hello :) wondering if someone could please help explain this question to me "Explain why ethanol will dissolve in water and in pentane?" - I know it's a simple question but I'm still struggling to understand the whole polar and non-polar concept! Thanks so much

Hey Liiz!

Sure thing! Ethanol will dissolve in water because of its hydrophilic (means water-liking) -OH group which allows it to interact with water molecules through hydrogen bonding. Ethanol will dissolve in pentane because of its non-polar hydrocarbon group which is able to interact with other non-polar substances such as pentane through dispersion forces.

This would usually be a 2 mark question and if you just write what I wrote above in the exam they are easy 2 marks to get! If you want me to expand on what I said, please dont hesitate to tell me! :)

Best Regards
Happy Physics Land
Post by: curious.egg on March 22, 2016, 08:33:03 pm
Hellooo!!

For the prac where we had to make a natural indicator, my school used purple cabbage and in the procedure, we added salt and ethanol when we were crushing it. Why was salt and ethanol added? Please help!
Post by: RuiAce on March 22, 2016, 08:48:41 pm
Attached is a question in chemical monitoring and management. I've done up to e) but am stuck on f). Could someone please help me c:

P.S Not entirely necessary but would prefer if someone completed the entire question so I can confirm my answers.

Thanks!
I have never seen the term "reaction efficiency" coined before in the HSC. Such a question will not be asked in this course.

a) Hydrogen and nitrogen introduced into the system. According to the reversible equation of the production of ammonia, as we initially has a concentration of ammonia of 0 mol L-1 it's unsurprising that ammonia begins to get produced. The system then reaches equilibrium.

N2(g) + 3 H2(g) ⇌ 2 NH3(g)

b) and c) A downward spike in the concentration of all 3 substances suggests that the volume of the system has been increased, which is equivalent to a decrease in pressure. As there are 4 moles of gas on the left and 0 on the right, LCP predicts that the equilibrium shifts to the left, which explains the gradual increase in the concentrations of nitrogen and hydrogen.

d) and e) An upward spike in only the concentration of hydrogen means that excess hydrogen was introduced into the system. Trivially, LCP predicts that the equilibrium will shift to the right to minimise this disturbance.

Post by: RuiAce on March 22, 2016, 08:50:07 pm
Hellooo!!

For the prac where we had to make a natural indicator, my school used purple cabbage and in the procedure, we added salt and ethanol when we were crushing it. Why was salt and ethanol added? Please help!

This is a highly specific question to which you will have to ask your teacher to determine what exactly were the impacts of the substances.

I performed this experiment using red cabbage and I did not require the addition of any of these reagents.
Post by: DrShellgon on March 22, 2016, 10:39:16 pm
Solids don't carry much kinetic energy for them to start reacting with other things by themselves. All that you do by adding a solid is somewhat alter the surface area for the solid to react; solid molecules aren't going to bump into other liquid/gaseous/aqueous substances just like that.

Now, what exactly do you mean by pure liquids? Do you mean something such as 18M sulfuric acid (H2SO4(l))?
__________________

In terms of dilutions.

Aqueous solution substances are called that because, they are, dissolved water. All water does is reduce the concentration of the substance (out of the definition of dilution).

Yep, by pure liquids, I meant adding reagants like the H2SO4 you mentioned, I'm just wondering, won't it react with a particular side of the equilibrium and thus cause a shift?

Also with adding just normal water to the equilibrium, am I correct to assume that since it's just water, the dilution it causes to both sides of the equilibrium is equal and thus it will overall have no effect on the equilibrium, similar to how a catalyst, when added, simply speeds up the rate of reaction and has no overall effect on the equilibrium?
Post by: Sine on March 23, 2016, 12:07:01 am
experimental procedure errors for a back titration?

is incorrect identification of end point okay (and If you have any others)
Post by: RuiAce on March 23, 2016, 07:58:56 am
Yep, by pure liquids, I meant adding reagants like the H2SO4 you mentioned, I'm just wondering, won't it react with a particular side of the equilibrium and thus cause a shift?

Also with adding just normal water to the equilibrium, am I correct to assume that since it's just water, the dilution it causes to both sides of the equilibrium is equal and thus it will overall have no effect on the equilibrium, similar to how a catalyst, when added, simply speeds up the rate of reaction and has no overall effect on the equilibrium?

Ok I had to do a bit of research to determine a formal answer.

The whole basis of Le Chatelier's Principle and all factors of dynamic equilibrium do not strictly relate to the moles of the substance, rather, the concentrations of the substances.

Effectively, gas concentrations can be measured as they will essentially be the moles of the substance over the volume of the vessel.
An aqueous substance's concentration can also be measured as it is the moles of the substance over the volume of the entire fluid they're dissolved in.

Solids and liquids, on the other hand, cannot be defined in terms of a concentration. You can add more moles of solid or liquid to a system, but this doesn't change their concentrations. The reason for this is, what is there to change the concentration with respect to?
- For solids, just adding more of them in really doesn't do much. They'll just sit there.
- For liquids, if you add more liquid what can you take the concentration of liquid with respect to? You increase the moles of the liquid, but you don't have something to take a volume with respect of, to find a concentration.

If your school elects to take up the industrial chemistry option topic, you will also be introduced to the equilibrium constant of a system, and find that in this constant (determines the ratio of reactants to products) no solid or liquid is included either.
Post by: RuiAce on March 23, 2016, 08:04:06 am
experimental procedure errors for a back titration?

is incorrect identification of end point okay (and If you have any others)
Back titration specifically is not something important in the HSC course.

Due to the nature of what it is compared to regular titration, the experimental procedure errors are similar to that of normal titration. This would include the incorrect identification of the endpoint as well as anything else (problematic method - equipment washed wrongly).
Post by: Maz on March 23, 2016, 09:41:18 pm
hey, can i please get some help with this question?
when you suffer from an acid stomach, the fluid in your stomach can contain hydrochloric acid at
a concentration of 2.5x10-4 mol L-1
to reduce the amount of acid in your stomach you are advised to take an antacid tablet that contains 5mg of aluminium hydroxide and 5mg of magnesium hydroxide. if at this time you have 1.5L of fluid in your stomach calculate the concentration of hydrogen ions and hydroxide ions in you stomach after you have taken the tablet.
if you took a second antacid tablet what would the hydrogen and hydroxide ion concentration be now?
i'd really appreciate the help...
thankyou :)
Post by: katherine123 on March 24, 2016, 06:45:09 pm
Hi can u check my 6 marks response? question is attached below

Le Chatalier’s principle applies in step 1 and 2 since reactions are in equilibrium, but it is not applied in step 3 since reaction goes to completion.
At step 1, low pressure and low temperature will cause reaction to shift towards the formation of NO(g). When heat is removed, reaction, being exothermic, will shift to the right to make up some of the reduced heat. When pressure is reduced, the reaction shifts right as 9 moles of gas react to form 10 moles of gas which results in increase in pressure.

The answer include “these conditions reduce the rate of reaction”  which I don’t really understand

At step 2, the production of NO2(g) is favoured at low temperature and high pressure. This is because the removal of heat will cause the exothermic reaction to shift right to counteract some of the lost heat, and increase in pressure cause reaction to shift right since 3 moles of gas reacts to form 2 moles of gas which reduce some of the increased pressure.
At step 3, high pressure and high temperature will favour the formation of HNO3. The increase in pressure also increases the collision between gas molecules and water, and increase in temperature will increase the average kinetic energy of particles which leads to more successful collisions, hence reaction will shift right.
I don’t get this bit included in the answer  “as this step is not an equilibrium, the highest yield and rate of formation of nitric acid will occur when reactants undergo more collisions per second. “

-Doesn’t change in pressure, temperature, concentration or volume have no effect on reactions that go to completion?
-Pressure only applies for gas molecules right?
Post by: Happy Physics Land on March 24, 2016, 07:54:52 pm
Hi can u check my 6 marks response? question is attached below

Le Chatalier’s principle applies in step 1 and 2 since reactions are in equilibrium, but it is not applied in step 3 since reaction goes to completion.
At step 1, low pressure and low temperature will cause reaction to shift towards the formation of NO(g). When heat is removed, reaction, being exothermic, will shift to the right to make up some of the reduced heat. When pressure is reduced, the reaction shifts right as 9 moles of gas react to form 10 moles of gas which results in increase in pressure.

The answer include “these conditions reduce the rate of reaction”  which I don’t really understand

At step 2, the production of NO2(g) is favoured at low temperature and high pressure. This is because the removal of heat will cause the exothermic reaction to shift right to counteract some of the lost heat, and increase in pressure cause reaction to shift right since 3 moles of gas reacts to form 2 moles of gas which reduce some of the increased pressure.
At step 3, high pressure and high temperature will favour the formation of HNO3. The increase in pressure also increases the collision between gas molecules and water, and increase in temperature will increase the average kinetic energy of particles which leads to more successful collisions, hence reaction will shift right.
I don’t get this bit included in the answer  “as this step is not an equilibrium, the highest yield and rate of formation of nitric acid will occur when reactants undergo more collisions per second. “

-Doesn’t change in pressure, temperature, concentration or volume have no effect on reactions that go to completion?
-Pressure only applies for gas molecules right?

Hey Katherine!

A very good answer, pretty much encompasses everything. You have provided the correct answers and correct reasoning. If I am to recommend anything, I would suggest putting in "According to Le Chatelier's principle" every time you refer to an equilibrium shift and at the end of your reasoning just put "... to minimise the disturbance". If I am to be picky, I would be careful with your final statement in step 3 where you stated "hence the reaction will shift right". Step three goes to completion, hence the reaction would always be going to the right, it would be considered incorrect to state "shift right" because this would indicate that you are considering this equation as an equilibrium reaction.

1. To answer your first concern, when you decrease your temperature (i.e. if you put the equilibrium mixture in a cold condition), the particles would be deprived of their kinetic energy and hence there will be less effective collisions. As we know from year 11 chemistry, rate of reaction is dependent upon the amount of effective collisions that take place, hence decrease in temperature would reduce rate of reaction. Less pressure (i.e. increase in volume) would provide a larger amount of space for gas molecules to flow around, hence decrease the chance of effective collision taking place (imagine yourself in a crowded party that takes place in a small space, there is a high chance you would walk into someone. Now imagine yourself in the same party with same amount of people in a much larger space, it is less likely that you would walk into someone). Hence decreased pressure would also cause reduced rate of reaction.

2. You have essentially said it in your answer without realising :D ! You have correctly state that rise in temperature and rise in pressure would both promote effective collisions and with more effective collisions occurring between reactants, more products will be formed and the rate of reaction will increase! Temperature rise and pressure certainly changes the position of equilibrium, but they apply universally to all reactions! You may like to have a look at your year 11 content on the factors that influence the rates of reactions. Think about when we havent learnt equilibrium reactions back in year 11, and how we applied those factors such as concentration, pressure and temperature to complete reactions.

3. In our HSC chemistry, we can assume that pressure only applies for gases, because it has the greatest influence on gases than any other states of matter (However in reality, pressure do have an impact upon liquids. This is what we call hydraulic pressure and is a dominant concept in mechanical and automotive industries. But in this case you dont have to worry about this too much.) Because we increased pressure, gas particles would be more concentrated in a smaller amount of space and hence it's not the water molecules that are colliding into gas particles, but it is the gas particles that are colliding into water molecules to generate the products. If you wish, you may also mention that there is some impact of pressure upon water molecules, but for the sake of HSC, you dont have to include it.

Hope my explanation was clear! If you have further questions please dont hesitate to ask! :D

Best Regards
Happy Physics Land
Post by: amandali on March 24, 2016, 08:05:42 pm
Hi i dont really understand how to determine suitable indicator to use for solutions.
Post by: Happy Physics Land on March 24, 2016, 08:57:42 pm
Hi i dont really understand how to determine suitable indicator to use for solutions.

Hey Amanda!

If we have a look at the indicator. It is able to distinguish between strong acids and weak acids (0.1M HCl reflects a pink colour, 0.001M HCl reflects a pale yellow colour), however it is unable to differentiate between neutral substances and bases (colour observed for distilled water is the same as the colour observed for bases). This means, that any reactions that would have an equivalence point of 7 or more than 7 would be unsuitable because there wont be any colour change to indicate whether neutrality has been achieved or basicity has been achieved. Hence we can first eliminated B) which is a strong acid + strong base reaction. This is going be neutral and hence the indicator wont be suitable. We can eliminate C) because this is a weak acid + weak base reaction, which will also yield a pH around 7 (be mindful however, it is difficult to conduct titration with weak acid + weak base due to a lack of distinct colour change, so in fact no indicator would be very suitable). We can also eliminate D) because this is a weak acid + strong base reaction which would yield an equivalence point above pH 7. As mentioned before, because there is no colour change to distinguish between neutral equivalence point and basic equivalence point, therefore D would also be unsuitable.

So this leaves us with A as our answer. However, before we put that down as our answer, we must also check whether we can determine the equivalence point of A using our indicator. Clearly, there is a colour change from strong acid to weak acid. Hence we are able to determine the equivalence point around pH 3 - 6 (i.e. weak acid pH range). Since A) is a strong acid + weak base reaction, its equivalence point would be around pH 3 - 6 (weak acid pH range) and hence A is our final answer.

Great question! Hope you understood my explanation! If you are still a little confused about some of my explanations please dont hesitate to ask! :)

Best Regards
Happy Physics Land
Post by: amandali on March 24, 2016, 09:46:35 pm
Whats the difference between PHB and PHB-V? Do i need to know how to draw the structure of these polymers ?
Post by: katherine123 on March 25, 2016, 11:23:18 am
Hi can u check my response
Is it too long and how do you allocate marks?

Question: 7 marks With reference to the underlying chemistry and with relevant equations, assess the impacts on society of two uses of ethanol

Ethanol is a renewable resource produced by fermentation of sugary crops such as sugar can and corn
C6H12O6(aq) --> 2C2H6O(aq) + 2CO2(g)
Carbon dioxide and water is released when ethanol is combusted hence it is a useful car fuel.
C2H6O(l) + O2(g)--> 2CO2(g) +3H2O(l)

Theoretically, the carbon dioxide released from combustion of ethanol balances with the CO2 absorbed during photosynthesis. Hence, ethanol is considered carbon neutral as it does not lead to increase in CO2 in the atmosphere, thus prevents global warming and has a significant positive impact on society.

However, fossil fuels are used during the farming process, distillation and distribution of ethanol. Hence, it does lead to a net increase in CO2 into the atmosphere. Moreover, production of ethanol requires the clearing of vast areas of arable land which can impact on food production, causes environmental problems like soil erosion and salinity and destructs animal habitats.

Weighing up the pros and cons, using ethanol as fuel has a negative impact on society as it does lead to increase in CO2 when the energy output of its production is taken into consideration which can elevate global warming and in addition it requires clearing of arable land which also leads to severe environmental problems.

Ethanol is also used as solvent for polar substances like water, and non-polar substances like oil and grease. It has non-polar region (-C2H3) which interacts with the non-polar substances via dispersion forces and also has a polar-region (OH-) which forms hydrogen bonds or dipole-dipole interact with polar substances.
(including a diagram  of ethanol bonding with polar/non-polar sub)

Therefore, ethanol is used as a solvent in cosmetic, perfumes and paints which create a positive impact on society by making lives easier with these products being available for use.

However, ethanol is derived from petroleum involving the process of cracking and dehydration. The mining of petroleum is environmentally damaging as it can cause problems like contamination of soil, erosion and oil leaks. Moreover, ethanol is volatile and flammable thus explosion may occur when it comes into contact with heat or flames if it is not stored properly which pose a severe risk to society.

Weighing up the pros and cons, ethanol is not good to be used as solvents as it poses a great risk to society due to its flammability which outweighs its positive impact of making lives easier.
Post by: katherine123 on March 25, 2016, 11:25:12 am
Hey Katherine!

A very good answer, pretty much encompasses everything. You have provided the correct answers and correct reasoning. If I am to recommend anything, I would suggest putting in "According to Le Chatelier's principle" every time you refer to an equilibrium shift and at the end of your reasoning just put "... to minimise the disturbance". If I am to be picky, I would be careful with your final statement in step 3 where you stated "hence the reaction will shift right". Step three goes to completion, hence the reaction would always be going to the right, it would be considered incorrect to state "shift right" because this would indicate that you are considering this equation as an equilibrium reaction.

1. To answer your first concern, when you decrease your temperature (i.e. if you put the equilibrium mixture in a cold condition), the particles would be deprived of their kinetic energy and hence there will be less effective collisions. As we know from year 11 chemistry, rate of reaction is dependent upon the amount of effective collisions that take place, hence decrease in temperature would reduce rate of reaction. Less pressure (i.e. increase in volume) would provide a larger amount of space for gas molecules to flow around, hence decrease the chance of effective collision taking place (imagine yourself in a crowded party that takes place in a small space, there is a high chance you would walk into someone. Now imagine yourself in the same party with same amount of people in a much larger space, it is less likely that you would walk into someone). Hence decreased pressure would also cause reduced rate of reaction.

2. You have essentially said it in your answer without realising :D ! You have correctly state that rise in temperature and rise in pressure would both promote effective collisions and with more effective collisions occurring between reactants, more products will be formed and the rate of reaction will increase! Temperature rise and pressure certainly changes the position of equilibrium, but they apply universally to all reactions! You may like to have a look at your year 11 content on the factors that influence the rates of reactions. Think about when we havent learnt equilibrium reactions back in year 11, and how we applied those factors such as concentration, pressure and temperature to complete reactions.

3. In our HSC chemistry, we can assume that pressure only applies for gases, because it has the greatest influence on gases than any other states of matter (However in reality, pressure do have an impact upon liquids. This is what we call hydraulic pressure and is a dominant concept in mechanical and automotive industries. But in this case you dont have to worry about this too much.) Because we increased pressure, gas particles would be more concentrated in a smaller amount of space and hence it's not the water molecules that are colliding into gas particles, but it is the gas particles that are colliding into water molecules to generate the products. If you wish, you may also mention that there is some impact of pressure upon water molecules, but for the sake of HSC, you dont have to include it.

Hope my explanation was clear! If you have further questions please dont hesitate to ask! :D

Best Regards
Happy Physics Land

Does yield apply for reactions in equilibrium   If not, why?
Post by: Happy Physics Land on March 25, 2016, 07:02:15 pm
Whats the difference between PHB and PHB-V? Do i need to know how to draw the structure of these polymers ?

Hey Amanda!

Sorry for the late reply! PHB and PHBV are both common biopolymers that possess contrasting physical properties! PHB is otherwise known as 3-hydroxybutyrate, which is stiff and brittle. Because of these properties, PHB have limited usage in commodities and hence we dont usually discuss about PHB. In contrast, PHBV, or biopol, is stronger and more flexible. Because of these desirable physical properties, PHBV have widespread use in medical implants (prosthetic devices) , specialty packaging and tissue repairing. The structure of these polymers are beyond the scope of chemistry syllabus and looking at past questions they have never asked for the structure of these, so no you wouldnt need to, but I guess you can look up the structure if you are interested and just in case they do ask you in an exam. What you definitely do need to know is the specific name of the bacteria that is involved in the production of PHB and PHBV, in this case the bacteria is called Cupriavidus Necator.

Good luck studying! :)

Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 25, 2016, 07:08:47 pm
Does yield apply for reactions in equilibrium   If not, why?

Hey Katherine!

Yes yield definitely do apply for reactions in equilibrium (as stated by le chatelier's principle, shifting of equilibrium to a certain side will favour the yield of that side), and it also applies to complete reactions as well! With the changes in temperature, changes in concentration and change in temperature, both equilibrium reactions and complete reactions will have different yields of products. You can refer to my previous answer for explanations on this! :)

Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 25, 2016, 07:37:51 pm
Hi can u check my response
Is it too long and how do you allocate marks?

Question: 7 marks With reference to the underlying chemistry and with relevant equations, assess the impacts on society of two uses of ethanol

Ethanol is a renewable resource produced by fermentation of sugary crops such as sugar can and corn
C6H12O6(aq) --> 2C2H6O(aq) + 2CO2(g)
Carbon dioxide and water is released when ethanol is combusted hence it is a useful car fuel.
C2H6O(l) + O2(g)--> 2CO2(g) +3H2O(l)

Theoretically, the carbon dioxide released from combustion of ethanol balances with the CO2 absorbed during photosynthesis. Hence, ethanol is considered carbon neutral as it does not lead to increase in CO2 in the atmosphere, thus prevents global warming and has a significant positive impact on society.

However, fossil fuels are used during the farming process, distillation and distribution of ethanol. Hence, it does lead to a net increase in CO2 into the atmosphere. Moreover, production of ethanol requires the clearing of vast areas of arable land which can impact on food production, causes environmental problems like soil erosion and salinity and destructs animal habitats.

Weighing up the pros and cons, using ethanol as fuel has a negative impact on society as it does lead to increase in CO2 when the energy output of its production is taken into consideration which can elevate global warming and in addition it requires clearing of arable land which also leads to severe environmental problems.

Ethanol is also used as solvent for polar substances like water, and non-polar substances like oil and grease. It has non-polar region (-C2H3) which interacts with the non-polar substances via dispersion forces and also has a polar-region (OH-) which forms hydrogen bonds or dipole-dipole interact with polar substances.
(including a diagram  of ethanol bonding with polar/non-polar sub)

Therefore, ethanol is used as a solvent in cosmetic, perfumes and paints which create a positive impact on society by making lives easier with these products being available for use.

However, ethanol is derived from petroleum involving the process of cracking and dehydration. The mining of petroleum is environmentally damaging as it can cause problems like contamination of soil, erosion and oil leaks. Moreover, ethanol is volatile and flammable thus explosion may occur when it comes into contact with heat or flames if it is not stored properly which pose a severe risk to society.

Weighing up the pros and cons, ethanol is not good to be used as solvents as it poses a great risk to society due to its flammability which outweighs its positive impact of making lives easier.

Hey Katherine!

Ok so let's start off breaking down the question and analysing the mark distribution:

Straight away, 2 marks for providing two equations for the two uses. (However, if your content isnt satisfactory, teacher may only award one mark for the equations because the equations are meant to be there to support your explanations. If your explanation isnt too desirable, it deprives the meaning of having an equation included).
1 mark for assessing (i.e. making a judgment)
2 marks for identifying the use of ethanol and explaining its impact on society (you may not be awarded any mark for just simply identifying the use, nor would you be awarded any mark for explaining impacts upon environment, but something you can relate to is that CO2 continuously contribute to greenhouse effect, and this will making the Earth environment inhospitable for human society.)
2 marks for identifying a second use of ethanol and explaining its impact on society.

What I do like about your response, is your frequent use of cause and effect language and your ability to make several judgments. These are evidences of you clearly understanding how to answer an explain/assess question! Well done! You have also included several social impacts and have also explained why these impacts are positive/negative. Another positive aspect of your response is the fact that you have provided the chemistry principles behind these uses, such as your successful identification of the -OH hydroxyl group and the hydrocarbon group. This answers the part "using underlying chemistry".

A fatal error (please dont be scared off by this harsh tone :) its still an articulate response!) in your writing is that you are overly focusing on environmental impacts. This is a question asking about social impacts, as soon as you mention the word environment, you are off-topic. If this question is on "society and environmental impacts", I would give 7/7 for your response, beyond all doubts. Even if you want to include environmental impacts, you will still need to ultimately relate it back to effect upon society, just like what you have done with the global warming concept. When you talked about arable land, I thought you would follow up with the idea "the use of land and agricultural products for the production of ethanol would create competition with human consumption" (this is a good idea for social impact if you wanna put it in). But instead you made a great point on environmental impact, but again it's not relevant to the question.

A small technicality error is your chemical formula for ethanol. You wrote C2H6O, but the more correct way of expressing it is C2H5OH. Sure, these two formulae have exactly the same number of hydrogen atoms, but we must not ignore the significance of the -OH hydroxyl group in ethanol, because that is what allows ethanol to dissolve in water! By writing C2H6O, you are ignoring the significance of -OH and it leaves a chance for the teacher to deduct you a mark there!

Overall, I would personally give a 5/7. Your teacher may give a 4-5/7. So essentially your response is a brand 5 range response, which is definitely good at this stage. If you apply my feedbacks, this would definitely be a band 6 response! :) Well done overall!

Best Regards
Happy Physics Land
Post by: atarz on March 25, 2016, 11:27:56 pm
I'm very confused... I have done two practice half-yearly tests from James Ruse. One said to use a stopper in the production of an ester while another said not to. Also I have seen esterification using thermometers. What should be used in esterification and why?
Post by: RuiAce on March 25, 2016, 11:51:40 pm
I'm very confused... I have done two practice half-yearly tests from James Ruse. One said to use a stopper in the production of an ester while another said not to. Also I have seen esterification using thermometers. What should be used in esterification and why?

The whole point of the condenser is to not only allow for high temperatures, but ALSO to avoid a buildup of pressure within the system. Otherwise, the system will explode. You will have glass everywhere.

The stopper, is a beautiful way to PROMOTE this buildup of pressure. Therefore it is very very very VERY bad for the process of reflux.

Now, perhaps a bit more importantly, if you look carefully at your diagram that is not a stopper. That is just the top of the condenser. I think you just got the diagram confused.

Note: The thermometer is optional. I've never refluxed with one but it's probably advantageous if you want your system to be heated to a specific temperature for reflux. It may promote safety, but I don't believe it's necessary.
Post by: atarz on March 26, 2016, 10:09:06 am
Thank you!
The whole point of the condenser is to not only allow for high temperatures, but ALSO to avoid a buildup of pressure within the system. Otherwise, the system will explode. You will have glass everywhere.

The stopper, is a beautiful way to PROMOTE this buildup of pressure. Therefore it is very very very VERY bad for the process of reflux.

Now, perhaps a bit more importantly, if you look carefully at your diagram that is not a stopper. That is just the top of the condenser. I think you just got the diagram confused.

Note: The thermometer is optional. I've never refluxed with one but it's probably advantageous if you want your system to be heated to a specific temperature for reflux. It may promote safety, but I don't believe it's necessary.
Post by: amandali on March 26, 2016, 04:38:27 pm
a) citric acid = weak, triprotic
b) acetic acid = weak, monoprotic
c)sulfuric acid=diprotic , strong
d) HCl = monoprotic strong

ive managed to rule out HCl and acetic acid  but im not sure why its citric
isnt citric really weak so its going to partially ionise producing much more less H+ than H2SO4
Post by: Happy Physics Land on March 26, 2016, 05:38:41 pm
a) citric acid = weak, triprotic
b) acetic acid = weak, monoprotic
c)sulfuric acid=diprotic , strong
d) HCl = monoprotic strong

ive managed to rule out HCl and acetic acid  but im not sure why its citric
isnt citric really weak so its going to partially ionise producing much more less H+ than H2SO4

Hey Amanda!

Well citric acid isnt REALLY weak, like l mean its still weak but its definitely not the weakest acid you will ever see. But then thats not the point in this question! Here, the amount of base required will depend on the stoichiometry of the reaction (i.e. the number of ionisable protons in the acid). Since citric acid has the most protons per mole (because its triprotic), it would require the most amount of base to change the colour of indicator.

But yes I do understand where your reasoning came from. Keep in mind however this question is about a neutralisation reaction, not so much about whether an acid is weak or strong. If we compare the balanced chemical equation for citric acid + NaOH (C6H8O7(aq) + NaOH(aq) --> NaC6H5O7(aq) + 3H2O(l)) and the balanced chemical equation for sulfuric acid + NaOH (H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2H2O(l)), we can see that 3 moles of NaOH is required to neutralise one mole of citric acid, and 2 moles of NaOH are required to neutralise one mole of citric acid. So I think this explains why the answer is citric acid not sulfuric acid more clearly for you.

If you still find it a bit hard to understand, please let me know! :)

Best Regards
Happy Physics Land
Post by: RuiAce on March 27, 2016, 01:44:56 am
a) citric acid = weak, triprotic
b) acetic acid = weak, monoprotic
c)sulfuric acid=diprotic , strong
d) HCl = monoprotic strong

ive managed to rule out HCl and acetic acid  but im not sure why its citric
isnt citric really weak so its going to partially ionise producing much more less H+ than H2SO4

Precisely what HPL was trying to say.

Because we're titrating with a strong base, the reaction goes to completion REGARDLESS of the strength of the acid.

So since the acids are both equal in volume AND equal in concentration, the triprotic one will take more NaOH to effectively be neutralised.
Post by: amandali on March 27, 2016, 09:25:17 am
Hi can u check my response?
- Is qualitative an indirect evidence and quantitative a direct evidence
- Am i allowed to omit the spectator ions for carbonate, iron reaction (below)
- is it enough to know the effects of acid rain as the only qualitative evidence
-For "discuss" questions  i dont need a concluding statement right ?

There has been an increase in the concentration of the oxides of nitrogen and sulfur in the atmosphere. Assess both the evidence to support this statement and the need to monitor these oxides  8 marks

Oxides of sulfur has increased in concentration from burning sulfide ores to extract metals CuS(s) + O2(g) -->Cu(s) + SO2(g) which releases SO2(g)
Whereas, acidic oxides of Nitrogen has increased mainly due to internal combustion engines and coal burning power stations.
N2(g) + O2(g) -->2NO(g)
2NO(g) + O2(g) -->2NO2(g)

These oxides the react with the water in the atmosphere:
SO2 (g)  + H2O(l) -->H2SO3(aq)
2NO2(g) + H2O (l)-->HNO2(aq)  + HNO3 (aq)
The evidence for the increase in these oxides comes from both direct of measurements of gas concentrations in the atmosphere and indirectly through effect of the increase in acid rain due to these oxides. The quantitative measurements are accurate and reliable due to the development of technologies in recent time. For eg. levels of SO2 and NO2 gases measured by analysing the gases trapped in bubbles of Antarctic ice which is a sample of the atmosphere from the past 200 years.
The indirect evidence comes from the effects of acid rain formed by these oxides. The acids H2SO3(aq) and HNO2(aq)/HNO3(aq) react and dissolves carbonates, which destructs historically important buildings and statues made of limestone and marble thus this becomes a record for the increase in levels of these oxides. 2H+(aq)+CaCO3(s)-->CO2(g) + H2O(l)+ Ca2+(aq) . Steel bridges and other steel structures also show evidence of damage due to acid rain since it reacts iron and dissolves it. 2H+(aq)+ Fe(s)-->H2(g)+Fe2+(aq). However, these evidences are less reliable as they could be resulted due to other oxides or factors.

It is vital to record the monitor and record increasing levels of pollution so that we will know when to act and reduce the levels of NOx and SO2 as they are respiratory irritants therefore can cause asthma and other respiratory diseases and thus have a devastating impact on population’s health. They also lead to formation of acid rain which will cause cuticle on leaves to strip off leading to degradation of forest.

Overall, the direct and indirect evidence both shows the increase in concentration of these oxides, however, they are not completely reliable as the concentration of oxides changes over time and they are not representative of all parts of the world.
Post by: Happy Physics Land on March 27, 2016, 01:05:28 pm
Hi can u check my response?
- Is qualitative an indirect evidence and quantitative a direct evidence
- Am i allowed to omit the spectator ions for carbonate, iron reaction (below)
- is it enough to know the effects of acid rain as the only qualitative evidence
-For "discuss" questions  i dont need a concluding statement right ?

There has been an increase in the concentration of the oxides of nitrogen and sulfur in the atmosphere. Assess both the evidence to support this statement and the need to monitor these oxides  8 marks

Oxides of sulfur has increased in concentration from burning sulfide ores to extract metals CuS(s) + O2(g) -->Cu(s) + SO2(g) which releases SO2(g)
Whereas, acidic oxides of Nitrogen has increased mainly due to internal combustion engines and coal burning power stations.
N2(g) + O2(g) -->2NO(g)
2NO(g) + O2(g) -->2NO2(g)

These oxides the react with the water in the atmosphere:
SO2 (g)  + H2O(l) -->H2SO3(aq)
2NO2(g) + H2O (l)-->HNO2(aq)  + HNO3 (aq)
The evidence for the increase in these oxides comes from both direct of measurements of gas concentrations in the atmosphere and indirectly through effect of the increase in acid rain due to these oxides. The quantitative measurements are accurate and reliable due to the development of technologies in recent time. For eg. levels of SO2 and NO2 gases measured by analysing the gases trapped in bubbles of Antarctic ice which is a sample of the atmosphere from the past 200 years.
The indirect evidence comes from the effects of acid rain formed by these oxides. The acids H2SO3(aq) and HNO2(aq)/HNO3(aq) react and dissolves carbonates, which destructs historically important buildings and statues made of limestone and marble thus this becomes a record for the increase in levels of these oxides. 2H+(aq)+CaCO3(s)-->CO2(g) + H2O(l)+ Ca2+(aq) . Steel bridges and other steel structures also show evidence of damage due to acid rain since it reacts iron and dissolves it. 2H+(aq)+ Fe(s)-->H2(g)+Fe2+(aq). However, these evidences are less reliable as they could be resulted due to other oxides or factors.

It is vital to record the monitor and record increasing levels of pollution so that we will know when to act and reduce the levels of NOx and SO2 as they are respiratory irritants therefore can cause asthma and other respiratory diseases and thus have a devastating impact on population’s health. They also lead to formation of acid rain which will cause cuticle on leaves to strip off leading to degradation of forest.

Overall, the direct and indirect evidence both shows the increase in concentration of these oxides, however, they are not completely reliable as the concentration of oxides changes over time and they are not representative of all parts of the world.

Hey Amanda!

1. Both qualitative and quantitative are direct evidences (not necessarily first-hand evidences though!). Qualitative requires you to describe this evidence. So for example in this case if you mentioned photochemical smog, you would need to account for this evidence and explain how this evidence affects monitoring oxide levels in the future. Quantitative evidence requires numbers, either from data that the question perhaps gives you in an exam or from data you've already known beforehand. Just making up some statistics here, for example you can say about 100000 tonnes of nitrogen oxides are contributed by internal combustion engines or 5000 tonnes of nitrogen oxides are measured to be released everyday or something like that.

2. With extended response questions like this, unless they ask you for a net ionic equation, you only need to include a balanced chemical equation. Of course, it is up to you whether you would want to delete spectator ions or not! For me, I usually wouldnt unless the spectator ions really causes confusion.

3. No definitely not! Know at least 3 evidences, because questions like "assess the evidence for nitrogen oxides/sulfur oxides" are extremely common in exams. You would need around 3-4 evidences and account for those evidences and what they suggest about what we should do in the future to produce less of those oxides. Some evidences are NOAA flask and gas chromatograph measurements, increase in coal burning power plants and photochemical smog. There! You already have 3 more evidences besides acid rain!

4. Personally I do, but it doesnt make such a big impact upon your final mark for the response if you dont include a concluding statement. You've gotta present both sides (i.e. positive and negative) of the question and then have one short last sentence that expresses your opinion (e.g. the benefits of _______ outweighs the disadvantages of ________ and hence _________ definitely benefits the society more).

Ok so I've had a look at your response now, and I will just point out some of the good stuff and some of those not-so-good stuff that you can improve on. But look, overall I reckon this would be a response somewhere between band 5 - band 6, and your teacher might choose to give you a high band 5. You have fluently illustrated what the question asks you to do and demonstrated a good understanding of those evidences and their impacts. You have in your establishing sentences accounted for the occurrence of your acidic rain and stated briefly the reason for concentrations of these oxides in atmosphere (to be brief with these things is good because they dont directly answer the question but you are providing examiners with backgrounds). I must also applaud you on your assessment of the evidences which is the question! Many candidates would forget or include very little judgments in their responses. What I often tell people to do is in the end just explicitly state the word judgment and state what your judgment is (i.e.  "Judgment: __________________________"). It sounds a bit clumsy but it tells the teacher you are answering the assess part of the question!

Ok now lets move onto the defects of your response. The biggest defect in your response is that your evidences are limited. You have mentioned two evidences: gas measurement and a whole tonne of info on acidic rain. This imbalance between the amount of explanation you put on gas measurement and the amount of explanation you put on acidic rain makes the examiner feel like "okay so this girl really knows her acidic rain stuff but she's just stating this gas measurement for the sake of stating it". I would recommend to either take out some information about acid rains and put more information into gas measurement. Or you can just simply add more information in gas measurement. To make your response stronger, you should really include another evidence such as photochemical smog.

Another defect is about your monitoring part of the question. You have used cause and effect language to explain the need for monitoring these acidic oxides but like what I said before, you are focusing a little too much on the evidence part and you are not balancing out your monitoring part. You can add one more point about the necessity to monitor these oxides (for example Nitrogen oxide's greenhouse effect is 100 times stronger than carbon dioxides) and you can also relate photochemical smog to your statement of acidic oxides being respiratory irritants.

Ok so out of 8 I would give a 6-7/8, and your teacher would most likely give you a mark like that as well (definitely not anything below 6). Overall its a very good piece of writing, well done! :D

Best Regards
Happy Physics Land
Post by: katherine123 on March 27, 2016, 01:32:15 pm
Is initiator and catalyst the same (organic peroxide for the production of LDPE)?

Analyse why ethylene is such as important starting material for the chemical industry, in your answer, include relevant chemical equations, and a description of new materials and fuels that can be prepared from ethylene. 7 marks

Ethene contains a reactive double bond which enables it to react readily with many substances forming polyethylene which is a starting material for plastics, a useful product which is widely used to make lives easier
[CH2CH2]n --> -[CH2-CH2-]-
The 2 types of polyethylene include LDPE and HDPE. HDPE is produced at a low gas pressure and temperature of 60 degrees with a Ziegler Natta catalyst. The catalyst attaches to the monomers which enable monomers to link and form a long chain. Once the activated chains collide, the chain growth stops and polyethylene is formed with an unbranched, crystalline structure which makes it rigid and brittle. Thus, it is used to make garbage bins, toys and garbage bags. LDPE is also produced in a similar way but under high gas pressure and temperature of 300 degrees with organic peroxide initiator. The chains formed are branched thus cannot pack closely together like HDPE which makes it soft and flexible. Hence, it is used to make glab wraps and plastic squeeze bottles.

It can also be converted to ethanol by heating it with steam at 300 degrees and high pressure using phosphoric catalyst as the catalyst. CH2CH2(g) + H20(l) --> CH3CH2OH(aq). Ethanol is a useful solvent for  perfume, cosmetics, and paints. Although the ethanol that is currently used as a fuel blend originates from fermentation of sugary crops, ethanol may be sourced from ethene in the future if a more economic method for its production is developed.
Post by: amandali on March 27, 2016, 04:41:17 pm
Hey Amanda!

1. Both qualitative and quantitative are direct evidences (not necessarily first-hand evidences though!). Qualitative requires you to describe this evidence. So for example in this case if you mentioned photochemical smog, you would need to account for this evidence and explain how this evidence affects monitoring oxide levels in the future. Quantitative evidence requires numbers, either from data that the question perhaps gives you in an exam or from data you've already known beforehand. Just making up some statistics here, for example you can say about 100000 tonnes of nitrogen oxides are contributed by internal combustion engines or 5000 tonnes of nitrogen oxides are measured to be released everyday or something like that.

2. With extended response questions like this, unless they ask you for a net ionic equation, you only need to include a balanced chemical equation. Of course, it is up to you whether you would want to delete spectator ions or not! For me, I usually wouldnt unless the spectator ions really causes confusion.

3. No definitely not! Know at least 3 evidences, because questions like "assess the evidence for nitrogen oxides/sulfur oxides" are extremely common in exams. You would need around 3-4 evidences and account for those evidences and what they suggest about what we should do in the future to produce less of those oxides. Some evidences are NOAA flask and gas chromatograph measurements, increase in coal burning power plants and photochemical smog. There! You already have 3 more evidences besides acid rain!

4. Personally I do, but it doesnt make such a big impact upon your final mark for the response if you dont include a concluding statement. You've gotta present both sides (i.e. positive and negative) of the question and then have one short last sentence that expresses your opinion (e.g. the benefits of _______ outweighs the disadvantages of ________ and hence _________ definitely benefits the society more).

Ok so I've had a look at your response now, and I will just point out some of the good stuff and some of those not-so-good stuff that you can improve on. But look, overall I reckon this would be a response somewhere between band 5 - band 6, and your teacher might choose to give you a high band 5. You have fluently illustrated what the question asks you to do and demonstrated a good understanding of those evidences and their impacts. You have in your establishing sentences accounted for the occurrence of your acidic rain and stated briefly the reason for concentrations of these oxides in atmosphere (to be brief with these things is good because they dont directly answer the question but you are providing examiners with backgrounds). I must also applaud you on your assessment of the evidences which is the question! Many candidates would forget or include very little judgments in their responses. What I often tell people to do is in the end just explicitly state the word judgment and state what your judgment is (i.e.  "Judgment: __________________________"). It sounds a bit clumsy but it tells the teacher you are answering the assess part of the question!

Ok now lets move onto the defects of your response. The biggest defect in your response is that your evidences are limited. You have mentioned two evidences: gas measurement and a whole tonne of info on acidic rain. This imbalance between the amount of explanation you put on gas measurement and the amount of explanation you put on acidic rain makes the examiner feel like "okay so this girl really knows her acidic rain stuff but she's just stating this gas measurement for the sake of stating it". I would recommend to either take out some information about acid rains and put more information into gas measurement. Or you can just simply add more information in gas measurement. To make your response stronger, you should really include another evidence such as photochemical smog.

Another defect is about your monitoring part of the question. You have used cause and effect language to explain the need for monitoring these acidic oxides but like what I said before, you are focusing a little too much on the evidence part and you are not balancing out your monitoring part. You can add one more point about the necessity to monitor these oxides (for example Nitrogen oxide's greenhouse effect is 100 times stronger than carbon dioxides) and you can also relate photochemical smog to your statement of acidic oxides being respiratory irritants.

Ok so out of 8 I would give a 6-7/8, and your teacher would most likely give you a mark like that as well (definitely not anything below 6). Overall its a very good piece of writing, well done! :D

Best Regards
Happy Physics Land

Is it sufficient if i add this to the gas measurement
Moreover, direct measurements are made by statutory bodies such as NSW EPA and scientists researching the atmosphere. NOx were found to be part of pollution generated over large cities due to use of internal combustion engines since industrial revolution

NOx produces photochemical smog after reacting in the presence of sunlight which is a toxic pollutant that causes respiratory diseases
Post by: jakesilove on March 27, 2016, 05:14:21 pm
Is initiator and catalyst the same (organic peroxide for the production of LDPE)?

Analyse why ethylene is such as important starting material for the chemical industry, in your answer, include relevant chemical equations, and a description of new materials and fuels that can be prepared from ethylene. 7 marks

Ethene contains a reactive double bond which enables it to react readily with many substances forming polyethylene which is a starting material for plastics, a useful product which is widely used to make lives easier
[CH2CH2]n --> -[CH2-CH2-]-
The 2 types of polyethylene include LDPE and HDPE. HDPE is produced at a low gas pressure and temperature of 60 degrees with a Ziegler Natta catalyst. The catalyst attaches to the monomers which enable monomers to link and form a long chain. Once the activated chains collide, the chain growth stops and polyethylene is formed with an unbranched, crystalline structure which makes it rigid and brittle. Thus, it is used to make garbage bins, toys and garbage bags. LDPE is also produced in a similar way but under high gas pressure and temperature of 300 degrees with organic peroxide initiator. The chains formed are branched thus cannot pack closely together like HDPE which makes it soft and flexible. Hence, it is used to make glab wraps and plastic squeeze bottles.

It can also be converted to ethanol by heating it with steam at 300 degrees and high pressure using phosphoric catalyst as the catalyst. CH2CH2(g) + H20(l) --> CH3CH2OH(aq). Ethanol is a useful solvent for  perfume, cosmetics, and paints. Although the ethanol that is currently used as a fuel blend originates from fermentation of sugary crops, ethanol may be sourced from ethene in the future if a more economic method for its production is developed.

Hey!

Yes, initiator and catalyst can be thought of as the same thing.

I like your answer; I think it goes into sufficient depth for most areas that you discuss. However, I would add something small about the PRODUCTION of ethene: Part of it's significance in the chemical industry is the ease by which it can be produced, particularly through catalytic cracking of fuel fractions. This is a really important reason as to why we love Ethene: not only is it great (as you've described) because of its reactivity, but also because of its potential abundance! Other than that, and perhaps rewording some of your sentences (read it outloud, and see if anything seems a bit rambly) I think you have covered lots of vital areas in answering the question.

Jake
Post by: Happy Physics Land on March 27, 2016, 06:36:15 pm
Is it sufficient if i add this to the gas measurement
Moreover, direct measurements are made by statutory bodies such as NSW EPA and scientists researching the atmosphere. NOx were found to be part of pollution generated over large cities due to use of internal combustion engines since industrial revolution

NOx produces photochemical smog after reacting in the presence of sunlight which is a toxic pollutant that causes respiratory diseases

Hey Amanda!

Yes if you add these your response would sound way stronger! Because now you are accounting for your evidences more and proposing a bit more new ideas! Well done! To get a full 8/8 I reckon you can just add in a few examples of how to monitor those oxides (e.g. look up if theres a legislation regarding the release of NOx or SOx pollutants, and then you can say "because of these evidences which demonstrate the negative consequences of these oxides, hence these emissions must be monitored. For example, the government body has participated in this monitoring process through this legislation _________, " This just makes the teacher realise that you have a great insight into the question and displays understandings towards those monitoring actions.

Im gonna use an analogy here (hehe :) no hates ). For me, I reckon originally your response was like a finely-made cake with icings sprinkled everywhere. Now after you apply the feedback and added in those stuff your response is like a finely-made cake that is beautifully decorated with a layer of evenly distributed icing. So now what you have to do is to just sprinkle some rainbow coloured sugar thingy on top!

Best Regards
Happy Physics Land
Post by: grace_joseph on March 27, 2016, 06:39:21 pm
Hi Jake,

Would you please be able to explain why this is the answer to the 2011 HSC paper q25 (attachment 1 and 2)? I don't really know why you need two equations, or even where the equations came from. I understand buffers but have never seen them proved using water before

Also, would you mind explaining question 15 of this paper to me when you could? (attachment 3) The answer is A :)

Thanks very much Jake :)
Post by: jakesilove on March 27, 2016, 07:07:10 pm
Hi Jake,

Would you please be able to explain why this is the answer to the 2011 HSC paper q25 (attachment 1 and 2)? I don't really know why you need two equations, or even where the equations came from. I understand buffers but have never seen them proved using water before

Also, would you mind explaining question 15 of this paper to me when you could? (attachment 3) The answer is A :)

Thanks very much Jake :)

Hey Grace!

Great question, as it really does seem like the equations were pulled out of nowhere! However, when you consider what a buffer is actually MEANT to do, the equations start to make sense!

A buffer is something that resists small changes in pH, which is vitally important for things like living organisms and cells. All the equations display are the acidic and basic nature of the buffer: in the first equation, you can clearly is the substance is a proton donor, whilst in the second you can clearly see that the substance is a proton acceptor. As such, the system can be both acidic and basic, and therefore resist small changes in pH! I hope that this makes sense: if you're ever unsure as to how to write out buffer equations, just add water to each part of the buffer and show the acidic and basic properties of the buffer!

As for Question 15, I don't think I can really answer it without the rest of the question. However, I assume that there are equal moles of strong and weak acid? In that case, remember that the important part of neutralisation reactions is the number of MOLES of the acid, as the number of moles of the base will need to equal that (in whatever ratio the chemical equation states) in order for neutralization to occur. For instance, a monoprotic WEAK acid of 0.1mol/L will require the same amount of NaOH to neutralise it as a monoprotic STRONG acid of 0.1mol/L, as the concentration is the important part in this reaction.

I hope this helps! Feel free to ask more questions so I can help clarify your understanding :)

Jake
Post by: grace_joseph on March 27, 2016, 08:02:29 pm
Hi Jake! Thank you very much for the buffer answer, yes I get it now :)

And sorry I forgot to add the first part of question 15! It's attached now but I still don't understand why a weak acid would require the same amount of base as a strong one, as when we do titrations I thought it was established that one drop of NaOH may overpower acetic acid but not have any effect on HCl?

Sorry! I really appreciate any help you can give, even if I still don't get it :D

Grace
Post by: amandali on March 27, 2016, 09:07:58 pm
Justify the continued use of Arrhenius definition of acids and bases, despite the development of the more sophisticated Bronsted-Lowry definition 3 marks
how do these 2 definitions describe neutralization differently

Water and ethanol are both used as solvents. Explain the diff and sim. in their solvent behaviour in terms of their molecular structures. Include a diagram 4 marks
I wrote my answer in a table (similarities+differences)
Is there anything else i can add to differences or is that sufficient
Similarities
•   Both water and ethanol are polar molecules
•   Both consists of polar hydroxyl group (-OH) which can attract and dissolve polar substances eg. HCl through dipole-dipole attraction, hydrogen bonding or dispersion forces.

differences
•   Ethanol has a non-polar region ie ethyl group (-CH2CH3) whereas water does not thus only ethanol can dissolve (some) non-polar substances eg. Hexane via dispersion forces. Water does not have non-polar regions therefore is a poor solvent for non-polar substances

Including a diagram of ethanol attract both polar and non-polar sub. and water attracting polar sub.
Post by: jakesilove on March 27, 2016, 10:36:47 pm
Hi Jake! Thank you very much for the buffer answer, yes I get it now :)

And sorry I forgot to add the first part of question 15! It's attached now but I still don't understand why a weak acid would require the same amount of base as a strong one, as when we do titrations I thought it was established that one drop of NaOH may overpower acetic acid but not have any effect on HCl?

Sorry! I really appreciate any help you can give, even if I still don't get it :D

Grace

Hey Grace!

When it comes to titrations, it is the Mol/L that is important. Think about it this way: let's pretend I titrate 25mL of monoprotic acid (0.1mol/L) with XmL of monoprotic base (0.1mol/L).

The equation is something like Acid + Base --> Salt + Water, and the ratio of Acid:Base will be 1:1.

Using our formulas, we know that 25mL of 0.1 mol/L acid will contain 0.0025 moles (0.1*0.025). Because it is a 1:1 reaction, we require 0.0025 moles of base, which will require a quantity of 25mL (as expected).

However, nowhere here did we discuss what specific acids or bases we used! The only important thing is the concentration: you can have a really concentrated weak acid, and a very dilute strong acid! It would be EASIER to have a strong, concentrated acid, but definitely not necessary. When it comes to Titrations, only the numbers matter.

Jake
Post by: RuiAce on March 27, 2016, 10:46:05 pm
Hey!

Yes, initiator and catalyst can be thought of as the same thing.

I like your answer; I think it goes into sufficient depth for most areas that you discuss. However, I would add something small about the PRODUCTION of ethene: Part of it's significance in the chemical industry is the ease by which it can be produced, particularly through catalytic cracking of fuel fractions. This is a really important reason as to why we love Ethene: not only is it great (as you've described) because of its reactivity, but also because of its potential abundance! Other than that, and perhaps rewording some of your sentences (read it outloud, and see if anything seems a bit rambly) I think you have covered lots of vital areas in answering the question.

Jake

Just popping in some chat Jake...

I didn't even realise that the peroxide initiator in LDPE and catalyst in HDPE worked the same way until AFTER the HSC!!
Post by: amandali on March 28, 2016, 01:40:52 am
Evaluate the success of the current usage of ethanol as an alternative car fuel obtained from sugar cane 4 marks

Currently, ethanol is used directly as a fuel in Brazil as there is enough arable land to grow sugary crops from which ethanol is derived. The advantage of ethanol is that it is renewable and abundant as it is produced from the fermentation of crops such as sugar cane, corn, wheats, C6H12O6 (aq) --> 2CH3CH2OH(aq) + 2CO2 (g)  , as opposed to the hydrocarbon fuels like octane which is produced from non-renewable fossil fuels.  Hence, ethanol has an unlimited use while hydrocarbon fuels are limited and are predicted to deplete within the next few decades. Moreover, ethanol combusts cleanly due to presence of oxygen in its molecule which enables a more complete combustion. This mean that less harmful respiratory irritants like C(s) and CO(g) that causes respiratory diseases are produced.
CH3CH2OH(l) + 3O2 (g) --> 2CO2 (g) + 3H2O(l) .

Meanwhile, it is not widely used countries like Australia as it lacks arable land to grow sugary crops and it is not economical to produce ethanol. The disadvantage of ethanol is that its production requires cultivating and clearing vast areas of arable land hence this will affect food production and give rise to environmental effects such as land erosion and destruction of animal habitats. An immense amount of energy is also required to manufacture fertilisers, plant and harvest the crop and distil ethanol thus it is not economically viable to do so.

Weighing the pros and cons, it is unlikely that ethanol will not be used as an alternative fuel unless an economic viable method for converting cellulose to glucose is discovered.
Post by: jakesilove on March 28, 2016, 11:18:40 am
Evaluate the success of the current usage of ethanol as an alternative car fuel obtained from sugar cane 4 marks

Currently, ethanol is used directly as a fuel in Brazil as there is enough arable land to grow sugary crops from which ethanol is derived. The advantage of ethanol is that it is renewable and abundant as it is produced from the fermentation of crops such as sugar cane, corn, wheats, C6H12O6 (aq) --> 2CH3CH2OH(aq) + 2CO2 (g)  , as opposed to the hydrocarbon fuels like octane which is produced from non-renewable fossil fuels.  Hence, ethanol has an unlimited use while hydrocarbon fuels are limited and are predicted to deplete within the next few decades. Moreover, ethanol combusts cleanly due to presence of oxygen in its molecule which enables a more complete combustion. This mean that less harmful respiratory irritants like C(s) and CO(g) that causes respiratory diseases are produced.
CH3CH2OH(l) + 3O2 (g) --> 2CO2 (g) + 3H2O(l) .

Meanwhile, it is not widely used countries like Australia as it lacks arable land to grow sugary crops and it is not economical to produce ethanol. The disadvantage of ethanol is that its production requires cultivating and clearing vast areas of arable land hence this will affect food production and give rise to environmental effects such as land erosion and destruction of animal habitats. An immense amount of energy is also required to manufacture fertilisers, plant and harvest the crop and distil ethanol thus it is not economically viable to do so.

Weighing the pros and cons, it is unlikely that ethanol will not be used as an alternative fuel unless an economic viable method for converting cellulose to glucose is discovered.

Hey Amanda,

I would add in some more aspects of ethanol as an ALTERNATIVE CAR FUEL, specifically things like the erosion of engines, specific heat capacity etc. that effects its current usage in Australia (limiting it to E10 etc.). Other than that, a really good response.

Jake
Post by: katherine123 on March 28, 2016, 12:50:39 pm
how do you know that there is change in pressure?  i know that change in concentration is indicated by sharp drop or increase

at 10min can it be increase in temperature as well (endothermic)  although its due to increase in CO concentration as indicated by the sharp increase
Post by: jakesilove on March 28, 2016, 01:15:04 pm
how do you know that there is change in pressure?  i know that change in concentration is indicated by sharp drop or increase

at 10min can it be increase in temperature as well (endothermic)  although its due to increase in CO concentration as indicated by the sharp increase

Hey!

You know that there is a change in pressure when there is no sharp drop, and the equilibrium clearly shifts to one side. For instance, at four minutes, there is clearly no sharp drop and yet there is an obvious shift to the products! This could either be because of temperature, concentration or pressure.

Because there is no sharp drop/increase, it isn't a change in concentration. As such, you would talk about the fact that it could be a change in temperature, a change in pressure, or both. Because there are more moles of gas on the side of the products, the only reason products would increase in concentration is if you DECREASE the pressure of the system (by Le Chatelier's). As it is endothermic, an increase in temperature will cause a shift to the right, and therefore this could explain the results at 4 minutes as well. There is no way to tell which of these options occurred, so you need to talk about both!

At 10 minutes, the reason for the change is (as you've identified) because of a removal of CO. This will cause a shift to the right, as per Le Chatelier's, giving the results on the graph. There is no clear change in temperature, just a classic change in concentration.
Post by: Maz on March 28, 2016, 06:53:18 pm
hey,
can i please get some help with this question? i got an  answer of 120g when the answer is actually 10.5g - big difference haha  :P
the question is
Calculate the mass of quicklime (calcium oxide), which will react completely with 250mL of 1.50mol L-1 hydrochloric acid.

thank you so much :)
Post by: grace_joseph on March 28, 2016, 06:55:34 pm
Hi Jake! Would you please be able to explain to me why the answer for 7 is A and why the answer for 11 is C?

Thank you so much!

Grace :)
Post by: lazydreamer on March 28, 2016, 07:19:38 pm
Hi Jake! Would you please be able to explain to me why the answer for 7 is A and why the answer for 11 is C?

Thank you so much!

Grace :)

In Q7, you need to look at how many hydrogen protons are donated by the acid to NaOH. So basically you need to know the protic nature of the acids. The answer is citric acid because it is triprotic, therefore 3mols of acid is needed for 1mol NaOH compared to the other mono/di-protic acids. Degree of ionisation is irrelevant to answering the question, i know, i've seen that Q before

For 11, you'd need to deduce that the compound formed is an alkene (only alkenes discolor bromine water immediately). Then you'd need to be able to link this with the dehydration of an alkanol, by heating with an acid, which produces an alkene. Therefore the answer is C, the alkanol
Post by: jakesilove on March 28, 2016, 07:26:00 pm
Hi Jake! Would you please be able to explain to me why the answer for 7 is A and why the answer for 11 is C?

Thank you so much!

Grace :)

Totally agree with Lazydreamer's answer! Let me know if you need any further clarification :)

Jake
Post by: jakesilove on March 28, 2016, 07:34:09 pm
hey,
can i please get some help with this question? i got an  answer of 120g when the answer is actually 10.5g - big difference haha  :P
the question is
Calculate the mass of quicklime (calcium oxide), which will react completely with 250mL of 1.50mol L-1 hydrochloric acid.

thank you so much :)

Hey!

My answer is below, I hope that it helps! Let me know if you need any further clarification. Always just follow the steps that I've outlined: Chemical formula, application of equations, solution!

(http://i.imgur.com/oBzntsh.png?1)

Jake
Post by: lazydreamer on March 28, 2016, 07:38:58 pm
Now i got a question

So in the Examples, why does the numbering start where it starts? Like why in 3 does it start closer to the methyl but in 4 it starts at the ethyl instead?
Thanks :D
Post by: RuiAce on March 29, 2016, 10:58:51 am
Now i got a question

So in the Examples, why does the numbering start where it starts? Like why in 3 does it start closer to the methyl but in 4 it starts at the ethyl instead?
Thanks :D

Firstly, the ethyl is written before the methyl because we want to maintain alphabetical order. E lies before M in the English the alphabet. But it seems like you already knew that.

We would like to choose, so that we always minimise the sum of the locant numbers in the molecule.

For Q3, if we count from left to right we will have
3-ethyl-5-methyl-hex-3-ene
3 + 5 = 8
If we count right to left we will have
4-ethyl-2-methyl-hex-3-ene
4 + 2 = 6

So we favour the latter.

For Q4, firstly keep in mind that there is an ethyl group on the right, because the double bond is what gives pentadiene it's name.
Now, using the same process

If we count from left to right we will have
If we count from right to left we will have

And here we have a problem. From maths, we know that 4+2+3=2+3+4. Oh dear!

Which is exactly the point I call you out for being a selective school student (or attempt to) and having to do harder questions.

The reason why the second one is favourable is that in the event that your sum of locants are the same (very unlikely to occur in the HSC), we again treat alphabetical order as priority. E lies before M, so the ethyl will get the lower number here.
Post by: RuiAce on March 29, 2016, 11:06:42 am
how do you know that there is change in pressure?  i know that change in concentration is indicated by sharp drop or increase

at 10min can it be increase in temperature as well (endothermic)  although its due to increase in CO concentration as indicated by the sharp increase

For questions like this there exists a rule of thumb

• No spike, and then things change -> Temperature change. If the nature of the reaction as exothermic or endothermic is given, you can work out if heated or cooled
• Spike in every substance -> Pressure change. Downward spike always means increase in pressure (or alternatively decrease in volume of the vessel)
• Spike in only one substance -> Concentration change. One of the things has been tampered with. Could be you introduced more or suddenly extracted a ton depending on spike up or down.
Post by: RuiAce on March 29, 2016, 11:10:55 am
In Q7, you need to look at how many hydrogen protons are donated by the acid to NaOH. So basically you need to know the protic nature of the acids. The answer is citric acid because it is triprotic, therefore 3mols of acid is needed for 1mol NaOH compared to the other mono/di-protic acids. Degree of ionisation is irrelevant to answering the question, i know, i've seen that Q before

For 11, you'd need to deduce that the compound formed is an alkene (only alkenes discolor bromine water immediately). Then you'd need to be able to link this with the dehydration of an alkanol, by heating with an acid, which produces an alkene. Therefore the answer is C, the alkanol
To build on:
It is irrelevant, but you should be aware that the reason why it is irrelevant is due to the titration with a strong base. The strong base being present, regardless of the strength of the acid, forces the reaction to go to completion no matter what.
Post by: indiago on March 29, 2016, 06:04:43 pm
Hello !!!

I have come across similar questions to this one but I can't seem to figure out how to do them. Please help !!!

pH 2.00 of 0.01M HCl.

Calculate the pH after 20 mL of 0.01 mol L-1 hydrochloric acid is diluted with 180 mL of water.

Thanks, India
Post by: lazydreamer on March 29, 2016, 11:02:01 pm
Firstly, the ethyl is written before the methyl because we want to maintain alphabetical order. E lies before M in the English the alphabet. But it seems like you already knew that.

We would like to choose, so that we always minimise the sum of the locant numbers in the molecule.

For Q3, if we count from left to right we will have
3-ethyl-5-methyl-hex-3-ene
3 + 5 = 8
If we count right to left we will have
4-ethyl-2-methyl-hex-3-ene
4 + 2 = 6

So we favour the latter.

For Q4, firstly keep in mind that there is an ethyl group on the right, because the double bond is what gives pentadiene it's name.
Now, using the same process

If we count from left to right we will have
If we count from right to left we will have

And here we have a problem. From maths, we know that 4+2+3=2+3+4. Oh dear!

Which is exactly the point I call you out for being a selective school student (or attempt to) and having to do harder questions.

The reason why the second one is favourable is that in the event that your sum of locants are the same (very unlikely to occur in the HSC), we again treat alphabetical order as priority. E lies before M, so the ethyl will get the lower number here.

HAHAHA quite the opposite, i was struggling with naming organic compounds so i went on that website to practice and came across something that made me more confused...but your explanation made it clearer so thanks! :D
Post by: RuiAce on March 30, 2016, 12:17:07 am
Hello !!!

I have come across similar questions to this one but I can't seem to figure out how to do them. Please help !!!

pH 2.00 of 0.01M HCl.

Calculate the pH after 20 mL of 0.01 mol L-1 hydrochloric acid is diluted with 180 mL of water.

Thanks, India

Logical consideration:

The dilution increases the volume from 20mL to 200mL. Note that 200 = 20 * 10
Because it's been diluted to a ten-fold value critically speaking, the pH goes up by 1. pH = 3.
This is because the pH scale is logarithmic (pH = -log10[H3O+])

Formal method:
According to the dilutions formula C1V1=C2V2
0.01 mol L-1 * 20 mL = [HCl]final * 200 mL
0.001 mol L-1 = [HCl]final

As HCl is both monoprotic and a strong acid
[H3O+]=[HCl]=0.001 mol L-1

So pH = -log10[H3O+] = -log10(0.001) = 3
Post by: katherine123 on March 30, 2016, 04:55:07 pm
During an investigation you are provided with:
- 3 volumetric flasks - each with volume capacity of 100ml, 205ml and 500ml
-2.2g anhydrous NaHCO3 molar mass of 84.01
You are asked to prepare a 0.250mol/L solution of NaHCO3

what i did was   i calculated the moles of NaHCO3 and find the volume required to have all the moles in the flask which turned out to be 105ml  so i thought 250 ml is the most suitable since it can contain all the volume of solution

2nd ques: NaOH absorbs moisture from air. If it was weighed and prepared for use as basic solution in X, what effect will this have?
a) calculated acid concentration will be correct
b) calculated acid concentration will be too low
c) calculated acid concentration will be too high
d) Ph at end point will be much lower than it should be
Post by: jakesilove on March 30, 2016, 06:09:02 pm
During an investigation you are provided with:
- 3 volumetric flasks - each with volume capacity of 100ml, 205ml and 500ml
-2.2g anhydrous NaHCO3 molar mass of 84.01
You are asked to prepare a 0.250mol/L solution of NaHCO3

what i did was   i calculated the moles of NaHCO3 and find the volume required to have all the moles in the flask which turned out to be 105ml  so i thought 250 ml is the most suitable since it can contain all the volume of solution

2nd ques: NaOH absorbs moisture from air. If it was weighed and prepared for use as basic solution in X, what effect will this have?
a) calculated acid concentration will be correct
b) calculated acid concentration will be too low
c) calculated acid concentration will be too high
d) Ph at end point will be much lower than it should be

Hey Katherine!

I agree that, for the first part, you would require 105mL to get the required concentration. However, remember that a volumetric flask has to be FILLED UP COMPLETELY in order to be accurate and effective. Therefore, your options are to use slightly less water (100mL), and therefore get a concentration slightly higher than required, or use much more water (250mL) and get a concentration much lower then required. Obviously, using the 100mL will get you closer to the desired value, and so we choose the first volumetric flask!

Okay, let's talk about the second question. If the NaOH has been measured to weigh MORE than it actually is (due to water), then in your calculations you will use a basic solution with a HIGHER concentration than it actually is. If you use LESS base than you think you are using to neutralise the SAME amount of acid, then the acid is going to be weaker than you expect (for instance, you might think it neutralises 0.5 moles of base, but it is actually only neutralising 0.45 moles of base. As C=n/V, the concentration will lower if n lowers). Therefore, the answer is c) as the actual concentration is lower than the calculated concentration.

Jake
Post by: Maz on March 31, 2016, 07:12:04 pm
hi
water from a domestic bore that sprayed onto a nearby fence was observed to leave orange brown deposits on it's surface. this indicated that the water contained iron compounds. the water was analysed for iron by titration in the following way: Aisr was bubbled through a 250 mL sample of water for several hours to convert all the dissolved iron into Fe3+ then boiled to precipitate all the calcium ions as calciu carbonate. the solid was removed by filtration then 10mL of standard 0.103mol L-1 NaOH was added to the bore water which precipitated all of the iron in the sample and left  some excess NaOH in the solution.
After the mixture was filtered it required 27.34mL of standard 0.0277 mol L-1 HCl acid for neutralisation. if the density of the bore water was 1.01g mL-1 calculate the concentration of iron in the bore water measured in parts per million...

so far i have just converted the HCl to moles giving: 0.000757318 moles...
i really don't know what to do after that...please can you help me?
Post by: jakesilove on March 31, 2016, 07:37:07 pm
hi
water from a domestic bore that sprayed onto a nearby fence was observed to leave orange brown deposits on it's surface. this indicated that the water contained iron compounds. the water was analysed for iron by titration in the following way: Aisr was bubbled through a 250 mL sample of water for several hours to convert all the dissolved iron into Fe3+ then boiled to precipitate all the calcium ions as calciu carbonate. the solid was removed by filtration then 10mL of standard 0.103mol L-1 NaOH was added to the bore water which precipitated all of the iron in the sample and left  some excess NaOH in the solution.
After the mixture was filtered it required 27.34mL of standard 0.0277 mol L-1 HCl acid for neutralisation. if the density of the bore water was 1.01g mL-1 calculate the concentration of iron in the bore water measured in parts per million...

so far i have just converted the HCl to moles giving: 0.000757318 moles...
i really don't know what to do after that...please can you help me?

Hey!

Unfortunately, I don't have time right now to do a full solution (although can write one up later, if no-one else has), however I can quickly explain the general method.

Knowing the moles of HCl, we can write out the reaction between HCL and NaOH to figure out the number of moles of NaOH that was in excess in solution (knowing that it is a 1:1 ratio).

Now that we have the EXCESS NaOH, we can figure out the amount of NaOH that reacted with Iron (figure out the total moles of NaOH added, and then subtract the excess).

We can write the reaction between Iron and NaOH. Knowing the number of moles of NaOH (from above) that reacts, we can figure out the number of moles of Iron!

Sorry that I can't go into much more detail right now.

Jake
Post by: g98 on March 31, 2016, 08:09:34 pm
Hi I know this is the Chem thread but I'm hoping someone out there knows something about Bio! Should you know the scientific names of animals and plants you use in examples? Thank You!!!
Post by: Rd2487 on April 01, 2016, 07:06:10 pm
Hi Jake
When you explain the salts nature how would you structure your response like lets say NH4CL?
So would you start saying that Cl is a weak conjugate anion of the strong acid HCl and hence it would not further hydrolyze.
Whereas the ammonium is a strong conjugate cation of the weak base ammonia hence it would further hydrolyze.
Now i get confused on this step after providing the reaction of NH4 with water. Would you say that because H30+ ions are produced in solution, it would make the salt acidic ??
Can you please provide a way to structure this response
Thanks
Post by: RuiAce on April 01, 2016, 07:55:16 pm
Hi Jake
When you explain the salts nature how would you structure your response like lets say NH4CL?
So would you start saying that Cl is a weak conjugate anion of the strong acid HCl and hence it would not further hydrolyze.
Whereas the ammonium is a strong conjugate cation of the weak base ammonia hence it would further hydrolyze.
Now i get confused on this step after providing the reaction of NH4 with water. Would you say that because H30+ ions are produced in solution, it would make the salt acidic ??
Can you please provide a way to structure this response
Thanks
As I have previously explained somewhere in this thread (not too sure where exactly).

The conjugate of something strong (HCl, NaOH, HNO3) is something extremely weak (respectively, Cl-, Na+, NO3-).
Whereas the conjugate of something weak (CH3COOH, NH3) is something that's also weak, but definitely not extremely weak (respectively CH3COOH-, NH4+)

However, when it comes to classifying salts as acidic/basic/neutral, one may actually safely assume substances such as Cl- and Na+ are neutral when writing exam responses. The reason for this is that the impact of these ions in the overall result is so negligible, it's almost non-existent.

Such a question would be about 3 marks at most (some may be lenient to give 2). Here is what I would say (not the BEST answer - I'm not currently doing the HSC)

The Cl- ion is a neutral anion. It's presence will not affect the acidity of the salt.
However, NH4+ is an acidic cation as it is able to react with water to form it's conjugate base:
NH4+ + H2O(l) ⇌ H3O+ + NH3(aq)
The acidic nature of ammonium chloride therefore occurs as a result of only the acidic ammonium ion impacting, as that of the chloride ion is negligible.
Post by: browny2409 on April 02, 2016, 01:49:05 pm
Hi,

When detailing the development of Acid-Base theories (Lavoisier, Davy, Arrhenius and Brøsted-Lowry) there was one HSC question which asked how each developed from the other

I am aware of what each theory, but could you help in clarifying exactly how each theory improved the former

Thanks
Nathan
Post by: jakesilove on April 02, 2016, 02:18:49 pm
Hi,

When detailing the development of Acid-Base theories (Lavoisier, Davy, Arrhenius and Brøsted-Lowry) there was one HSC question which asked how each developed from the other

I am aware of what each theory, but could you help in clarifying exactly how each theory improved the former

Thanks
Nathan

Hey Nathan!

Essentially, all you need to know is that each subsequent theory was developed as a result of an unexplained finding! For instance, Lavoisier believed that acids contained Oxygen molecules (as you know). However, Davy noticed that HCl acted as an acid, and therefore obviously Lavoisier's theory was inadequate! As such, Davy developed his own theory.

Each development has similar examples that you can use, however I wouldn't bother. The only question you could really get either a multiple choice, asking you to determine which theory is outlined in the question, or a 'describe Arrhenius's theory of Acids and Bases' etc.

How many marks was that question you found? I would say you just need to outline the general aspects of each theory, and saying that they built on each other as new scientific information came to light. You're not expected to know examples of each: merely that they become increasingly complex,and increasingly accurate as they tend to encapsulate more and more actual acids.

Jake
Post by: RuiAce on April 02, 2016, 03:13:56 pm
Hi,

When detailing the development of Acid-Base theories (Lavoisier, Davy, Arrhenius and Brøsted-Lowry) there was one HSC question which asked how each developed from the other

I am aware of what each theory, but could you help in clarifying exactly how each theory improved the former

Thanks
Nathan

During the HSC I wrote a full 8 marker on acid-base theories. I was able to find it somehow.

Past theories of acids and bases tend to purely rely on what was believed to be the chemical composition of all acids and their properties. Lavoisier's theory stated that acids were substances which contained hydrogen, as he found that when compounds involving oxygen dissolved in water, they produced acidic properties, such as CO2, NO2 and P4O10. This also meant that some acids were recognised to be acids, e.g. H2SO4. However, this is now known to be significantly flawed, as some substances containing oxygen are in fact basic (e.g. Na2CO3, Na2O) and why some other acids did not contain oxygen (e.g. HCl). Davy then proposed that acids were substances that contained hydrogen through the discovery of HCl, through means of dissolving gaseous hydrogen chloride into water and producing a highly acidic solution. Davy also found that metals could displace the hydrogen in acids: Metal + Acid -> Salt + Hydrogen gas. Yet, this was also flawed because other substances such as CH4 are not acidic, yet also contain hydrogen.
Arrhenius then developed a theory that was, at the time, much more revolutionary and a more accurate definition of acids, and bases. Arrhenius showed that acids were substances that when dissolved in water, ionised to form H+ ions, and similarly for bases albeit OH- ions. Arrhenius was also able to initiate the definitions of relative strengths of acids through his theory from an understanding of degrees of ionisation, and also demonstrated how acids and bases can undergo a neutralisation process: Acid + Base -> Salt + Water. This is reasonably accurate, and is sometimes taught as a simplified model of how acids and bases work, however it did not cater for other bases such a Na2CO3 and NH3.
The most accurate definition of acids and bases was proposed by Johannes Bronsted and Thomas Lowry. The Bronsted-Lowry theory defines acids and bases in terms of their chemical behaviour, that is, acids were substances that could donate a proton/protons, whereas bases could accept them. This theory focused much more on the properties of acids and bases rather than their chemical composition and an understanding of what conjugate acids and bases were. By consideration of the ability to donate/accept H+, it was then finally understood why substances were acidic, basic or neutral. (It also led into the identification of amphiprotic species with reference conjugate acids and bases.)
Post by: RhyannahHamer on April 03, 2016, 05:46:09 pm
Hey there,
I'm doing notes under the syllabus right now, and I don't think I've got any sufficient notes on the following dot point:
'describe the difference between a strong and a weak acid in terms of an equilibrium between the intact molecule and its ions'
If you could help me out, it'd be much appreciated.
Thanks :)
Post by: jakesilove on April 03, 2016, 10:07:59 pm
Hey there,
I'm doing notes under the syllabus right now, and I don't think I've got any sufficient notes on the following dot point:
'describe the difference between a strong and a weak acid in terms of an equilibrium between the intact molecule and its ions'
If you could help me out, it'd be much appreciated.
Thanks :)

Hey!!

Basically, for that dot point I would have a few things up your sleeve.

Definitions of strong and weak acids

Always start by defining things. A strong acid is, by definition, an acid that ionises completely in solution. A weak acid is just any acid that does NOT ionise completely in solution (and usually only ionises to a tiny extent, eg. 1-10%).

Formulas

But what do the above statements even mean? The easiest way to describe strong and weak acids in terms of equilibrium is to use equations. So, let's do that!

Let's start by looking at a strong acid. I always like to use HCl, because its the one used most often in the HSC.

We write the reaction of HCl and Water to show ionisation.

(http://i.imgur.com/CM4B2wb.png?1)

Now, we know that HCl is a strong acid, and so ionises completely. In terms of equilibrium, this means that the reaction lies entirely to the right!

Now, let's look at weak acids or weak bases. You can use whatever weak acid or base you want.

First, we write out the equation.

(http://i.imgur.com/mc7JMgy.png?1)

Now, we know that Ammonia is a weak base. Therefore, in terms of equilibrium, we can say that the reaction lies largely to the left!

That's really all you need to know in terms of acids/bases and equilibrium. Hope it helps!

Jake
Post by: Maz on April 05, 2016, 01:10:31 am
hey
i've tried to do this titration question a couple of times- i just don't know how to do it :(
the question:
9.32g of 'coudy ammonia' was dissolved in water and made up to 250mL in a volumetric flask. 20mL portions of this solution were tritrated with 0.98 mol L^-1 HCl. An average of 25.8 mL HCl was required for the neutralisation using methyl orange indicator. Calculate the percentage by mass of ammonia in the 'cloudy ammonia'.
i'd really appreciate it if you could help me out...
thankyou  :)
Post by: browny2409 on April 06, 2016, 09:38:43 am

Just another question, when summarising the chemistry of the fermentation process (Production of Materials), do we need to mention acid hydrolysis and enzyme digestion, because i have heard of these terms but don't have them in my notes - all i refer to is the invertase enzyme and the zymase enzyme brought about by the yeast

Thanks
Nathan
Post by: jakesilove on April 06, 2016, 09:52:51 am
hey
i've tried to do this titration question a couple of times- i just don't know how to do it :(
the question:
9.32g of 'coudy ammonia' was dissolved in water and made up to 250mL in a volumetric flask. 20mL portions of this solution were tritrated with 0.98 mol L^-1 HCl. An average of 25.8 mL HCl was required for the neutralisation using methyl orange indicator. Calculate the percentage by mass of ammonia in the 'cloudy ammonia'.
i'd really appreciate it if you could help me out...
thankyou  :)

Hey!

I'm so sorry for the late response, I totally missed your question! This is a tough one, that requires multiple applications of the titration formulas. My working is below, I hope it helps!

(http://i.imgur.com/amSlGC0.png?1)
(http://i.imgur.com/RDtuK5Q.png?1)

Jake
Post by: jakesilove on April 06, 2016, 09:59:14 am

Just another question, when summarising the chemistry of the fermentation process (Production of Materials), do we need to mention acid hydrolysis and enzyme digestion, because i have heard of these terms but don't have them in my notes - all i refer to is the invertase enzyme and the zymase enzyme brought about by the yeast

Thanks
Nathan

Hey Nathan,

Whenever I tutor students, and whenever I answered questions on this topic in my HSC, I always recommend one method for exactly this problem: Mention the terms, but that's it.

An important step in the process is going from Cellulose to Glucose. At the moment, this process is expensive and time consuming, and requires either acid hydrolysis and enzyme digestion. However, you definitely don't need to know more information about either of those processes. I would definitely mention that they are expensive and time consuming, and name-drop the terms, but that's it.

Not only would you get full marks by mentioning it, but as it is really an 'addition' to the course the marker will look really favorably on you for going above and beyond, despite the fact that maybe you don't understand the terms you are using (as I definitely don't either!).
Post by: Neutron on April 06, 2016, 10:21:39 am
Hey guys!

Got my Chemistry half yearly today and I've got a stupid question D: This is the question:

Which of the following aqueous solutions has a pH greater than 7?
a) Sodium ethanoate
b) Sodium chloride
c) Ammonium nitrate
d) Ammonium chloride

What's a good way to tell if you're unsure of the compound? Thanks!

Neutron
Post by: jakesilove on April 06, 2016, 10:32:41 am
Hey guys!

Got my Chemistry half yearly today and I've got a stupid question D: This is the question:

Which of the following aqueous solutions has a pH greater than 7?
a) Sodium ethanoate
b) Sodium chloride
c) Ammonium nitrate
d) Ammonium chloride

What's a good way to tell if you're unsure of the compound? Thanks!

Neutron

Hey!

EDIT: I was mistaken when answering this question: see below posts for the correct answer!

I don't think that's a stupid question at all! I think the most difficult part of the HSC Chemistry curriculum are sections where you just kind of need to memorise content, or rules, to get marks in a question.

Let's approach the multiple choice you've posed. First, I think it is fair to eliminate B straight away, as we know that NaCl is a neutral salt. It kind of has to be: we wouldn't want to be putting acidic/basic substances or our chips.

Next, I would try to recall that anything with an -oate on the end is know as an 'alkanoic acid' (from the Esterifaction part of your curriculum). Knowing that, I think it would be fair to eliminate A as well, since it will be acidic.

As for the last two, I honestly have no idea how to decide which is basic. I hope someone else in the forums can help me out, but I can't think of a way to distinguish between them in terms of any HSC level chemistry. I'm really sorry about that: at least at this stage you have a 50/50 chance of being right!

Again, so sorry about my terrible response.

Jake
Post by: Neutron on April 06, 2016, 11:53:45 am
Ahh it's okay, but strangely enough the answer was A? :/ No idea, also, what's a good amphoteric substance to use as an example because I originally chose Aluminium Oxide but I don't seem to be able to find what it forms when reacted with NaOH? Thanks :D

Neutron
Post by: IntelxD on April 06, 2016, 12:03:45 pm
Hey guys!

Got my Chemistry half yearly today and I've got a stupid question D: This is the question:

Which of the following aqueous solutions has a pH greater than 7?
a) Sodium ethanoate
b) Sodium chloride
c) Ammonium nitrate
d) Ammonium chloride

What's a good way to tell if you're unsure of the compound? Thanks!

Neutron

Sodium ethanoate will dissociate into sodium ions and ethanoate ions in an aqueous solution. Ethanoate (also known as acetate) is the weak conjugate base of ethanoic acid. Thus, ethanoate will react with the H+ ions, lowering [H+] and slightly increasing the pH of the solution above 7. Ammonium in option c and d is a weak acid and so will have the opposite effect. You should also note that ions such as chloride, sodium and nitrate do in fact have acidic or basic properties. They however are so weak that their effect on the pH of the solution is negligible and usually produces opposing effects (eg, NaCl) leaving the pH unchanged.
Post by: helloworld on April 06, 2016, 12:25:24 pm
Does anyone have answers to thushans notes for Chem on here?
Post by: HopefulLawStudent on April 06, 2016, 12:50:42 pm
Hey. This is the HSC Chem thread. This question is specific to VCE Chem (cuz I think Thushan only does notes/stuff for VCE). So if you pop on over to the VCE chem board (Chemistry), you'll probably be more likely to get an answer to your question cuz VCE chem kids are less likely to hang around over in this section of the forum.
Post by: helloworld on April 06, 2016, 01:14:22 pm
Thanks!
Post by: RuiAce on April 06, 2016, 05:43:55 pm
Hey guys!

Got my Chemistry half yearly today and I've got a stupid question D: This is the question:

Which of the following aqueous solutions has a pH greater than 7?
a) Sodium ethanoate
b) Sodium chloride
c) Ammonium nitrate
d) Ammonium chloride

What's a good way to tell if you're unsure of the compound? Thanks!

Neutron

We're looking for the basic salt. Because ammonium nitrate and ammonium chloride are acidic, they're automatically out. Why did I eliminate them in a matter of 1 second?

NO3- and Cl- are the conjugates of strong acids. That means they're extremely weak; virtually neutral. So they don't impact.

And it just so happens that substances such as Na+ and K+ are neutral cations. Whereas NH4+ is actually weakly acidic.

(Note, if I had ammonium acetate that would be different. That'd be around neutral.)

And then, like mentioned, sodium chloride is neutral. This is trivial, but it's not the first thing I notice due to the difference between sodium ions and ammonium ions.

Now, I think Jake made a mistake. Whilst ethanoic acid (aka acetic acid) is acidic, the ethanoate (acetate) ion is consequently basic. It is the conjugate to acetic acid, and is thus what causes the basic properties. Just like Intel mentioned.
Post by: jakesilove on April 06, 2016, 10:06:20 pm
Now, I think Jake made a mistake.

You're absolutely right, I'm sorry about that! Your responses all look fantastic, thanks for helping out!

Jake
Post by: Johny1234567 on April 10, 2016, 08:29:00 am
Draw separate concentration-time graphs for the following situations relating to the Haber Reaction.
a) The reactants are placed into the reaction vessel and are allowed to react for a longer period of time.
b) The reaction first reacts to equilibrium, and then the reaction vessel is heated.
c) The formation of ammonia with and without a catalyst, assuming all other conditions are the same. You do not need to show how hydrogen/ nitrogen vary on this graph.
Post by: RuiAce on April 10, 2016, 09:01:42 am
Draw separate concentration-time graphs for the following situations relating to the Haber Reaction.
a) The reactants are placed into the reaction vessel and are allowed to react for a longer period of time.
b) The reaction first reacts to equilibrium, and then the reaction vessel is heated.
c) The formation of ammonia with and without a catalyst, assuming all other conditions are the same. You do not need to show how hydrogen/ nitrogen vary on this graph.
a) As the Haber process does effectively take a HUGE amount of time to reach equilibrium, show this by making the process seemingly unable to obtain equilibrium in graph 1, but then achieving it in graph 2.
b) Recall that the equation 3 H2(g) + N2(g) ⇌ 2 NH3(aq) with ∆H=-92 kJ mol-1
The forward reaction is exothermic. Hence, if we introduce heat into the system, according to Le Chatelier's principle the system will attempt to readjust it's equilibrium to counter the change by shifting it to the left.
From a graph that's already attained equilibrium, suddenly the concentrations of the reactants (hydrogen and nitrogen) must go up a bit, and that of ammonia will go down.
c) Catalyst speeds up the rate of reaction. Hence, you achieve the same equilibrium, but you achieve it faster. This just means that on the graph, say the equilibrium is at 3. You increase up to 3 more quickly, but you end up with the same horizontal line anyway.

I will let someone else post relevant diagrams

P.S. I haven't been helping much these days cause I want to focus on mid sems
Post by: Johny1234567 on April 10, 2016, 11:05:20 am
ty, one more question:
Calculate the yield of the Haber reaction that used 30 kg of Nitrogen gas in excess hydrogen gas to form 1.2 kg of ammonia.
Post by: jakesilove on April 10, 2016, 11:22:15 am
ty, one more question:
Calculate the yield of the Haber reaction that used 30 kg of Nitrogen gas in excess hydrogen gas to form 1.2 kg of ammonia.

Hey!

Below is my solution. Hope it helps!

(http://i.imgur.com/ykfGQYA.png?1)

Jake
Post by: chloe9756 on April 14, 2016, 04:20:07 pm
hey, this is a really simply question but i can't manage to get my head around a few things:

The solubilities at 25C of three white barium salts, in grams per 100mL of water are:
Barium Sulfate 0.00025
Barium Nitrate 10.1
Barium Iodide 220

15g of a white powder was prepared by mixing 5.0g of each of these three barium salts. Your task to to obtain a sample of pure barium sulphate and a sample of virtually pure barium nitrate. Write the method for the separation process. Explain at each stage what happens to the mixture.

Firstly, when the question says to obtain PURE barium sulphate and VIRTUALLY PURE barium nitrate, how should that change the process you choose to use. Also, the answers specifically says that you must add more than 55mL of water, why is this so? And it also says after separating barium sulphate through filtration, barium nitrate will stay dissolved but barium iodide will crystallise out. Why is this so, doesn't barium iodide have a higher solubility. Thank you :)
Post by: katherine123 on April 20, 2016, 03:16:56 am
Help with this ques thanks
0.132g of pure carboxylic acid (R-COOH) was dissolved in 25ml of water and titrated with 0.120mol/L NaOH solution
A volume of 14.80ml was required to reach the endpoint of the titration.
The carboxylic acid could be:
A)HCOOH
B)CH3COOH
C)C2H5COOH
D)C3H7COOH
ans:C

question 2: Oxalic acid C2O4H2
A 25.0ml solution of oxalic acid reacts completely with 15.0ml of 2.50mol/L of NaOH
what is the concentration of the oxalic acid solution?
ans: 0.750M
Post by: mmadeleine on April 23, 2016, 06:29:55 pm
Hi, just a bit stuck on this titration question - I don't really understand how to do it so any help would be appreciated.  :)

"The electrolyte in car batteries is sulfuric acid. A student decided to determine the concentration of this acid in a well-charged car battery by taking exactly 2 mL by pipette and titrating it with 1.16 mol/L sodium hydroxide solution. 17.1 mL was needed to reach the equivalence point. Calculate the molarity of the sulfuric acid in the battery."

Thanks!
Post by: jakesilove on April 23, 2016, 07:22:08 pm
Hi, just a bit stuck on this titration question - I don't really understand how to do it so any help would be appreciated.  :)

"The electrolyte in car batteries is sulfuric acid. A student decided to determine the concentration of this acid in a well-charged car battery by taking exactly 2 mL by pipette and titrating it with 1.16 mol/L sodium hydroxide solution. 17.1 mL was needed to reach the equivalence point. Calculate the molarity of the sulfuric acid in the battery."

Thanks!

Hey!

Below is the general approach for all titration questions. I would recommend becoming quite comfortable with the method. Hope this helps!

(http://i.imgur.com/LBOLUSM.png?1)

Jake
Post by: Johny1234567 on April 24, 2016, 02:03:46 am
What is the effect on solubility of adding concentrated sodium carbonate to the carbon dioxide equilibrium.

Na2CO3 + H3O^+ -> CO2(g) + 2Na^+ + H2O
So, since sodium carbonate reacts with hydronium, the disturbance is that it decreases the hydronium concentration. This means it needs to shift right to increase hydronium concentration, correct? While this would increase CO2(aq) thus increasing solubility, there would be an increase in CO2(g) as well due to the reaction as stated above. Would the increase in CO2(g) also be treated as a disturbance, thus leading to an increase in solubilty. Or, does it simply decrease carbon dioxide solubity since it's a gas. Need some help please !
Post by: RuiAce on April 24, 2016, 08:51:05 am
What is the effect on solubility of adding concentrated sodium carbonate to the carbon dioxide equilibrium.

Na2CO3 + H3O^+ -> CO2(g) + 2Na^+ + H2O
So, since sodium carbonate reacts with hydronium, the disturbance is that it decreases the hydronium concentration. This means it needs to shift right to increase hydronium concentration, correct? While this would increase CO2(aq) thus increasing solubility, there would be an increase in CO2(g) as well due to the reaction as stated above. Would the increase in CO2(g) also be treated as a disturbance, thus leading to an increase in solubilty. Or, does it simply decrease carbon dioxide solubity since it's a gas. Need some help please !

The carbon dioxide equilibrium is just this:
H2CO3(aq) ⇌ H2O(l) + CO2(g)

Therefore I believe either you started the question wrongly, or the question was given misleadingly.

Introduction of sodium carbonate (which is a basic substance) will automatically cause it to react with the carbonic acid. According to LCP, because the concentrations of carbonic acid falls down, the equilibrium will shift to the left to minimise this disturbance. This implies that more carbon dioxide will be converted to carbonic acid, and thus solubility should increase.
Post by: RuiAce on April 24, 2016, 08:54:00 am
hey, this is a really simply question but i can't manage to get my head around a few things:

The solubilities at 25C of three white barium salts, in grams per 100mL of water are:
Barium Sulfate 0.00025
Barium Nitrate 10.1
Barium Iodide 220

15g of a white powder was prepared by mixing 5.0g of each of these three barium salts. Your task to to obtain a sample of pure barium sulphate and a sample of virtually pure barium nitrate. Write the method for the separation process. Explain at each stage what happens to the mixture.

Firstly, when the question says to obtain PURE barium sulphate and VIRTUALLY PURE barium nitrate, how should that change the process you choose to use. Also, the answers specifically says that you must add more than 55mL of water, why is this so? And it also says after separating barium sulphate through filtration, barium nitrate will stay dissolved but barium iodide will crystallise out. Why is this so, doesn't barium iodide have a higher solubility. Thank you :)

Making an assumption that this is indeed preliminary chemistry.

If the solubility of barium sulfate is as poor as THAT compared to the other substances, then filtration is automatically the best way to go. This is because you will need TONS of water for all of that barium sulfate to dissolve, whereas the other substances only require few. Thus, filtration should be used to seperate barium sulfate without any further thought.

Filtration only separates the barium sulfate. The barium nitrate and barium iodide remain dissolved in solution. I have absolutely no clue where they gathered that barium iodide will crystallise out.
Post by: RuiAce on April 24, 2016, 09:08:06 am
Help with this ques thanks
0.132g of pure carboxylic acid (R-COOH) was dissolved in 25ml of water and titrated with 0.120mol/L NaOH solution
A volume of 14.80ml was required to reach the endpoint of the titration.
The carboxylic acid could be:
A)HCOOH
B)CH3COOH
C)C2H5COOH
D)C3H7COOH
ans:C

question 2: Oxalic acid C2O4H2
A 25.0ml solution of oxalic acid reacts completely with 15.0ml of 2.50mol/L of NaOH
what is the concentration of the oxalic acid solution?
ans: 0.750M

For Q2, oxalic acid is indeed, diprotic.
2 NaOH(aq) + C2O4H2(aq) → Na2C2O4(aq) + 2 H2O(l)

Hence the mole ratio is given
n(NaOH) = 2n(C2O4H2)
Thus, using n = CV
[NaOH] * V(NaOH) = 2* [C2O4H2] * V(C2O4H2)

2.50 mol L-1 * 15.0 mL = 2 * [C2O4H2] * 25.0 mL
[C2O4H2] = 0.75 mol L-1 as the answers requested
______________________________________

Q1 is one of those typical rigorous questions that the HSC puts in, where you essentially have to do up to 4 calculations to find the answer.
We wish to determine if methanoic acid (formic acid), ethanoic acid (acetic acid), propanoic acid or butanoic acid was used in the titration. We firstly note that ALL of the carboxylic acids are monoprotic.

Acid + NaOH(aq) → Sodium salt + H2O(l)

In every case, the mole ratio is 1 to 1.
Thus, we have
n(Acid) = n(NaOH)

We will apply n=CV to the sodium hydroxide, but n=m/M to the acid:
m/M = CV
0.132g / M = 0.0148 L * 0.120 mol L-1
M = 74.32432... g mol-1

Now that we know the molar mass, we can simply use the periodic table to find out which acid is the correct acid.

It turns out that propanoic acid is correct:
M = 2*12.01 + 5*1.008 + 12.01 + 16.00 + 16.00 + 1.008 = 74.078

The tiny amount of inaccuracy should've been anticipated since the start.
Post by: Johny1234567 on April 24, 2016, 12:50:49 pm
The carbon dioxide equilibrium is just this:
H2CO3(aq) ⇌ H2O(l) + CO2(g)

Therefore I believe either you started the question wrongly, or the question was given misleadingly.

Introduction of sodium carbonate (which is a basic substance) will automatically cause it to react with the carbonic acid. According to LCP, because the concentrations of carbonic acid falls down, the equilibrium will shift to the left to minimise this disturbance. This implies that more carbon dioxide will be converted to carbonic acid, and thus solubility should increase.
I thought the CO2 equilibrium consisted of 4 equations in the following sequence:
1. CO2(g) ⇌ CO2(aq)
2. CO2(aq) + H2O(l) ⇌ H2CO3(aq)
3. H2CO3(aq) ⇌ H+(aq) + HCO3-(aq)
4. HCO3-(aq) ⇌ H+(aq) + CO3^2-(aq)
Post by: RuiAce on April 24, 2016, 02:55:07 pm
The first step is trivially assumed. Writing that out is pointless.

We are talking about the carbon dioxide equilibrium, whereby the second equation (with carbon dioxide as a gas) is predominant. The third and fourth equations technically do occur as a result of Brønsted-Lowry acid base theory.

However, their presence is negligible to this question, as the carbonic acid equilibrium is only specific to the interaction of carbon dioxide with water. This produces a substance that is mostly acidic, not basic.

Worthwhile mention: the hydrogen carbonate ion buffer makes greater use of the generation of the carbonate ion, and the carbonic acid equilibrium
Post by: Johny1234567 on April 24, 2016, 06:03:36 pm
Thank you :)
Post by: katherine123 on April 26, 2016, 10:56:20 pm
esters are polar but why are they non-water soluble?
Post by: RuiAce on April 27, 2016, 06:56:00 am
esters are polar but why are they non-water soluble?

Whilst the components of esters are quite appreciably polar, that is, the alkanol and alkanoic acid seperate, the ester itself is only a tiny bit polar. The polarity of the ester occurs as as the carboxylic acid group (-COOH) does contain one extra polar oxygen.

However, this is the only site of polarity. That OH part was dismantled when the ester was formed (note that water is a product of esterification). As this is the only site, the solubility of the substance will not be that great as soon as the chains start becoming longer in length. Much more of the molecule consists of carbon chains which are non-polar and do not readily dissolve into water.

(Some of the very short chained esters still somewhat dissolve into water.)
Post by: Happy Physics Land on April 27, 2016, 02:04:24 pm
Whilst the components of esters are quite appreciably polar, that is, the alkanol and alkanoic acid seperate, the ester itself is only a tiny bit polar. The polarity of the ester occurs as as the carboxylic acid group (-COOH) does contain one extra polar oxygen.

However, this is the only site of polarity. That OH part was dismantled when the ester was formed (note that water is a product of esterification). As this is the only site, the solubility of the substance will not be that great as soon as the chains start becoming longer in length. Much more of the molecule consists of carbon chains which are non-polar and do not readily dissolve into water.

(Some of the very short chained esters still somewhat dissolve into water.)

Essentially there is still partial solubility but as the chain length increases this solubility involving -COOH carboxylic group can be considered negligible
Post by: RuiAce on April 27, 2016, 08:17:27 pm
Essentially there is still partial solubility but as the chain length increases this solubility involving -COOH carboxylic group can be considered negligible
An ester doesn't have a -COOH functional group that's an alkanoic acid
Post by: JellyBeanz on April 27, 2016, 08:26:51 pm
An ester doesn't have a -COOH functional group that's an alkanoic acid

I think he's referring to the COO section when an ester is formed through the condensation reaction between an alkanoic acid and alkanol.
Post by: RuiAce on April 27, 2016, 09:07:53 pm
I think he's referring to the COO section when an ester is formed through the condensation reaction between an alkanoic acid and alkanol.

The "COO" group (obviously it isn't really one) loses most of it's impact mainly cause it's right in the middle of the molecule. Partial solubility is definitely present but if it's more towards the middle it is slightly limited from impacting as much as if it were on the end.

If it were on the end, an entire tail (or head) would be dissolved. Kinda the principles behind how soaps work (industrial chemistry option for the HSC). But being in the middle it will hardly dissolve a bit as the molecule would have to bend too much for the solubility to be appreciable.
Post by: vaish_vj on April 30, 2016, 12:30:18 pm
Hey!
My second 3/4 chem sac is regarding aspirin and oil of wintergreen (methyl salicylate).
Is they key structural difference between the two the fact that aspirin lacks a distinct hydroxyl functional group or the positioning of the methyl functional group on aspirin?
Post by: jakesilove on April 30, 2016, 04:07:21 pm
Hey!
My second 3/4 chem sac is regarding aspirin and oil of wintergreen (methyl salicylate).
Is they key structural difference between the two the fact that aspirin lacks a distinct hydroxyl functional group or the positioning of the methyl functional group on aspirin?

Hey! I think you're looking for the VCE Chemistry question board! Sorry, won't be able to help you out here :)

Jake
Post by: HighTide on May 02, 2016, 07:19:25 am
Hey!
My second 3/4 chem sac is regarding aspirin and oil of wintergreen (methyl salicylate).
Is they key structural difference between the two the fact that aspirin lacks a distinct hydroxyl functional group or the positioning of the methyl functional group on aspirin?
Apologies for invading the HSC thread... the structural difference would be more towards the functional groups present. Methyl salicylate has a hydroxyl group whilst aspirin has a carboxyl group. But also, if you draw it out, you'll see another difference in the components to form the ester.
Post by: Maz on May 02, 2016, 07:45:32 pm
Hey
i have a titration assessed practical coming up soon
the assessment is the we have to find the percentage of household ammonia in a cleaning product by tritrating it against HCl... with the solution diluted to 100mL and an aliquot of 10mL
I've done the calculation part to one...but we are also given density...and i don't quite know where the density fits into it?
the HCl has a concentration of 0.1082
Average volume of HCl titrate used
= 4.85 cm3
No. of moles of HCl reacted = 0.1082 x (4.85 x10-3)
= 5.248 x 10-4 mol

From equation, mole ratio of HCl ≡ NH3   is 1:1,
no. of moles of NH3 reacted = 5.248 x 10-4 mol
No. of moles of NH3 in volumetric flask (diluted) = (100 ÷ 10) × 5.248 x 10-4 mol
= 5.248 x 10-3 mol
No. of mole of NH3 in 10mL (undiluted)   = 5.248 x 10-3 mol
Concentration of NH3   = (5.248 x 10-3) ÷ (10 x 10-3) = 5.248 x 10-1 moldm-3

Molar mass of NH3 = 14.01 + 3 (1.0079)
= 17.0337 g mol-1

Therefore, weight-volume percentage of NH3 in window cleaner
= ((17.0337 x 5.248 x 10-1)/ 103) × 100%
= 0.894%

Percentage of ammonia in the window cleaner is: 0.894%
found the percentage of ammonia...but didnt use the density given...could you please tell me where i went wrong?
i'd really appreciate the help
Thankyou  :)
Post by: RuiAce on May 02, 2016, 09:10:49 pm
Hey
i have a titration assessed practical coming up soon
the assessment is the we have to find the percentage of household ammonia in a cleaning product by tritrating it against HCl... with the solution diluted to 100mL and an aliquot of 10mL
I've done the calculation part to one...but we are also given density...and i don't quite know where the density fits into it?
the HCl has a concentration of 0.1082
Average volume of HCl titrate used
= 4.85 cm3
No. of moles of HCl reacted = 0.1082 x (4.85 x10-3)
= 5.248 x 10-4 mol

From equation, mole ratio of HCl ≡ NH3   is 1:1,
no. of moles of NH3 reacted = 5.248 x 10-4 mol
No. of moles of NH3 in volumetric flask (diluted) = (100 ÷ 10) × 5.248 x 10-4 mol
= 5.248 x 10-3 mol
No. of mole of NH3 in 10mL (undiluted)   = 5.248 x 10-3 mol
Concentration of NH3   = (5.248 x 10-3) ÷ (10 x 10-3) = 5.248 x 10-1 moldm-3

Molar mass of NH3 = 14.01 + 3 (1.0079)
= 17.0337 g mol-1

Therefore, weight-volume percentage of NH3 in window cleaner
= ((17.0337 x 5.248 x 10-1)/ 103) × 100%
= 0.894%

Percentage of ammonia in the window cleaner is: 0.894%
found the percentage of ammonia...but didnt use the density given...could you please tell me where i went wrong?
i'd really appreciate the help
Thankyou  :)

Density = Mass / Volume

So if they give you the density and you have the volume, you can use it to find the mass

Take note that n = m/M, not n = V/M. Also take note that only WATER has a density of 1. For water, m = V
Post by: Maz on May 02, 2016, 09:21:02 pm
Density = Mass / Volume

So if they give you the density and you have the volume, you can use it to find the mass

Take note that n = m/M, not n = V/M. Also take note that only WATER has a density of 1. For water, m = V

so using that method could you bypass all the parts where in my solution i have worked it out through moles...
would u just do density x av titre (volume)= mass?
Post by: amandali on May 03, 2016, 12:13:08 am
1. How to explain "why can multiple indicators be used in titration of strong acid?"

2. You are given a titration graph of weak acid + strong base with curve ending at the basic region
is the reason for that due to weak acid producing strong conjugate base which reacts with H2O to produce OH- that increases pH
Post by: fluffchuck on May 03, 2016, 07:00:00 pm
Hey guys,

Do both ionic and covalent compounds contain intermolecular and intramolecular bonds? Thanks!!
Post by: Elizawei on May 03, 2016, 10:49:35 pm
Hey guys,

Do both ionic and covalent compounds contain intermolecular and intramolecular bonds? Thanks!!

Yup, they both do  ;)

EDIT: SORRY didn't realise this was the HSC chem thread, oops  :P
Post by: katherine123 on May 03, 2016, 11:46:30 pm
Explain the effect of liquefying the ammonia on the yield of reaction  2 marks

N2(g) + 3H2(g) <--> 2NH3(g)
[NH3(g)] decreases as NH3(g) --> NH3(l)   so does it shift to right because the concentration of NH3(g) has decreased  or because the reaction is under high pressure so it will shift to reduce the number of gas molecules to reduce pressure
Post by: amandali on May 03, 2016, 11:58:36 pm
Can you check my response? thanks

a) since pressure is inversely proportional to volume, the decrease in volume will increase the pressure. According to Le chatalier's principle, the equilibrium will shift right so that there will be fewer total gaseous molecules in the system since for every 2  molecules that react only 1 is formed which reduces pressure. This shift will cause an increase in concentration of Z(g) and also heat released as forward reaction is exothermic.

b) catalyst does not affect the position of equilibrium because it lowers the activation energy which increases the rate of both forward and reverse reactions equally. Therefore, equilibrium state is reached sooner but the yield of product is not increased.
Post by: Rishi97 on May 04, 2016, 10:37:54 am
Can you check my response? thanks

a) since pressure is inversely proportional to volume, the decrease in volume will increase the pressure. According to Le chatalier's principle, the equilibrium will shift right so that there will be fewer total gaseous molecules in the system since for every 2  molecules that react only 1 is formed which reduces pressure. This shift will cause an increase in concentration of Z(g) and also heat released as forward reaction is exothermic.

b) catalyst does not affect the position of equilibrium because it lowers the activation energy which increases the rate of both forward and reverse reactions equally. Therefore, equilibrium state is reached sooner but the yield of product is not increased.

Good responses :)
Just a small fix up is needed for response 2. The questions asks about the effect of a catalyst on the equilbrium mixture. However, you started off talking about the position of equilbrium straight away. This is correct but maybe just structure it a bit differently. So start of by explaining the role of a catalyst and then apply it to the situation/question given. For eg. a catalyst is a substance that increases the rate of reaction by lowering the activation energy. The catalyst increases the rate of both the forward and reverse reactions. Thus, the equilbrium state is reached sooner but the amount of yield is not affected.
Hope this makes sense :) great effort just need a bit of fixing up...but if your teacher is not a harsh marker, you may have still gotten the marks :)
Post by: meretz.etai on May 05, 2016, 11:26:50 am
Hey Jake,

Im having some problems understanding the idea equilibrium, can help explain?

Post by: Happy Physics Land on May 08, 2016, 01:24:20 pm
Hey Jake,

Im having some problems understanding the idea equilibrium, can help explain?

Hey meretz!

Clearly I'm not Jake, but I would be more than happy to help you out here! In year 11, all the equations that we have learnt about are all complete reactions, which means that the reaction only goes one way because the reaction would proceed until all the reactants are used up. A good example is combustion reaction (2C8H18(l) + 25O2(g) ----> 18H2O(l) + 16CO2(g). So if you think about it, when you burn a fuel, do you ever see the products H2O and CO2 reacting to form the original fuel again? Definitely not. In contrast an equilibrium reaction is basically any reactions that do not go to complete and therefore the reaction can be reversed. For instance, the Haber Process N2(g) + H2(g) --> / <--- NH3(g) is a reversible process. In this case, when nitrogen reacts with hydrogen to form ammonia, some ammonia decomposes back into nitrogen and hydrogen as well. The amount of reactants/products forms are governed by Le Chatelier's principle which talk about the impact of temperature, pressure and concentration upon the equilibrium reaction. Be mindful that this does not affect complete reactions because it is a ONE-WAY reaction and therefore the products cannot decompose to form reactants again.

When we talk about chemical equilibrium, we are referring to a state of a close chemical system in which the concentrations of both reactants and products do not change with time and the rate of the forward reaction is equal to the rate of the reverse reaction. For instance, consider a sealed bottle of soda water. There is dynamic equilibrium happening because carbon dioxide is dissolving in the liquid and carbon dioxide being form as gas in the air above the liquid simultaneously. The equilibrium is maintained unless concentration changes, pressure changes or temperature changes which will cause a disturbance.

Post by: hannahboardman98 on May 11, 2016, 07:39:14 pm

A particular brand of vinegar claimed to 4.3% acetic acid. 10mL of the vinegar was diluted to make a 100mL solution. The claim was tested using a titration where 16mL of 0.1M sodium hydroxide was needed to neutralise 25mL of the 10% vinegar solution. Calculate the molarity (moles/L) and the percentage (w/v, ie - g/100mL = % ) of acetic acid in the original vinegar. Comment on the 4.3% claim.
Post by: jakesilove on May 12, 2016, 02:17:22 pm

A particular brand of vinegar claimed to 4.3% acetic acid. 10mL of the vinegar was diluted to make a 100mL solution. The claim was tested using a titration where 16mL of 0.1M sodium hydroxide was needed to neutralise 25mL of the 10% vinegar solution. Calculate the molarity (moles/L) and the percentage (w/v, ie - g/100mL = % ) of acetic acid in the original vinegar. Comment on the 4.3% claim.

Hey! I've attached my answers below. These sorts of questions can be tough, as they require a whole bunch of steps and a lot of maths. Once you've done a few of them, however, you should be able to work through others!

(http://i.imgur.com/Umatd5x.png?1)

Jake
Post by: Loki98 on May 12, 2016, 05:20:52 pm
Hey,
I have a titration practical coming up to determine the ethanoic acid content in a vinegar sample.
In the marking criteria it asks for "specific equations for the reactions that occur during the lab," but i only know of the neutralisation reaction. What could the other possible equations be?
Also for the conclusion, it asks to "identify area(s) of the lab most likely to cause experimental error and propose solutions to problems encountered in the investigation." I have asked my teacher about errors such as Naoh being hygroscopic and altering the results, the end point not being the same as the equivalence point and how colour changes of the indicator are not instant, I have also asked about the indicator changing the pH of the solution and also how each individual can have a unique colour perception when determining the end point. But, my teacher has rejected all of those suggestions and also said no human errors and equipment errors can be written. What other possible errors and improvements are there? Am i misreading something from the criteria?
Also, what should i write about the validity and accuracy of the experiment?
-Thanks
Post by: jakesilove on May 12, 2016, 06:12:04 pm
Hey,
I have a titration practical coming up to determine the ethanoic acid content in a vinegar sample.
In the marking criteria it asks for "specific equations for the reactions that occur during the lab," but i only know of the neutralisation reaction. What could the other possible equations be?
Also for the conclusion, it asks to "identify area(s) of the lab most likely to cause experimental error and propose solutions to problems encountered in the investigation." I have asked my teacher about errors such as Naoh being hygroscopic and altering the results, the end point not being the same as the equivalence point and how colour changes of the indicator are not instant, I have also asked about the indicator changing the pH of the solution and also how each individual can have a unique colour perception when determining the end point. But, my teacher has rejected all of those suggestions and also said no human errors and equipment errors can be written. What other possible errors and improvements are there? Am i misreading something from the criteria?
Also, what should i write about the validity and accuracy of the experiment?
-Thanks

Hey!

I'm sure that the marking criteria will really only require you to know neutralisation reactions for a Titration prac like this. I wouldn't worry about other areas of Chemistry for that :)

I'll have to think a bit more about your error question, because those sound like all the important errors to me. The dotpoint must mean something else; I feel like the 'area(s) of the lab' part makes it a bit weird and ambiguous, and I don't really know what the suggest. Surely, somehow, it is talking in a more general sense about the actual physical features of the lab. Maybe thinks like the floor being at a slope, causing your reading on the Burrette to be skewed? Air pressure? Humidity? Light causing increased reaction time? Temperature? I really don't know what the suggest, because it feels like you nailed all the important things and we're just pulling at strings.

For validity, you want to talk about whether you actually measured what you were trying to measure. In real terms, that means you want to have controlled all of the variables in your experiment. Things like correctly washing your equipment, keeping the room at a constant temperature, the hydroscopicness of NaOH etc (all of which change the result of your experiment) fall into this category.

For accuracy, you want to talk about your measuring devices. How accurate was your scale, did you account of paralax error, how accurate are your eyes. Essentially, it asks to what extent your measured values are close to the real value.

I'm sorry I couldn't be more helpful for the substance of your question! Maybe ask your teacher to explain the dotpoint; you not understanding it isn't your fault, its theirs.

Jake
Post by: Happy Physics Land on May 12, 2016, 06:43:56 pm
Hey,
I have a titration practical coming up to determine the ethanoic acid content in a vinegar sample.
In the marking criteria it asks for "specific equations for the reactions that occur during the lab," but i only know of the neutralisation reaction. What could the other possible equations be?
Also for the conclusion, it asks to "identify area(s) of the lab most likely to cause experimental error and propose solutions to problems encountered in the investigation." I have asked my teacher about errors such as Naoh being hygroscopic and altering the results, the end point not being the same as the equivalence point and how colour changes of the indicator are not instant, I have also asked about the indicator changing the pH of the solution and also how each individual can have a unique colour perception when determining the end point. But, my teacher has rejected all of those suggestions and also said no human errors and equipment errors can be written. What other possible errors and improvements are there? Am i misreading something from the criteria?
Also, what should i write about the validity and accuracy of the experiment?
-Thanks

Hey Loki!

Reading your description, I'm assuming that you are conducting a titration involving acetic acid and NaOH? Ok, then let's get started!

When the teacher asks you for equations, you need to know 3 equations: (for convenience sake lm not gonna write in states, but you should write them in an exam)
1. The Balanced chemical equation (CH3COOH + NaOH ----> NaCH3COO + H2O)
2. The net ionic equation (CH3COO- + Na+ ----> NaCH3COO)
3. Hydrolysis equations of salt (CH3COO- + H2O ---> CH3COOH + OH-)

Equation 1 and 2 are pretty straight forwards. Equation 3 is also significant because it shows why the endpoint would be slightly basic. Common exam questions would ask you for whether the endpoint is basic, acid or neutral and would ask you to identify a suitable indicator. If your teacher wants to differentiate between band 5 students from band 6 students, they would ask you why the endpoint would be basic. In this case, referring to equation 3 from above, since hydroxide ion is given off, the endpoint would hence be basic.

With your second question identifying the possible errors in the experiment, I would personally have written a human error or equipment error (it's kinda weird how your teacher is completely rejecting these errors). But look if your teacher says none of these can be included, then we have to find some other flaws in the experiment. One thing you can perhaps talk about is impurities from the atmosphere entering the conical flask or the burette and this can cause slight differences. Another thing maybe carrying out titration at the wrong temperature, which would affect experimental outcome because some indicators are quote sensitive to surrounding temperature. I'm a little confused by the meaning of "area(s) of the lab", because Im not too sure whether your teacher is talking about the experiment or the physical places in the lab that would cause error. Can you please enquire the teacher and come back to us on that? That would be helpful.

With the validity and accuracy part its quite easy, I will list some things for you to consider and you can write it out in a formal manner:

Validity:
- the extent to which the experiment tests the hypothesis
- Is there only one independent variable involved? If yes, then the experiment is valid
- Does your methods answer the aim? If yes, then the experiment is valid
- Is there a very significant difference between your result and your hypothesis? If yes, then the experiment is invalid. (This may sound like a point for accuracy, but accuracy is only concerned with relatively small deviations from your hypothesis. A significant deviation would mean that you are doing something wrong)

Accuracy:
- Did you thoroughly clean your equipments? Did you do it for multiple times? If yes, accurate
- Are your instruments correctly calibrated? If yes, accurate
- Is your weighing instrument presenting a precise quantity? (i.e. the value it displays does not fluctuate) If yes, accurate
- Are you biased in your reading of the scale? If yes, inaccurate
- Is the scale on your pipette and burette correct? If yes, accurate
- Are you reading from the bottom of meniscus (i.e. parallax error)? If yes, accurate
Then talk about how to minimise the impact of these inaccuracies if they exist.

Hope my answer helped! Remember to come back to us after you have enquired the teacher about that second point! :D

Best Regards
Happy Physics Land

- Are all other variables controlled? If yes, then the experiment
Post by: Happy Physics Land on May 12, 2016, 06:45:29 pm
Hey!

I'm sure that the marking criteria will really only require you to know neutralisation reactions for a Titration prac like this. I wouldn't worry about other areas of Chemistry for that :)

I'll have to think a bit more about your error question, because those sound like all the important errors to me. The dotpoint must mean something else; I feel like the 'area(s) of the lab' part makes it a bit weird and ambiguous, and I don't really know what the suggest. Surely, somehow, it is talking in a more general sense about the actual physical features of the lab. Maybe thinks like the floor being at a slope, causing your reading on the Burrette to be skewed? Air pressure? Humidity? Light causing increased reaction time? Temperature? I really don't know what the suggest, because it feels like you nailed all the important things and we're just pulling at strings.

For validity, you want to talk about whether you actually measured what you were trying to measure. In real terms, that means you want to have controlled all of the variables in your experiment. Things like correctly washing your equipment, keeping the room at a constant temperature, the hydroscopicness of NaOH etc (all of which change the result of your experiment) fall into this category.

For accuracy, you want to talk about your measuring devices. How accurate was your scale, did you account of paralax error, how accurate are your eyes. Essentially, it asks to what extent your measured values are close to the real value.

I'm sorry I couldn't be more helpful for the substance of your question! Maybe ask your teacher to explain the dotpoint; you not understanding it isn't your fault, its theirs.

Jake

Man the one time that I actually get to answer a chem question and you beat me to it
Post by: RuiAce on May 12, 2016, 08:28:01 pm
Man the one time that I actually get to answer a chem question and you beat me to it

I'm just too overloaded with work right now or I'd address every chemistry question. For now, I'm focusing on my specialty (even though it's not my top interest).
Post by: cajama on May 12, 2016, 09:03:33 pm
Hi!
I'm not sure if this has been asked before but I do three sciences; Chemistry, Biology and Physics and I often get my scientists mixed up between the subjects (especially with Chemistry and Biology). What information of the scientist is sufficient for me to know?
Post by: jakesilove on May 12, 2016, 09:31:46 pm
Hi!
I'm not sure if this has been asked before but I do three sciences; Chemistry, Biology and Physics and I often get my scientists mixed up between the subjects (especially with Chemistry and Biology). What information of the scientist is sufficient for me to know?

Hey!

I can't really help you with Biology, but in terms of learning scientist information in general, you don't need to know very much detail. If it's an important dot-point, tested often (eg. Physicist in the Space section of the curriculum), then you should know at least some history, their contribution to the scientific field, and maybe some extra facts the throw in and sound smart. This applies similarly to the Haber process for Chemistry. When it comes to less important dotpoints, almost never tested (eg. that one weird one about a chemist at the start of Chemical Monitoring and Management), you don't need to worry too much. Know what sort of science they're involved with, maybe some chemical processes they use.

Overall, you don't need to know very much. Mainly, know context and contribution. There aren't many dotpoints requiring you to know about the contribution of specific scientists, so hopefully as you learn each one a bit better you won't get them as mixed up!

This all being said, there are some dotpoints that are person-related and require quite a bit of knowledge (as sometimes you can get a 5-7 marker on them). For instance, Westinghouse vs Edison. Basically just look through some past papers and decide how significant that person is; often, you won't need to know very much.

Jake
Post by: cajama on May 12, 2016, 10:15:25 pm
Hey!

I can't really help you with Biology, but in terms of learning scientist information in general, you don't need to know very much detail. If it's an important dot-point, tested often (eg. Physicist in the Space section of the curriculum), then you should know at least some history, their contribution to the scientific field, and maybe some extra facts the throw in and sound smart. This applies similarly to the Haber process for Chemistry. When it comes to less important dotpoints, almost never tested (eg. that one weird one about a chemist at the start of Chemical Monitoring and Management), you don't need to worry too much. Know what sort of science they're involved with, maybe some chemical processes they use.

Overall, you don't need to know very much. Mainly, know context and contribution. There aren't many dotpoints requiring you to know about the contribution of specific scientists, so hopefully as you learn each one a bit better you won't get them as mixed up!

This all being said, there are some dotpoints that are person-related and require quite a bit of knowledge (as sometimes you can get a 5-7 marker on them). For instance, Westinghouse vs Edison. Basically just look through some past papers and decide how significant that person is; often, you won't need to know very much.

Jake

Ahh thanks!  ;D
Post by: Happy Physics Land on May 12, 2016, 10:40:33 pm
Hi!
I'm not sure if this has been asked before but I do three sciences; Chemistry, Biology and Physics and I often get my scientists mixed up between the subjects (especially with Chemistry and Biology). What information of the scientist is sufficient for me to know?

Hey Cajama:

Adding onto Jake's point, in chemistry there is actually a syllabus dotpoint which demands for students to know about the works of different chemists and the importance of their collaboration. To satisfy this dotpoint you would need to be familiarised with the various fields of chemistry such as analytical chemists, industrial chemists, environmental chemists, and radiography chemists. It is also recommended that you look up the techniques each one would use in their individual specialties. For example, analytical chemists use the technique of AAS (Atomic Absorption Spectroscopy) to qualitatively detect the presence of lead in water. It is also recommended that you know the name of a chemist that works for a particular field (This actually came up in a HSC exam a few years ago) and you would be required to talk about the details of the works that relate to the chemist.
Post by: smiley2101 on May 15, 2016, 12:16:37 pm
why is the answer B? thank you!!!!! :)
Post by: RuiAce on May 15, 2016, 03:41:05 pm
why is the answer B? thank you!!!!! :)

Using n=m/M, for every 1 gram of substance x we have y moles of it:

x is CO, y = 1/28.01 mol
x is CH4, y = 1/16.042 mol
x is C2H2, y = 1/26.036 mol
x is C2H6, y = 1/30.068 mol

So since the heat q released of reaction is just enthalpy change * moles:

CO: 1/28.01 * 233 = 8.318 kJ
CH4: 1/16.042 * 890 = 55.479 kJ
C2H2: 1/26.036 * 1300 = 49.93 kJ
C2H2: 1/30.068 * 1560 = 51.88 kJ

Post by: ProfLayton2000 on May 16, 2016, 06:00:23 pm
(Hydrogen Iodide) there,

So in the acidic env. topic we were told that phosphoric acid is weak, nitric acid is strong etc. Is there a definite list of strong/weak acids/bases we need to know for the HSC course?
Post by: jakesilove on May 16, 2016, 06:56:04 pm
(Hydrogen Iodide) there,

So in the acidic env. topic we were told that phosphoric acid is weak, nitric acid is strong etc. Is there a definite list of strong/weak acids/bases we need to know for the HSC course?

Hey!

Yep there certainly is! The acids that you need to know are strong acids are Nitric acid, Sulfuric acid and Hydrochloric acid. You can assume that any other acid you are given is weak (there are very few strong acids!).

Similarly, when it comes to strong bases you only really need to know that Sodium Hydroxide is weak.

Hope that helps!
Post by: RuiAce on May 16, 2016, 09:01:07 pm
Hey!

Yep there certainly is! The acids that you need to know are strong acids are Nitric acid, Sulfuric acid and Hydrochloric acid. You can assume that any other acid you are given is weak (there are very few strong acids!).

Similarly, when it comes to strong bases you only really need to know that Sodium Hydroxide is weak.

Hope that helps!
Whoops.

@ProfLayton2000

Anyway, as Jake has pointed out, the main ones that you need to know are:

Strong acids:
HNO3
HCl
H2SO4

However, it should be quite obvious as to why HBr and HI are also strong. (What may be surprising instead, is that HF is in fact, weak. This occurs due to the fact fluorine is the most electronegative element.)

Strong bases are simple. Group I hydroxides and that is it.
Post by: Neutron on May 16, 2016, 09:14:54 pm
Hey guys!

I was learning about coordinate covalent bonds today and I was wondering how you can tell which pairs form a product that does contain a coordinate covalent bond? Sorry if this seems simple but here's an example of a question:

Which of the following pairs form a product containing a coordinate covalent bond?
a) Ca and 2H+
b) H2O and H+
c) Ag+ and Cl-
d) NH4+ and OH-

Thank you! you're all amazing :D

Neutron
Post by: RuiAce on May 16, 2016, 09:35:27 pm
Hey guys!

I was learning about coordinate covalent bonds today and I was wondering how you can tell which pairs form a product that does contain a coordinate covalent bond? Sorry if this seems simple but here's an example of a question:

Which of the following pairs form a product containing a coordinate covalent bond?
a) Ca and 2H+
b) H2O and H+
c) Ag+ and Cl-
d) NH4+ and OH-

Thank you! you're all amazing :D

Neutron

The hydronium H3O+ is known to contain a coordinate covalent bond.

Otherwise, the process of elimination works as follows:
c) is definitely wrong as AgCl is an example of an ionic compound
a) is an impossibility.
d) is wrong because NH4+ and OH- will react in a neutralisation reaction to produce H2O and NH3
Therefore the answer must be b)
Post by: katherine123 on May 17, 2016, 09:38:47 pm
1. what is the method to identify that it's CaSO4(s)
the ans: include adding HNO3 and i dont get the purpose of that    isnt acid added just for like metal carbonates

don't we normally identify what ions are present in the solution rather than what type of compound it is

2. do we differentiate Fe(OH)2 and Fe(OH)3 because Fe(OH)2 goes from white to brown and then Fe(OH)3 stays consistently brown

3.the solubility rules differ across different sources
i got mine from the matrix website but i dont know if its a good idea

Post by: RuiAce on May 17, 2016, 10:06:21 pm
1. what is the method to identify that it's CaSO4(s)
the ans: include adding HNO3 and i dont get the purpose of that    isnt acid added just for like metal carbonates

don't we normally identify what ions are present in the solution rather than what type of compound it is

2. why does Fe(OH)2 (s) turn from white to brown

Not enough info

If these are your cation/anion tests, you need to give the whole entire question.

(For Q1 however, if that IS the entire question, then CaSO4 can be VERIFIED by an acid test, but not IDENTIFIED.)
Post by: RuiAce on May 19, 2016, 10:36:38 pm
Still insufficient information after the edit for Q1. Nitric acid will react with both carbonates and sulfates. But it doesn't distinguish between which is what.

Q2: The hydroxide test to determine which iron ion (lol) is present is a bit ambiguous.

In the laboratory, sometimes what happens is that Fe2+ shows up in a colour it's not meant to show up. Fe3+ is technically meant to stay brown as you stated, but I think when I did the test for Fe2+ I actually ended up with green.

Q3: ALWAYS keep in mind that multiple cation have MORE THAN ONE acceptable test pathway to determine its nature. Some sources refer to "secondary tests" because they provide further evidence to what they call a "primary test"
Post by: Johny1234567 on May 20, 2016, 11:09:44 pm
What is the half-equations illustrating the reaction between iodide ions and dilute sulfuric acid?
I know how to do it with concentrated sulfuric acid, but the aqueous symbol for dilute throws me off since I'm not quite sure about the oxidation half. Help appreciated :)
Post by: Sheikie on May 21, 2016, 04:31:21 pm
hey, I was wondering what is the best way to study for chemistry pracs?
Post by: RuiAce on May 21, 2016, 06:19:25 pm
What is the half-equations illustrating the reaction between iodide ions and dilute sulfuric acid?
I know how to do it with concentrated sulfuric acid, but the aqueous symbol for dilute throws me off since I'm not quite sure about the oxidation half. Help appreciated :)
This is a bit confusing. Which galvanic cell are you interested in? Or what's the full question.

Alternatively, give the half equations to concentrated sulfuric acid for a reference point for me

hey, I was wondering what is the best way to study for chemistry pracs?

Studying for pracs is a matter of:
- Knowing the aim
- Knowing the method
- Only knowing results if its relevant (however this is often more than not)
- Knowing limitations to the experiment
- Identifying controls/variables, as well as relating validity/reliability/discussion
Post by: jakesilove on May 21, 2016, 06:47:15 pm
hey, I was wondering what is the best way to study for chemistry pracs?

Hey!

I think that, generally, there is little to study when it comes to chemistry pracs. Most of the time, you aren't being assessed on specific, actual knowledge of the course; rather, on your analytical skills, your ability to form a discussion and sometimes your ability to do calculations.

Other than being comfortable with a titration, in case you are asked to do one, I think that the main this to worry about is being able to write balanced chemical formulas, use equations from throughout the course, and understanding what accuracy, reliability and validity are. This latter point is especially important, in case you are asked (as often occurs) to write out a discussion for your prac.

You need to be able to write a risk assessment (risk, precaution, response) and be able to graph any results you find.

I think that's about it; I didn't elaborate on each of the above notes because I figured that you probably understand most of them, but if there is any part of this answer that I can expand on (accuracy, reliability and validity for example) please let me know! Happy to help out however I can.

Jake
Post by: RuiAce on May 21, 2016, 07:06:01 pm
Hey!

I think that, generally, there is little to study when it comes to chemistry pracs. Most of the time, you aren't being assessed on specific, actual knowledge of the course; rather, on your analytical skills, your ability to form a discussion and sometimes your ability to do calculations.

Other than being comfortable with a titration, in case you are asked to do one, I think that the main this to worry about is being able to write balanced chemical formulas, use equations from throughout the course, and understanding what accuracy, reliability and validity are. This latter point is especially important, in case you are asked (as often occurs) to write out a discussion for your prac.

You need to be able to write a risk assessment (risk, precaution, response) and be able to graph any results you find.

I think that's about it; I didn't elaborate on each of the above notes because I figured that you probably understand most of them, but if there is any part of this answer that I can expand on (accuracy, reliability and validity for example) please let me know! Happy to help out however I can.

Jake

Kinda need both in my opinion. I distinctively recall panicking when a past paper question said "outline a experiment to produce an ester in a laboratory"
Post by: jakesilove on May 21, 2016, 07:14:39 pm
Kinda need both in my opinion. I distinctively recall panicking when a past paper question said "outline a experiment to produce an ester in a laboratory"

I totally agree in terms of needing to know the actual pracs etc. when it comes to examination, but I assume the question was about an in-class practical task? A prac exam?

Jake
Post by: RuiAce on May 21, 2016, 08:56:52 pm
I totally agree in terms of needing to know the actual pracs etc. when it comes to examination, but I assume the question was about an in-class practical task? A prac exam?

Jake

Oh touche. It does give off that vibe.

I generally just assume the final exam though because what you have to memorise for that overlaps into what you have to know for a practical exam anyway
Post by: therealqwerty on May 23, 2016, 08:13:22 pm
Can you explain how to do this question, thanks  :)
Post by: RuiAce on May 23, 2016, 08:52:51 pm
Can you explain how to do this question, thanks  :)
Should be Q and R I reckon.

Q is the most reactive because not only does it burn in oxygen and react with acid, but it even reacts with water. (It might be aluminium)

R is the least reactive because it just reacts slowly with oxygen

S is in the middle

So the difference in Eo values for Q and R should be the greatest
Post by: jakesilove on May 23, 2016, 09:19:41 pm
Should be Q and R I reckon.

Q is the most reactive because not only does it burn in oxygen and react with acid, but it even reacts with water. (It might be aluminium)

R is the least reactive because it just reacts slowly with oxygen

S is in the middle

So the difference in Eo values for Q and R should be the greatest

Yep I agree! You always want to look for a reaction between a substance highest on the standard reduction potential table, and a substance lowest on the standard reduction potential table. As this table is also a marker of reactivity, you want a substance that is very reactive, and a substance that is not very reactive, to make the best possible galvanic cell.

Jake
Post by: therealqwerty on May 24, 2016, 07:48:01 am
Should be Q and R I reckon.

Q is the most reactive because not only does it burn in oxygen and react with acid, but it even reacts with water. (It might be aluminium)

R is the least reactive because it just reacts slowly with oxygen

S is in the middle

So the difference in Eo values for Q and R should be the greatest

Thanks :)
Post by: angela99 on May 24, 2016, 06:53:07 pm
my teacher constantly tells me that my prac sections in the exam are my weakest.. im not really sure how to study the pracs we do in class or how to write notes on them
any tips or layouts?
Post by: RuiAce on May 24, 2016, 07:28:09 pm
my teacher constantly tells me that my prac sections in the exam are my weakest.. im not really sure how to study the pracs we do in class or how to write notes on them
any tips or layouts?

Studying for pracs is a matter of:
- Knowing the aim
- Knowing the method
- Only knowing results if its relevant (however this is often more than not)
- Knowing limitations to the experiment
- Identifying controls/variables, as well as relating validity/reliability/discussion
Hey!

I think that, generally, there is little to study when it comes to chemistry pracs. Most of the time, you aren't being assessed on specific, actual knowledge of the course; rather, on your analytical skills, your ability to form a discussion and sometimes your ability to do calculations.

Other than being comfortable with a titration, in case you are asked to do one, I think that the main this to worry about is being able to write balanced chemical formulas, use equations from throughout the course, and understanding what accuracy, reliability and validity are. This latter point is especially important, in case you are asked (as often occurs) to write out a discussion for your prac.

You need to be able to write a risk assessment (risk, precaution, response) and be able to graph any results you find.

I think that's about it; I didn't elaborate on each of the above notes because I figured that you probably understand most of them, but if there is any part of this answer that I can expand on (accuracy, reliability and validity for example) please let me know! Happy to help out however I can.

Jake
...
Post by: ProfLayton2000 on June 02, 2016, 09:01:27 pm
How do you assess the validity of an experiment?
Post by: jakesilove on June 02, 2016, 09:32:11 pm
How do you assess the validity of an experiment?

Hey there!

For validity, you're essentially asking yourself did the experiment test what you were trying to test. In other words, did the set up actually answer the aim of the experiment?

This often takes into account things like whether or not you have controlled your variables, whether there may be any impurities in your substances, whether you put the right things in the right place and whether you used the correct kind of glassware. An example would be the experiment where you test the specific heat capacity of specific substances (ethanol etc.). Most of the heat DOESN'T get transferred into the water; most of it is lost to the environment! So, you're not REALLY testing what you're trying to test, greatly limiting the validity of the experiment.

If you have any specific questions, regarding specific pracs, let me know! Hope this helped :)

Jake
Post by: browny2409 on June 04, 2016, 03:15:32 pm
Hi, there is a dotpoint about the reliability of the sulphate fertiliser prac - what exactly else would you talk about for reliability other then repeatability?

Thanks
Nathan
Post by: RuiAce on June 04, 2016, 04:00:28 pm
Hi, there is a dotpoint about the reliability of the sulphate fertiliser prac - what exactly else would you talk about for reliability other then repeatability?

Thanks
Nathan

Just kidding :P But not really - in a first hand investigation, repetition is the main requirement for reliability to be achieved. The more you repeat the experiment, the more reliable it is.

As for a dot point, yes. Refer to Chemical Monitoring and Management section 3. The dot points are #3 and #4 of the third column.

(Since your results involve a numeric analysis, if your figures are all over the place then that's clearly not as reliable as figures such as 70g, 72g and 69.5g. So if validity or accuracy crumbles, in this case reliability may also be damaged. This is probably the reason why reliability has a seperate dot point for this practical.)
Post by: relativity1 on June 06, 2016, 02:35:15 pm
couple of questions
-how to improve accuracy/reliability of the molar heat of combustion prac
-why weak acids have lower molar heats of neutralization than strong acids (something to do with ionisation i think)
Post by: Adriaclya on June 06, 2016, 03:12:30 pm
Hi guys, i am really confused about moles, molar masses and such. Can someone please enlighten me?
Thanks!
Post by: RuiAce on June 06, 2016, 05:31:59 pm
couple of questions
-how to improve accuracy/reliability of the molar heat of combustion prac
-why weak acids have lower molar heats of neutralization than strong acids (something to do with ionisation i think)

Reliability is trivial. Keep repeating the experiment.
Accuracy can be a bit harder though. For this prac, accuracy ties down to validity a lot. Most of the time your molar heat of combustion prac will be quite noticeably inaccurate because heat loss to the surroundings is so hard to perfectly control. It's almost impossible to create the perfect environment in the laboratory to perform a fully accurate experiment here.

Hi guys, i am really confused about moles, molar masses and such. Can someone please enlighten me?
Thanks!
That doesn't really give me much direction as to where you should be headed.

Avogadro's number: 6.022 * 1023 = NA

1 mol of a substance = Avogadro's number times the number of molecules of that substance.
I.e. 1 mol of O2 is the same as saying you have  6.022 * 1023 oxygen molecules present.

The fundamental equation of chemistry is n=m/M where m is the mass of a substance present, and M is the molar mass of said substance. The molar mass can be extracted straight off the periodic table.
Post by: Johny1234567 on June 12, 2016, 01:52:39 am
Analyse the industrial consideratinos of the Solvay Process. (7 marks)

Critically analyse the environmental concerns posed by the Solvay Process and the way in which these are addressed. (7 marks)
Post by: ProfLayton2000 on June 15, 2016, 05:05:23 pm
Hi there,

"Chemists are involved in monitoring and managing the environment. They deal with the reactants, products and conditions of reactions that occur in the environment around us. To do this they identify and measure chemicals around us.  Discuss equipment and methods used by chemists to monitor our environment" (7 marks)

What do you suggest I include in my answer (and could you recommend a structure )?
Post by: jakesilove on June 15, 2016, 08:06:37 pm
Hi there,

"Chemists are involved in monitoring and managing the environment. They deal with the reactants, products and conditions of reactions that occur in the environment around us. To do this they identify and measure chemicals around us.  Discuss equipment and methods used by chemists to monitor our environment" (7 marks)

What do you suggest I include in my answer (and could you recommend a structure )?

Hey!

This is an awfully vague, slightly absurd question to be asked. Still, for a 7 marker, it's a beast of a thing so it's important that we are able to tame it! I would probably recommend talking about at least two big hitter issues, discussing what they are, how scientists monitor it and why scientists monitor it.

The most obvious choices are the production of CFCs (and depletion of Ozone) and polluting waterways. In terms of structure, I would start with a quick introduction (Scientists monitor the environment and use chemical equations etc. etc. to protect etc. etc.). Then, throw a subheading "Ozone" in. Talk about what Ozone is, why it's important, what CFCs are, why they were significant, how we detected them, change over time etc. MAKE SURE TO TALK ABOUT EQUIPMENT used to monitor this process: This is specifically in the question, and forgetting to do so could lose you marks. Then, I'd do the same thing with waterways. Be fairly brief; you want to get across the fact that chemicals are scary, and can have huge impacts on our environment, and that monitoring them is the only way to be sure we don't all die. At the same time, mention specific information if you can (dates, equipment, processes). It's possible that talking about two separate things is a bit much; perhaps one would be sufficient. Either way, as long as you get the point across you'll get the marks! If you feel like writing up a response, I'd be happy to take a look at it :)

Let me know if I can elaborate on any of this or answer any other questions!

Jake
Post by: katherine123 on June 17, 2016, 02:37:13 am
Post by: Jakeybaby on June 17, 2016, 08:15:30 am

It's irrelevant to the fact that one is weak and one is strong. It's due to both acids having the same concentration, therefore will both be titrated by the base at the same rate. The equivalence point will be the same as the same about of base will be needed to reach equivalence.
Post by: jakesilove on June 17, 2016, 09:06:37 am
It's irrelevant to the fact that one is weak and one is strong. It's due to both acids having the same concentration, therefore will both be titrated by the base at the same rate. The equivalence point will be the same as the same about of base will be needed to reach equivalence.

Absolutely on point! Think about it this way: When you're doing the maths (ie. C=n/v, moles=mass/molar mass etc.), the 'strong/weak' nature of the acid doesn't come into play at all. All you care about is the concentration, and the molar ratio, which is exactly the same between the two substances! Thanks for the response Jakey, quality stuff.

Jake
Post by: nimasha.w on June 17, 2016, 11:25:14 pm
hi! i'm stuck on trying to find a ppt test to distinguish between Ca2+ and Ba2+
Post by: Jakeybaby on June 17, 2016, 11:51:04 pm
hi! i'm stuck on trying to find a ppt test to distinguish between Ca2+ and Ba2+
Sorry, could I ask what a ppt test is?
If you could use a flame test, this would determine the difference, as Ca2+ would produce a 'brick-red' coloured flame whereas Ba2+ would produce a 'yellow-bright green' coloured flame.

Sorry, realised what ppt. means.
You could add NaOH to determine between the two if a precipitate test is the way to go.

Ca2+ will produce a white precipitate, whereas Ba2+ will not produce a precipitate due Ba(OH)2 being soluble.
Post by: RuiAce on June 17, 2016, 11:53:17 pm
hi! i'm stuck on trying to find a ppt test to distinguish between Ca2+ and Ba2+

The one that precipitates is the Ca2+

(This is assuming you have to use precipitation. Otherwise a flame test is really easy.)
Post by: m_chan98 on June 19, 2016, 11:43:03 am
Hi! So I was wondering whether someone could provide just a bit of a summary of what solubility rules I need to know? I've consulted a few different sources but each of them are different. Also, I understand that the compounds are soluble, but soluble in what?? Thanks!!!  :)
Post by: jakesilove on June 19, 2016, 12:00:48 pm
Hi! So I was wondering whether someone could provide just a bit of a summary of what solubility rules I need to know? I've consulted a few different sources but each of them are different. Also, I understand that the compounds are soluble, but soluble in what?? Thanks!!!  :)

That's a seriously good question! I memorised the following solubility rules, which are really easy to remember once you start using the pneumonic device outlined below.

(http://i.imgur.com/wF8DyMR.png?1)

Those are the main solubility rules. I think that there are some others (eg. something to do with Silver Nitrate precipitating a white powder specifically) which discuss the COLOUR of the precipitate, but the actual precipitation is all outlined above. I wouldn't be learning more rules than what I've posted above.

As for your second questions; I assume it's just in water. I've tried doing some research, but everything comes up blank. However, you definitely don't need to think about that!

Let me know if I can be  more specific/helpful regarding any of this,

Jake
Post by: amandali on June 19, 2016, 12:56:14 pm
can you check my answer ?

CH3COOH(aq) + H2O(l) <---> H3O+(aq) +CH3COO-(aq)
The addition of OH- from NaOH will react with H3O+ which consequently reduces the concentration of H3O+ . This will cause the equilibrium to shift right to increase the concentration of H3O+. Hence, a buffer system is established which minimises pH increase.

The answer given has the equation the other way around so is this one correct as well?
Post by: jakesilove on June 19, 2016, 01:00:29 pm
can you check my answer ?

CH3COOH(aq) + H2O(l) <---> H3O+(aq) +CH3COO-(aq)
The addition of OH- from NaOH will react with H3O+ which consequently reduces the concentration of H3O+ . This will cause the equilibrium to shift right to increase the concentration of H3O+. Hence, a buffer system is established which minimises pH increase.

The answer given has the equation the other way around so is this one correct as well?

Hey! Your equation is definitely fine; as it's an equilibrium, you can write it whichever way you want. The only thing I would add is a brief explanation of acidity/basidity. Whist it may be obvious to you that an increase in H30+ is an increase in H+, which increased acidity (decreasing pH), perhaps throw in a single sentence saying that. This is just to make sure you give a really comprehensive response!

Jake
Post by: RuiAce on June 19, 2016, 02:41:37 pm
can you check my answer ?

CH3COOH(aq) + H2O(l) <---> H3O+(aq) +CH3COO-(aq)
The addition of OH- from NaOH will react with H3O+ which consequently reduces the concentration of H3O+ . This will cause the equilibrium to shift right to increase the concentration of H3O+. Hence, a buffer system is established which minimises pH increase.

The answer given has the equation the other way around so is this one correct as well?

Also be sure to explicitly say "Le Chatelier's principle" whenever you're referring to equilibrium behaviour.
Post by: jakesilove on June 19, 2016, 02:46:34 pm
Also be sure to explicitly say "Le Chatelier's principle" whenever you're referring to equilibrium behaviour.

Ah yes of course: every single equilibrium answer should have something like "By Le Chatelier's principle (which says that a system undergoing a change with shift in equilibrium to minimise that change)...." etc.
Post by: smiley2101 on June 22, 2016, 04:34:39 pm

Hey guys!! Why is the answer A and not B?Thanks so much !
Post by: RuiAce on June 22, 2016, 05:13:42 pm
In the reaction, 2 moles of H2 react with 1 mole of CO.

This means that the moles of hydrogen gas are used up at TWICE the rate the moles of CO are.

Hence the dotted curve represents the concentration of H2 and the blocked curve represents that of CO, because the concentration of H2 should be lower if it goes down more quickly.

Because the dotted curve is the one that spikes up, we know that H2 was what was added. Not CO.
Post by: anotherworld2b on June 24, 2016, 08:48:31 pm
Hello :)
I was wondering could I ask for help in regards to writing a scientific report (investigation) on molar mass?
I wasn't sure if I could ask this here :/

Design an investigation to determine the composition and molar mass of an unknown white substance
I dissolved the unknown bicarbonate (given info its a bicarbonate) with hydrochloric acid then boiled the substance
using a Bunsen burner to isolate the element. I later used a flame test to identify the substance as sodium bicarbonate

I wasn't sure what to put in the introduction of my report and the depth and what I would need to write to explain the concepts used
I am also confused about how to determine the composition and molar mass of the sodium bicarbonate
I tried to calculate the mass and mole of substances but I'm not sure if I am approaching it properly.
HELP is greatly appreciated TT ^ TT

1. Mass of tube + NaHCO3

51.6g

2. Mass of tube

47.6g

3. Mass of NaHCO3

4g

4. Moles of NaHCO3

0.0459 gmol -1

5. Mass of tube + NaCl

50.7g

6. Mass of NaCl

3.1g

7. Moles of NaCl

0.0530 gmol -1

Mass of heating basin + water

71.5g

Mass of water

20.8g

Mass of carbon dioxide

23.9g

Ratio: Moles NaCl /Moles NaHCO3

0.0530 : 0.0459

Mass of NaCo3 = 51.6 - 47.6 = 4g (used in experiment)

Mass of NaHCO3 = 22.98977 + 1.00794 + 12.0107 + (15.9994x3)

= 22.98977 + 1.00794 + 12.0107 + 47.9982

= 84.00661

Mole of NaHCO3 = 4/ 84.00661 = 0.0459

Post by: jakesilove on June 25, 2016, 02:08:10 pm
Hello :)
I was wondering could I ask for help in regards to writing a scientific report (investigation) on molar mass?
I wasn't sure if I could ask this here :/

Design an investigation to determine the composition and molar mass of an unknown white substance
I dissolved the unknown bicarbonate (given info its a bicarbonate) with hydrochloric acid then boiled the substance
using a Bunsen burner to isolate the element. I later used a flame test to identify the substance as sodium bicarbonate

I wasn't sure what to put in the introduction of my report and the depth and what I would need to write to explain the concepts used
I am also confused about how to determine the composition and molar mass of the sodium bicarbonate
I tried to calculate the mass and mole of substances but I'm not sure if I am approaching it properly.
HELP is greatly appreciated TT ^ TT

1. Mass of tube + NaHCO3

51.6g

2. Mass of tube

47.6g

3. Mass of NaHCO3

4g

4. Moles of NaHCO3

0.0459 gmol -1

5. Mass of tube + NaCl

50.7g

6. Mass of NaCl

3.1g

7. Moles of NaCl

0.0530 gmol -1

Mass of heating basin + water

71.5g

Mass of water

20.8g

Mass of carbon dioxide

23.9g

Ratio: Moles NaCl /Moles NaHCO3

0.0530 : 0.0459

Mass of NaCo3 = 51.6 - 47.6 = 4g (used in experiment)

Mass of NaHCO3 = 22.98977 + 1.00794 + 12.0107 + (15.9994x3)

= 22.98977 + 1.00794 + 12.0107 + 47.9982

= 84.00661

Mole of NaHCO3 = 4/ 84.00661 = 0.0459

Hey!

I have to admit, I'm not 100% sure what this investigation is actually asking you to do. Looks like you're having particular trouble with the introduction, depth, and explanation, so I'll just generally talk about that here. Hope it helps!

You don't really need an introduction for a scientific report. You want to have an aim, a hypothesis, equipment, a method, results (potentially calculations) and a discussion. Don't worry too much about having an introduction; just make sure to include the above listed features.

In terms of depth, just make sure your method includes every step that you used, including recording data, washing glassware, repetition etc. You don't really need to explain WHY you do anything until the discussion.

In the discussion, definitely talk about your method. Particularly, outline the solubility rules/flame tests that you knew to be true, and why you used them in this specific example. You basically want to go through accuracy, reliability and validity in every discussion. Talk about the flaws of the experiment, where sources of error lie, that kind of thing. This really comes down to your interpretation of the question, and I don't think I have enough information to properly give you advice for that.

Just make sure to go through the key components of a standard scientific paper. Make sure to include tables, graphs etc. with all the correct units, and make sure that your method is in past tense. Those sorts of things just make everything seem super professional. Your calculations seem right, but that's only a tiny part of an experimental report. Just as an example, check out the report someone else posted here!

Let me know if you have other specific questions; sorry that I have been extremely vague and general with my response, I'm just not sure what you have or haven't done yet.

Jake
Post by: Maz on June 25, 2016, 09:19:45 pm
Hey,
I have a fuel cell test on Monday and I was just wondering if you had any idea what kind of questions they generally ask. It is on the hydrogen-oxygen and methanol fuel cell specifically :)

Also,
Would you happen to know if there are any variations in electrolytes used (if any) and chemical reasons for these variations?

I'd really appreciate any ideas
Thankyou :)
Post by: Happy Physics Land on June 25, 2016, 11:23:18 pm
Hey,
I have a fuel cell test on Monday and I was just wondering if you had any idea what kind of questions they generally ask. It is on the hydrogen-oxygen and methanol fuel cell specifically :)

Also,
Would you happen to know if there are any variations in electrolytes used (if any) and chemical reasons for these variations?

I'd really appreciate any ideas
Thankyou :)

Hey mq123! Thanks for asking (it's been a long time since I lasted posted in the forums)

I didnt really do fuel cell for my secondary investigation on batteries, I will try my best to help you here but if you dont seem to benefit from it Im extremely sorry and I hope someone more professional can be here to provide better advices.

In direct methanol fuel cell:
- Fundamental mechanism involves the exchange of protons
- Main fuel being methanol, which offers the advantage of being portable, energy-dense and chemically stable under normal temperature and pressure
- Methanol is reacted to hydrogen through steam reforming
- Common operation temperature 50 - 120 degrees celsius
- Know about how methanol cross-over reduces battery efficiency
- CO2 produces at the anode, negatively impacting the environment
- Anode equation: CH3OH + H2O --> CO2+6H[/sup]+[/sup] + 6e-
- Cathode equation: 6H+ + 6e- --> 3H2
- Overall cell equation: CH3OH + H2O --> CO2 + 3H2
- Know about the structure and purpose of: Oxygen flow field, anode/cathode backing, anode/cathode current collector
- Requires a continuous source of fuel and oxygen to sustain the chemical reaction, hence usually used in an ambient condition

In Hydrogen-Oxygen Cell:
- Anode equation: 2H2 --> 4H+ + 4e-
- Cathode equation: O2 + 4e- + 4H+ --> 2H2O
- Overall cell equation: 2H2 + O2 --> 2H2O
- Energy is released in the form of heat and electricity
- Unused H2 is circulated/reused for maximum efficiency
- Proton exchange reaction
- Know the purpose of proton exchange membrane
- Know the catalyst used at both the cathode and anode

Common exam questions will often involve:
- asking about the chemistry behind the functioning of battery (I.e. need to state anode and cathode equations, overall equation and the working process that leads to redox reaction)
- asking about the impact of the battery on society and environment
- asking about cost and compare this cost to the more common dry cell batteries

Hope this helps! :)

Best regards
Happy Physics Land
Post by: anotherworld2b on June 28, 2016, 01:58:26 pm
Hi I was wondering could I get help in calculating the molar mass and composition of sodium bicarbonate?
Could I get a reply as soon as possible? I have to hand in my assignment tomorrow  :'(
We used 4g of NaHCO3 and got 3.1g of NaCl

NaHCO3(s) +  HCl(aq) → NaCl (s) + H2O(l) + CO2(g)

Info I have: ( not entirely sure its correct)
1. Mass of tube + NaHCO3
51.6g
2. Mass of tube
47.6g
3. Mass of NaHCO3
4g
4. Moles of NaHCO3
0.0459 gmol -1
5. Mass of tube + NaCl
50.7g
6. Mass of NaCl
3.1g
7. Moles of NaCl
0.0530 gmol -1
Mass of heating basin + water
71.5g
Mass of water
20.8g
Mass of carbon dioxide
23.9g
Ratio: Moles NaCl /Moles NaHCO3
0.0530 : 0.0459

Post by: amandali on June 28, 2016, 08:38:09 pm
can you check my response thanks

In this option you studied one natural product that was not a fossil fuel. Describe the issues associated with shrinking world supplies of this natural product, and evaluate progress being made to solve the problems identified. 7 marks
A natural product is one that is used with little or no modification. An example include raw rubber which is a naturally occurring addition polymer formed from the polymerisation of 2-methyl-1,3-butadiene (isoprene) and obtained from sap of rubber trees. Many issues arise as its supply starts to shrink thus a replacement is required. In the early 20th century, the demand for rubber outstripped supply as rubber trees can only produce a certain amount of rubber each year. Rubber was needed for tyres for military vehicles thus its limited supply adversely affected countries’ performance at war. In ww2, Japan had control of rubber producing areas which threatened the supply of natural rubber in other countries. Moreover, automobile industry was growing and car manufacturers needed rubber. Due to the rising in demand for rubber, a first replacement for rubber is produced which is Styrene Butadiene rubber (SBR) formed from monomers of 1,3-butadiene and styrene. It is vulcanised with short Sulfur chains forming cross-links between polymer chains. It is more favourable than raw rubber as it has the improved properties ie. more durable, more resistant to chemical attack, and stronger. The progress made solved the problems as it increased the supply of rubber such that its demand is being met and cost is able to be maintained low.
Post by: jakesilove on June 28, 2016, 11:14:13 pm
Hi I was wondering could I get help in calculating the molar mass and composition of sodium bicarbonate?
Could I get a reply as soon as possible? I have to hand in my assignment tomorrow  :'(
We used 4g of NaHCO3 and got 3.1g of NaCl

NaHCO3(s) +  HCl(aq) → NaCl (s) + H2O(l) + CO2(g)

Info I have: ( not entirely sure its correct)
1. Mass of tube + NaHCO3
51.6g
2. Mass of tube
47.6g
3. Mass of NaHCO3
4g
4. Moles of NaHCO3
0.0459 gmol -1
5. Mass of tube + NaCl
50.7g
6. Mass of NaCl
3.1g
7. Moles of NaCl
0.0530 gmol -1
Mass of heating basin + water
71.5g
Mass of water
20.8g
Mass of carbon dioxide
23.9g
Ratio: Moles NaCl /Moles NaHCO3
0.0530 : 0.0459

Hey,

Unfortunately, I can't do calculations for you, and I won't be able to get to this properly tonight. I definitely can't do your assignments for you. All I can recommend is having a proper think about what you're actually trying to find, and utilising the equations at your disposal to the best of your ability.

Good luck,

Jake
Post by: jakesilove on June 29, 2016, 10:22:36 am
can you check my response thanks

In this option you studied one natural product that was not a fossil fuel. Describe the issues associated with shrinking world supplies of this natural product, and evaluate progress being made to solve the problems identified. 7 marks
A natural product is one that is used with little or no modification. An example include raw rubber which is a naturally occurring addition polymer formed from the polymerisation of 2-methyl-1,3-butadiene (isoprene) and obtained from sap of rubber trees. Many issues arise as its supply starts to shrink thus a replacement is required. In the early 20th century, the demand for rubber outstripped supply as rubber trees can only produce a certain amount of rubber each year. Rubber was needed for tyres for military vehicles thus its limited supply adversely affected countries’ performance at war. In ww2, Japan had control of rubber producing areas which threatened the supply of natural rubber in other countries. Moreover, automobile industry was growing and car manufacturers needed rubber. Due to the rising in demand for rubber, a first replacement for rubber is produced which is Styrene Butadiene rubber (SBR) formed from monomers of 1,3-butadiene and styrene. It is vulcanised with short Sulfur chains forming cross-links between polymer chains. It is more favourable than raw rubber as it has the improved properties ie. more durable, more resistant to chemical attack, and stronger. The progress made solved the problems as it increased the supply of rubber such that its demand is being met and cost is able to be maintained low.

Hey! I didn't do this area of Chemistry, but I think your answer is spot on! The only thing I would mention is that the last part of the question asks you to 'evaluate'. When you talk about SBR, you've mentioned its benefits, but are there any detrimental aspects? Impact on the environment, cost effectiveness, timeline etc? It is good to mention something like this, even if you decide that, overall, the process is beneficial.

Jake
Post by: zoeh on June 29, 2016, 05:30:06 pm
Hi,
I just have a few questions about electrolysis because it wasn't really explained well in class…
When the electrodes used are inert how does this impact on the reaction? Is the anode and cathode still the 2 electrodes? And where does the oxidation occur? For example if graphite electrodes were used in copper sulfate electrolyte would copper reduce at the cathode and what would be oxidised?
thanks so much
Post by: jakesilove on June 29, 2016, 09:04:01 pm
Hi,
I just have a few questions about electrolysis because it wasn't really explained well in class…
When the electrodes used are inert how does this impact on the reaction? Is the anode and cathode still the 2 electrodes? And where does the oxidation occur? For example if graphite electrodes were used in copper sulfate electrolyte would copper reduce at the cathode and what would be oxidised?
thanks so much

Hey! Basically, if an electrode is inert, it is non-reactive (bit of a tautology there, sorry). It won't oxidise or reduce at any point in the reaction. Generally, Graphite or Platinum electrodes are used to signify that the electrode is to be inert. However, something else needs to be happening for this cell to move forward.
An inert electrode acts only as a location for electrons to move from one place to another, and thus for some kind of oxidation/reduction to occur. For instance, if you had a half-cell with a platinum electrode in water, the platinum doesn't actually do anything; it is the water that is oxidised or reduced! This goes for any solution in which an inert electrode is placed (the standard one used in the HSC, I believe, is bubbling Hydrogen gas through a liquid. The gas attaches to the Platinum electrode, is oxidised or reduced by the inflow/outflow of electrons, and changes states accordingly).

I hope that this clears things up for you! If you see an inert electrode, you need to look for something else to move the reaction forward. Then, compare this other substance to the other half cell, using your table of reduction potentials, in order to determine which side oxidises and which side reduces (as with any standard galvanic cell).

Let me know if I can clear any of this/anything else up!

Jake

Post by: zoeh on June 29, 2016, 10:58:24 pm
Hey! Basically, if an electrode is inert, it is non-reactive (bit of a tautology there, sorry). It won't oxidise or reduce at any point in the reaction. Generally, Graphite or Platinum electrodes are used to signify that the electrode is to be inert. However, something else needs to be happening for this cell to move forward.
An inert electrode acts only as a location for electrons to move from one place to another, and thus for some kind of oxidation/reduction to occur. For instance, if you had a half-cell with a platinum electrode in water, the platinum doesn't actually do anything; it is the water that is oxidised or reduced! This goes for any solution in which an inert electrode is placed (the standard one used in the HSC, I believe, is bubbling Hydrogen gas through a liquid. The gas attaches to the Platinum electrode, is oxidised or reduced by the inflow/outflow of electrons, and changes states accordingly).

I hope that this clears things up for you! If you see an inert electrode, you need to look for something else to move the reaction forward. Then, compare this other substance to the other half cell, using your table of reduction potentials, in order to determine which side oxidises and which side reduces (as with any standard galvanic cell).

Let me know if I can clear any of this/anything else up!

Jake
Thanks so much that clears it up! :)
Post by: conic curve on June 29, 2016, 10:58:35 pm
What are the most important things in prelim chem (other than chemical equations and moles) which are important in HSC chem (and will reappear in HSC chem)?

Thanks
Post by: anotherworld2b on June 30, 2016, 09:51:32 am
I'm sorry if my question seemed like I was taking advantage of this thread.
I was unsure how to calculate molar mass, moles and ect and what formula I needed to use.
I am still quite confused about molar mass and moles calculations in general.
Could I have help in understanding when to use what formula and where to use it?
Post by: jakesilove on June 30, 2016, 08:31:16 pm
What are the most important things in prelim chem (other than chemical equations and moles) which are important in HSC chem (and will reappear in HSC chem)?

Thanks

There is very little actual crossover between the courses. In fact, I'd hesitate to say that there is absolutely none, whatsoever. What is important is a good understanding of chemical reactions (ie. Neutralization reactions, that kind of thing), a working knowledge of chemical terms (solutions, mixtures, moles, gas etc.) and not much else. Don't worry too much about the content you are learning now, and its relation to the HSC course. Just learn it for your exams, and then move on to the HSC curriculum later this year!

Jake
Post by: jakesilove on June 30, 2016, 08:44:15 pm
I'm sorry if my question seemed like I was taking advantage of this thread.
I was unsure how to calculate molar mass, moles and ect and what formula I needed to use.
I am still quite confused about molar mass and moles calculations in general.
Could I have help in understanding when to use what formula and where to use it?

That's not a problem :)

Okay, so let's do a brief refresher of moles, molar mass and mass, and how to calculate each from the other.

A mole is a collection of atoms (in fact, it is an exact number of atoms, Avagadro's number. This isn't important in the HSC, but knowing this makes understanding the concepts a lot easier). 2 moles is two times Avagadro's number of atoms, 1/2 a mole is 50% Avagadro's number etc. As you will imagine, each element's atom will have a different mass. So, if a single atom of Nitrogen was heavier than a single atom of Helium, you would expect a mole of Nitrogen to weigh more than a single mole of Helium (despite them having the same number of atoms!). The molar mass of a substance is the mass of one mole of the element.

So, let's put this all together into some calculations. The formula for relating all of this is

$Moles = \frac{Mass}{Molar Mass}$

So, if we had 3g of Carbon, which has a molar mass of approximately 12g, then we have 1/4 moles of Carbon!

Easy.

Now, there's some more complicated maths that I think your assignment required you to do. Say you had a reaction between two substances. You can balance the equation easily, by adding numbers out the front of each reactant and product to make sure the same number of elements/atoms are on each side. A 2 signifies 2 moles, a 1 signifies 1 mole etc. Now, say you had a reaction that looks something like this:

$2A+B -> C$

We know that for any two moles of A, we need 1 mole of B to react, which will produce 1 mole of C. Therefore, if we know we have 2g of A, we can use the above formula to figure out how many moles that is (using the molar mass). We then know we need HALF as many moles of B, and can figure out the relevant mass of B (using it's molar mass). This is stuff that's really important to the HSC syllabus; I'm not sure what level you're at but hopefully this sounds familiar.

That's the basics of the whole thing. I could go into much more depth, but that really is a bulk of the skills in this course. If what I've written above doesn't make sense, or you're having trouble with anything, please feel free to post specific questions because at the moment it's hard to gauge where you're having problems. If none of this is making any sense, I would talk to your teacher about having a few lessons with them. It's a big topic area, and it's important you understand what's going on.

Hope that this helped, at least a little.

Jake

Post by: MysteryMarker on July 02, 2016, 09:03:54 pm
Hey guys, just got a past trial paper question that i don't really understand. I don;t know what the answer is but i've been told it is A. Can anyone explain to me how this is so, and if possible, maybe through the aid of a lewis electron dot diagram? Thanks Guys.

Which Reaction involved the formation of a coordinate covalent bond?

a)   H+ + OH- → H2O
b)   Mg + 2H+ →Mg2+ + H2
c)   C2H4 + H2 → C2H6
d)   CH4 + Br2 → CH3Br + HBr

I originally thought the answer was b as the magnesium atom technically 'provides' both electrons to form the bonding pair for the hydrogen atom.

Cheers broskies.

Post by: RuiAce on July 03, 2016, 07:31:06 pm
Hey guys, just got a past trial paper question that i don't really understand. I don;t know what the answer is but i've been told it is A. Can anyone explain to me how this is so, and if possible, maybe through the aid of a lewis electron dot diagram? Thanks Guys.

Which Reaction involved the formation of a coordinate covalent bond?

a)   H+ + OH- → H2O
b)   Mg + 2H+ →Mg2+ + H2
c)   C2H4 + H2 → C2H6
d)   CH4 + Br2 → CH3Br + HBr

I originally thought the answer was b as the magnesium atom technically 'provides' both electrons to form the bonding pair for the hydrogen atom.

Cheers broskies.
B is wrong because if you react an acid with a metal you get a salt plus the hydrogen gas. E.g.
Mg(s) +2 HCl(aq) → MgCl2(aq) + H2(g)

C is obviously wrong because alkene → alkane cannot possibly become a coordinate bond

D is wrong mainly because it's just the formation of a CFC (or rather, BCFC here).

The answer is A. It should have been established clearly that the species H3O+, that is, the hydronium ion features a coordinate bond between one of the hydrogens and the oxygen. This is because the acid H+ isn't actually present. What's actually present is indeed, the hydronium ion. The more accurate way equation A is written is this:

H3O+ + OH- ⇌ 2 H2O(l)
Post by: trixqwe on July 03, 2016, 10:59:48 pm
Hi there! I'm a little stuck on how to answer this question and I would appreciate any help  :)

Compare an 'amphiprotic' substance and an 'amphoteric' substance. Use examples and equations to support your answer.

Post by: RuiAce on July 04, 2016, 09:30:42 am
Hi there! I'm a little stuck on how to answer this question and I would appreciate any help  :)

Compare an 'amphiprotic' substance and an 'amphoteric' substance. Use examples and equations to support your answer.

Technically amphiprotic substances are all amphoteric, but not vice versa. Just like how all squares are rectangles.

Amphoteric just means it exhibits both acidic and basic properties. You can take Al2O3 and compare two reactions. When aluminium oxide reacts with an acid, it's exhibiting basic properties. When it reacts with a base, it's exhibiting acidic properties. (Don't remember the equations right now, but either Al2O3 or ZnO is included in the Jacaranda textbook.)

Amphiprotic specifically relates to B-L theory. That is, an acid is a proton donor, whereas a base is a proton acceptor. Amphiprotic substances are both. An example is your HCO3- ion (or if that's too inadequate, just consider NaHCO3).
It's easy to show that if you react sodium hydrogen carbonate with hydrochloric acid, you get water, carbon dioxide and sodium carbonate. Whereas if you react it with sodium hydroxide, you get sodium carbonate and water.
Post by: katherine123 on July 06, 2016, 02:27:30 am
I did part i but not sure how to part ii)    what are the negative points?
i)
1.add 150ml of water into measuring cylinder A and 100ml of water into measuring cylinder B
2.transfer 10ml of water using 10ml pipette from measuring cylinder A to B (forward process)
3.Transfer 5ml of water using 5ml pipette from measuring cylinder B to A (backward process)
4.Repeat step 2 and 3 (one transfer cycle) until the volume of water remain constant in both measuring cylinder (equilibrium)

ii)Information collected is highly valid given that it is a qualitative model of an equilibrium reaction. In this procedure, it can be seen that the process occurs 2 directions (water transferred from measuring cylinder A to B or B to A) which is similar to an equilibrium. It is also observed that the volume of water in both measuring cylinder remain constant after transferring, like an equilibrium  with no macroscopic changes. Moreover, no water is removed from the system which is similar to equilibrium which is a closed system
Post by: MysteryMarker on July 06, 2016, 11:53:35 am
Hey guys, just got a question on eutrophication and the methods used to determine the concentration of phosphates and nitrates/nitrogen.

The method for determining organic nitrogen within waterways is to convert it into ammonia. The nitrogen concentration is then found through acid-base titration. Is this in enough detail for exams or should i know about how they convert the nitrogen into ammonia? (heat in sulphuric acid and stuff.)
Post by: katherine123 on July 06, 2016, 04:45:36 pm
im quite confused about the electrolysis of molten, concentrated and aqueous of NaCl

what info does the polarity of electrode give?
Post by: conic curve on July 06, 2016, 04:51:01 pm
Sometimes with examples in chemical equations, I get confused with the states. Like sometimes h2o it is in liquid state and sometimes it is in gaseous state. How do you know when it's in liquid state and when it's in a gaseous state (using h2o as an example)?

Thanks
Post by: RuiAce on July 06, 2016, 11:04:58 pm
I did part i but not sure how to part ii)    what are the negative points?
i)
1.add 150ml of water into measuring cylinder A and 100ml of water into measuring cylinder B
2.transfer 10ml of water using 10ml pipette from measuring cylinder A to B (forward process)
3.Transfer 5ml of water using 5ml pipette from measuring cylinder B to A (backward process)
4.Repeat step 2 and 3 (one transfer cycle) until the volume of water remain constant in both measuring cylinder (equilibrium)

ii)Information collected is highly valid given that it is a qualitative model of an equilibrium reaction. In this procedure, it can be seen that the process occurs 2 directions (water transferred from measuring cylinder A to B or B to A) which is similar to an equilibrium. It is also observed that the volume of water in both measuring cylinder remain constant after transferring, like an equilibrium  with no macroscopic changes. Moreover, no water is removed from the system which is similar to equilibrium which is a closed system
Providing the question could help here, thanks. I am not sure where "negative points" comes from just by looking at the answer.
Hey guys, just got a question on eutrophication and the methods used to determine the concentration of phosphates and nitrates/nitrogen.

The method for determining organic nitrogen within waterways is to convert it into ammonia. The nitrogen concentration is then found through acid-base titration. Is this in enough detail for exams or should i know about how they convert the nitrogen into ammonia? (heat in sulphuric acid and stuff.)
Back when I did the course, all I memorised was "Kjeldahl method" and nothing about the specifics at all. It is all that I felt was necessary.
You may find more useful information from another person.
im quite confused about the electrolysis of molten, concentrated and aqueous of NaCl

what info does the polarity of electrode give?
The polarity of the electrode determines which half-equation is occurring at the electrode. (Always remember that the polarity of the cathode/anode flip for an electrolytic cell; instead of a positive cathode and negative anode, we consider a negative cathode and positive anode.)

Recall the mnemonic RED-CAT AN-OX.
This means that the reduction occurs at the cathode, whereas the oxidation occurs at the anode.

We analyse three cases:

Case 1: Molten NaCl.
The only species present are Na+ and Cl-. This means that sodium ions will be reduced to sodium metal, and chloride ions will be oxidised to produce chloride gas.
+'ve anode: 2 Cl- -> Cl2(g) + 2e-
-'ve cathode: Na+ + e- -> Na(s)

Case 2: Aqueous solution of NaCl.
The two species present are as above, and H2O as well now. We use our data sheet to determine what is more likely to occur.

-'ve cathode: -2.71V is required to reduce sodium ions back to sodium. But -0.83V is required to reduce water into hydrogen gas and base (hydroxide ions). Clearly, it will take considerably less voltage for the latter equation to occur. Hence, provide water is present, that will ALWAYS be the reaction.
2 H2O(l) + 2 e- -> 2 OH- + H2(g)

+'ve cathode: -1.36V is required to oxidise chloride ions to chloride gas. -1.23V is required to oxidise water into acid (hydrogen ions) and oxygen gas. There is a voltage difference, however it's quite small, so we cannot say for sure which reaction will occur. We seperate up our cases:

Subcase 1: Concentrated aqueous solution (generally >2M, although some sources say >1M)
There is a good abundance of chloride ions present, hence they will be oxidised.
2 Cl- -> Cl2(g) + 2e-

Subcase 2: Dilute aqueous solution (generally <0.1M)
Not many chloride ions are present at all. The oxidation of water will be more dominant here.
2 H2O(l) -> 4 H+ + O2(g) + 4 e-

(Note: The electrolysis of dilute NaCl solution is basically the electrolysis of water.)

Sometimes with examples in chemical equations, I get confused with the states. Like sometimes h2o it is in liquid state and sometimes it is in gaseous state. How do you know when it's in liquid state and when it's in a gaseous state (using h2o as an example)?

Thanks

Whilst a bit of this will require your knowledge of chemistry (e.g. aluminium reacts specifically with STEAM), more often than not it should be implied as to what state water is in. You may choose to provide examples to voice your confusion, which I can walk through.
Post by: vm1997 on July 07, 2016, 12:56:32 pm
Few questions on Water quality management

"Describe the tests for any factors that need to be specifically monitored in these areas/with these activities to ensure that we have safe drinking water" I can only name a couple of tests that need to be done for this.
Briefly, what tests can be done for this and what's the chemistry behind them?

"Identify the organisations that monitor the wwaterways in NSW and describe the types of chemistry that is used in the monitoring process"  I know a few suc as SAC, but WHAT TESTS DO THEY DO??
Post by: RuiAce on July 07, 2016, 06:35:42 pm
Few questions on Water quality management

"Describe the tests for any factors that need to be specifically monitored in these areas/with these activities to ensure that we have safe drinking water" I can only name a couple of tests that need to be done for this.
Briefly, what tests can be done for this and what's the chemistry behind them?

"Identify the organisations that monitor the wwaterways in NSW and describe the types of chemistry that is used in the monitoring process"  I know a few suc as SAC, but WHAT TESTS DO THEY DO??
The HSC expects you to know each of the following tests and how we can use it to determine overall water quality.

- Concentration of common ions: Easy, just use your cation/anion tests and gravimetric analysis.
- Total dissolved solids: This is just an arbitrary way of saying how much solid is there in the water, measured in ppm. We could use gravimetric analysis but it's not that good since the amount of solid present generally isn't much. So we use the fact that the things dissolved are ions and will conduct electricity - thus we use a conductivity meter
- Hardness: This measures the concentration of Mg2+ and Ca2+. We use EDTA titration here.
- Turbidity: The cloudiness of water (on a macroscopic level) is sort of how clear it is. We take measurements of turbidity using a "turbidity tube". We need to know how turbid the water is to assess the ease of aquatic plants in photosynthesizing as an example.
- Acidity: This is obvious. And we can use a pH probe where possible.
- Dissolved oxygen: This is how much oxygen there is in the water. We use Winkler titration to find out this value. Harmful anaerobic bacteria grow when DO is low.
- Biochemical oxygen demand: By definition, "Biochemical Oxygen Demand is the DO necessary completely breakdown organic matter in the water through aerobic bacteria." To measure BOD, we take a sample of water and record the initial DO. We then leave the sample in the dark for 5 days before taking another value. The difference is the BOD.

I 100% rote-learnt the catchment dot point though. You can find information on that (and also more information on the tests) here. https://www.scribd.com/doc/49281194/The-Student-s-Guide-to-HSC-Chemistry
Post by: ssarahj on July 08, 2016, 06:35:11 pm
You can find information on that (and also more information on the tests) here. https://www.scribd.com/doc/49281194/The-Student-s-Guide-to-HSC-Chemistry

cheers for the link to that guide, it looks super helpful!!!  :)
Post by: katherine123 on July 11, 2016, 11:26:57 pm
Providing the question could help here, thanks. I am not sure where "negative points" comes from just by looking at the answer.Back when I did the course, all I memorised was "Kjeldahl method" and nothing about the specifics at all. It is all that I felt was necessary.

Post by: amandali on July 11, 2016, 11:55:23 pm
In this option you studied one natural product that was not a fossil fuel. Describe the issues associated with shrinking world supplies of this natural product, and evaluate progress being made to solve the problems identified. 7 marks

since it says  "evaluate progress being made"   does the last sentence suffice the negative point of using synthetic rubber or is it okay to just focus on the positives
Post by: RuiAce on July 12, 2016, 08:57:24 am
If by negative points you just mean how it might not be invalid, well one thing could be that it doesn't accurately display the rate at which equilibrium is achieved. Two weaker arguments may be that the model also only demonstrates how an equilibrium is achieved, but we don't know if the forward reaction is forcibly going twice as fast (or something) as the reverse reaction. Nor do we know if the final ratio of products to reactants is right.

In this option you studied one natural product that was not a fossil fuel. Describe the issues associated with shrinking world supplies of this natural product, and evaluate progress being made to solve the problems identified. 7 marks

since it says  "evaluate progress being made"   does the last sentence suffice the negative point of using synthetic rubber or is it okay to just focus on the positives

Yeah. From memory when I wrote about rubber, with the negatives of SBR I just wrote about how it's still being made out of petrochemicals, and why that's bad (which is something you should know from the production of materials topic).

These are not my notes, but basically they were all I memorised for section 1 of industrial chemistry.

(One thing that you will find in these notes at the very end is how progress is being made to develop an alternative to styrene-butadiene rubber, but made out of biomass. This is an example of even further progress to counter the drawbacks of SBR, and may or may not be worth mentioning in your 7-mark response. At this point though, because I'm too out of shape I won't critique your response for now, sorry.)
Post by: conic curve on July 12, 2016, 10:17:41 am
Could someone here please explain to me the important concepts in the metals module (i.e. Moles, spectator ions, half equations, limiting and excess, ionic equations, etc) I struggle to understand the maths behind this and require a lot of help (it'd be great if you could provide examples as well)

CHeers
Post by: anotherworld2b on July 12, 2016, 07:59:02 pm
I tried to balance question e but im not sure why i got it wrong

Post by: RuiAce on July 12, 2016, 08:28:50 pm
I tried to balance question e but im not sure why i got it wrong
Line 2 is correct.

C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)

You can use WolframAlpha to show you the answer to balancing out the equations. (Just make sure to use an equal sign.)

Could someone here please explain to me the important concepts in the metals module (i.e. Moles, spectator ions, half equations, limiting and excess, ionic equations, etc) I struggle to understand the maths behind this and require a lot of help (it'd be great if you could provide examples as well)

CHeers
Most of this stuff is quite fundamental to chemistry. Make sure you have read through your textbook carefully enough.

In SI base units, the measurement of the mole is used to measure the quantity of particles that exist. This is opposed to the mass, which measures how much matter there IS in a particle.

1 mol is defined as the number of carbon atoms in exactly 12 grams of carbon solid. The correct value is approximately 6.022 * 1023 which you can find on your data sheet. This number is famously known as Avogadro's number.

NA =  6.022 * 1023 mol-1

This leads to the fundamental relationship:
$n = \frac{N}{N_A}$
where n = moles of a substance (measured in mol)
N = the actual number of atoms there are

A number's quantity of particles is directly linked to the mass of the substance present. The relationship is given as

$n=\frac{m}{M}$
where:

n - moles of a substance present (measured in mol)
m - the mass of a substance given (measured in g for the sake of chemistry)
M - molar mass (measured in g mol-1)

By definition, the molar mass is just the mass divided by the quantity present. This is essentially the above formula rearranged (M=n/m). For historical reasons, it has units g mol-1 instead of kg mol-1.

However, every individual element has a different atomic number and atomic mass. The molar mass will differ depending on what species we have.

The number the molar mass (for monatomic elements) is just its atomic mass. (There is a conversion between atomic mass and molar mass but this we will ignore). Hence, the molar mass for, say, carbon, is 12.01 g mol-1. The molar mass of O is 16.00 g mol-1.

The molar mass of compounds is just that of the individual elements added together. E.g. the molar mass of PbSO4 equals
M(Pb) + M(S) + 4 * M(O)
= 207.2 + 32.07 + 4*16
= 303.27 g mol-1

As for the rest? Asking for all that in the one go is THOROUGHLY exhaustive and you will need to provide questions to set a basis. In the future, please also ask one or two questions (especially if they're too related to the same thing) at a time. Just like with the trigonometry questions - all of those at once was quite overloading.
Post by: Sahar8642 on July 12, 2016, 08:46:29 pm
Hey,
Need Help for this q.
Its worth 5 marks
A sample of lemon juice is to be analysed in the laboratory. A student took 25.00 mL of the juice
and diluted it to 250.00 mL. Exactly 25.00 mL of the diluted lemon juice is titrated with standard
0.1045 mol L-1
sodium hydroxide solution using phenolphthalein as the indicator. An average titre
of 24.05 mL of sodium hydroxide was required
Assuming that the lemon juice contained only citric acid (molar mass = 192.1 g/mol), calculate
the concentration in mol L-1 of citric acid in the undiluted lemon juice.

Thank You!
Post by: RuiAce on July 12, 2016, 08:58:49 pm
Hey,
Need Help for this q.
Its worth 5 marks
A sample of lemon juice is to be analysed in the laboratory. A student took 25.00 mL of the juice
and diluted it to 250.00 mL. Exactly 25.00 mL of the diluted lemon juice is titrated with standard
0.1045 mol L-1
sodium hydroxide solution using phenolphthalein as the indicator. An average titre
of 24.05 mL of sodium hydroxide was required
Assuming that the lemon juice contained only citric acid (molar mass = 192.1 g/mol), calculate
the concentration in mol L-1 of citric acid in the undiluted lemon juice.

Thank You!
We always start with an equation. Even if you forget the formula for citric acid (2-hydroxypropane-1,2,3-tricarboxylic-acid) you should be well aware of how it is triprotic.

The reaction goes to completion as we have a strong base.
C3H4OH(COOH)3(aq) + 3 NaOH(aq) -> C3H4OH(COO)3Na(aq) + 3 H2O(l)

We may determine the (average) moles of NaOH solution.
n(NaOH)
= CV
= 0.1045 mol L-1 * 0.02500 L
= 2.6125 * 10-3 mol

1 mol of NaOH reacts with 1/3 mol of citric acid.

n(C3H4OH(COOH)3) = n(NaOH)/3 = 8.7083333333... * 10-4 mol

This gives us the moles of citric acid used in the reaction. We know that 25 mL of citric acid was used in the reaction. Hence the concentration of the citric acid is

C2
= n/V
=  (8.7083333333... * 10-4 mol) / (0.025 L)
= 0.03483333... mol L-1

Notice that this is the concentration of citric acid AFTER dilution. To obtain the concentration BEFORE dilution, we use the concentrations formula C1V1 = C2V2

C1 * 25.00 mL = 0.03483333... mol L-1 * 250.00 mL

C1 = 0.34833333... mol L-1

= 3.483 * 10-1 mol L-1 (4 sig. fig.)

(There may be a slight calculation error somewhere but the method should be good.)
Post by: anotherworld2b on July 12, 2016, 11:28:45 pm
How would you do these questions?
Post by: kristine.faelnar on July 12, 2016, 11:32:00 pm
Hi guys!!
So I was making my notes for my option topic (Forensics) and I came across this syllabus point and I have no bloody idea what I'm doing. I don't really understand it and it hasn't been gone through much in class.

"Perform a first-hand investigation and gather first-hand information to identify the range of solvents that may be used for chromatography and suggest mixtures that may be separated and identified by the use of these solvents."

Post by: conic curve on July 12, 2016, 11:47:12 pm
Hi guys!!
So I was making my notes for my option topic (Forensics) and I came across this syllabus point and I have no bloody idea what I'm doing. I don't really understand it and it hasn't been gone through much in class.

"Perform a first-hand investigation and gather first-hand information to identify the range of solvents that may be used for chromatography and suggest mixtures that may be separated and identified by the use of these solvents."

I believe that is a practical
Post by: kristine.faelnar on July 12, 2016, 11:53:10 pm
I have never done it before, so I have no idea what to do for that :(
I tried finding notes online as well and I couldn't find anything
Post by: RuiAce on July 13, 2016, 07:49:52 am
How would you do these questions?
These were...a bit awkward to read... sideways and upside-down photo

Q4: Not really important at all. One thing to note about the ammonium ion is just the coordinate covalent bond from the nitrogen to the fourth hydrogen ion. Yes, the coordinate bond is to a hydrogen ion (aka. a proton), not a hydrogen atom. This hydrogen ion basically gives that extra +1 charge.

Whereas for hydroxide, the hydroxide ion is formed when water undergoes self ionisation. What happens is that one of the hydrogens in H2O essentially "comes off", but it doesn't take its electron with it. The hydrogen and oxygen were initially covalently bonded, but the hydrogen comes off and leaves its electron with oxygen. This causes the -1 charge.

I have no idea how this relates to the fertiliser practical. Maybe the rest of the textbook could help here.
Post by: conic curve on July 13, 2016, 08:49:59 am
I have never done it before, so I have no idea what to do for that :(
I tried finding notes online as well and I couldn't find anything

Just wait until your teacher does it in class and then you compose the practical
Post by: RuiAce on July 13, 2016, 10:22:34 am
I have never done it before, so I have no idea what to do for that :(
I tried finding notes online as well and I couldn't find anything
If you haven't done a practical yet purely because you're not that far into the syllabus yet, you cannot be asked it in the exam. If you do, take it up to the teacher or faculty coordinator.

(Unfortunately I have no clue about forensic chemistry.)
Post by: RuiAce on July 13, 2016, 10:42:13 am
Did you do Industrial chemistry?
Yes

Posts like these aren't necessary. Please delete it.

Post by: conic curve on July 13, 2016, 11:10:52 am
Yes

Posts like these aren't necessary. Please delete it.

Oh I was just asking ahaha  :D

Anyways if anyone here is generous enough could you please explain to me spetator ions, ionic equations, half equations, OILRIG, etc all with examples (if it's too much do 1-2 at a time). I seriously suck at those and I want to continue chemistry in year 12 (I heard it comes up in year 12 BTW and that's the main reason why I'm worried)

Thanks so much  :)
Post by: HighTide on July 13, 2016, 11:24:47 am
Oh I was just asking ahaha  :D

Anyways if anyone here is generous enough could you please explain to me spetator ions, ionic equations, half equations, OILRIG, etc all with examples (if it's too much do 1-2 at a time). I seriously suck at those and I want to continue chemistry in year 12 (I heard it comes up in year 12 BTW and that's the main reason why I'm worried)

Thanks so much  :)
Alright, although this is a question, you should probably have a go at these first. A simple google search, or using a textbook would give you the answers.

Spectator ions don't participate in the reaction. So if have like a compound like sodium sulfate and barium nitrate, the precipitate would be barium sulphate, the sodium ions and nitrate ions are spectator ions as they are not involved in the main reaction, but they are in the solution.
Ionic equations are kind of self explanatory. If you have a specific question about it, I can elaborate.
Half equations: One for oxidation and another for reduction since both occur spontaneously.
OILRIG is just oxidation is losing electrons. Reduction is gaining electrons.

Alright examples you don't really need at this point. But since you're keen on learning this (it does indeed come up in year 12), you should check out videos from Khan Academy, or other youtube sites because they introduce you to the basics. These basics will suffice for year 11 and year 12. If you have any specific question after that, I'd be happy to answer it.
Post by: conic curve on July 13, 2016, 02:12:49 pm
Alright, although this is a question, you should probably have a go at these first. A simple google search, or using a textbook would give you the answers.

Spectator ions don't participate in the reaction. So if have like a compound like sodium sulfate and barium nitrate, the precipitate would be barium sulphate, the sodium ions and nitrate ions are spectator ions as they are not involved in the main reaction, but they are in the solution.
Ionic equations are kind of self explanatory. If you have a specific question about it, I can elaborate.
Half equations: One for oxidation and another for reduction since both occur spontaneously.
OILRIG is just oxidation is losing electrons. Reduction is gaining electrons.

Alright examples you don't really need at this point. But since you're keen on learning this (it does indeed come up in year 12), you should check out videos from Khan Academy, or other youtube sites because they introduce you to the basics. These basics will suffice for year 11 and year 12. If you have any specific question after that, I'd be happy to answer it.

How come spectator ions don't take place in the reaction?

Yeah I know OILRIG, I just tend to get confused with examples. Usually does it require the periodic table?

Post by: WLalex on July 13, 2016, 03:19:10 pm

Yeah I know OILRIG, I just tend to get confused with examples. Usually does it require the periodic table?

No it does not, or not to my knowledge, require the periodic table. Basically when looking at an equation you are trying to determine the characteristics of the reactants ad products in determining which has lost (oxidation) and gained (reduction) electrons....HINT: look at the change in states/charges

As a rule of thumb, most of the time that is...the metal e.g. Mg will be oxidised

Alex
Post by: RuiAce on July 13, 2016, 03:23:19 pm
Regarding spectator ions:

Spectator ions are present because they're a part of the species. However, after the reaction is complete, nothing has happened to them. For example:

CuSO4(aq) + Zn(s) -> Cu(s) + ZnSO4(aq)

In the reaction, copper (II) ions turned to copper solid
Zinc solid turned to zinc ions

The sulfate ion did nothing. The species present are copper(II) sulfate and zinc sulfate, but the sulfate itself never got altered. It remained as SO42-.

We call this a spectator ion due to the fact that it is there just to ensure the reaction happens. It doesn't participate in the reaction.

It isn't a condition either because it doesn't HAVE to be there for the reaction to happen.
____________

As to why they don't take part in the reaction? Note that the definition of spectator ion is something THAT does not take part. It's just there because it's there in the mixture.
Post by: wesadora on July 13, 2016, 08:57:22 pm
I'm having trouble with a following two questions in one of my school's past trial exams:

Spoiler
"A student used this equipment [experimental setup for alkanol's heat of combustion prac: i.e. retort stand, clamp, spirit burner, water, thermometer, conical flask] to heat 250g of water. The mass of the spirit burner, which contained ethanol, decreased from 296.52g to 295.95g.

Given the heat of combustion of ethanol is 29.7kJ/g, calculate the maximum possible change in the temperature of the water."

--> this question I'm confused about the kJ/g...seeing ∆H = Q/n for kJ/mol, i just thought of doing ∆H=Q/m as it was in kJ/g (kilojule per gram), and solving for Q, as: 29.7 = Q/0.57 (0.57 is mass of ethanol burned)...yeah i got lost :(

Spoiler
"Ozone concentrations are measured in Dobson units (DU). DU are the standard way to express ozone concentration in the stratosphere. A concentration of one DU means there would be 2.7x1020 ozone molecules in a layer of air that was one square metre in area and 0.01mm thick.

A baseline value of 220 DU is chosen as the starting point for an ozone hole in the stratosphere since total ozone values of less than 200 DU were not found in historic obsevations over Antarctica over 1979.
Which of the following concentrations, in moles per cubic metre (molm'3), is most nearly equivalent to 220 DU?"

A) 0.05 molm-3
B) 10 molm-3
C) 5000 molm-3
D) 10 000 molm-3

--> this one i am just lost in.

Post by: RuiAce on July 13, 2016, 09:12:37 pm
I'm having trouble with a following two questions in one of my school's past trial exams:

Spoiler
"A student used this equipment [experimental setup for alkanol's heat of combustion prac: i.e. retort stand, clamp, spirit burner, water, thermometer, conical flask] to heat 250g of water. The mass of the spirit burner, which contained ethanol, decreased from 296.52g to 295.95g.

Given the heat of combustion of ethanol is 29.7kJ/g, calculate the maximum possible change in the temperature of the water."

--> this question I'm confused about the kJ/g...seeing ∆H = Q/n for kJ/mol, i just thought of doing ∆H=Q/m as it was in kJ/g (kilojule per gram), and solving for Q, as: 29.7 = Q/0.57 (0.57 is mass of ethanol burned)...yeah i got lost :(

Spoiler
"Ozone concentrations are measured in Dobson units (DU). DU are the standard way to express ozone concentration in the stratosphere. A concentration of one DU means there would be 2.7x1020 ozone molecules in a layer of air that was one square metre in area and 0.01mm thick.

A baseline value of 220 DU is chosen as the starting point for an ozone hole in the stratosphere since total ozone values of less than 200 DU were not found in historic obsevations over Antarctica over 1979.
Which of the following concentrations, in moles per cubic metre (molm'3), is most nearly equivalent to 220 DU?"

A) 0.05 molm-3
B) 10 molm-3
C) 5000 molm-3
D) 10 000 molm-3

--> this one i am just lost in.
If they give you the enthalpy change per gram, then ∆H=-Q/m (don't forget the negative) is correct.

Q = mc∆T = 16.929 kJ

Continue.
_________________________________
This question tests how well you know your units.
$1 DU = 2.7\times 10^{20}\text{ molecules, PER volume of }1m^2 \times 0.01mm$
$\text{So }1 DU = 2.7\times 10^{20}\text{ molecules, per volume of }10^{-5}m^3$
$\text{Which is equivalent to saying }2.7\times 10^{25}\text{ molecules, per volume of }1mm$
$\therefore 1DU = 2.7\times 10^{25}\text{ molecules }m^{-3}$
$\text{Recall from preliminary that the fundamental definition of moles is }n=\frac{N}{N_A}\\ \text{where }N_A=6.022\times 10^{23}mol^{-1} \qquad \text{(Avogadro's number)}$
$\frac{2.7 \times 10^{25}}{6.022 \times 10^{23} mol^{-1}}=44.8835...mol$
$\text{Therefore we just found out that }1DU = 44.8835...mol\, mm^{-3}$
$220 DU = 9863.8326...mol \, mm^{-3}$
Tell me if I am wrong.
Post by: wesadora on July 13, 2016, 09:17:38 pm
That's correct, answer was D. I'll have a look at the working and redo the first question and see what I get.

Thanks Rui, also 1 more quick question:
I just came across a question using crystalline oxalic acid (aka oxalic acid dihydrate) as a primary standard. This comes in the form (COOH)2.2H2O

I'm confused because in class we used it and my teacher said to include the 2H2O in the molar mass, while i SWEAR my chem tutor said not to. And this makes a BIG difference in the calculation (obviously) ._.
Post by: RuiAce on July 13, 2016, 09:18:53 pm
That's correct, answer was D. I'll have a look at the working and redo the first question and see what I get.

Thanks Rui, also 1 more quick question:
I just came across a question using crystalline oxalic acid (aka oxalic acid dihydrate) as a primary standard. This comes in the form (COOH)2.2H2O

I'm confused because in class we used it and my teacher said to include the 2H2O in the molar mass, while i SWEAR my chem tutor said not to. And this makes a BIG difference in the calculation (obviously) ._.
I'd say to include the molar mass of the dihydrate. I will ask around because I haven't done this type of question in ages.

This post will be deleted.
________

A response I got

Well I guess if crystalline oxalic acid (just (COOH)2 without the hydrates) was used, adding it to water to get those extra dihydrates shouldn't impact the experimental results, so I think he shouldn't include it in his calculations. Because, according to me, adding the oxalic acid to the water just causes the acid to essentially dissolve in the water and pick up that dihydrate (similar to changing from a solid to aqueous subscript) and so it should be treated as just a change of state.

But I'm not completely sure.
Post by: ProfLayton2000 on July 13, 2016, 10:51:39 pm
Hi there,

So on the HSC Chem 2011 paper q29b: "Why does the neutralisation of any strong acid in an aqueous solution by any strong base always result in a heat of reaction approximately -57kJ mol^-1

Could you please explain why it is what it is?
Post by: RuiAce on July 13, 2016, 10:58:45 pm
Hi there,

So on the HSC Chem 2011 paper q29b: "Why does the neutralisation of any strong acid in an aqueous solution by any strong base always result in a heat of reaction approximately -57kJ mol^-1

Could you please explain why it is what it is?
Because:
a) The reaction goes to completion between two completely ionised substances
b) The reaction H3O+ + OH- -> 2 H2O(l) has ∆H=-57 kJ mol-1 (approximately).

That's all you need. See if you can realise how that just builds up to an answer.
Post by: anotherworld2b on July 14, 2016, 12:44:28 am
I was wondering for chemcial equations how do you know the state of each substance/compound whether it is aqueoud, liquid, gas or solid?
Post by: jakesilove on July 14, 2016, 10:16:29 am
I was wondering for chemcial equations how do you know the state of each substance/compound whether it is aqueoud, liquid, gas or solid?

Hey!

To some extent, it's just intuition after you've done a hell of a lot of practice. The only liquid that you really deal with is water: Assume water is liquid, unless it is clearly steam (in which case it is a gas). Anything else, in like a 'liquid' form, but containing metal ions etc. will always be aqueous. That means any acidic solution, any basic solution, any ions in solution, will be aqueous. Gas, again, is usually obvious from the question or from having done heaps of past papers. Things like hydrogen gas, oxygen, nitrogen (all diatomic) will be in a gaseous form. I can't give much more advice than that, except to say DO HEAPS OF PAST PAPERS and this will all because clear. Think about what is actually happening in the reaction, and you'll be sweet :)

Jake
Post by: RuiAce on July 14, 2016, 12:09:06 pm
I was wondering for chemcial equations how do you know the state of each substance/compound whether it is aqueoud, liquid, gas or solid?
On top of what Jake said, some are trivial whilst others are logic.

E.g. The water you produce in neutralisation between an acid and base has to be a liquid. You're mixing two liquids together.

As opposed to the combustion of ethanol where you're achieving such high temperatures and water must have been boiled already.

An example of what I call a special case I know is that aluminium refuses to react with liquid water, but rather steam.
Post by: katherine123 on July 15, 2016, 08:22:01 pm
can you check my response attached below thanks

when talking about conditions for certain process involving equilibrium  do i have to talk about what happens if (temperature, pressure) is too low or too high   or do i just need to talk about the compromise temperature (the temperature is not too high or low thus favours the forward reaction)?
Post by: conic curve on July 15, 2016, 08:37:44 pm
When going down a group in the periodic table, a new electron shell is added to the atom. Why is this the case, is this "by definition"

Also this would increase (what I said above) the atomic radius as the outermost electrons are placed further from the nucleus. Why is this the case?
Post by: zsteve on July 15, 2016, 08:43:28 pm
When going down a group in the periodic table, a new electron shell is added to the atom. Why is this the case, is this "by definition"

Also this would increase (what I said above) the atomic radius as the outermost electrons are placed further from the nucleus. Why is this the case?

The periodic table was constructed in such a way that patterns in the elements are made obvious and systematic (to a limited degree of course, trends break down as you go down a group).
Every time a new electron shell is added, the number of valence electrons can be thought of as being 'reset', hence elements in the same group have the same number of valence electrons and have similar (although by no means identical) properties.

As more electrons are added, they need to occupy higher energy levels which are thus more distant from the nucleus. This is a consequence of orbital theory which restricts the number of electrons in a particular energy state. The lowest energy state in an element (1s) can hold 2e-, once this is filled, it goes to the next orbital (2s), and so on, with the electrons becoming progressively more distant from the nucleus.

Across a period, as you gain more electrons (and consequently more protons), the atomic radius actually decreases because the pull on each valence electron becomes greater (due to more protons)
Post by: conic curve on July 15, 2016, 08:46:09 pm
Another question but why did Mendeleev leave gaps in his periodic table?
Post by: zsteve on July 15, 2016, 08:50:08 pm
Another question but why did Mendeleev leave gaps in his periodic table?

That was purely due to the fact that the 'next' element actually skipped a property, and Mendeleev believed that there was a missing element which hadn't been discovered which belonged there. Later on, that element was discovered.
Post by: Happy Physics Land on July 16, 2016, 06:44:57 pm
Another question but why did Mendeleev leave gaps in his periodic table?

Because not all the elements can be found in the order as we see them now - some are more difficult to synthesise or extract than other due to the technological deficiency of the time.
Post by: Happy Physics Land on July 16, 2016, 07:00:06 pm
can you check my response attached below thanks

when talking about conditions for certain process involving equilibrium  do i have to talk about what happens if (temperature, pressure) is too low or too high   or do i just need to talk about the compromise temperature (the temperature is not too high or low thus favours the forward reaction)?

Hey Katherine,

Your response is certainly very detailed and encompasses the essential things you need to know. When you get asked about equilibrium questions involving ammonia, you usually talk about why there is a compromise. I see you have already included those into your response so thats very good.

A couple of things that would be beneficial for you to add:
- You want to be more specific with the type of catalyst you are using. Iron oxide is ok, but l think its better to say Fe3O4 (magnetite). This shows the teacher you really know your stuff.
- With temperature, you dont want very high temperature also because it can damage the catalyst. Since catalysts are quite expensive to replace and relatively vulnerable, you want to preserve them in good conditions by using lower temperatures.
- You must also add the liquidification of ammonia as a part of your conditions. By this l mean constantly turning ammonia gas into liquid to remove gaseous ammonia from the equilibrium and hence the equilibrium will shift right, favouring the production of more ammonia. You touched on this in your last condition but its very important to mention LIQUIDIFICATION of ammonia.

Other than those points, I think this is a very good table to study for ammonia section of chemical monitoring and management. Kudos to you, you have done a very good job!

Best Regards
Happy Physics Land
Post by: RuiAce on July 17, 2016, 08:24:40 am
Hey Katherine,

Your response is certainly very detailed and encompasses the essential things you need to know. When you get asked about equilibrium questions involving ammonia, you usually talk about why there is a compromise. I see you have already included those into your response so thats very good.

A couple of things that would be beneficial for you to add:
- You want to be more specific with the type of catalyst you are using. Iron oxide is ok, but l think its better to say Fe3O4 (magnetite). This shows the teacher you really know your stuff.
- With temperature, you dont want very high temperature also because it can damage the catalyst. Since catalysts are quite expensive to replace and relatively vulnerable, you want to preserve them in good conditions by using lower temperatures.
- You must also add the liquidification of ammonia as a part of your conditions. By this l mean constantly turning ammonia gas into liquid to remove gaseous ammonia from the equilibrium and hence the equilibrium will shift right, favouring the production of more ammonia. You touched on this in your last condition but its very important to mention LIQUIDIFICATION of ammonia.

Other than those points, I think this is a very good table to study for ammonia section of chemical monitoring and management. Kudos to you, you have done a very good job!

Best Regards
Happy Physics Land
Yeah, I agree with the use of "magnetite" over iron oxide. Btw liquefaction is the scientific word for condensation.

However, I do think 400deg C (and 200 atm pressure) is fine in itself. At this temperature, the catalyst is safe. The more important thing is that the catalyst remains pure as was mentioned, because if the catalyst is impure, then the exposed surface decreases, thus limiting the impact of the catalyst. Pretty sure that's all we mean by an impure catalyst.
____________

@Katherine: Regarding impurities, it is good that you mentioned things such as O2 which should ALWAYS be removed to avoid violent explosions in the reaction vessel, even inert gases such as argon should be monitored because they still slow down the rate of reaction. They just happen to be a "counter-catalyst" - they slow both the forward AND reverse rate. (Which is why they're not as significant in my opinion but still important)
Post by: katherine123 on July 17, 2016, 08:52:57 pm

2013 There is a compromise between maximising yield and minimising the environmental impact of industrial processes. Justify this statement with reference to the production of sulfuric acid. 7 marks
The production of sulfuric acid is achieved through the mining and contact process which involves the maximisation of yield with environmental considerations throughout the process.
Mining: Sulfur is extracted through Frasch process in which super-heated steam and high-pressure air is forced into an underground sulfur deposit, melting the Sulfur and forcing it to the surface where it solidifies. Although this process produces high purity sulfur, it causes land subsidence and requires the removal of trees for land, thus damaging natural ecosystems. Mining also requires energy which is obtained by burning of fossil fuels which releases CO2 thus contributing to global warming and enhanced greenhouse effect.

Contact process step 1: Sulfur obtained by Frasch process is heated in dry air in a combustion furnace, forming SO2
S(s) +O2(g) --> SO2(g) + heat. Heat produced via this exothermic reaction is used to generate steam to produce electricity thus minimising environmental impacts as it reduce the reliance on fossil fuels thus minimising the amount of CO2 released.

Contact process step 2: SO2 is passed through a catalytic converter and several beds of V2O5 catalyst where it is converted into SO3. 2SO2(g) + O2(g) -->2SO3(g) +heat. The conditions of the reaction are manipulated to maximise yield of SO3 with environmental considerations. The forward reaction is exothermic thus the increase in temperature will force the equilibrium to favour backward reaction in order to reduce some of the increased heat according to Le Chatelier’s principle (LCP) which would reduce the yield of SO3. If the temperature is too high, the catalyst will be damaged which makes it less efficient and if temperature is too low, the rate of reaction will decrease which leads to the decrease in yield of SO3. Thus, a compromise temperature of approx. 400 degrees is maintained to ensure temperature is low enough to encourage forward reaction and high enough for particles to have sufficient kinetic energy to undergo reaction. Heat released via the exothermic reaction is used to generate electricity and unreacted gases are recycled back into the converter to reduce the necessity to mine for sulfur thus minimising environmental impacts of enhanced greenhouse effect and land subsidence. High pressure will favour the forward reaction (LCP) in order to reduce some of the increase in pressure thus increasing yield of SO3. However, low pressure of 1-2atm is used instead as it will be dangerous and costly if it is too high. V2O5 catalyst lowers the activation energy which increases the rate of reaction and enables lower temperature to be used, thus lowering the energy requirement and cost, and moreover reducing the production of energy from burning of fossil fuels which releases CO2 which minimises environmental impact. The concentration of oxygen is maintained high to force equilibrium to the right (LCP) in order to reduce some of the increased concentration of Oxygen thus increasing yield of SO3.

Contact process step 3: SO3(g) is dissolved in concentrated  H2SO4 to produce oleum:   SO3(g) + H2SO4(l) -->H2S2O7(l). Water is then added to H2S2O7 to form concentrated H2SO4: H2S2O7(l) + H2O(l) -->2H2SO4(l).

Do I need to include catalyst even though it doesn’t increase the yield
and do i need to be more specific about the compromise temperature:
-   Initial gas stream is at 1000 degrees after combustion of sulfur.
-   Gas stream is cooled to about 550 degrees and passed over bed of V2O5 catalyst which produces 70% conversion of SO2 to SO3
-   Gas stream is further cooled to 400 degrees and passed over a second catalytic bed, producing 97% conversion
-   After conversion of SO3 to H2SO4 (via oleum), remaining gas stream is passed over final bed of V2O5 to produce 99.7% conversion

Assess the impact of atomic absorption spectroscopy (AAS) on the scientific understanding of the effects of trace elements 4 marks
Before AAS was developed there were no other techniques sensitive enough to measure low concentrations of trace elements. Therefore, scientists were not able to detect and determine concentration of trace elements. However, after the development of AAS, which measures intensity of certain wavelengths of light from the electromagnetic spectrum (different element absorb specific wavelengths of light), the concentration of trace element can be determined very accurately as low as 0.01 ppm. The lamp used in AAS emits specific wavelengths that corresponds to the element being measured thus absorption of element is not affected by presence of other elements. This allows scientist to measure concentration and rectify deficiency of specific trace element (eg. animal health could not be maintained due to Cobalt deficiencies in pasture in coastal SW Australia) by providing fertilisers or dietary supplements to animals to enhance plant growth and animal and human health. Thus, AAS has enabled scientists to detect and measure concentration of trace elements which led to the increase in scientific understanding of the effects of trace elements.
Post by: RuiAce on July 17, 2016, 09:35:39 pm
I don't think anyone's done industrial chem so I'll add some input for this one despite loss of experience.

2013 There is a compromise between maximising yield and minimising the environmental impact of industrial processes. Justify this statement with reference to the production of sulfuric acid. 7 marks
The production of sulfuric acid is achieved through the mining and contact process which involves the maximisation of yield with environmental considerations throughout the process.
Mining: Sulfur is extracted through Frasch process in which super-heated steam and high-pressure air is forced into an underground sulfur deposit, melting the Sulfur and forcing it to the surface where it solidifies. Although this process produces high purity sulfur, it causes land subsidence and requires the removal of trees for land, thus damaging natural ecosystems. Mining also requires energy which is obtained by burning of fossil fuels which releases CO2 thus contributing to global warming and enhanced greenhouse effect.

Contact process step 1: Sulfur obtained by Frasch process is heated in dry air in a combustion furnace, forming SO2
S(s) +O2(g) --> SO2(g) + heat. Heat produced via this exothermic reaction is used to generate steam to produce electricity thus minimising environmental impacts as it reduce the reliance on fossil fuels thus minimising the amount of CO2 released.

Contact process step 2: SO2 is passed through a catalytic converter and several beds of V2O5 catalyst where it is converted into SO3. 2SO2(g) + O2(g) -->2SO3(g) +heat. The conditions of the reaction are manipulated to maximise yield of SO3 with environmental considerations. The forward reaction is exothermic thus the increase in temperature will force the equilibrium to favour backward reaction in order to reduce some of the increased heat according to Le Chatelier’s principle (LCP) which would reduce the yield of SO3. If the temperature is too high, the catalyst will be damaged which makes it less efficient and if temperature is too low, the rate of reaction will decrease which leads to the decrease in yield of SO3. Thus, a compromise temperature of approx. 400 degrees is maintained to ensure temperature is low enough to encourage forward reaction and high enough for particles to have sufficient kinetic energy to undergo reaction. Heat released via the exothermic reaction is used to generate electricity and unreacted gases are recycled back into the converter to reduce the necessity to mine for sulfur thus minimising environmental impacts of enhanced greenhouse effect and land subsidence. High pressure will favour the forward reaction (LCP) in order to reduce some of the increase in pressure thus increasing yield of SO3. However, low pressure of 1-2atm is used instead as it will be dangerous and costly if it is too high. V2O5 catalyst lowers the activation energy which increases the rate of reaction and enables lower temperature to be used, thus lowering the energy requirement and cost, and moreover reducing the production of energy from burning of fossil fuels which releases CO2 which minimises environmental impact. The concentration of oxygen is maintained high to force equilibrium to the right (LCP) in order to reduce some of the increased concentration of Oxygen thus increasing yield of SO3.

Contact process step 3: SO3(g) is dissolved in concentrated  H2SO4 to produce oleum:   SO3(g) + H2SO4(l) -->H2S2O7(l). Water is then added to H2S2O7 to form concentrated H2SO4: H2S2O7(l) + H2O(l) -->2H2SO4(l).

Do I need to include catalyst even though it doesn’t increase the yield
and do i need to be more specific about the compromise temperature:
-   Initial gas stream is at 1000 degrees after combustion of sulfur.
-   Gas stream is cooled to about 550 degrees and passed over bed of V2O5 catalyst which produces 70% conversion of SO2 to SO3
-   Gas stream is further cooled to 400 degrees and passed over a second catalytic bed, producing 97% conversion
-   After conversion of SO3 to H2SO4 (via oleum), remaining gas stream is passed over final bed of V2O5 to produce 99.7% conversion
I reckon for questions like this the focus is mainly on the Contact process for the production of sulfuric acid rather than the Frasch process for extraction of sulfur. This is because step 2 is where the equilibrium reaction happens. However, as you mentioned, the Frasch process does indeed come with environmental considerations such as land subsidence. Another point I would include here is that if the superheated steam leaks you are potentially causing thermal pollution which may lead to global warming. Also, the sulfur collected must be stored ASAP to ensure that it doesn't oxidise by itself and leak SO2 into the atmosphere.

Whilst the 1000deg C for step 1 isn't really much of a hassle, including it is better than not as it shows you have better understanding of the actual Contact process. This reinforces your point that sufficient heat energy will indeed be produced to be converted to electricity.

With the catalyst beds for step 2, generally I only include the first and fourth with their relative percentages. But yes I would DEFINITELY include that one because the idea is 30% or 3% emission is unacceptable by environmental guidelines, but 0.3% is okay. The cooling...well yes I would include it. And you've already integrated it in there somehow as well.
Post by: anotherworld2b on July 18, 2016, 02:21:14 am
I was wondering if i could get help with these questions  XD
Post by: RuiAce on July 18, 2016, 09:55:54 am
I was wondering if i could get help with these questions  XD
Can we please not have so many questions waved around at once?

For the first one, you know that complete combustion involves the production of carbon dioxide and water.

C6H12(l) + O2(g) -> CO2(g) + H2O(g)

Use logic to determine what order to balance in:
C: Reactants - 6. Products - 1. Therefore 6 in front of CO2
H: Reactants - 12. Products - 2. Therefore 6 in front of H2O
O: Products - 6*12 + 6*6 = 18. Reactants - 2. Therefore 9 in front of O2

C6H12(l) + 9 O2(g) -> 6 CO2(g) + 6 H2O(g)

The second one is more open-ended, as there exists more than one equation for incomplete combustion. The products must include at least one of C and CO, however they can include any of C, CO or CO2 at once.
__________
Answer to Q5 is incomplete because CO is there
__________
Reactivity is a chemical property as it is essentially a measure of how likely it will form a new substance. Note that physical properties do not involve formation of a new substance.

For the net ionic equations, try writing complete equations first. (Remember: Metal + Acid -> Salt + Hydrogen)
Post by: katherine123 on July 18, 2016, 12:37:01 pm
6 marks
Chemists can assist in reversing or minimising the environmental problems caused by technology and the human demand for products and services. With reference to this statement, assess the need for chemists to collaborate when monitoring the environmental impact of a named electrochemical cell.

Therefore, there is a great need for chemists to collaborate to minimise the damage caused by this technology as some chemists monitor the environment while others use the results of this to develop environmental friendly alternatives.
Post by: EEEEEEP on July 18, 2016, 12:56:16 pm
6 marks
Chemists can assist in reversing or minimising the environmental problems caused by technology and the human demand for products and services. With reference to this statement, assess the need for chemists to collaborate when monitoring the environmental impact of a named electrochemical cell.

Therefore, there is a great need for chemists to collaborate to minimise the damage caused by this technology as some chemists monitor the environment while others use the results of this to develop environmental friendly alternatives.

>> Good • Demonstrates a thorough knowledge and understanding of the identifiedroles of chemists and the environmental impact of a correctly named electrochemical cell
>> GOOD • Assesses the need for collaboration between chemists
>> Good • Demonstrates coherence and logical progression of ideas with correct scientific terminology used
Post by: onepunchboy on July 19, 2016, 03:49:57 am
Can someone give me a basic run down on how aas works. Ive read the textbook but still very confusing to me
Post by: jakesilove on July 19, 2016, 09:20:13 am
I have a 25% assessed report to write up, 1000 +/- 200 words about biopolymers, my chosen PFA. I have attached an image of the issue I have been presented with.

The actual task is requires a critical discussion of the science relevant to the PFA, all in a concise report with an introduction, background information, the actual synopsis and a relevant conclusion.

I just want to know what you would recommend in regards to the format and structure of my report, as well as what you think is vital to include in order to maximise my marks.

Hey!

For a report like this, it is important to include a number of things. I would recommend subheading by topic, and using tables and graphs where possible (noting that graphs won't increase your word count!).

I think history is important. When did you biopolymer start being produced? How efficient was it? When was it discovered? For what purpose was it produced. Then, it is easy to cite advances in technology that has brought it to the stage it is at today. You should clearly indicate how these developments have been useful to both the production and use of the polymer; if the production has become more environmentally friendly/renewable, that is important as it 1) protects the environment and 2) allows for a sustainable, replenishing supply of the polymer!

Make sure to include chemistry. What is actually happening, what catalyst is being used, what does the reaction look like etc. Try to include chemical terminology as much as possible, as this brings the assignment away from a 'society and the environment' answer and towards a more chemical sound, scientific one.

The aspects that you've written in your question are of course crucial. Use that structure, but within "background information" I would include history or discover, development and uses (just as examples). You need to think about what is specifically important for your biopolymer, after doing fairly extensive research.

It's always hard to tell where marks lie for an assignment like this. All I can recommend is doing lots of research, including what you think is important, and if you go way over word count trying to take out stuff that doesn't directly lend itself to the question. Include lots of statistics to impress the marker (what is the reaction yield? What year did developments take place? Who discovered the process?).

Good luck! Let me know if you have any more specific questions, noting that we can't do your assignment for you of course!

Jake
Post by: jakesilove on July 19, 2016, 09:24:19 am
Can someone give me a basic run down on how aas works. Ive read the textbook but still very confusing to me

Hey! You really only need to know what AAS is in a very basic sense; textbooks often go into great depth, and the fact is you just don't need to know that. So, here's what I learnt for my HSC.

AAS is a technique to identify absolutely minuscule (ppm) amounts of heavy metals. First, a calibration curve must be calculated, in which a sample is tested that DOES NOT have any of the heavy metal present. The liquid will be 'aspirated' or 'vaporized', by being fed into a flame. Then, light is shone on one side of the flame, and detected on the other. From this, a curve is established (based on the readings on the detector). Then, the liquid which contains some heavy metal is tested. The same process is undergone (aspiration, taking readings from the detector). However, as there are now heavy metal particles floating around in the flame, some on the light will be reflected/refracted. This will lead to a different curve being present, which can be compared to the original curve. We can therefore establish the concentration of the heavy metal!

Obviously it's much more complicated that this. You also need to think about the effects of AAS (allowing us to detect small quantities = good, as we can protect society and the environment) as well as potential issues (only detect heavy metals, destroys sample, expensive to run etc.). Let me know if you need any further help!

Jake
Post by: RuiAce on July 19, 2016, 09:28:10 am
Can someone give me a basic run down on how aas works. Ive read the textbook but still very confusing to me
AAS:

0. Remember the diagram off by heart. This is one of those scenarios where remembering the diagram actually helps, as mentioned in Jake's lecture.

1. The hollow cathode lamp shines into the flame
1. a) Every individual element has it's own absorption spectrum. The hollow cathode lamp must be shining light corresponding to the absorption spectrum of each element. Hence, every element has its own cathode lamp.
2. A standardised solution (this just means that its concentration is KNOWN) is fed into the flame.
2. a) This is used to calibrate the AAS machine.
3. Light passing through the flame will get absorbed by the metal ions and reemitted. I'm not sure if you need to know about the physics here for the HSC chemistry course, but I can explain it later if you wish so.
4. A collimator focuses the light
5. The light reaches a prism and is dispersed.
6. The photomultiplier picks up the light. Calibration is done.
7. Repeat steps 2 to 6 with the solution of unknown concentration.

Absorption is a measure of the amount of light absorbed relative to the reference beam. It is proportional to the concentration of the metal sample.

(Lel. Jake beat me here.)
Post by: onepunchboy on July 19, 2016, 11:19:22 am
Thanks for the explanation guys i understand aas now ayy :)
Post by: anotherworld2b on July 19, 2016, 12:02:46 pm
For a lab report we were assigned to do we had to identify an unknown bicarbonate.

I was wondering if the substance was sodium bicarbonate and you react it with Hydrochloric.

Would it be correct to assume that this reaction will follow the basic reaction

Acid + Metal Carbonate -> salt + carbon dioxide + water

I heard that bicarbonates wont react with HCL? Is that true?

In the method we then boiled the solution that is produced to get the sodium salt. Using the salt we conducted a flame test to validate the metal.

I was wondering what are the chemistry concepts behind this method?
Post by: RuiAce on July 19, 2016, 05:55:27 pm
For a lab report we were assigned to do we had to identify an unknown bicarbonate.

I was wondering if the substance was sodium bicarbonate and you react it with Hydrochloric.

Would it be correct to assume that this reaction will follow the basic reaction

Acid + Metal Carbonate -> salt + carbon dioxide + water

I heard that bicarbonates wont react with HCL? Is that true?

In the method we then boiled the solution that is produced to get the sodium salt. Using the salt we conducted a flame test to validate the metal.

I was wondering what are the chemistry concepts behind this method?
No it definitely reacts.

You can use HCl to test for the presence of HCO3- OR CO32-. However you can't distinguish between which one is there as easily.

Note that test for HCO3- is not explicitly stated in the syllabus.
Post by: tanishap on July 19, 2016, 07:23:35 pm
Hey Jake!

I came to the repeat Chem lecture last week at UTS and you showed us this totally boss document that was literally 3 pages of all the reactions we had to memorise, but I forgot where to find it. Can you just let me know where I can find that chemistry bible please? :))

Thanks heaps! :)
Post by: jakesilove on July 19, 2016, 07:26:04 pm
Hey Jake!

I came to the repeat Chem lecture last week at UTS and you showed us this totally boss document that was literally 3 pages of all the reactions we had to memorise, but I forgot where to find it. Can you just let me know where I can find that chemistry bible please? :))

Thanks heaps! :)

You can find it right HERE! Go mental :)

Jake
Post by: marynguyen18 on July 19, 2016, 10:58:06 pm
hey Jake I went to your ATAR Notes lecture at UTS and i'm still really confused with the HSC 2013 question 23, is there any chance you could re explain it?
Post by: onepunchboy on July 19, 2016, 11:19:14 pm

Hey can someone explain the answer to this question, i put d but its wrong hehe
Thanks
Post by: RuiAce on July 19, 2016, 11:35:11 pm

Hey can someone explain the answer to this question, i put d but its wrong hehe
Thanks
Have you tried calculating the resultant pH for A and B?

The concentration of H+ is going to go down heaps if you react it with a base.

(Can't see the value on the top right corner to do the calculations)

The dilution just raises the pH by 1
Post by: onepunchboy on July 19, 2016, 11:37:21 pm
Oh right its 0.01 mol/L
Post by: RuiAce on July 19, 2016, 11:45:18 pm
Original pH = 2

D) The dilution factor by 10 will change the pH into 3

Take B)

HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)

nHCl initial = 0.1L * 0.01 mol L-1 = 10-3 mol
nNaOH initial = 0.01L * 0.1mol L-1 = 10-3 mol

Everything's going to be reacted. The final solution will be neutral with a pH of 7

(You may want to test out the CaCO3 reaction as well just incase the change in pH is greater. But I doubt it will be.)
Post by: amandali on July 20, 2016, 04:33:24 pm
are these the correct and main steps for Solvay process?
Post by: RuiAce on July 20, 2016, 05:07:30 pm
are these the correct and main steps for Solvay process?
Yep I believe those are correct. Try to have all of those equations memorised.
Post by: conic curve on July 21, 2016, 10:58:34 am
How do you do well in the water module?

I'm finding that module really hard and I don't want to drop chemistry in the future. I feel that I don't have that level of understanding in chemistry

Also

1. why does alcohol dissolve in both polar and non polar?
2. Why does water dissolve covalent compounds?

Thanks  ;D
Post by: RuiAce on July 21, 2016, 11:05:07 am
How do you do well in the water module?

I'm finding that module really hard and I don't want to drop chemistry in the future. I feel that I don't have that level of understanding in chemistry

Also

1. why does alcohol dissolve in both polar and non polar?
2. Why does water dissolve covalent compounds?

Thanks  ;D
The first question is more of a production of materials question that draws on knowledge for water.

As a rule of thumb, likes dissolve likes

The hydroxyl group (-OH) can be involved with extensive hydrogen bonding between other O, N and F molecules. Because the -OH is also polar, it also engages in dipole-dipole interactions with other polar molecules.

But the long carbon chain is non-polar. The linear non-polar chain is what will dissolve with other non-polar substances such as alkanes, through dispersion forces.

(Dispersion forces are as a result of one side of the molecule temporarily becoming more polar than the other, thereby forming a dipole. Dipole-dipole interactions happen when there is a permanent dipole going on. Hydrogen bonding is only separate from dipole-dipole interactions because it's much stronger; about 1/10 the strength of an actual covalent bond)

Q2 is wrong. Water dissolves certain covalent compounds. It does not dissolve stuff like methane or propane.
The covalent compounds it dissolves are all polar because water IS a polar molecule
Post by: conic curve on July 21, 2016, 03:59:45 pm
The first question is more of a production of materials question that draws on knowledge for water.

As a rule of thumb, likes dissolve likes

The hydroxyl group (-OH) can be involved with extensive hydrogen bonding between other O, N and F molecules. Because the -OH is also polar, it also engages in dipole-dipole interactions with other polar molecules.

But the long carbon chain is non-polar. The linear non-polar chain is what will dissolve with other non-polar substances such as alkanes, through dispersion forces.

(Dispersion forces are as a result of one side of the molecule temporarily becoming more polar than the other, thereby forming a dipole. Dipole-dipole interactions happen when there is a permanent dipole going on. Hydrogen bonding is only separate from dipole-dipole interactions because it's much stronger; about 1/10 the strength of an actual covalent bond)

Q2 is wrong. Water dissolves certain covalent compounds. It does not dissolve stuff like methane or propane.
The covalent compounds it dissolves are all polar because water IS a polar molecule

TBH I find that the water module is a hard module and things are very difficult for me to grasp, especially things like VSPER theory, electron pair geometry, spatial geometry, etc

Is this "like dissolves like" something by definition/convention or is there a reason as to why "like dissolves like"

It says here that water is a polar molecule because of it's shape: http://www.triangularwave.com/a1b1.htm and that one side is positively charged and one side is negatively charged. Why is this? (again is this "by definition")
Post by: RuiAce on July 21, 2016, 04:40:36 pm
TBH I find that the water module is a hard module and things are very difficult for me to grasp, especially things like VSPER theory, electron pair geometry, spatial geometry, etc

Is this "like dissolves like" something by definition/convention or is there a reason as to why "like dissolves like"

It says here that water is a polar molecule because of it's shape: http://www.triangularwave.com/a1b1.htm and that one side is positively charged and one side is negatively charged. Why is this? (again is this "by definition")
Yes that's correct.

If instead, water were a linear molecule like CO2, there would not be a region for high electron density. Recall that electrons like to exist in pairs. However, like charges are obviously going to repel each other - no pair of electrons wants to hug another pair. So that extra 2 electrons in oxygen (that are not covalently bonded) will make advantage of the molecule being bent, and cluster at the top of the oxygen in H2O instead.

It is more easily shown if I drew a diagram tbh though.

Water does take a bit of effort to make sense out of yeah.
Post by: conic curve on July 21, 2016, 04:46:55 pm
Yes that's correct.

If instead, water were a linear molecule like CO2, there would not be a region for high electron density. Recall that electrons like to exist in pairs. However, like charges are obviously going to repel each other - no pair of electrons wants to hug another pair. So that extra 2 electrons in oxygen (that are not covalently bonded) will make advantage of the molecule being bent, and cluster at the top of the oxygen in H2O instead.

It is more easily shown if I drew a diagram tbh though.

Water does take a bit of effort to make sense out of yeah.

Try explaining with this picture: https://www.google.com.au/search?hl=en&biw=1366&bih=667&site=webhp&tbm=isch&q=water+molecule+structure&sa=X&ved=0ahUKEwivhsaR-oPOAhXCHZQKHddwB_0QhyYIIw&dpr=1#imgrc=J0EWfEKYgTnznM%3A

(if this helps obviously)

Edit: I'll redo them

http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/water.gif
http://witcombe.sbc.edu/water/images/chemistrydipolescicx.jpg
Post by: RuiAce on July 21, 2016, 04:53:39 pm
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps4wqhzana.png)
Post by: Jakeybaby on July 21, 2016, 07:03:11 pm
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps4wqhzana.png)
Oh wow, I love it
Post by: anotherworld2b on July 22, 2016, 01:24:13 am
How would you do q5c and d?
I got the answers for a and b but i didnt know how to do c and d.

Help is greatly appreciated
Post by: RuiAce on July 22, 2016, 01:32:22 am
How would you do q5c and d?
I got the answers for a and b but i didnt know how to do c and d.

Help is greatly appreciated
Consider the original balanced equations to know what is going on first.

c) This is a metal and acid reaction. We know that:
Metal + Acid -> Salt + Hydrogen gas

So treating our species appropriately:

Zn(s) + H2SO4(aq) -> ZnSO4(aq) + H2(g)

Have any charges changed?
Zn: It has went from zinc solid to the zinc ion
SO4: The sulfate ion is as is
H: The hydrogen ion has become hydrogen gas.

So we have Zn(s) + 2 H+ -> Zn2+ + H2(g)

c) This is a precipitation reactants. But it's made easier because they give us the reactants and the product.
Pb(NO3)2 + 2 KNO3(aq) + PbI2(s)

Have any changed occurred?
Pb: It stayed as Pb2+ but it precipitated into the product PbI2(s) which cannot be split!
NO3: The nitrate ion is as is
K: The potassium ion is as is
I: It stayed as I- but it precipitated into the product PbI2(s) which cannot be split!

So we have Pb2+ + 2 I- -> PbI2(s)
_____________

An alternate way to approach these questions is to write out the complete ionic equation based off the balanced equation and then cancel out any spectator ions.

For c) this would be

Zn(s) + 2 H+ + SO42- -> Zn2+ + SO42- + H2(g)
Post by: conic curve on July 22, 2016, 01:16:14 pm
I need help with the water topic since I find it quite difficult

Q. A soluble molecular substance such as sugar (e.g. sucrose C11H22O11) is mixed with water

a. Identify the forces that mist break for it to dissolve

I was not sure of this and was thinking about intermolecular or intramolecular but I seriously don't know.

b. Describe the attraction that then develops between the free water and sugar molecules

I'm not sure of this. Is it Hydrogen bonding?

Thanks guys  ;)
Post by: RuiAce on July 22, 2016, 01:25:52 pm
I need help with the water topic since I find it quite difficult

Q. A soluble molecular substance such as sugar (e.g. sucrose C11H22O11) is mixed with water

a. Identify the forces that mist break for it to dissolve

I was not sure of this and was thinking about intermolecular or intramolecular but I seriously don't know.

b. Describe the attraction that then develops between the free water and sugar molecules

I'm not sure of this. Is it Hydrogen bonding?

Thanks guys  ;)
Draw out the molecular structure of sucrose. It is full of polar OH bonds everywhere. The fact that they are polar means that sucrose molecules will engage in dipole-dipole interactions (and maybe a small amount of hydrogen bonding) with each other. (Do you see why?)

If water is to dissolve sucrose, the intermolecular forces between sucrose molecules must be broken as water has to start interacting intermolecularly with sucrose instead. (No intramolecular bonds should be broken or we no longer have sucrose).

Because the polarity of water is much stronger than that of sucrose, it will overcome the intermolecular forces between sucrose molecules and form extensive hydrogen bonding with sucrose molecules instead, thereby dissolving the substance.

Process of thought:
a) Water has to break something. What is the only logical thing to break? In fact, what even is there to break is a better question
b) It is hydrogen bonding. Just draw out the shape of sucrose and water and you should be able to realise how the polar -OH groups on sucrose can interact with water.
Post by: amandali on July 22, 2016, 01:48:53 pm

dont know how to do q17 b
Post by: RuiAce on July 22, 2016, 02:00:42 pm

dont know how to do q17 b
Please give the answers to a next time if you could do it. Would make life easier :) (especially since I might've gotten it wrong).
\begin{align*}i) \quad \frac{[SO_3]^2}{[SO_2]^2[O_2]}&=174\\ \therefore \frac{5^2}{[O_2]}&=174\\ [O_2]&=0.143678 \dots... mol \, L^{-1}\\ &\approx 1.4 \times 10^{-1}\, mol L^{-1}\end{align*}
$\text{Similarly for ii) }[O_2]=\frac{9^2}{174}=0.465517\dots mol L^{-1} \approx 4.7 \times 10^{-1} \, mol \, L^{-1}$
Because the equilibrium constant is a decently large number, at equilibrium there should be more of the products than the reactants.

Firstly, our calculations implicitly verify this as in the event the concentration of oxygen is greater, there would indeed be more of SO2 being converted into SO3 (90% instead of 50%).

How has Le Chatelier's principle been obeyed here? I reckon it's actually quite simple. We are saying that if there's more O2 (the reactant), then the forward reaction should be favoured. The equilibrium shifts towards the right, converting SO2 to SO3 in the process. If the concentration of O2 is greater, indeed we should anticipate that the forward reaction will occur to a greater extent.
Post by: marynguyen18 on July 22, 2016, 09:25:58 pm
Hey i was just wondering how would you structure a 4-7 mark question on Cellulose?
Post by: conic curve on July 22, 2016, 09:27:07 pm
Hey i was just wondering how would you structure a 4-7 mark question on Cellulose?

It really depends on what the question is and what the marking criteria is
Post by: marynguyen18 on July 22, 2016, 09:34:08 pm
What if the question on Cellulose is related to it as being an alternative to petrochemicals and its potential as a raw material for chemicals used in industry?
Post by: RuiAce on July 22, 2016, 09:41:57 pm
What if the question on Cellulose is related to it as being an alternative to petrochemicals and its potential as a raw material for chemicals used in industry?
Think about why it can be an alternative to petrochemicals. This goes back down to the fact that natural production of ethylene comes from fossil fuels = bad for reasons you should already know.

Cellulose is a renewable resource. If you want to produce ethanol, all you have to do is ferment it and done. And you can easily get more cellulose from dead plant matter.

(But you should always analyse the negatives of this process, e.g. lots of arable land needed and the energy required to trigger fermentation is massive.)

Essentially, the part about being an alternative to petrochemicals is a process of outweighing the advantages and disadvantages against burning fossil fuels. The raw materials part can be related to things such as

- Ethanol: Very powerful solvent used in a variety of scenarios
- Polyethylene: A common polymer that we use today for plastics

Also, make sure to give a judgement if it ever asks you to evaluate/assess.
Post by: marynguyen18 on July 22, 2016, 09:46:18 pm
Thank you so much! Also when you make a judgement does it mean that you have to say what you think would be a good alternative?
Post by: RuiAce on July 22, 2016, 09:49:01 pm
You make a judgment based off the question.

E.g. if the question is about the potentials of cellulose in being a replacement, you would recollect the info you have and say hence it is most likely that it is a good/bad alternative. Or something like overall, it is (say, adequate) in that it is able to...

Your judgment doesn't have to be "correct" - it just has to make sense and link to the question.

Post by: conic curve on July 22, 2016, 09:50:30 pm
Thank you so much! Also when you make a judgement does it mean that you have to say what you think would be a good alternative?

Judgement really means to give your own opinion

Anyways can anyone give me tips on how to answer hydrocarbon questions because I really suck at that

Thanks
Post by: RuiAce on July 22, 2016, 09:59:18 pm
Judgement really means to give your own opinion

Anyways can anyone give me tips on how to answer hydrocarbon questions because I really suck at that

Thanks
Before commencing, your prefixes of meth-, eth-, prop-, but-, pent- etc. should be known off by heart first.

The alkane series only has single bonds. The key characteristic of the alkenes series is that somewhere in there, there is a double bond.

A complete structural formula includes the bonds. 1-pentene is basically saying that there is a double bond, coming off carbon #1.

C=C-C-C-C

with H's coming off the carbons. (Note: if you have a double bond, do not draw in too many H's)

A condensed formula just writes out what's there. For 1-pentene it would be
CH2CHCH2CH2CH3

For heptane this would be
CH3CH2CH2CH2CH2CH2CH3 (count the carbons! The prefix matters!)

For 1-heptene this would be
CH3CH2CHCHCH2CH2CH3

This one might not be obvious. It helps to draw out the complete structural formula whenever you're stuck.

Molecular formulae are easy. Alkanes have formula CnH2n+2, whereas alkenes have formula C2H2n

I'm leaving stuff unanswered because I want you to do some work yourself.
Post by: conic curve on July 22, 2016, 10:03:37 pm
Before commencing, your prefixes of meth-, eth-, prop-, but-, pent- etc. should be known off by heart first.

The alkane series only has single bonds. The key characteristic of the alkenes series is that somewhere in there, there is a double bond.

A complete structural formula includes the bonds. 1-pentene is basically saying that there is a double bond, coming off carbon #1.

C=C-C-C-C

with H's coming off the carbons. (Note: if you have a double bond, do not draw in too many H's)

A condensed formula just writes out what's there. For 1-pentene it would be
CH2CHCH2CH2CH3

For heptane this would be
CH3CH2CH2CH2CH2CH2CH3 (count the carbons! The prefix matters!)

For 1-heptene this would be
CH3CH2CHCHCH2CH2CH3

This one might not be obvious. It helps to draw out the complete structural formula whenever you're stuck.

Molecular formulae are easy. Alkanes have formula CnH2n+2, whereas alkenes have formula C2H2n

I'm leaving stuff unanswered because I want you to do some work yourself.

Yes I know that but when it comes to answering them, I get stuck

Could you at least give me hints on them?

What does it mean by like 3-methy-2-difluro-1-ene (or something similar to that)?
Post by: RuiAce on July 22, 2016, 10:13:30 pm
Yes I know that but when it comes to answering them, I get stuck

Could you at least give me hints on them?

What does it mean by like 3-methy-2-difluro-1-ene (or something similar to that)?
Should've been more specific about methyl groups and that stuff, yeah.

Firstly, let's clear up a nomenclature: 1-propene and prop-1-ene mean the same thing. The convention is to use the latter (very retarded change they implemented a few years ago, but we use it).

Let's fix your question into something more manageable.
3-methy-1,2-difluro-hept-1-ene

Getting a CFC (or rather HCFC) with a methyl group present is pretty darn rare but we can still do it.

We start by identifying the prefix: hept. This means there are 7 carbons

Note that it's hept-1-ene. Hence, there is a double bond coming off carbon 1

C=C-C-C-C-C-C

We are told that a methyl group comes off carbon number 3. We can add that in.

C
C=C-C-C-C-C-C

We are told that a fluorine comes off carbon 1, AND carbon 2. We can draw them in wherever appropriate

Cl     C
C=C-C-C-C-C-C
Cl

Then just chuck all the appropriate hydrogens in there last.
Post by: conic curve on July 22, 2016, 10:20:02 pm
Should've been more specific about methyl groups and that stuff, yeah.

Firstly, let's clear up a nomenclature: 1-propene and prop-1-ene mean the same thing. The convention is to use the latter (very retarded change they implemented a few years ago, but we use it).

Let's fix your question into something more manageable.
3-methy-1,2-difluro-hept-1-ene

Getting a CFC (or rather HCFC) with a methyl group present is pretty darn rare but we can still do it.

We start by identifying the prefix: hept. This means there are 7 carbons

Note that it's hept-1-ene. Hence, there is a double bond coming off carbon 1

C=C-C-C-C-C-C

We are told that a methyl group comes off carbon number 3. We can add that in.

C
C=C-C-C-C-C-C

We are told that a fluorine comes off carbon 1, AND carbon 2. We can draw them in wherever appropriate

Cl     C
C=C-C-C-C-C-C
Cl

Then just chuck all the appropriate hydrogens in there last.

Thanks so much  :D
Post by: anotherworld2b on July 23, 2016, 07:07:15 pm
I've tried to answer q6 b and c but i got them wrong and i dont know why
Post by: RuiAce on July 23, 2016, 07:35:14 pm
I've tried to answer q6 b and c but i got them wrong and i dont know why
H+ + OH- -> H2O(l)

You are safe to cancel out the numbers because this is a net IONIC form we're talking about.
_________________________________

Your working out for c) is wrong in the products. Here are proposed amendments.

1. Seperate CH3COOH(aq) into H+ and CH3COO- for better clarity in the reactants first
2. You wrote acetic acid when it's not there. It's only the acetate ion. Thus, replace the mistaken acetic acid with CH3COO-.

You should get MgO(s) + 2 H+ -> Mg2+ + H2Ol
Post by: anotherworld2b on July 23, 2016, 11:09:18 pm
I got the answer for q6 b but the answer for q6 is different.

I was also wondering how would you do q 8? I dont understand the question itself.

H+ + OH- -> H2O(l)

You are safe to cancel out the numbers because this is a net IONIC form we're talking about.
_________________________________

Your working out for c) is wrong in the products. Here are proposed amendments.

1. Seperate CH3COOH(aq) into H+ and CH3COO- for better clarity in the reactants first
2. You wrote acetic acid when it's not there. It's only the acetate ion. Thus, replace the mistaken acetic acid with CH3COO-.

You should get MgO(s) + 2 H+ -> Mg2+ + H2Ol
Post by: RuiAce on July 23, 2016, 11:19:04 pm
I got the answer for q6 b but the answer for q6 is different.

I was also wondering how would you do q 8? I dont understand the question itself.
CH3COOH(aq) is acetic acid. I disagree with the answer because the fact it is acetic acid means that it is definitely dissociated.

Unless the question is being pedantic. Their answer isn't "completely" unjustified because some unsplit CH3COOH is in there as well. So I can see where they're coming from. But the question should've made that clearer - I don't like these ambiguous questions.
___________________

Q8 is combining equations. You have to do some maths here.

a) Equation 2 has just one CO in the reactants. But equation 1 has two COs in the products
So we have to multiply equation 2 by two to merge the equations:

2 CO(g) + 2 FeO(s) -> 2 Fe(l) + 2 CO2(g)

Substitute equation 1 in there to get:

C(s) + CO2(g) + 2 FeO(s) -> 2 Fe(l) + 2 CO2(g)

Cancel out the duplicated carbon dioxide:

C(s) + 2 FeO(s) -> 2 Fe(l) + CO2(g)

It's sort of like simultaneous equations: Perform (1) + 2*(2)

I'll let you have a try at the other two first. If you're still stuck, however (especially with 8c - that one is trickier), just post again
Post by: Aliceyyy98 on July 24, 2016, 12:44:05 am
Hello!

I have a quick question about how coordinate covalent bonds work? Could someone please explain to me, perhaps using examples that would be great :) thanks in advance!

Cheers
Post by: RuiAce on July 24, 2016, 01:09:36 am
Hello!

I have a quick question about how coordinate covalent bonds work? Could someone please explain to me, perhaps using examples that would be great :) thanks in advance!

Cheers
The basic idea is that whereas in your normal covalent bonds, two atoms share an electron with each other, in a coordinate covalent bond one atom shares BOTH of the electrons with the other.

E.g. The first one you must know of is ozone. In O2, there's a double bond between the two oxygen molecules (there are two covalent bonds, so both atoms share two electrons with each other). However in O3, that third oxygen gets TWO electrons from one of the other oxygen atoms.

It's more easily demonstrated with a Lewis dot structure diagram.
Post by: Aliceyyy98 on July 24, 2016, 01:17:19 am
Hi thank you for you answer, just to add on, how would you know if a molecule has a covalent coordinate bond or not without a diagram or anything :(

The basic idea is that whereas in your normal covalent bonds, two atoms share an electron with each other, in a coordinate covalent bond one atom shares BOTH of the electrons with the other.

E.g. The first one you must know of is ozone. In O2, there's a double bond between the two oxygen molecules (there are two covalent bonds, so both atoms share two electrons with each other). However in O3, that third oxygen gets TWO electrons from one of the other oxygen atoms.

It's more easily demonstrated with a Lewis dot structure diagram.
Post by: RuiAce on July 24, 2016, 01:22:35 am
Hi thank you for you answer, just to add on, how would you know if a molecule has a covalent coordinate bond or not without a diagram or anything :(
A pro-tip is to try drawing them all with covalent bonds first. If you can't successfully FORCE them to have a full outer shell (8 electrons) using only normal covalent bonds, there's probably a coordinate covalent bond somewhere.

However, the following three are ones that you just need to know that there is a coordinate covalent bond:
- Ozone O3: From one oxygen to the third oxygen
- Hydronium ion H3O+: From the oxygen to the third hydrogen
- Ammonium ion NH4+: From the nitrogen to the fourth hydrogen
Post by: angiezhang9 on July 24, 2016, 10:00:56 am
Hey :)

I don't understand this multiple choice question from the 2014 Catholic Trials.

An experiment was performed to determine the volume of carbon dioxide gas released during the fermentation of a glucose solution. The reaction took place over several days in a flask containing yeast that was immersed in a water bath at 28 degrees and stoppered with cotton wool. Which of the following factors would most significantly compromise the validity of this experiment?

A) The evaporation of water
B) The temperature of the water bath
C) Measurement error associated with determining the mass lost
D) Lack of oxygen due to the presence of the cotton wool plug

Thanks heaps! :)
Post by: RuiAce on July 24, 2016, 10:13:55 am
Hey :)

I don't understand this multiple choice question from the 2014 Catholic Trials.

An experiment was performed to determine the volume of carbon dioxide gas released during the fermentation of a glucose solution. The reaction took place over several days in a flask containing yeast that was immersed in a water bath at 28 degrees and stoppered with cotton wool. Which of the following factors would most significantly compromise the validity of this experiment?

A) The evaporation of water
B) The temperature of the water bath
C) Measurement error associated with determining the mass lost
D) Lack of oxygen due to the presence of the cotton wool plug

Thanks heaps! :)
One again, CSSA is giving very dodgy questions.

D) is obviously wrong - The whole process of fermentation is dependent on anaerobic conditions. By removing oxygen, we are promoting the validity of the experiment.
C) is asking for the wrong thing - A measurement error affects the accuracy of the experiment, not the validity.
B) can be a bit more tricky. Obviously the best conditions for fermentation is at 37°C, but the water bath is at 28°C. This does not promote the fermentation of glucose and is also a candidate for why the experiment is invalid.

However, we can't just use this our basis. To consider the validity, look at the aim.

What we want to determine, is the volume of carbon dioxide released IN fermentation. The fermentation itself? It just has to proceed! Note that at some temperature like 28°C the reaction will still proceed, just that it will be hindered by a slow process.

Therefore B is out. But why is the answer A?

This is my proposal. When you have an aim, the idea is that you keep all other factors fixed and leave behind only one variable. As the water evaporates (especially given how slow fermentation is), the impact of the water bath is altered. Potentially the temperature of the water bath changes, but not only that, the entire glucose solution is no longer under the same, consistent 28°C anymore! It could be under the 10°C or 40°C air! Because we don't fix all our other factors, the experiment's validity is more heavily impacted here. I'll let someone else give a better answer.
Post by: Happy Physics Land on July 24, 2016, 12:02:14 pm
Hey :)

I don't understand this multiple choice question from the 2014 Catholic Trials.

An experiment was performed to determine the volume of carbon dioxide gas released during the fermentation of a glucose solution. The reaction took place over several days in a flask containing yeast that was immersed in a water bath at 28 degrees and stoppered with cotton wool. Which of the following factors would most significantly compromise the validity of this experiment?

A) The evaporation of water
B) The temperature of the water bath
C) Measurement error associated with determining the mass lost
D) Lack of oxygen due to the presence of the cotton wool plug

Thanks heaps! :)

Hey Angie!

A is the correct option simply because all we are trying to measure is the volume of CO2 produced. If water evaporates (especially since we are carrying out the experiment under 28 degrees here, so a large chance that some water would have evaporated) then we obtain a measurement that includes a combined mass loss of CO2 and H2O. Validity means "are we measuring what we want to measure" and this is clearly not the case if water has also evaporated. Keep in mind, WE ONLY WANT TO MEASURE THE MASS OF CARBON DIOXIDE.

Explaining this one step further. When you conducted this experiment what you would have done is to measure the mass of fermentation mixture before fermentation has taken place, and the mass of fermentation mixture after fermentation has taken place. The loss in mass after fermentation you would assume to be CO2 (You would probably have carried out the experiment with a limewater flask connected to the fermentation mixture as well). But keep in mind that even at room temperature, water evaporation still occurs, meaning that whilst CO2 is being evaporated, water is too! So when you measure the lost weight, thats a combined weight of evaporated CO2 and WATER. So evidently we are not measuring JUST the mass of CO2 and therefore the experiment becomes invalid.

So, how do we improve validity? We can do one of the two things:

1. Keep a carton of water under the same temperature as your fermentation mixture. Make sure the volume of the water is the same as the volume of your fermentation mixture. Measure the loss in the amount of water at the same time when you measure the loss in mass of the fermentation mixture, subtract this from the measured weight loss of the fermentation mixture. Why are we doing this? Because we want to see how much water has evaporated under the same conditions and in an equal amount of time as fermentation takes place. This way we can measure only the mass of CO2 produced.

2. This is an easier method. Simply connect your fermentation mixture to a conical flask filled with limewater. When CO2 is released it reacts with Ca(OH)2 (lime) to form a milky liquid CaCO3. By measuring the increase in the weight of the limewater and divide this weight by the molar mass of CO2 we can work out how many moles of CO2 has been produced and therefore calculate the volume of CO2 produced. Make sure when you do this experiment that everything is enclosed to prevent CO2 from escaping into the air. Why do we not worry about water evaporation in this case? Because it doesnt affect anything! It does not react or enter the limewater and therefore all we are measuring here is the amount of CO2 produced!

A bit long (quite long tbh) for a multiple choice answer, but I thought I would just make everything clearer for you. :)

Best Regards
Happy Physics Land
Post by: angiezhang9 on July 24, 2016, 02:08:32 pm
Hey Angie!

A is the correct option simply because all we are trying to measure is the volume of CO2 produced. If water evaporates (especially since we are carrying out the experiment under 28 degrees here, so a large chance that some water would have evaporated) then we obtain a measurement that includes a combined mass loss of CO2 and H2O. Validity means "are we measuring what we want to measure" and this is clearly not the case if water has also evaporated. Keep in mind, WE ONLY WANT TO MEASURE THE MASS OF CARBON DIOXIDE.

Explaining this one step further. When you conducted this experiment what you would have done is to measure the mass of fermentation mixture before fermentation has taken place, and the mass of fermentation mixture after fermentation has taken place. The loss in mass after fermentation you would assume to be CO2 (You would probably have carried out the experiment with a limewater flask connected to the fermentation mixture as well). But keep in mind that even at room temperature, water evaporation still occurs, meaning that whilst CO2 is being evaporated, water is too! So when you measure the lost weight, thats a combined weight of evaporated CO2 and WATER. So evidently we are not measuring JUST the mass of CO2 and therefore the experiment becomes invalid.

So, how do we improve validity? We can do one of the two things:

1. Keep a carton of water under the same temperature as your fermentation mixture. Make sure the volume of the water is the same as the volume of your fermentation mixture. Measure the loss in the amount of water at the same time when you measure the loss in mass of the fermentation mixture, subtract this from the measured weight loss of the fermentation mixture. Why are we doing this? Because we want to see how much water has evaporated under the same conditions and in an equal amount of time as fermentation takes place. This way we can measure only the mass of CO2 produced.

2. This is an easier method. Simply connect your fermentation mixture to a conical flask filled with limewater. When CO2 is released it reacts with Ca(OH)2 (lime) to form a milky liquid CaCO3. By measuring the increase in the weight of the limewater and divide this weight by the molar mass of CO2 we can work out how many moles of CO2 has been produced and therefore calculate the volume of CO2 produced. Make sure when you do this experiment that everything is enclosed to prevent CO2 from escaping into the air. Why do we not worry about water evaporation in this case? Because it doesnt affect anything! It does not react or enter the limewater and therefore all we are measuring here is the amount of CO2 produced!

A bit long (quite long tbh) for a multiple choice answer, but I thought I would just make everything clearer for you. :)

Best Regards
Happy Physics Land

Thanks so much for clearing this up. It makes sense now :) RuiAce and Happy Physics Land
Post by: Happy Physics Land on July 24, 2016, 03:02:33 pm

Thanks so much for clearing this up. It makes sense now :) RuiAce and Happy Physics Land

No worries angie, what we do here the best is to clear things up! :)
Post by: anotherworld2b on July 25, 2016, 01:10:39 am
CH3COOH(aq) is acetic acid. I disagree with the answer because the fact it is acetic acid means that it is definitely dissociated.

Unless the question is being pedantic. Their answer isn't "completely" unjustified because some unsplit CH3COOH is in there as well. So I can see where they're coming from. But the question should've made that clearer - I don't like these ambiguous questions.
___________________

Q8 is combining equations. You have to do some maths here.

a) Equation 2 has just one CO in the reactants. But equation 1 has two COs in the products
So we have to multiply equation 2 by two to merge the equations:

2 CO(g) + 2 FeO(s) -> 2 Fe(l) + 2 CO2(g)

Substitute equation 1 in there to get:

C(s) + CO2(g) + 2 FeO(s) -> 2 Fe(l) + 2 CO2(g)

Cancel out the duplicated carbon dioxide:

C(s) + 2 FeO(s) -> 2 Fe(l) + CO2(g)

It's sort of like simultaneous equations: Perform (1) + 2*(2)

I'll let you have a try at the other two first. If you're still stuck, however (especially with 8c - that one is trickier), just post again

For the carbon dioxide in q8 a how do you know to cancel it out? Also does it matter which equation you subsitute?

I am not sure how to do part c  :-[
Post by: RuiAce on July 25, 2016, 09:17:50 am
For the carbon dioxide in q8 a how do you know to cancel it out? Also does it matter which equation you subsitute?

I am not sure how to do part c  :-[
I didn't. It appeared on both the products AND the reactants once I did simultaneous equations, so that's when I knew to cancel them out.

(It's because I'm not sure whether we want Cl to cancel out, or ClO to cancel out.)
Post by: leila_ameli on July 25, 2016, 01:35:03 pm
heyy
I need a some help with this question:

Bond energy is the energy needed to break the chemical bond between two atoms. The bond energies for each of the two allotropes of oxygen are shown in the table.
Oxygen                                       Ozone
Bond Energy (kJ mol-1)                         498                                             364
Account for the difference in bond energy for these two allotropes of oxygen.
Post by: Aliceyyy98 on July 25, 2016, 01:48:44 pm
Hi guys,

Could someone please the concept of buffers? with an example and preferably equations to illustrate? thanks!
Post by: Maz on July 25, 2016, 04:01:01 pm
Hi guys,

Could someone please the concept of buffers? with an example and preferably equations to illustrate? thanks!
Hey human
Hopefully, I can help a bit...
Buffers are solutions with the ability to resist pH change when small quantities of acid/base are added to them. Buffers are generally a solution which contains a weak acid-base conjugate pair (a weak acid and its conjugate base or vice-versa). These acids and bases and co-exist without neutralising each other, as neutralisation generally occurs in an acid-base reaction, yet they can still react to neutralise any strong acid or strong base added to the buffer.
An example of a buffer solution could include a 2mol ethanoic acid solution, and 2 mol of sodium ethanoate; then making the solution u to 1L. The buffer contains a high concentration of both weak acid CH3COOH and its conjugate weak base, CH3COO-.
The equilibrium reaction is shown below:
CH3COOH (aq) + H20 (l)   <---->   CH3COO- +H3O+(aq)
Buffer behaviour can be predicted via Le Chateliers Principle (LCP). Eg, if you add a base to the buffer will neutralise some of the H3O+ present, causing its concentration to fall. This falling concentration causes the equilibrium reaction to shift right, as explained via LCP, (forward reaction favoured) o replace some of the lost hydronium concentration, and preventing too much of a significant fall. This is the property of a buffer being able to resist some changes in pH. If you flood the buffer with acid/base this will not work, and a pH change will result (each buffer has a buffer capacity)
Post by: jakesilove on July 25, 2016, 04:25:06 pm
heyy
I need a some help with this question:

Bond energy is the energy needed to break the chemical bond between two atoms. The bond energies for each of the two allotropes of oxygen are shown in the table.
Oxygen                                       Ozone
Bond Energy (kJ mol-1)                         498                                             364
Account for the difference in bond energy for these two allotropes of oxygen.

Hey!

Essentially, all I would be talking about to answer this question is the reactivity of each alloptrope. As Ozone has a lower bond energy, it will be easier to break/react. Oxygen will be more difficult to break/react. Clearly, Ozone is more reactive than Oxygen. We can explain this by saying that coordinate covalent bonds are easier to break than normal covalent bonds. This explains why Ozone only really forms in higher levels of the atmosphere, where even more reactive Oxygen free radicals are present to react. I think this is as much as an answer would require in the HSC!

Jake
Post by: MysteryMarker on July 25, 2016, 09:03:03 pm
Industrial Chemistry Question:

During the electrolysis of dilute NaCl, why is the water oxidised instead of the Chlorine ions? Because in the concentrated NaCl solution, the chlorine ions do get oxidised, however in the very dilute solution they do not. Just curious as to why this is.

Cheers.
Post by: marynguyen18 on July 25, 2016, 09:11:07 pm
Hey i was wondering if anyone has studied/currently studying shipwrecks that could help me understand the difference between a galvanic cell and an electrolytic cell, because I'm getting confused with the charge of the cathode and anode are.
Post by: RuiAce on July 25, 2016, 09:23:15 pm
Industrial Chemistry Question:

During the electrolysis of dilute NaCl, why is the water oxidised instead of the Chlorine ions? Because in the concentrated NaCl solution, the chlorine ions do get oxidised, however in the very dilute solution they do not. Just curious as to why this is.

Cheers.
Compare the Eo values of both oxidations:

2 Cl- ⇌ Cl2(g) + 2 e-                  Eo=-1.36 V
2 H2O(l) ⇌ O2(g) + 4 H+ + 4 e-   Eo=-1.23 V

Notice how the two values are very similar. This is opposed to the reduction, where the voltage required to cause sodium to reduce is significantly larger to water reducing.

Because the voltage is so similar, the EMF supplied will seek out the more abundant substance as about a similar amount of voltage is consumed anyway for either reaction. In concentrated sodium chloride solution, there is actually a sufficient amount of chloride ions for their oxidation to be favoured. In dilute solution, there really isn't that many, and water will be oxidised anyway.
Hey i was wondering if anyone has studied/currently studying shipwrecks that could help me understand the difference between a galvanic cell and an electrolytic cell, because I'm getting confused with the charge of the cathode and anode are.
Fortunately this overlaps into industrial chemistry.

Galvanic cells and electrolytic cells function oppositely.

In a galvanic cell, a chemical reaction takes place producing a voltage, thereby chemical energy is converted to electrical energy
In an electrolytic cell, electricity is passed through a cell to reverse the effect of the galvanic cell, thereby electrical energy is converted to galvanic energy.
(The EMF value for a galvanic cell is positive, whereas that of an electrolytic cell is negative.)

Effectively, electrolytic cells can be used as battery rechargers.

The mnemonic RED-CAT AN-OX still takes play. Reduction happens at the cathode and oxidation happens at the anode.

However, the polarity reverses in an electrolytic cell.

For a galvanic cell, the anode is negative and the cathode is positive.
Cathode: Cu2+ + 2 e- -> Cu(s)
Anode: Zn(s) -> Zn2+ + 2 e-

For an electrolytic cell, the anode is positive and the cathode is negative.
Cathode: Zn2+ + 2 e- -> Zn(s)
Anode: Cu(s) -> Cu2+ + 2 e-

Deleted last bit. Forgot the why. JAKE!
Post by: marynguyen18 on July 25, 2016, 09:37:25 pm
thank you so much! but with the half equations why is that some of them are either producing H+ or OH- if that makes sense?
Post by: RuiAce on July 25, 2016, 09:39:01 pm
thank you so much! but with the half equations why is that some of them are either producing H+ or OH- if that makes sense?
Not quite

All I can respond to out of that is that one of them is an oxidation whilst the other is a reduction
Post by: onepunchboy on July 25, 2016, 09:56:57 pm

Hey guys im not sure to do q 15 hehe thanks
Post by: MysteryMarker on July 25, 2016, 10:03:00 pm
Compare the Eo values of both oxidations:

2 Cl- ⇌ Cl2(g) + 2 e-                  Eo=-1.36 V
2 H2O(l) ⇌ O2(g) + 4 H+ + 4 e-   Eo=-1.23 V

Notice how the two values are very similar. This is opposed to the reduction, where the voltage required to cause sodium to reduce is significantly larger to water reducing.

Because the voltage is so similar, the EMF supplied will seek out the more abundant substance as about a similar amount of voltage is consumed anyway for either reaction. In concentrated sodium chloride solution, there is actually a sufficient amount of chloride ions for their oxidation to be favoured. In dilute solution, there really isn't that many, and water will be oxidised anyway.Fortunately this overlaps into industrial chemistry.

Cheers man, another question for industrial chemistry:

Within a diaphragm cell, why is it more important to stop the migration of hydroxide ions into the anode compartment then it is to stop the migration of chloride ions into the cathode compartment?

Post by: anotherworld2b on July 25, 2016, 10:51:34 pm
I didn't. It appeared on both the products AND the reactants once I did simultaneous equations, so that's when I knew to cancel them out.

(It's because I'm not sure whether we want Cl to cancel out, or ClO to cancel out.)
Post by: Happy Physics Land on July 25, 2016, 11:30:10 pm
Cheers man, another question for industrial chemistry:

Within a diaphragm cell, why is it more important to stop the migration of hydroxide ions into the anode compartment then it is to stop the migration of chloride ions into the cathode compartment?

Hey Mystery Marker!

Both are equally just as important as each other. By preventing OH- from entering the anode we are ensuring that all of the Na+ from the brine are reacting with OH-. This is important because after all, it takes energy to reduce water to produce hydroxide and therefore any waste of OH- (i.e. those that dont end up participating in the formation of NaOH) is undesirable as it is wasting the energy input. We also dont want Cl- to migrate into cathode compartment because we dont like the formation of NaCl in our final product, all we want is NaOH. A major concern that may answer your question is that we can get Na+ and Cl- very easily, just from natural sea water (brine) but to get OH- we need quite an intensive amount of energy (diaphragm cell needs about 2750kwh energy). So it's less of a concern to form NaCl than to not form NaOH.

Best regards
Happy Physics Land
Post by: RuiAce on July 25, 2016, 11:40:03 pm

Lol i made a dumb. Of course the final products must stay there.

In (1) and (2) there is exactly 1 mol of Cl(g) there. Thus, we can immediately perform (1) + (2)

CFCl3(g) + Cl(g) + O3(g) -> CFCl2(g) + Cl(g) + ClO(g) + O2(g)

Cancel out the Cl on both sides:
CFCl3(g) + O3(g) -> CFCl2(g) + ClO(g) + O2(g)

This new equation, and equation 3, both have exactly 1 mol of ClO present. Thus we can perform (3) + the above

CFCl3(g) + O3(g) + ClO(g) + O(g) -> CFCl2(g) + ClO(g) + O2(g) + Cl(g) + O2(g)

Cancel out the ClO:

CFCl3(g) + O3(g) + O(g) -> CFCl2(g) + 2 O2(g) + Cl(g)

Hey guys im not sure to do q 15 hehe thanks
This is just your ordinary rigorous calculation.

Write out relevant equations:
A) K2CO3 + 2 HCl -> CO2 + H2O + 2 KCl
B) KHCO3 + HCl -> CO2 + H2O + KCl
C) Na2CO3 + 2 HCl -> CO2 + H2O + 2 KCl
D) NaHCO3 + HCl -> CO2 + H2O + KCl

Clearly, 1 mol of reactant _ yields 1 mol of CO2 in all four equations.

Now, determine the moles of each substance present using n=m/MM:
A) nCO2 = nK2CO3 = 7.23 * 10-3 mol
B) nCO2 = nKHCO3 = 1.05 * 10-2 mol
C) nCO2 = nNa2CO3 = 9.43 * 10-3 mol
D) nCO2 = nNaHCO3 = 1.19 * 10-2 mol

So since n = V/VM, clearly the answer must be D.

And now for the smart way

However obvious it was that the moles of the reactant and the moles of CO2 were the same varies for each person. But without even writing out the full equation you should've might've been able to see it. This is because CO32- is being decomposed down to CO2 by HCl, which has no carbons in it!

Then, because m=1, we have that the moles is INVERSELY proportional to the molar mass. Thus the one with the smallest molar mass (NaHCO3) will yield the largest moles.
Post by: Aliceyyy98 on July 26, 2016, 12:57:50 am
Hey human
Hopefully, I can help a bit...
Buffers are solutions with the ability to resist pH change when small quantities of acid/base are added to them. Buffers are generally a solution which contains a weak acid-base conjugate pair (a weak acid and its conjugate base or vice-versa). These acids and bases and co-exist without neutralising each other, as neutralisation generally occurs in an acid-base reaction, yet they can still react to neutralise any strong acid or strong base added to the buffer.
An example of a buffer solution could include a 2mol ethanoic acid solution, and 2 mol of sodium ethanoate; then making the solution u to 1L. The buffer contains a high concentration of both weak acid CH3COOH and its conjugate weak base, CH3COO-.
The equilibrium reaction is shown below:
CH3COOH (aq) + H20 (l)   <---->   CH3COO- +H3O+(aq)
Buffer behaviour can be predicted via Le Chateliers Principle (LCP). Eg, if you add a base to the buffer will neutralise some of the H3O+ present, causing its concentration to fall. This falling concentration causes the equilibrium reaction to shift right, as explained via LCP, (forward reaction favoured) o replace some of the lost hydronium concentration, and preventing too much of a significant fall. This is the property of a buffer being able to resist some changes in pH. If you flood the buffer with acid/base this will not work, and a pH change will result (each buffer has a buffer capacity)

Thanx a bunch!
Post by: Alexander23 on July 26, 2016, 08:56:38 am
I've been told that the addition of an inert gas has no effect on a reaction at equilibrium. I was wondering why it has no effect on equilibrium direction, even if you are adding in this non reaction gas in a closed system which seems to decrease volume and hence increase pressure.

Also, I was hoping for a clearer explanation as to why temperature is the only factor which effects the K expression of an equilibrium reaction.

Thanks.   :)
Post by: RuiAce on July 26, 2016, 09:14:02 am
I've been told that the addition of an inert gas has no effect on a reaction at equilibrium. I was wondering why it has no effect on equilibrium direction, even if you are adding in this non reaction gas in a closed system which seems to decrease volume and hence increase pressure.

Also, I was hoping for a clearer explanation as to why temperature is the only factor which effects the K expression of an equilibrium reaction.

Thanks.   :)
Adding an inert gas is essentially doing the opposite of the effect of a catalyst - it slows down the rate equilibrium is attained however does nothing to the equilibrium itself.

This is because by adding a catalyst, you are not hindering just ONE of the forward/reverse reactions. You are actually hindering BOTH of the forward/reverse reactions.

If you change the pressure by pumping in more equilibrium mixture you are causing an imbalance in the concentrations of products and reactants. This is the same as varying the volume of the vessel, because the amount of space you have will promote either the forward or reverse reaction to allow the moles of gas to be reasonable. (Recall: Avogadro's law states that at equal temperatures and pressures, systems will rather having an equal number of molecules of gases.)

Now, temperature is an extrinsic factor to the position of the equilibrium. All other factors such as pressure and varying concentrations are intrinsic. When you alter the pressure by pumping equilibrium mixture, change the volume of the vessel or increase/decrease concentrations of just one substance, as a consequence of Le Chatelier's Principle the system will try to bring back the equilibrium it started with.

Note that K is just a rough indicator of where the equilibrium actually is to begin with! The concentrations of your reactants/products readjust so that the reaction quotient Q returns to K.

When you alter the temperature, however, you didn't actually alter the concentrations of the reactants and products. Instead, you manipulated the fact that the reaction is exothermic or endothermic. From LCP we do know that the equilibrium does indeed adjust if you supply/take out heat, however in doing so you're forcing the ratio of products and reactants to change themselves WITHOUT actually having added/taken away any in advance! Hence, K will actually change for different temperatures
Post by: liiz on July 26, 2016, 02:55:45 pm
Hey there, just wondering if someone would be please able to help me with this question. I find calculation questions super hard so any explanation would be awesome, thankyou!!  ;D ;D"Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation P2O5(s)  + 3H2O(l) → 2H3PO4 (aq) Phosphoric acid reacts with sodium hydroxide according to the following equation. H3PO4 (aq) + 3NaOH(aq) → Na3PO4 (aq) + 3H2O(l). A student reacted 1.42 g of phosphorus pentoxide with excess water. What volume of 0.30 mol L-1 sodium hydroxide would be required to neutralise all the phosphoric acid produced?"
Post by: sweetcheeks on July 26, 2016, 03:52:49 pm
Hey there, just wondering if someone would be please able to help me with this question. I find calculation questions super hard so any explanation would be awesome, thankyou!!  ;D ;D"Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation P2O5(s)  + 3H2O(l) → 2H3PO4 (aq) Phosphoric acid reacts with sodium hydroxide according to the following equation. H3PO4 (aq) + 3NaOH(aq) → Na3PO4 (aq) + 3H2O(l). A student reacted 1.42 g of phosphorus pentoxide with excess water. What volume of 0.30 mol L-1 sodium hydroxide would be required to neutralise all the phosphoric acid produced?"

Initially we start with 1.42g of phosphorus pentoxide. First we must find the amount of moles of phosphorus pentoxide. Each molecule of phosphorus pentoxide produces two mole of phosphoric acid. The phosphoric acid produced will then react with Sodium hydroxide in a 1:3 ratio, that is you need 3 molecules of NaOH per molecule of phosphoric acid.

Here are the steps
1. Find moles of phosphorus pentoxide, using the formula n=m/M where n=number moles m=grams M=molecular weight
2. Ratio this to moles of Phosphoric acid produced (2:1)
3. Substitute the mole of phosphoric acid into the second equation and ratio to NaOH to find the amount (in mole) of NaOH required. Using the formula n/C=v (rearranged C=n/v), where n=moles of NaOH, C= concentration of the NaOH solution and V is the volume in litres.

It helps to draw a diagram when doing questions like this. Also, write out both equations and work in columns under the substances of interest (i.e. when finding moles of phosphorus pentoxide, work underneath it on the formula.)

Is there anything in particular that you struggle with when it comes to these type of questions?
Post by: RuiAce on July 26, 2016, 05:01:10 pm
Hey there, just wondering if someone would be please able to help me with this question. I find calculation questions super hard so any explanation would be awesome, thankyou!!  ;D ;D"Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation P2O5(s)  + 3H2O(l) → 2H3PO4 (aq) Phosphoric acid reacts with sodium hydroxide according to the following equation. H3PO4 (aq) + 3NaOH(aq) → Na3PO4 (aq) + 3H2O(l). A student reacted 1.42 g of phosphorus pentoxide with excess water. What volume of 0.30 mol L-1 sodium hydroxide would be required to neutralise all the phosphoric acid produced?"
Because the equations are given to you, there's one less thing to do. Because normally when you work between reactants/products and other reactants/products you always need a chemical equation.

You also always work in moles. According to n=m/MM

nP2O5 = 1.42/(2*30.97 + 5*16.00) = 0.010004227... mol

Look at the equation and compare the mole ratio. 1 mol of diphosphorus pentoxide produces 2 moles of phosphoric acid

nH3PO4 = 2 nP2O5 = 0.02000845427... mol

So we now know the moles of phosphoric acid we must neutralise.

Look at the second equation for the neutralisation. Clearly three moles of NaOH is required to neutralise one mole of H3PO4

nNaOH = 3 nH3PO4 = 0.06002536282... mol

We are given the concentration of NaOH. Because we know that C=n/V

V = n/C

VNaOH =  0.06002536282... / 0.30
=  0.06002536282... L

The lowest amount of significant figures used in this question is three (from the mass of diphosphorus pentoxide. So rounding to 3 s.f.

V ≈ 6.0 * 10-2 L

Key strategy: Break the question apart and do not try to do multiple things at once. Chemistry calculations are always one-way.

It helps to draw a diagram when doing questions like this. Also, write out both equations and work in columns under the substances of interest (i.e. when finding moles of phosphorus pentoxide, work underneath it on the formula.)
Post by: sweetcheeks on July 26, 2016, 05:22:31 pm
@RuiAce a diagram that shows the steps of the process. It helps visualize what is going on. for example, you draw the phosphorus pentoxide being weighed out, being added to water, the solution being reacted with NaOH. Add the information underneath to help keep track of what is going on and what where calculations need to be made. It works especially great for things like titrations involving samples and dilutions. My teacher swears by this and always did it when he worked in the field.
Post by: RuiAce on July 26, 2016, 05:30:43 pm
@RuiAce a diagram that shows the steps of the process. It helps visualize what is going on. for example, you draw the phosphorus pentoxide being weighed out, being added to water, the solution being reacted with NaOH. Add the information underneath to help keep track of what is going on and what where calculations need to be made. It works especially great for things like titrations involving samples and dilutions. My teacher swears by this and always did it when he worked in the field.
A flow chart?
Post by: MysteryMarker on July 26, 2016, 05:35:04 pm
Industrial Chemistry Question:

During the production of soap in a school laboratory, why is saturated salt solution added to aid in the precipitation of soap from the mixture?

Thanks guys.
Post by: RuiAce on July 26, 2016, 05:46:48 pm
Industrial Chemistry Question:

During the production of soap in a school laboratory, why is saturated salt solution added to aid in the precipitation of soap from the mixture?

Thanks guys.
Haven't done this in ages so I don't remember much at all but I'm pretty sure you do this both in the laboratory and in industrial preparation.

The idea is to manipulate a precipitation procedure known as 'salting out'. The solubility of the soap is forced down because the other ions (Na+ and Cl-), which are more soluble, dissolve in their place. It makes it easier for us to collect the soap by using this method.
Post by: leila_ameli on July 26, 2016, 08:20:30 pm
Hey!

Essentially, all I would be talking about to answer this question is the reactivity of each alloptrope. As Ozone has a lower bond energy, it will be easier to break/react. Oxygen will be more difficult to break/react. Clearly, Ozone is more reactive than Oxygen. We can explain this by saying that coordinate covalent bonds are easier to break than normal covalent bonds. This explains why Ozone only really forms in higher levels of the atmosphere, where even more reactive Oxygen free radicals are present to react. I think this is as much as an answer would require in the HSC!

Jake

thankyouu!!
Post by: liiz on July 26, 2016, 08:47:03 pm
Initially we start with 1.42g of phosphorus pentoxide. First we must find the amount of moles of phosphorus pentoxide. Each molecule of phosphorus pentoxide produces two mole of phosphoric acid. The phosphoric acid produced will then react with Sodium hydroxide in a 1:3 ratio, that is you need 3 molecules of NaOH per molecule of phosphoric acid.

Here are the steps
1. Find moles of phosphorus pentoxide, using the formula n=m/M where n=number moles m=grams M=molecular weight
2. Ratio this to moles of Phosphoric acid produced (2:1)
3. Substitute the mole of phosphoric acid into the second equation and ratio to NaOH to find the amount (in mole) of NaOH required. Using the formula n/C=v (rearranged C=n/v), where n=moles of NaOH, C= concentration of the NaOH solution and V is the volume in litres.

It helps to draw a diagram when doing questions like this. Also, write out both equations and work in columns under the substances of interest (i.e. when finding moles of phosphorus pentoxide, work underneath it on the formula.)

Is there anything in particular that you struggle with when it comes to these type of questions?
Ah, awesome I get you. Those steps really helped to guide me through working it out :) And the diagram suggestion is really awesome too!! I've never thought to do diagrams like that before, and as a more visual learner it's something I can definitely incorporate now!! Thanks so much for your help :)) I think I just struggle to simply comprehend what the question is actually asking me and relate it to the formulas I know and what info they've given. I guess just doing heaps of practice, I'll (hopefully) get there haha
Post by: liiz on July 26, 2016, 08:49:14 pm
Because the equations are given to you, there's one less thing to do. Because normally when you work between reactants/products and other reactants/products you always need a chemical equation.

You also always work in moles. According to n=m/MM

nP2O5 = 1.42/(2*30.97 + 5*16.00) = 0.010004227... mol

Look at the equation and compare the mole ratio. 1 mol of diphosphorus pentoxide produces 2 moles of phosphoric acid

nH3PO4 = 2 nP2O5 = 0.02000845427... mol

So we now know the moles of phosphoric acid we must neutralise.

Look at the second equation for the neutralisation. Clearly three moles of NaOH is required to neutralise one mole of H3PO4

nNaOH = 3 nH3PO4 = 0.06002536282... mol

We are given the concentration of NaOH. Because we know that C=n/V

V = n/C

VNaOH =  0.06002536282... / 0.30
=  0.06002536282... L

The lowest amount of significant figures used in this question is three (from the mass of diphosphorus pentoxide. So rounding to 3 s.f.

V ≈ 6.0 * 10-2 L

Key strategy: Break the question apart and do not try to do multiple things at once. Chemistry calculations are always one-way.
Thanks Ruiace for your input too!! I was able to do it from Sweetcheeks and then check my working with yours which was really helpful!! Awesome tip as well to not do multiple things at once - will stick by it a bit more now!! Thanks again :))
Post by: MysteryMarker on July 27, 2016, 04:22:56 pm
Assess the significance of the industrial development of the Haber process to society at the beginning on the 1900s. Include a relevant chemical equation in your answer. 6 marks.

Hey guys, just wondering what would the recommended structure be for this/ some main points i would need to discuss to get the 6 marks.

Thanks
Post by: jakesilove on July 27, 2016, 11:13:52 pm
Assess the significance of the industrial development of the Haber process to society at the beginning on the 1900s. Include a relevant chemical equation in your answer. 6 marks.

Hey guys, just wondering what would the recommended structure be for this/ some main points i would need to discuss to get the 6 marks.

Thanks

Hey!

Firstly, holy crap. What a terrifying question. Still, definitely something that you can work through, and get 6 marks for. Let's look at the subheadings that I would recommend you used.

History

You need to explain WHY the Haber process was developed, by putting it in context. Talk about the importance of Ammonia is soil, which was naturally sourced from Chile. The British blockaded Germany following the outbreak of WWI (1914-8), and they could not longer receive Ammonia. Their crops perished, and hundreds of thousands of Germans died on the homefront. Haber was a chemist who developed a process to produce Ammonia.

Chemistry

Here, put the Chemical equation (Which I can't be bothered typing out), and the reaction conditions that yield the highest result (plus the catalyst). YOU WILL LOSE A MARK FOR LEAVING OUT STATES!

Significance

Here, you need to make an assessment. The Haber process allowed for Ammonia-rich soil to grow crops, saving thousands of starving Germans. It was also used to produce bombs. Both of these, potentially, INCREASED the length of WWI, in which over 17 million people were killed. Thus, the development of the process at this specific period of time could be conceived as vastly negative.

You're assessment doesn't need to be exactly what I said above. However, notice the specific details I've thrown in (dates of WWI, people killed, Chile etc.). I've done that to impress the marker, because this is a beast of a question, and figuring out where to get 6 marks is tough. Make sure to remember to impress the marker!

That's what I think you should include; feel free to throw in some more suggestions! You need to be ready for shitty questions like this, so know each dot point in depth. Hopefully, however, you don't come across a beast like this in your Trials :)

Jake
Post by: Maz on July 28, 2016, 12:58:04 pm
Hey guys, :)
My class recently did an experiment on the reactivity of alcohols. We reacted primary, secondary and tertiary alcohols with dichromate, permanganate, and sodium...I have formed the reaction redox equations. We have a test on it on Friday and I was wandering if you had any idea about some of the questions that may come up/any questions that I should really study?
Also one more thing please...For household cleaning would it be better to have a primary alchohol, or secondary? In a primary one there is more solubility, but less reactivity. In a secondary one there would be slightly less solubility, and a bit more reactivity. I'm thinking secondary...but I'm not really sure?
Any help would be really appreciated :)

Thankyou,
Maryam
Post by: jakesilove on July 28, 2016, 02:39:58 pm
Hey guys, :)
My class recently did an experiment on the reactivity of alcohols. We reacted primary, secondary and tertiary alcohols with dichromate, permanganate, and sodium...I have formed the reaction redox equations. We have a test on it on Friday and I was wandering if you had any idea about some of the questions that may come up/any questions that I should really study?
Also one more thing please...For household cleaning would it be better to have a primary alchohol, or secondary? In a primary one there is more solubility, but less reactivity. In a secondary one there would be slightly less solubility, and a bit more reactivity. I'm thinking secondary...but I'm not really sure?
Any help would be really appreciated :)

Thankyou,
Maryam

Hey MQ,

Unfortunately, none of that is part of the HSC curriculum. I've done some Chemistry at Uni, but never heard of that terminology. I'm afraid I'm not going to be able to help you out, I'm really sorry! I hope you can do some research and find the answers you seek :)

Jake
Post by: HighTide on July 28, 2016, 04:07:56 pm
Hey guys, :)
My class recently did an experiment on the reactivity of alcohols. We reacted primary, secondary and tertiary alcohols with dichromate, permanganate, and sodium...I have formed the reaction redox equations. We have a test on it on Friday and I was wandering if you had any idea about some of the questions that may come up/any questions that I should really study?
Also one more thing please...For household cleaning would it be better to have a primary alchohol, or secondary? In a primary one there is more solubility, but less reactivity. In a secondary one there would be slightly less solubility, and a bit more reactivity. I'm thinking secondary...but I'm not really sure?
Any help would be really appreciated :)

Thankyou,
Maryam
Interesting question. I was given to think that primary alcohols were also more reactive due to their exposure? And secondary, and tertiary alcohols were progressively more stable. From what we've learned at uni, detergents and all that rely on the ability to form micelles. So one side would be hydrophobic and this would attach to the dirt, the other part would be hydrophilic and form hydrogen bonds, so technically more soluble would be preferred? This is just an idea and this is for detergents and stuff like that. I'm not 100% sure about household cleaning, but perhaps it's a similar idea.
Hopefully |zxn| or jyce may be able to shed light on this.

Post by: anotherworld2b on July 29, 2016, 01:30:13 am
I was wondering if i could get help in understandong dipole dipole forces, dispersion forces and hydrogen bonding. We went through these forces extremely briefly and quickly at school so i am bit confused about what there are and their purpose
Post by: wesadora on July 29, 2016, 01:52:14 pm
Two questions are attached. I don't even understand what w/v is, and I'm just very confused on the whole dissolved oxygen calculations in general. I suppose I understand the theory behind it (i.e. more dissolved oxygen in a waterway = good, as it is plentiful for aquatic life to live healthily etc. etc.) but don't understand these calcs. Thanks :)
Post by: jakesilove on July 29, 2016, 02:54:49 pm
I was wondering if i could get help in understandong dipole dipole forces, dispersion forces and hydrogen bonding. We went through these forces extremely briefly and quickly at school so i am bit confused about what there are and their purpose

(http://images.slideplayer.com/1/261608/slides/slide_22.jpg)

Hey! I'll just give a brief summary of these forces.

Hydrogen Bonding

Hydrogen is attracted to very electronegative elements. So, if there is hydrogen present in the chemical structure of a compound, and there are also atoms of Oxygen, Flourine or Nitrogen (FON!), there will be strong intermolecular forces between them. See the image above, it makes it pretty clear.

Dipole-Dipole

This is similar to Hydrogen bonding, but less strong. Some elements have positive charges (eg. metals) and some have negative charges (eg. halons). These will naturally be attracted, and that's what dipole-dipole forces are. If a compound has a 'negative' side, it will be attracted to the 'positive' side of another compound. This is only the case if the compound is polar (ie. has a positive and negative side).

Dispersion forces

These forces occur is all compounds. As electrons move around the nucleus, sometimes they will be in certain areas with a higher density than elsewhere. As such, there will be a greater negative charge in that region. This essentially creates a 'more positive' area, and a 'more negative' area, which line up and are attracted to each other.

Hope this helps! Let me know if I can expand; they aren't too difficult as concepts, hopefully the image helps.

Jake
Post by: jakesilove on July 29, 2016, 03:02:56 pm
Two questions are attached. I don't even understand what w/v is, and I'm just very confused on the whole dissolved oxygen calculations in general. I suppose I understand the theory behind it (i.e. more dissolved oxygen in a waterway = good, as it is plentiful for aquatic life to live healthily etc. etc.) but don't understand these calcs. Thanks :)

Two questions are attached. I don't even understand what w/v is, and I'm just very confused on the whole dissolved oxygen calculations in general. I suppose I understand the theory behind it (i.e. more dissolved oxygen in a waterway = good, as it is plentiful for aquatic life to live healthily etc. etc.) but don't understand these calcs. Thanks :)

Cool question!

So, we know that at 25 degrees, 8mg/L of Oxygen will be dissolved. This can be read straight from the graph. However, there are 10.0L of water, therefore we know that 10*8 = 80mg of Oxygen is dissolved in the liquid.

Now, we need a volume of Oxygen, not a weight. First, let's convert from mg to g.

$80mg=0.08g$

We can now convert this into moles.

$n=\frac{0.008}{2*16}=0.00025 moles$

We know that moles of gas occupy a fixed volume at a certain temperature: specifically, 24.79L per mole.

$0.00025*24.79=0.0062L$

So the answer is A, 62mL. I think I may have screwed up the number of zeros somewhere (there should be one less), but that is the general structure of the answer. Hope that helps!

Jake
Post by: wesadora on July 29, 2016, 03:36:19 pm
Thanks Jake! I getcha! :) not bad actually haha
here's the other one i was confused about, sorry thought I attached two.

and a quick question while my brain doesn't forget to ask: Does ozone have a higher MP/BP than Oxygen because of the stronger IMFs (i.e. dipole dipole AND two sites for hydrogen bonding? or is it just the dipole dipole?

CHEEEEEEEEEEEEEEEEERS m8 :)
Post by: conic curve on July 29, 2016, 03:47:43 pm
I need help with a practical question (since I suck at them)

A student performed an experiment to determine the molar heat of solution of ammonium chloride. However, their experimental value was observed to be lower than the standard theoretical value. Which of the following could best explain the observation?

a)Wind gusts reducing the heat transfer from the flame to the water
b)Inclusion of the mass of salt in the caliorimetry experiment
c)Incomplete combustion of the salt, leading to less heat transfer
d) Over stirring of the solution leading to heat generation

Also how do you prepare for prac exams at school?
Post by: RuiAce on July 29, 2016, 03:59:05 pm
Thanks Jake! I getcha! :) not bad actually haha
here's the other one i was confused about, sorry thought I attached two.

and a quick question while my brain doesn't forget to ask: Does ozone have a higher MP/BP than Oxygen because of the stronger IMFs (i.e. dipole dipole AND two sites for hydrogen bonding? or is it just the dipole dipole?

CHEEEEEEEEEEEEEEEEERS m8 :)
There's no hydrogen in ozone for there to be hydrogen bonding. It's the bent structure that promotes dipole-dipole interactions.

I did that question recently for my student. Analyse it systematically as always through determining moles.

nNa2SO3 = CV = 8.00*10-3 L * 0.0100 mol L-1 = 8 * 10-5 mol.

Observe the mole ratio: 1/4 mol of O2 reacts with 1 mol of Na2SO3
nO2 = 2 * 10-5 mol

Utilise your units. The question asks for weight/volume. Convert the moles of O2 into a mass:
mO2 = nMM = 2*10-5 mol * 32.00 g mol-1 = 6.4*10-4 g

(w/v) is a measure of how many grams there are of a substance, in how many millilitres of solution. As a percentage, %(w/v) therefore measures how many grams are present in how many one hundred millilitres of a solution.

%(w/v) = 6.4*10-4 g / 50.00mL * 100% = 1.28*10-3%

Further examples can be found here http://www.ausetute.com.au/wtvol.html
I need help with a practical question (since I suck at them)

A student performed an experiment to determine the molar heat of solution of ammonium chloride. However, their experimental value was observed to be lower than the standard theoretical value. Which of the following could best explain the observation?

a)Wind gusts reducing the heat transfer from the flame to the water
b)Inclusion of the mass of salt in the caliorimetry experiment
c)Incomplete combustion of the salt, leading to less heat transfer
d) Over stirring of the solution leading to heat generation

Also how do you prepare for prac exams at school?
Which one would be most logical?

If you ask me, complete combustion will be quite easy for a molecule with a relatively low molar mass so C is out. D cannot even contribute because the temperature change in q=mc∆T and moles reacted will cancel out in finding a value for ∆H (do you see why?). And if anything, the specific heat capacity will be what changes the value calculated, not the mass of the salt in the experiment.

How to prepare for practical experiments at school? If you're lucky your teacher will give up time to allow you to do the experiment again at lunch. Otherwise, you should've already done the experiment once in class and be prepared to utilise the same techniques. Go over the methods, go over any possible discussion questions, remember what you require and get into it.
Post by: conic curve on July 29, 2016, 04:20:34 pm
There's no hydrogen in ozone for there to be hydrogen bonding. It's the bent structure that promotes dipole-dipole interactions.

I did that question recently for my student. Analyse it systematically as always through determining moles.

nNa2SO3 = CV = 8.00*10-3 L * 0.0100 mol L-1 = 8 * 10-5 mol.

Observe the mole ratio: 1/4 mol of O2 reacts with 1 mol of Na2SO3
nO2 = 2 * 10-5 mol

Utilise your units. The question asks for weight/volume. Convert the moles of O2 into a mass:
mO2 = nMM = 2*10-5 mol * 32.00 g mol-1 = 6.4*10-4 g

(w/v) is a measure of how many grams there are of a substance, in how many millilitres of solution. As a percentage, %(w/v) therefore measures how many grams are present in how many one hundred millilitres of a solution.

%(w/v) = 6.4*10-4 g / 50.00mL * 100% = 1.28*10-3%

Further examples can be found here http://www.ausetute.com.au/wtvol.htmlWhich one would be most logical?

If you ask me, complete combustion will be quite easy for a molecule with a relatively low molar mass so C is out. D cannot even contribute because the temperature change in q=mc∆T and moles reacted will cancel out in finding a value for ∆H (do you see why?). And if anything, the specific heat capacity will be what changes the value calculated, not the mass of the salt in the experiment.

How to prepare for practical experiments at school? If you're lucky your teacher will give up time to allow you to do the experiment again at lunch. Otherwise, you should've already done the experiment once in class and be prepared to utilise the same techniques. Go over the methods, go over any possible discussion questions, remember what you require and get into it.

No not quite. It would be great if you could explain further  :D

I don't think my teacher would allow that. I wish she could though  :'(
Post by: RuiAce on July 29, 2016, 04:39:15 pm
No not quite. It would be great if you could explain further  :D

I don't think my teacher would allow that. I wish she could though  :'(
Regarding that part.

∆H = q/n = q*MM/msalt = mH2OC ∆T * MM / msalt

If we over stir, the change in temperature deltaT could be increased, however the mass used on the bottom also goes up because we combust more of the substance. The overall effects cancel out and do nothing to the value of DeltaH.
Post by: conic curve on July 29, 2016, 05:27:25 pm
Regarding that part.

∆H = q/n = q*MM/msalt = mH2OC ∆T * MM / msalt

If we over stir, the change in temperature deltaT could be increased, however the mass used on the bottom also goes up because we combust more of the substance. The overall effects cancel out and do nothing to the value of DeltaH.

I'm confused with this:

Nacl(aq) -->na+(aq)+cl-(aq)

K2so4(aq)-->2k+(aq)+so4 2-(aq)

For the second one does it have to do with the charges or what?

For the first one I don't get how they are in aqueous state and that when they dissolve, they are still in aqueous state

H2so4(aq)-->2h+(aq)+so4 2-(aq)

Des this have to do with the charges or what?
Post by: RuiAce on July 29, 2016, 05:29:30 pm
I'm confused with this:

Nacl(aq) -->na+(aq)+cl-(aq)

K2so4(aq)-->2k+(aq)+so4 2-(aq)

For the second one does it have to do with the charges or what?

For the first one I don't get how they are in aqueous state and that when they dissolve, they are still in aqueous state

H2so4(aq)-->2h+(aq)+so4 2-(aq)

Again does this have to do with the charges or what?
Your question is not making sense here. If they're dissolved their already in aqueous state. Aqueous means dissolved in solution.

(Splitting NaCl(aq) into Na+ and Cl- does nothing. They were always dissolved)
Post by: conic curve on July 29, 2016, 05:38:05 pm
Your question is not making sense here. If they're dissolved their already in aqueous state. Aqueous means dissolved in solution.

(Splitting NaCl(aq) into Na+ and Cl- does nothing. They were always dissolved)

I meant why is it represented that way, with the plus and minus charges?
Post by: RuiAce on July 29, 2016, 05:49:45 pm
I meant why is it represented that way, with the plus and minus charges?
NaCl(s) is just solid table salt.

NaCl(aq) is a shorthand for Na+(aq) + Cl-(aq).

If a substance featuring an ionic bond is not dissolved, the fact that the sodium ions and chloride ions are attracted to each other will allow positive and negative (opposite) charges to attract. This forms a sturdy lattice of alternating NaCl molecules.

If a substance featuring an ionic bond is dissolved, a ton of polar water molecules will be attracted to the charged ions. The interaction between the sodium and chloride ions breaks apart due to the interference of the water. There is no longer a true NaCl left being held together; instead a ton of Na+ and Cl- ions are present in the water.
Post by: conic curve on July 29, 2016, 07:33:41 pm
I'm stuck with the following chemical equations

Zn(s)-->Zn2+(aq)+Zn-. Why is there a "2" there?
Cu2+ + 2e- --->Cu(s) Is this based on charges in the periodic table?

How would you find the net ionic equation of this as an example?

Ca(s)+ZnSO4(aq)--->CaSO4(aq)+Zn(s)
Post by: RuiAce on July 29, 2016, 07:49:53 pm
I'm stuck with the following chemical equations

Zn(s)-->Zn2+(aq)+Zn-. Why is there a "2" there?
Cu2+ + 2e- --->Cu(s) Is this based on charges in the periodic table?

How would you find the net ionic equation of this as an example?

Ca(s)+ZnSO4(aq)--->CaSO4(aq)+Zn(s)
You don't use the periodic table to determine charges on transitional elements.

They will always specify if the iron ion has a 2+ or 3+ charge. Copper, lead and tin generally have a 2+ charge when unstated. If they say copper has a 1+ charge, or lead/tin has a 4+ charge, then go with that.

The zinc ion has a 2+ charge
The silver ion has a 1+ charge

These are common ones that you need to just know. Just like how you need to know sulfates have a 2- charge.
__________________________

Analyse the species separately watching out for where an electron transfer has occurred:
Ca: It went from solid metal to becoming a calcium ion in calcium sulfate
Zn: It went from being an ion in zinc sulfate to just being a solid metal
SO4: Nothing happened to sulfate

So you have Ca(s) + Zn2+ -> Ca2+ + Zn(s)
Post by: onepunchboy on July 30, 2016, 11:04:40 am

Hi guz , i dont really get this graph question
Post by: RuiAce on July 30, 2016, 12:22:59 pm

Hi guz , i dont really get this graph question
Note how we're talking about rates here.

We know that at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. I'll assumed you deciphered this part out yourself since you picked A.

We are saying that the equilibrium shifts to the right. This means that the forward reaction is being favoured. Thus, the rate A+B->C is happening is going to be much greater than the rate C->A+B is happening. This is why the answer is B.

Of course, the two graphs collide once the equilibrium has been established.
Post by: conic curve on July 31, 2016, 09:26:54 am
In prelim chem, is it possible that they'll assess you on groundwater, even though it's a minor part of the syllabus? (I've never seen a question on groundwater before)
Post by: jakesilove on July 31, 2016, 09:50:53 am
In prelim chem, is it possible that they'll assess you on groundwater, even though it's a minor part of the syllabus? (I've never seen a question on groundwater before)

I don't remember anything about the prelim syllabus, but if groundwater is in there, then yes they can assess you on it. If you think that it is unlikely to be tested, just learn it in vague detail, in the event that a question does come up.
Post by: senara on July 31, 2016, 01:08:39 pm
this is a really generic question but in the case we have to describe an experiment, what exactly do we need to include? because different papers have different marking guidelines :/
Post by: RuiAce on July 31, 2016, 01:15:26 pm
this is a really generic question but in the case we have to describe an experiment, what exactly do we need to include? because different papers have different marking guidelines :/
The question gives you the aim, so the method that you present them has to address the aim. (Diagram is optional - situationally preferrable)

What do you mean different papers have different marking guidelines? The method is different for every individual experiment. Unless you can post two questions the exact same and their marking guidelines for comparison.
Post by: senara on July 31, 2016, 01:43:36 pm
The question gives you the aim, so the method that you present them has to address the aim. (Diagram is optional - situationally preferrable)

What do you mean different papers have different marking guidelines? The method is different for every individual experiment. Unless you can post two questions the exact same and their marking guidelines for comparison.

No I mean, some requires safety precautions others variables, discussion etc. and for example a 5 mark question do we need to include everything, for e.g. aim, hypothesis, variables, method, safety...... and so and how would that differ from a 7 mark question (if that makes sense...)
Post by: RuiAce on July 31, 2016, 01:46:53 pm
No I mean, some requires safety precautions others variables, discussion etc. and for example a 5 mark question do we need to include everything, for e.g. aim, hypothesis, variables, method, safety...... and so and how would that differ from a 7 mark question (if that makes sense...)
Look at the question. If it just says to outline then all they want is a method.

Discussion is almost always asked as a separate question, e.g. assess the validity of your experiment, or e.g. comment on why the calculated Eo value is less than the theoretical.

I do not recall having to include the safety in an experiment unless it actually says to outline the safety. You'll have to provide examples of this. (In fact, I've never heard of a question about an experiment being a 7 marker)

Generally, there's not enough marks for the examiners to give away for you to write a full scientific report on an experiment. Most of the time they just want to see the method or a discussion question as a follow-up.
Post by: senara on July 31, 2016, 01:59:22 pm
Look at the question. If it just says to outline then all they want is a method.

Discussion is almost always asked as a separate question, e.g. assess the validity of your experiment, or e.g. comment on why the calculated Eo value is less than the theoretical.

I do not recall having to include the safety in an experiment unless it actually says to outline the safety. You'll have to provide examples of this. (In fact, I've never heard of a question about an experiment being a 7 marker)

Generally, there's not enough marks for the examiners to give away for you to write a full scientific report on an experiment. Most of the time they just want to see the method or a discussion question as a follow-up.

Ahh okay.. Thank you :)
Post by: jakesilove on July 31, 2016, 03:06:18 pm
Look at the question. If it just says to outline then all they want is a method.

Discussion is almost always asked as a separate question, e.g. assess the validity of your experiment, or e.g. comment on why the calculated Eo value is less than the theoretical.

I do not recall having to include the safety in an experiment unless it actually says to outline the safety. You'll have to provide examples of this. (In fact, I've never heard of a question about an experiment being a 7 marker)

Generally, there's not enough marks for the examiners to give away for you to write a full scientific report on an experiment. Most of the time they just want to see the method or a discussion question as a follow-up.

100% agree with Rui!
Post by: bethjomay on July 31, 2016, 03:47:32 pm
Hey atar notes! This isn't a specific past paper question, but I was wondering if anyone could give me an accurate/concise version of the reactions involved in CFC's depleting ozone in the atmosphere? I can't seem to find any two sources which agree! Thanks!
Post by: conic curve on July 31, 2016, 04:36:14 pm
For this extended response; why they are more metals available to use now than 200 yrs ago, what would I have to talk about other than "recently developed technology" which made it cheaper and easier for us to extract?
Post by: Sssssrr on August 01, 2016, 09:58:29 pm
hey, could some please explain what equilibrium constants actually are
many thanks
Post by: RuiAce on August 01, 2016, 10:00:43 pm
hey, could some please explain what equilibrium constants actually are
many thanks
What it actually is? Well it is a measure of just where exactly does the equilibrium lie for a reversible reaction. I.e. by how much the products or reactants dominate at equilibrium.
Post by: conic curve on August 01, 2016, 10:03:16 pm
hey, could some please explain what equilibrium constants actually are
many thanks

The equilibrium constant of a chemical reaction is the value of the reaction quotient when the reaction has reached equilibrium. An equilibrium constant value is independent of the analytical concentrations of the reactant and product species in a mixture, but depends on temperature and on ionic strength. Known equilibrium constant values can be used to determine the composition of a system at equilibrium.
Post by: RuiAce on August 01, 2016, 10:04:22 pm
Equilibrium constants are separated into a homogenous equilibrium and heterogenous equilibrium

A homogeneous equilibrium has everything present in the same phase. The usual examples include reactions where everything is a gas, or everything is present in the same solution

A heterogeneous equilibrium has things present in more than one phase. The usual examples include reactions involving solids and gases, or solids and liquids.
All of this isn't actually a part of the syllabus, fyi.
Post by: jakesilove on August 01, 2016, 10:13:04 pm
All of this isn't actually a part of the syllabus, fyi.

I think maybe it's part of Industrial Chem? Not entirely sure
Post by: RuiAce on August 01, 2016, 10:15:36 pm
I think maybe it's part of Industrial Chem? Not entirely sure
It's not. Relating heterogeneous/homogeneous isn't.

Source: Personal experience :P
Post by: jakesilove on August 01, 2016, 10:17:02 pm
It's not. Relating heterogeneous/homogeneous isn't.

Source: Personal experience :P

Ah sorry I see I see, I didn't actually see your definition post :)
Post by: onepunchboy on August 02, 2016, 07:48:08 pm
Im not sure how to convert into ppm. Can someone crash course this q for me and maybe ppm in general? Thanks
Post by: RuiAce on August 02, 2016, 07:52:03 pm
Im not sure how to convert into ppm. Can someone crash course this q for me and maybe ppm in general? Thanks
To say the least
$\text{Concentration in parts per million is a measure of concentration in }\frac{\textit{mass of thing relevant}}{\textit{mass of entire object}}\\ \text{or alternatively }\frac{\textit{mass of thing relevant}}{\textit{volume of entire liquid}}$
The units "ppm" is decomposed to mg/kg or mg/L

I.e. 1 ppm = 1mg/1kg
OR 1 ppm = 1mg/1L

To determine concentration in ppm, convert the numerator into a mass in mills. Also convert the bottom to a mass in kilograms, or a mass in litres.

To go from molarity (moles per litre) you would leave the denominator the same (possibly equal to 1), and use m=nMM on the numerator and convert the grams to mills.

(Your question can't be done yet as I don't know what the actual precipitate is.)
Post by: onepunchboy on August 02, 2016, 07:58:45 pm
To say the least
$\text{Concentration in parts per million is a measure of concentration in }\frac{\textit{mass of thing relevant}}{\textit{mass of entire object}}\\ \text{or alternatively }\frac{\textit{mass of thing relevant}}{\textit{volume of entire liquid}}$
The units "ppm" is decomposed to mg/kg or mg/L

I.e. 1 ppm = 1mg/1kg
OR 1 ppm = 1mg/1L

To determine concentration in ppm, convert the numerator into a mass in mills. Also convert the bottom to a mass in kilograms, or a mass in litres.

To go from molarity (moles per litre) you would leave the denominator the same (possibly equal to 1), and use m=nMM on the numerator and convert the grams to mills.

(Your question can't be done yet as I don't know what the actual precipitate is.)
Silver chloride hehe
Post by: RuiAce on August 02, 2016, 09:33:23 pm
Silver chloride hehe
The percentage composition of chloride ions in silver chloride (by molar masses) is

35.45/(107.9+35.45) * 100% = 24.7296826...%

Thus, the mass of chloride ions present in the sample is
mCl- = 3.65g * 24.7296826.../100 = 0.9026334147...g

Convert this mass into milligrams: 902.6334147... mg

The mass of the water is in mL. We want to convert this to L: 0.05 L

Concentration in ppm = 902.6334147...mg / (0.05 L) = 18052.66829... ppm = 1.80*104 ppm (3 s.f.)

That's quite a large concentration. Unsurprising because 3.65g precipitate out of 50mL is massive in the context of ppm but still dubious. Will let someone else check my working here.
Post by: jakesilove on August 02, 2016, 09:39:02 pm
The percentage composition of chloride ions in silver chloride (by molar masses) is

35.45/(107.9+35.45) * 100% = 24.7296826...%

Thus, the mass of chloride ions present in the sample is
mCl- = 3.65g * 24.7296826.../100 = 0.9026334147...g

Convert this mass into milligrams: 902.6334147... mg

The mass of the water is in mL. We want to convert this to L: 0.05 L

Concentration in ppm = 902.6334147...mg / (0.05 L) = 18052.66829... ppm = 1.80*104 ppm (3 s.f.)

That's quite a large concentration. Unsurprising because 3.65g precipitate out of 50mL is massive in the context of ppm but still dubious. Will let someone else check my working here.

Was just about to post my working, and got the same solution. Unless we're both doing the same thing wrong, pretty sure that checks out!
Post by: conic curve on August 02, 2016, 09:47:39 pm
I'm not getting the idea of this question:

The concentration of mercury in a solution is 0.005g (% w/w). Express this concentration in ppm
Post by: jakesilove on August 02, 2016, 09:53:59 pm
I'm not getting the idea of this question:

The concentration of mercury in a solution is 0.005g (% w/w). Express this concentration in ppm

Assuming that this just means 0.005% of the solution was Mercury (I can't think of another way to answer this question), we can convert it straight to ppm. Parts per million are exactly that; if there were a million 'parts', how many would be Mercury? Well, 0.005% of them. You can plug this straight into a calculator, and find that the answer is 50ppm
Post by: conic curve on August 03, 2016, 07:03:30 am
Assuming that this just means 0.005% of the solution was Mercury (I can't think of another way to answer this question), we can convert it straight to ppm. Parts per million are exactly that; if there were a million 'parts', how many would be Mercury? Well, 0.005% of them. You can plug this straight into a calculator, and find that the answer is 50ppm

Thanks Jake

Need help:

The haber process is an industrial process used to produce ammonia (NH3) from hydrogen and nitrogen gas in the air. The reaction shown below details this process

(it is attached)

Identify three factors that can increase the forward reaction rate and explain how this occurs (6 marks)
Post by: HighTide on August 03, 2016, 07:46:22 am
Thanks Jake

Need help:

The haber process is an industrial process used to produce ammonia (NH3) from hydrogen and nitrogen gas in the air. The reaction shown below details this process

(it is attached)

Identify three factors that can increase the forward reaction rate and explain how this occurs (6 marks)
You should have a go at these. Read the principles here: http://www.chemguide.co.uk/physical/equilibria/lechatelier.html
Hopefully you have a go at it first, then read the spoiler.
Spoiler
Increase reaction rate by:
1. Increase temperature--> increase kinetic energy = more collisions  at right orientation in a given time
2. Increase pressure --> Decrease volume= less room= more collisions at right orientation in a given time
3. Add a catalyst--> Decrease activation energy = increase proportion of particles which can surpass activation energy = faster
Post by: conic curve on August 03, 2016, 08:02:36 am
You should have a go at these. Read the principles here: http://www.chemguide.co.uk/physical/equilibria/lechatelier.html
Hopefully you have a go at it first, then read the spoiler.
Spoiler
Increase reaction rate by:
1. Increase temperature--> increase kinetic energy = more collisions  at right orientation in a given time
2. Increase pressure --> Decrease volume= less room= more collisions at right orientation in a given time
3. Add a catalyst--> Decrease activation energy = increase proportion of particles which can surpass activation energy = faster

Le chaletier (if that;s how you spell it) is not part of the year 11 syllabus

In order to compare the reactivity of three different metals, Macy conducted an experiment where in which she placed the metals in different electrolyte solutions and recorded observations that occurred over time.

1. When metal A was placed in a 0.1M nitrate solution of Metal B there was no reaction observed
2. When metal B was placed in a 0.1M nitrate solution of Metal C a precipitate was seen to deposit on the surface of metal A
3. Metal B was then placed in a 0.1M nitrate solution of Metal C. Macy noticed that a precipitate was deposited on the surface of metal B

Arrange the metals in order of decreasing reactivity and explain the reasoning behind your arrangement (4 marks)
Post by: RuiAce on August 03, 2016, 08:45:14 am
Le chaletier (if that;s how you spell it) is not part of the year 11 syllabus

In order to compare the reactivity of three different metals, Macy conducted an experiment where in which she placed the metals in different electrolyte solutions and recorded observations that occurred over time.

1. When metal A was placed in a 0.1M nitrate solution of Metal B there was no reaction observed
2. When metal B was placed in a 0.1M nitrate solution of Metal C a precipitate was seen to deposit on the surface of metal A
3. Metal B was then placed in a 0.1M nitrate solution of Metal C. Macy noticed that a precipitate was deposited on the surface of metal B

Arrange the metals in order of decreasing reactivity and explain the reasoning behind your arrangement (4 marks)
If I remember correctly whilst you don't explicitly learn LCP in prelim the same rules are applied and you still need to know what he said.

Now, precipitation reactions in this context are intended to demonstrate metal displacement reactions.
The more reactive metal will displace the less reactive metal and become ionised.

E.g. Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
Zinc being more reactive than copper displaces it.

In #1, metal A did not displace metal B. So metal A is not as reactive as metal B
#2 I think you had a typo. You started talking about metal A out of nowhere
In #3, metal B did displace metal C. So metal B is more reactive than metal C
Post by: wesadora on August 03, 2016, 05:16:39 pm
For anyone that does Shipwrecks...I HAVE A QUESTION.
I am aware rusting does not occur when pH is above 9, but why does a more acidic environment (lower pH) accelerate rusting?
The explanation given to me was that the reduction reaction O2 + 2H2O + 4e-<--> 4OH- exists in equilibrium - so, when H+ ions (from acid) are present it reacts with the OH- produced, and thus by LCP removing the product (OH) will cause the equilibrium to shift to the right and consequently accelerating rusting.

HOWEVER, my teacher and many other sources say that the reaction at the cathode is not reversible and i'm....confused.
cheers :)
Post by: onepunchboy on August 03, 2016, 05:31:15 pm

Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?
Post by: wesadora on August 03, 2016, 05:48:35 pm
'The pH of 5 x 10-3M H2SO4 is'?

apparently it's 2. I'm getting 2.3 from -log10[H+], but apparently you're supposed to do -log10[2 x 5 x 10-3]. why? :/
Post by: jakesilove on August 03, 2016, 06:07:47 pm
For anyone that does Shipwrecks...I HAVE A QUESTION.
I am aware rusting does not occur when pH is above 9, but why does a more acidic environment (lower pH) accelerate rusting?
The explanation given to me was that the reduction reaction O2 + 2H2O + 4e-<--> 4OH- exists in equilibrium - so, when H+ ions (from acid) are present it reacts with the OH- produced, and thus by LCP removing the product (OH) will cause the equilibrium to shift to the right and consequently accelerating rusting.

HOWEVER, my teacher and many other sources say that the reaction at the cathode is not reversible and i'm....confused.
cheers :)

I'm sorry I can't answer your question in any more detail, because the explanation you were given is how I understood the process to work! Personally, I would just use that LCP explanation in an exam. Hopefully someone doing Shipwrecks now can help us out here!
Post by: jakesilove on August 03, 2016, 06:11:29 pm
'The pH of 5 x 10-3M H2SO4 is'?

apparently it's 2. I'm getting 2.3 from -log10[H+], but apparently you're supposed to do -log10[2 x 5 x 10-3]. why? :/

Hey! Sulfuric acid is DIPROTIC. That means it donates 2 hydrogen ions (two protons), instead of just one (like HCl does). This means that, for one more of Sulfuric acid, two moles of hydrogen ions are donated. Therefore, you DOUBLE the concentration before plugging it into the formula. If it was triprotic (eg. Citric acid), you would TRIPLE the concentration! Hope that makes sense!
Post by: jakesilove on August 03, 2016, 06:17:41 pm

Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?

To be completely honest, whilst I can't really answer your question directly, I can answer the question by elimination. I think that's the best way to deal with this question anyway.

Firstly, by looking to the Polymer, we know that one monomer should be 5 carbon lengths long, and one should be 4 carbon lengths long. This immediately rules out A or D, leaving us with B or C.

Now, let's get of an OH from one monomer, and a H from another monomer. We know that a H2O is condensed out of the monomers somewhere, so this is a fair thing to do. According to B, this would result in one of two things: either BOTH monomers still have an OH group, or one doesn't an an OH group, and one has two. Looking to the Polymer, we can't see a single OH in sight. As such, B doesn't make any sense; we should see some left over.

Well damn. Actually, C has two OH in each monomer. Maybe that means two Water molecules are eliminated? I'm stumped, sorry I'm not great at visualising Chemistry stuff like this. Maybe someone can come help us out?
Post by: wesadora on August 03, 2016, 06:27:58 pm

Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?

Pretty much, you NEED two OH groups to form the (C-O-C carboxylic bond AND have excess water (H2O) on both sides of the monomer.
By process of elimination, you first need two monomers with 4 and 5 carbons, and thus the answer is either B or C.
The reason the answer is C, is because each monomer given has two OH groups. The simplified structural formula has a [CH2CH3], which on the 5 carbon monomer in question C is the two left-most most carbons leftover, after the OH groups connect.

This is really hard to explain just using words and it would be so much easier for you to visualise if I could physically point things out with my finger but....yeah ._.
Post by: RuiAce on August 03, 2016, 06:34:14 pm

Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?
The question's answered so I'm going to give a piece of advice here.

Try to work backwards. I always figured out which two MONOMERS would give the POLYMER in question
Post by: Loki98 on August 04, 2016, 12:55:17 pm
How would i do this question?
The molar heat of combustion of 1-propanol is 2021kJmol-1. What mass of 1-propanol must be combusted to boil 300mL of water if the water has an initial temperature of 20°C and only absorbs 75% of the heat released?

Post by: RuiAce on August 04, 2016, 01:22:37 pm
How would i do this question?
The molar heat of combustion of 1-propanol is 2021kJmol-1. What mass of 1-propanol must be combusted to boil 300mL of water if the water has an initial temperature of 20°C and only absorbs 75% of the heat released?
Because the density of water is just 1 g mL-1, we know that the mass of water present is 300g.

Because only 75% of the heat is released, our enthalpy is

q = 75% * mc∆T = 0.75 * 300g * 4.18 J g-1 K-1 * 20 K = 18810 J

So q = 18.81 kJ

Using the fact that ∆H * n = -q

-2021 kJ mol-1 * n = -18.81 kJ
n = 0.00930727362... mol

Lastly, using the fact that m = n*MM

m = (0.00930727362...mol) * (60.094 g mol-1) = 0.5590320831... g

Then round to 2 s.f.
Post by: Loki98 on August 04, 2016, 01:42:01 pm
Thx for the help RuiAce :D
Post by: wesadora on August 04, 2016, 02:31:09 pm
For part b), i keep getting 18ppm as my answer, but apparently the answer is 4.5 x 10-3 ppm.
plz explain :'(

my working was just 0.45mg/25mL * 40/40 (to get it into micrograms/L) but I feel there's more to it xD
Post by: jakesilove on August 04, 2016, 02:53:14 pm
For part b), i keep getting 18ppm as my answer, but apparently the answer is 4.5 x 10-3 ppm.
plz explain :'(

my working was just 0.45mg/25mL * 40/40 (to get it into micrograms/L) but I feel there's more to it xD

Hey!

I hope the below answer makes enough sense. Let me know if you need more details!

(http://i.imgur.com/9q8Xfjq.jpg?1)

Jake
Post by: wesadora on August 04, 2016, 03:01:18 pm
Hey!

I hope the below answer makes enough sense. Let me know if you need more details!

Jake

Legend. :)
Also- in a chemistry exam, when do you have to/don't have to define LCP? My logic says do it the very first time it's required to state it, and then i can just leave it and say 'by LCP' for the rest of the exam later in the paper whenever i need to...but have chem trial tomorrow and wanna make sure HAHA!
Cheers:)
Post by: wesadora on August 04, 2016, 03:04:14 pm
hold on....why is the treated water sample "0.450 x 10-3"mg/25mL in the first line of your working? I thought that frmo the previous question before and just reading off the graph it's .450 mg/25mL (450 x -3)
Post by: jakesilove on August 04, 2016, 03:06:43 pm
Legend. :)
Also- in a chemistry exam, when do you have to/don't have to define LCP? My logic says do it the very first time it's required to state it, and then i can just leave it and say 'by LCP' for the rest of the exam later in the paper whenever i need to...but have chem trial tomorrow and wanna make sure HAHA!
Cheers:)

You need to define it in every single question to do with LCP. It's not enough to prove to the marker once that you do actually know it; it is vital to any question about equilibrium! Just because you show the marker you can do concentration calculations in one question, doesn't mean you can skip the next calculation question :P For the same reason, ALWAYS define LCP (as painful as that may be)
Post by: Loki98 on August 04, 2016, 04:36:45 pm
Could someone please help me out with this dot point, "Identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature." How to i determine which salts dissolved in water, form acidic, basic or neutral solutions?
Post by: RuiAce on August 04, 2016, 04:41:33 pm
Could someone please help me out with this dot point, "Identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature." How to i determine which salts dissolved in water, form acidic, basic or neutral solutions?
By comparing what ions built up to give that salt.

Ions such as Na+, K+, Cl- and NO3- are all neutral cations. Their conjugate acid/base is so strong that they are only a tiny bit basic/acidic and don't impact.

As opposed to something like NH4+, which is slightly acidic because NH3 is just decently basic.
Or something like CH3COO- which is slightly basic because CH3COOH is just somewhat acidic.
hold on....why is the treated water sample "0.450 x 10-3"mg/25mL in the first line of your working? I thought that frmo the previous question before and just reading off the graph it's .450 mg/25mL (450 x -3)
Milli does not mean micro

1000 micrograms in a milligram
Post by: conic curve on August 04, 2016, 05:20:33 pm
NiCl2(aq)+NaOH(aq)-->Ni(OH)2(s)+NaCl(aq)

I don't understand the complete ionic equation and net ionic equation behind this?

AgNO3(aq)+cl-(aq)-->AgCl(s)

I don't get the complete ionic equation in this
Post by: RuiAce on August 04, 2016, 05:54:46 pm
NiCl2(aq)+NaOH(aq)-->Ni(OH)2(s)+NaCl(aq)

I don't understand the complete ionic equation and net ionic equation behind this?

AgNO3(aq)+cl-(aq)-->AgCl(s)

I don't get the complete ionic equation in this
No tidy typing cause I'm on my phone

With your complete ionic equation all you do is separate the aqueous substances into their components really. It's only for the net ionic equation where you have to throw out your spectator ions.

However, one very important thing to note is that both of your given equations are unbalanced.

First one:
NiCl2(aq) + 2 NaOH(aq) -> Ni(OH)2(s) + 2 NaCl(aq)

In complete ionic:
Ni(2+) + 2 Cl(-) + 2 Na(+) + 2 OH(-) -> Ni(OH)2(s) + 2 Na(+) +2 Cl(-)

Cancel out all the spectators for net ionic
Ni(2+) + 2 OH(-) -> Ni(OH)2(s)
Post by: anotherworld2b on August 04, 2016, 10:02:09 pm
Hi was wondering if i could get help understanding what this question is asking. Valency?
Post by: Skidous on August 04, 2016, 10:10:09 pm
It's talking about the number of covalent bonds each element can make, it relates to valency in that it involves the sharing of electrons

So for Cl it has a valency of 7, so it needs an extra electron to become stable (like the noble gases) in which case it forms a covalent bond with another Cl atom, they share that one covalent bond and thus share a valence electron to become stable.

Then you go through and figure out the number of covalent bonds needed to make be valence shells of each element full

For example
O has 6 valence, therefore 2 covalent bonds are needed to fill the valence shell
H has 1 valence and thus forms one covalent bond to fill the valence shell (it only occupies the first shell)

Hope this helps
Skidous
Post by: RuiAce on August 04, 2016, 10:13:05 pm
It's talking about the number of covalent bonds each element can make, it relates to valency in that it involves the sharing of electrons

So for Cl it has a valency of 7, so it needs an extra electron to become stable (like the noble gases) in which case it forms a covalent bond with another Cl atom, they share that one covalent bond and thus share a valence electron to become stable.

Then you go through and figure out the number of covalent bonds needed to make be valence shells of each element full

For example
O has 6 valence, therefore 2 covalent bonds are needed to fill the valence shell
H has 1 valence and thus forms one covalent bond to fill the valence shell (it only occupies the first shell)

Hope this helps
Skidous
The question asks for the maximum amount. This makes me question as to whether they mean maximum amount so that the atoms are "happy", or the actual maximum amount.

Why do I say this? Oxygen technically only needs (and ever wants) 2 covalent bonds to be happy. Sulfur, on the other hand, can take 2, 4 or even 6.

I feel that this was a badly worded question due to its lack of clarity.
Post by: wesadora on August 04, 2016, 10:13:54 pm
is the answer C because citric acid only has a 1% ionisation strength compared to hydrochloric acid's 100%? What's the reasoning/how do you work this out?
Post by: RuiAce on August 04, 2016, 10:16:36 pm
is the answer C because citric acid only has a 1% ionisation strength compared to hydrochloric acid's 100%? What's the reasoning/how do you work this out?
pH is 2? Shouldn't the answer then be B because [H3O+] = 10-pH = 10-2 = 0.010
Calling in Jake...

However yes, ionisation strength definitely plays a part.
Post by: jakesilove on August 04, 2016, 10:27:18 pm
pH is 2? Shouldn't the answer then be B because [H3O+] = 10-pH = 10-2 = 0.010
Calling in Jake...

However yes, ionisation strength definitely plays a part.

Yes, the answer is definitely B, not C. Like Rui shows, we know that the concentration of HCl to be 0.010 moles per liter. That equation works directly for all strong acids/bases, so it can't be wrong. This means that the only possible answer is B. We also happen to know that Citric acid is much, much weaker, and so requires a concentration that is much stronger (eg. double) that HCl to produce the same pH. B therefore continues to make sense, and so is the only possible answer.

Jake
Post by: anotherworld2b on August 04, 2016, 10:29:45 pm
The question asks for the maximum amount. This makes me question as to whether they mean maximum amount so that the atoms are "happy", or the actual maximum amount.

Why do I say this? Oxygen technically only needs (and ever wants) 2 covalent bonds to be happy. Sulfur, on the other hand, can take 2, 4 or even 6.

I feel that this was a badly worded question due to its lack of clarity.

For part c wouldnt N have a valence of 5 eletrons? So it needs 3 more electrons? But the answer says 5?
Post by: RuiAce on August 04, 2016, 10:31:42 pm
For part c wouldnt N have a valence of 5 eletrons? So it needs 3 more electrons? But the answer says 5?
Just as I suspected.

They want to know what is the maximum amount of covalent bonds that can form. That is always the amount of electrons in the valence shell.

For nitrogen, this is 5.

(In reality will it form 5? Never. But that's not the point of the question.)
Post by: anotherworld2b on August 04, 2016, 10:35:55 pm
Just as I suspected.

They want to know what is the maximum amount of covalent bonds that can form. That is always the amount of electrons in the valence shell.

For nitrogen, this is 5.

(In reality will it form 5? Never. But that's not the point of the question.)

Oh okay  ;D
Thank you i think it get it now  :D
Post by: anotherworld2b on August 04, 2016, 10:51:42 pm
I struggle with drawing lewis structures and was wondering if i could get some tips kn how to draw them
An example im struggling with is
SO4 ^2-
Post by: Jakeybaby on August 04, 2016, 11:18:20 pm
I struggle with drawing lewis structures and was wondering if i could get some tips kn how to draw them
An example im struggling with is
SO4 ^2-
Initially, we will count the number of valence electrons:

We know that Sulfur has 6, as does Oxygen.
Therefore:
Number of valence electrons = Valence Electrons(sulfur) + 4*Valence Electrons(oxygen) + 2(negative 2 charge shows 2 electrons are present)
Number of valence electrons = 6 + 6(4) + 2
= 32

Drawing the ion:
Spoiler
(https://i.gyazo.com/2136bc0cf3fafac2cf5727c7fd3ce3f8.png)

We need to ensure that the outer atoms have octets, therefore, we need to draw 6 dots around each of the oxygen atoms, and counting the number of valence electrons present for each atom: it can be seen that all the oxygens have 8, as does the centre sulfur atom.
Spoiler

But hang on, we're not done:
The formal charges need to be looked at:
Formal Charge = Valence Electrons - Non-Bonding Valence Electrons - Bonding Electrons/2

Therefore:
Formal ChargeOxygen = 6 - 6 - 2/2 = -1
Formal ChargeSulfur = 6 - 0 - 8/2 =  +2

We want the formal charges to be as close to 0 as possible on each atom, hence, to do this, we can take 2 valence electrons from 2 seperate oxygen atoms, to make some double bonds:
Spoiler
(https://i.gyazo.com/c0b736b4e5e2d6b55b21a4a1dddd3f5b.png)
Now calculating the formal charges again:
Formal ChargeSulfur = 6 - 0 - 12/2 =  0
Formal ChargeBlack Oxygen = 6 - 6 - 2/2 =  -1
Formal ChargeBlue Oxygen = 6 - 4 - 4/2 =  0
Therefore, as we cannot get the formal charges any closer to 0, we have the lewis structure of the sulfate ion.

Spoiler
(https://i.gyazo.com/ccb02b725768741f26fba536685f1825.png)
Post by: RuiAce on August 04, 2016, 11:41:33 pm
I think the Lewis structure for the sulfate ion is not in the HSC course.

The concept of "formal charges" is not taught in the HSC nor preliminary courses. At most, the octet rule is.
Post by: sire123 on August 06, 2016, 03:06:01 am
Can you guys explain to me why for the bromine water prac, it's either Br2 or HOBr? Can you show both word and structural equations for the reactions that take place with cyclohexene + bromine water, and cyclohexene + bromine water?
Post by: RuiAce on August 06, 2016, 08:34:39 am
Can you guys explain to me why for the bromine water prac, it's either Br2 or HOBr? Can you show both word and structural equations for the reactions that take place with cyclohexene + bromine water, and cyclohexene + bromine water?
For your first question, it does not matter whether we use Br2 or HOBr. However, for reference sake, this is what actually happens.

Generally speaking, rather than using liquid bromine we mix the bromine into the water. A reaction takes place:
Br2(l) + H2O(l) -> HBr(aq) + HOBr(aq)

Note that the hydrobromic acid does nothing.

(Not too sure but I think bromine water is preferred over actual bromine because it oxidises more easily.)
____________________

To understand the reactions that take place you need to consider what exactly are substitution and addition reactions. Substitution reactions are exactly that - certain atoms effectively displace other atoms. Addition reactions are when bonds are broken and the atoms form new bonds.

So say you went from ethane to ethanol through reaction with water. Then two of the hydrogens are going to have to be substituted for the H and OH from water.
If you went form ethene to ethanol though, the double bond breaks and the water molecule just adds on. The H goes to one carbon whereas the OH goes to the other.

Now, cyclohexene has formula C6H10. We can't really assign it a number as to where the double bond is (unlike say, 1-hexene), but we know that there is a double bond there. It has a high electron density and is very reactive, so when the HOBr is introduced it will readily add on.

(Note that HOBr is what has the brown or whatever colour. The bromine water is what gets decolourised, not the cyclohexene.)

C6H10(l) + HOBr(aq) -> C6H10BrOH(l)

On the contrary, cyclohexane only contains single bonds. For a substitution reaction to take place on top of any of those single bonds we need to provide much more energy. In fact, this can only be done so through adding UV radiation.

C6H12(l) + HOBr(aq) -(UV)-> C6H10BrOH(l) + H2(g)

Under laboratory conditions, you're not going to find that much UV at all. So you will see the cyclohexene decolourise it, and not the cyclohexane. That is all.

Structural diagrams can be found at EasyChem
Post by: sire123 on August 06, 2016, 10:14:51 am
Wow cheers Rui! I have a couple more questions.

Well first one is a 6 marker "Discuss the process of fermenting glucose into ethanol. Include a chemical equation and a chemical structure of ethanol." How would you tackle and structure this q? Is it asking for the prac, and all the conditions needed to make this work...

And what is the difference between Biopol, polydroxybutanoate, and polyhydroxybutyrate? Are they all the same?
Post by: jakesilove on August 06, 2016, 01:14:53 pm
Wow cheers Rui! I have a couple more questions.

Well first one is a 6 marker "Discuss the process of fermenting glucose into ethanol. Include a chemical equation and a chemical structure of ethanol." How would you tackle and structure this q? Is it asking for the prac, and all the conditions needed to make this work...

And what is the difference between Biopol, polydroxybutanoate, and polyhydroxybutyrate? Are they all the same?

Hey Sire!

Let's start with that 6 marker: "Discuss the process of fermenting glucose into ethanol. Include a chemical equation and a chemical structure of ethanol."

Okay, clearly we need a chemical equation and the chemical structure of ethanol. I would probably smash that out right at the start, so that you don't forget!

(http://i.imgur.com/FAmGeIU.gif?1)
Make sure to include Yeast as a catalyst as well! As for the chemical structure of Ethanol:
(http://www.bbc.co.uk/staticarchive/728c1c64297ffeacfb59d01b4f2d71ec3a6e4b98.gif)

Easy! Okay, let's try to answer the actual question. We don't really need to be talking about the prac specifically (ie. you don't need to write a method). Instead, I would be listing as much information as possible. I would be writing something like this.

The fermentation of Glucose to Ethanol can be described chemically by the equation given above. However, to achieve the greatest yield of Ethanol, certain conditions must be satisfied.

- Enzymes to convert the Glucose to Ethanol (specifically, Yeast)
- Moderate temperatures (about 37 degrees Celsius)
- A mixture of Glucose in solution
- Anaerobic conditions (ie. low oxygen)
- An Ethanol concentration <15% (or else the Yeast will die)

Great! So we've identified a whole bunch of important factors, we've explained how the process works, we're nearly done! Finally, I would talk about the fact that if we want to USE the ethanol, we need to use multiple iterations of fractional distillation. This is because the Ethanol/Glucose solution will only have a concentration of about 15%. As such, we need to purify it! This allows it to be used.

That should be enough to satisfy the 6 marks! You can always add some extra info, but I don't think you need more than that :)

As for your Biopol/polydroxybutanoate/polyhydroxybutyrate conundrum, basically Biopol is the brand name of PHB (like Panadol is the brand name, but Paracetamol is the drug in it!). I don't think that polydroxybutanoate exists, but polyhydroxybutyrate is the full chemical name of PHB (I would use this last term).

Let me know if you need any more help! Great questions :)

Jake
Post by: jakesilove on August 06, 2016, 07:23:57 pm
I'm just posting a couple of questions for another user; feel free to write up some solutions, otherwise I'll get to an answer shortly!

(http://i.imgur.com/kW0MFep.jpg)
(http://i.imgur.com/aoAn3Dx.jpg)
(http://i.imgur.com/fAjyAVa.jpg)
(http://i.imgur.com/qGYKHAU.jpg)

Jake
Post by: jakesilove on August 06, 2016, 07:31:49 pm
I'm just posting a couple of questions for another user; feel free to write up some solutions, otherwise I'll get to an answer shortly!

(http://i.imgur.com/kW0MFep.jpg)
(http://i.imgur.com/aoAn3Dx.jpg)
(http://i.imgur.com/fAjyAVa.jpg)
(http://i.imgur.com/qGYKHAU.jpg)

Jake

For the first question, you need to start by stating Le Chatelier's Principle. Something along the lines of "When a system undergoes some kind of change, the equilibrium will shift to try to minimise that change". Then, identify that the equilibrium is exothermic (as delta H is negative). As such, an increase in temperature will shift the equilibrium to the left (towards the reactants). Elaborate on this a little, and this will explain the results observed!
Post by: jakesilove on August 06, 2016, 07:35:03 pm
I'm just posting a couple of questions for another user; feel free to write up some solutions, otherwise I'll get to an answer shortly!

(http://i.imgur.com/kW0MFep.jpg)
(http://i.imgur.com/aoAn3Dx.jpg)
(http://i.imgur.com/fAjyAVa.jpg)
(http://i.imgur.com/qGYKHAU.jpg)

Jake

The third question is to do with Hydrogen bonds and dipole-dipole interaction. Identify that Ethane/Butane etc. will only have dispersion inter-molecular forces. These are the weakest inter molecular forces. However, the identified compounds will have dipole-dipole interactions (as the hydroxide groups are electronegative) and hydrogen bonding (between hydrogen and Oxygen). Dipole-dipole interactions are stronger than dispersion, and Hydrogen bonding is stronger still. Therefore, as the intermolecular bonding is much stronger, the boiling point will be much higher (more energy required to break the bonds). This explains the boiling points!
Post by: jakesilove on August 06, 2016, 07:38:55 pm
I'm just posting a couple of questions for another user; feel free to write up some solutions, otherwise I'll get to an answer shortly!

(http://i.imgur.com/kW0MFep.jpg)
(http://i.imgur.com/aoAn3Dx.jpg)
(http://i.imgur.com/fAjyAVa.jpg)
(http://i.imgur.com/qGYKHAU.jpg)

Jake

For the final question, let's break down the two marks into 1) Improving validity and 2) Improving reliability.

Validity

This method does not take into account the fact that water is evaporating as the carbon dioxide is being released. A more valid experiment would be filling a similar can with water, and weighing the water evaporation over time simultaneously. Then, the water evaporation can be subtracted from the changes over time, and the carbon dioxide can be isolated.

Reliability

This one is easy; just do it multiple times! Potentially, take more measurements as well. Nothing too tricky here
Post by: sire123 on August 06, 2016, 08:09:49 pm
Can you do the second image as well? And continuing that q it says "Justify the conditions used during he Haber synthesis of ammonia - High temperature (reactor) and High pressure". But isn't 400 degrees celsius and 200 atm not that high?
Post by: RuiAce on August 06, 2016, 08:42:42 pm
Can you do the second image as well? And continuing that q it says "Justify the conditions used during he Haber synthesis of ammonia - High temperature (reactor) and High pressure". But isn't 400 degrees celsius and 200 atm not that high?
Why not? If you ask me 400 degrees would burn me and 200 atm would easily crush me in a matter of seconds as well. (Keep in mind 1atm is everyday pressure)

The temperature of 400deg C is a moderately high temperature chosen as a compromise to balance out between the need to increase the rate of reaction (high temperatures lead to increased kinetic energy among particles) and forcing the equilibrium to the right (Because the production of ammonia is exothermic, by LCP lower temperatures favour the yield)

Pressure is self explanatory: High pressures favour the yield by LCP since there are fewer moles of gas in the products. However if we excessively increase the pressure though, we risk the glassware breaking and losing everything we have.
Post by: sire123 on August 06, 2016, 08:57:06 pm
Ah u have a point there haha! Nah i just read somewhere it's quite an intermediate temperature and pressure. Then what does it mean Justify use for high temperature (reactor)? Is it talking about the reactants
Post by: RuiAce on August 06, 2016, 09:02:40 pm
Because since LCP predicts that low temperatures shift the equilibrium to the right, having a higher temperature is counter-intuitive.
Post by: amandali on August 07, 2016, 04:04:30 pm

how to know which one is oxidant and reductant
Post by: RuiAce on August 07, 2016, 04:42:40 pm

how to know which one is oxidant and reductant
Consult your table of standard reduction potentials and observe the relevant Eo values to determine if a reaction will occur.

In beaker 1, zinc is higher on the potentials sheet than acid. In fact, this is just your ordinary metal + acid -> salt + hydrogen gas reaction.

Zn(s) + H2SO4(aq) -> ZnSO4(aq) + H2(g)

Remember that the oxidant (aka oxidising agent) is the one that gets reduced. Using OIL-RIG, we know that reduction is gain, so we identify the species that gained electrons.

The correct answer is H+, because two hydrogen ions gained an electron in the formation of hydrogen gas.

Equation 2 works similarly, however is an example of metal displacement:
Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
Post by: Skidous on August 07, 2016, 08:06:24 pm

how to know which one is oxidant and reductant

I know this isn't an answer but I want to know how you got your picture in the post like that
Post by: RuiAce on August 07, 2016, 08:13:14 pm
I know this isn't an answer but I want to know how you got your picture in the post like that
You either upload it to imgur or something and copy/paste the [img] code, or you upload it where you write the post - attachments are uploaded below the box you type your message
In her instance, however, she uploaded through Tapatalk.
Post by: anotherworld2b on August 08, 2016, 12:43:05 am
Hi :D
I have some more chem questions if its okay  :)

I tried q3b but i was unable to get the right answer

For q13 i am not sure how to approach the question
Post by: RuiAce on August 08, 2016, 09:19:37 am
Hi :D
I have some more chem questions if its okay  :)

I tried q3b but i was unable to get the right answer

For q13 i am not sure how to approach the question
$\text{Percentage composition is just a measure of how much of a certain species is present in a molecule.}\\ \text{We can use molar masses here because we know the relevant species.}\\ \text{Note that there are 2 moles of potassium, 1 mole of sulfur and 4 moles of oxygen.}$
MM_{K_2SO_4}=39.10\times 2+32.07 + 4 \times 16.00 =174.27 g \, mol^{-1}\\ \begin{align*}K&= \frac{39.10\times 2}{174.27} \times 100=44.87\dots \\ S&= \frac{32.07}{174.27}\times 100 = 18.20\dots \\ O&=\frac{16.00\times 4}{174.27}\times 100=36.72 \dots\end{align*}
$\textit{LaTeX is being a jerk so I omitted the percent signs}$
Q13 has gone way outside what could possibly be asked in the HSC. However, it still seems like nothing more than gravimetric analysis so this is my attempt.

a) Weight loss - According to the data given the hydrated copper sulfate definitely decomposed, but because we're talking about 200oC here, the water most likely got evaporated off.
b) 1.67/4.77 * 100% = 35%(approx) - use mass, not molar mass to find percentage composition here.
c) Error is not in the HSC course.
Post by: sire123 on August 09, 2016, 01:26:29 pm
A q asks "Write the full ground state electron configuration of chromium showing:
- principle shell notation
- sub-shell notation
- orbital notation
How do u do it?
Post by: senara on August 11, 2016, 08:58:35 pm
Hi Just wondering do we need to know how to draw dry cell and vanadium redox?
Post by: jakesilove on August 11, 2016, 10:30:34 pm
Hi Just wondering do we need to know how to draw dry cell and vanadium redox?

Hey! It is definitely knowing how to draw your battery, if it is at all simple. Specifically, the Dry Cell is quite easy to draw, so I would recommend having that up your sleeve. Vanadium Redox is slightly more complicated, so you probably don't need to have that down. In reality, you might never use either, so it's like an additional "LOOK HOW MUCH I KNOW!" point that you can use to score some marks
Post by: RuiAce on August 11, 2016, 10:37:46 pm
Hi Just wondering do we need to know how to draw dry cell and vanadium redox?
Food for thought: I literally drew out the Gratzell cell in the exam last year to help get my marks. (Dry cell wasn't examined for us.)
Post by: Loki98 on August 13, 2016, 05:23:55 pm
The colourless aqueous solution of a white solid sample formed a precipitate with an acidified aqueous solution of lead (II) nitrate. what is the possible identity of the original solution of this sample?
a) copper (II) sulfate
b) Barium chloride
c) Sodium sulfate
d) silver chloride
Could you please explain how you disprove the incorrect options.
Thx =]
Post by: RuiAce on August 13, 2016, 05:45:35 pm
The colourless aqueous solution of a white solid sample formed a precipitate with an acidified aqueous solution of lead (II) nitrate. what is the possible identity of the original solution of this sample?
a) copper (II) sulfate
b) Barium chloride
c) Sodium sulfate
d) silver chloride
Could you please explain how you disprove the incorrect options.
Thx =]
This answer is incomplete. I'll ask that someone else finish it off because I'm not sure how to incorporate the acidification. I've started it off in the meantime

Since we know that all nitrates are soluble, there is no way that the precipitate formed could've involved the nitrate ion.

Hence, we consider substances involving the lead ion Pb2+ which are insoluble. Problem is that both PbSO4 AND PbCl2 are insoluble!

Begin a process of elimination.
D) is the first one to go. This is because AgCl is also insoluble. Since silver chloride is insoluble, it cannot have been the aqueous solution.
A) is the second one to go. The presence of the Cu2+ ion means that the solid would've been blue, when clearly it is white.

The last bit is the hard one.
Post by: MysteryMarker on August 13, 2016, 06:27:30 pm
Kinda stuck on this question,

CH3NH2(aq) + H2O <----> OH- + CH3NH3+\

If the methyl amine (CH3NH2(aq)) reacts with HCl to produce the salt methyl ammonium chloride. Write the equation for this neutralisation reaction. Explain, with the aid of an equation, why this salt does no produce a neutral solution when dissolved in water.

Cheers.
Post by: RuiAce on August 13, 2016, 08:21:55 pm
Kinda stuck on this question,

CH3NH2(aq) + H2O <----> OH- + CH3NH3+\

If the methyl amine (CH3NH2(aq)) reacts with HCl to produce the salt methyl ammonium chloride. Write the equation for this neutralisation reaction. Explain, with the aid of an equation, why this salt does no produce a neutral solution when dissolved in water.

Cheers.
Given that the equation is reversible due to the presence of ⇌, we know that methyl amine is weakly basic and its conjugate; the methyl ammonium ion is weakly acidic.

Suppose we react with hydrochloric acid. Because we have a strong acid, the equation loses its nature as reversible. So it just becomes this.
CH3NH2(aq) + HCl(aq) -> CH3NH3Cl(aq)

If we dissolve this salt in water, it will dissociate.
CH3NH3Cl(s) ->  CH3NH3+ + Cl-

The chloride ion is neutral. Because HCl is a strong acid, Cl- is so basic to the point it basically doesn't react.
However, the weakly acidic methyl ammonium ion will just do the original equation's thing in reverse (i.e. donate a proton):

CH3NH3+ + H2O(l) -> H3O+ + CH3NH2(aq)

So in fact, the salt formed is slightly acidic. Not neutral.
Post by: MysteryMarker on August 13, 2016, 09:46:22 pm
Given that the equation is reversible due to the presence of ⇌, we know that methyl amine is weakly basic and its conjugate; the methyl ammonium ion is weakly acidic.

Suppose we react with hydrochloric acid. Because we have a strong acid, the equation loses its nature as reversible. So it just becomes this.
CH3NH2(aq) + HCl(aq) -> CH3NH3Cl(aq)

If we dissolve this salt in water, it will dissociate.
CH3NH3Cl(s) ->  CH3NH3+ + Cl-

The chloride ion is neutral. Because HCl is a strong acid, Cl- is so basic to the point it basically doesn't react.
However, the weakly acidic methyl ammonium ion will just do the original equation's thing in reverse (i.e. donate a proton):

CH3NH3+ + H2O(l) -> H3O+ + CH3NH2(aq)

So in fact, the salt formed is slightly acidic. Not neutral.

For the reaction between methyl amine and hydrochloric acid, shouldn't this be a neutralisation reaction as we established that methyl amine is a weak base and we know that hydrochloric acid is a strong acid? So shouldn't water also be a product in this reaction?

Post by: RuiAce on August 13, 2016, 09:51:03 pm
For the reaction between methyl amine and hydrochloric acid, shouldn't this be a neutralisation reaction as we established that methyl amine is a weak base and we know that hydrochloric acid is a strong acid? So shouldn't water also be a product in this reaction?
I didn't know you always had to produce water in neutralisation:

NH3 + HCl -> NH4Cl
Post by: MysteryMarker on August 14, 2016, 10:27:21 am
I didn't know you always had to produce water in neutralisation:

NH3 + HCl -> NH4Cl

OHHHHHHH rekt. Thanks man, i understand it now :P

Got another question:

The correct statement about biopolymers if:

B)All biopolymers can be manufactured synthetically in the laboratory by condensation reactions.
C)Synthetic biopolymers are being produced from living organisms and are replacing polymers made from petrochemicals
D)All natural biopolymers are made by condensation reactions involving glucose monomers.

Cheers.
Post by: jakesilove on August 14, 2016, 11:45:30 am
OHHHHHHH rekt. Thanks man, i understand it now :P

Got another question:

The correct statement about biopolymers if:

B)All biopolymers can be manufactured synthetically in the laboratory by condensation reactions.
C)Synthetic biopolymers are being produced from living organisms and are replacing polymers made from petrochemicals
D)All natural biopolymers are made by condensation reactions involving glucose monomers.

Cheers.

Hey,

The answer here is definitely C. Just do this by elimination, in terms of what you have actually studied. Biopolymers may be biodegradable, they may not be, however you aren't supposed to know EVERY biopolymer out there! So, the answer can't be A. Again, you would need to know the reaction of EVERY biopolymer, some of which may be condensation reactions, most of which would be addition reactions, for the answer to be B. So, it isn't. Same goes for D. However, the REASON you learn about biopolymers is that we need to find alternatives to petrochemicals (the focus of the whole of Production of Materials). Plus, BIOpolymers are obviously produced by living organisms. So, the answer is C.

Jake
Post by: RuiAce on August 14, 2016, 11:58:35 am
Hey Jake, please finish off the question that I only half completed at the bottom of the last page. I still can't show that sodium sulfate (prediction) would be the correct one
Post by: jakesilove on August 14, 2016, 12:10:01 pm
Hey Jake, please finish off the question that I only half completed at the bottom of the last page. I still can't show that sodium sulfate (prediction) would be the correct one

On it!
Post by: jakesilove on August 14, 2016, 12:17:40 pm
This answer is incomplete. I'll ask that someone else finish it off because I'm not sure how to incorporate the acidification. I've started it off in the meantime

Since we know that all nitrates are soluble, there is no way that the precipitate formed could've involved the nitrate ion.

Hence, we consider substances involving the lead ion Pb2+ which are insoluble. Problem is that both PbSO4 AND PbCl2 are insoluble!

Begin a process of elimination.
D) is the first one to go. This is because AgCl is also insoluble. Since silver chloride is insoluble, it cannot have been the aqueous solution.
A) is the second one to go. The presence of the Cu2+ ion means that the solid would've been blue, when clearly it is white.

The last bit is the hard one.

So, for reasons unknown to me, and irrelevant to the HSC curriculum, Lead (II) Chloride is slightly soluble in Acidic conditions. It is also slightly soluble in water (but only a tiny bit!). Either way, the BETTER answer is going to be C, because Lead Sulfate is not soluble at all. This isn't something that you're supposed to know. Typical James Ruse going wayyy beyond the curriculum.

Jake
Post by: MysteryMarker on August 14, 2016, 12:57:38 pm
Hey,

The answer here is definitely C. Just do this by elimination, in terms of what you have actually studied. Biopolymers may be biodegradable, they may not be, however you aren't supposed to know EVERY biopolymer out there! So, the answer can't be A. Again, you would need to know the reaction of EVERY biopolymer, some of which may be condensation reactions, most of which would be addition reactions, for the answer to be B. So, it isn't. Same goes for D. However, the REASON you learn about biopolymers is that we need to find alternatives to petrochemicals (the focus of the whole of Production of Materials). Plus, BIOpolymers are obviously produced by living organisms. So, the answer is C.

Jake

Thanks man, totally understand it now. Just got a few more questions if thats alright.

1. Sodium hydrogen carbonate is a common laboratory chemical. Explain why arrhenius acid/base definition is unable to account for the acid/base properties of this species whereas the BL theory can. Include chemical equations. (5marks)

For this question would i just state Arrhenius's definition + BL definition, and then show that HCO3- is amphiprotic. By its reaction with water, showing it accepting a hydrogen ion and donating a hydrogen ion, and as it can do both Arhennius would be unable to define this as a base/acid?

2. What is meant by the Eo of an electrode in an electrochemical cell. (2marks)

3. For the complete combustion of a fuel, is the water produced in the reaction gaseous or liquid?

Cheers.
Post by: RuiAce on August 14, 2016, 01:07:38 pm
Thanks man, totally understand it now. Just got a few more questions if thats alright.

1. Sodium hydrogen carbonate is a common laboratory chemical. Explain why arrhenius acid/base definition is unable to account for the acid/base properties of this species whereas the BL theory can. Include chemical equations. (5marks)

For this question would i just state Arrhenius's definition + BL definition, and then show that HCO3- is amphiprotic. By its reaction with water, showing it accepting a hydrogen ion and donating a hydrogen ion, and as it can do both Arrhenius would be unable to define this as a base/acid?

Also, note that Arrhenius thought bases had to ionise to produce OH-. You can't do that with NaHCO3 either.

2. What is meant by the Eo of an electrode in an electrochemical cell. (2marks)

3. For the complete combustion of a fuel, is the water produced in the reaction gaseous or liquid?

Cheers.
1. That does show how it's a B-L acid because according to B-L theory, acids are proton donors.

Svante Arrhenius' definition stated that acids ionised in solution (water) to produce H+. When you dissolve your sample of sodium hydrogen carbonate you don't produce that; you just produce the sodium ion and the hydrogen carbonate ion.

NaHCO3(s) -> Na+ + HCO3-

So it fails to be an Arrhenius acid.

2. Your table of standard reduction potentials lists several Eo values for individual half equations.

The EMF produced in an electrochemical cell is just the sum of the two relevant half equations. It is thus, a measure of the voltage produced, from you making that electrochemical cell. (Remember: In a Galvanic cell, chemical energy is converted into electrical energy.)

3. Gas. The temperatures in combustion are extremely high.
Post by: MysteryMarker on August 14, 2016, 04:25:59 pm
Classify the salt formed from HCO3- as acidic, basic or neutral and justify its classification. Include a chemical equation to illustrate your answer. (2marks)

Kinda confused for this question, cause i know that HCO3- is amphiprotic so wouldn't its acidity/basicity depend on the solvent? Or would the salt just be neutral because assuming the solvent is water, it will produce an equal amount of hydrogen and hydroxide ions?

Cheers.
Post by: RuiAce on August 14, 2016, 04:33:02 pm
Classify the salt formed from HCO3- as acidic, basic or neutral and justify its classification. Include a chemical equation to illustrate your answer. (2marks)

Kinda confused for this question, cause i know that HCO3- is amphiprotic so wouldn't its acidity/basicity depend on the solvent? Or would the salt just be neutral because assuming the solvent is water, it will produce an equal amount of hydrogen and hydroxide ions?

Cheers.
Actually, for the hydrogen carbonate ion it exhibits basic properties with water.

HCO3- + H2O(l) ⇌ H2CO3(aq) + OH-

I think this is because water is a stronger acid than hydrogen carbonate. Or alternatively hydrogen carbonate is a stronger base than water. It's also found in the human body as a buffer system in this way.

I'm not entirely sure how this works though.
Post by: MysteryMarker on August 14, 2016, 06:15:07 pm
Ahh, so HCO3- acts as a base. Would this be because it is the strong conjugate base of the weak acid H2CO3?

Another question i have is justify the use of AAS in the mining industry, and in safeguarding the environment.

Just curious as to what points i should bring up for 'mining industry'.

Cheers.
Post by: Happy Physics Land on August 14, 2016, 08:18:54 pm
Ahh, so HCO3- acts as a base. Would this be because it is the strong conjugate base of the weak acid H2CO3?

Another question i have is justify the use of AAS in the mining industry, and in safeguarding the environment.

Just curious as to what points i should bring up for 'mining industry'.

Cheers.

Hey MysteryMarker!

To your enquiry about the conjugate base/acid pair, yes and no. HCO3- is a strong base since H2CO3. However, it is only strong compared to water, meaning that when reacting with water, it is able to accept a hydrogen proton from water. But compared to NaOH which is a very strong base, HCO3- is less strong as a base. In our blood stream, HCO3- and H2CO3 act as buffer pairs and if HCO3- is a strong base then it wouldnt have the buffer effect since strong bases result in irreversible reactions. So when you talk about HCO3-, say its a strong base RELATIVE TO WATER.

In regards to AAS, I will just jot down a few dotpoints here for you to consider:
- Before AAS there was no way to detect the existence of trace elements in waterways (trace elements = extremely small quantities)
- AAS has helped industries to monitor heavy metal pollution into nearby waterways or atmosphere
- For example, if you do industrial chemistry, you can talk about the mercury cell which leeches the neurotoxin mercury into waterways and this can bioaccumulate and affect the health of animals going up the food chain
- If we wanna talk about mining industry, we can say that AAS helps to detect lead or mercury vapours that may rise from underground deposit caverns and these can really threaten workers' health
- And almost if you want to see the purity of the metal ore that is dug up from the ground, AAS would be useful for detecting small amounts of impurities

I think its quite hard to talk about the use of AAS in mining industry without being a mining engineer but generally AAS would just be used for detecting heavy metal existences and purity of materials. In HSC, they would ask for applications of AAS, but very unlikely to ask about its relation with mining industry. More likely they would be asking about AAS' use in water quality examination or just its general uses.

Best Regards
Happy Physics Land
Post by: Happy Physics Land on August 14, 2016, 08:22:04 pm
Let's just proclaim the good news:

I AM WELL AND TRULY BACK BOYS
Post by: jakesilove on August 15, 2016, 08:30:10 am
Let's just proclaim the good news:

I AM WELL AND TRULY BACK BOYS

The prodigal son has returned!
Post by: anotherworld2b on August 15, 2016, 09:53:58 am
Hi was wondering if I could get help with chromatography. I wasn't sure where to get information for these aspects. :-X

•Justify the use of a particular chromatography technique by considering
•the properties of the substances being separated,
•the amount of substance available for analysis
•the sensitivity of the equipment

Post by: Loki98 on August 15, 2016, 01:18:56 pm
Industrial Chemistry Questions:
What kind of molecule would a micelle be classified as?
How do you explain soap acting as a emulsifier with reference to the formation of micelles?
Post by: Happy Physics Land on August 15, 2016, 02:00:41 pm
Industrial Chemistry Questions:
What kind of molecule would a micelle be classified as?
How do you explain soap acting as a emulsifier with reference to the formation of micelles?

Hey Loki!!!

Ohhhh okay we don't usually classify micelle as a molecule, rather I would say its just a mixture. A micelle is essentially a spherical structure where the hydrophobic hydrocarbon chain from soap is adsorbed (i.e. dissolved, attached) into grease/oil through dispersion force and the negatively charged hydrophilic carboxylate head interacts with water molecules through ion-dipole forces. This is how I would define as a micelle, and I attached an image below to show you what a micelle look like. Now why is it a mixture? Because nothing is reacted, nothing is consumed and its just simply soap molecules attached onto grease.

(http://i.imgur.com/CHXaBMK.png)

Ok so I will kinda explain emulsification to you in numbered steps so its clearer:
1. Soap molecules are dissolved in water
2. Hydrophilic head of soap ion interacts with water molecule through ion-dipole, hydrocarbon chain interacts with and dissolves into grease through dispersion forces
3. Surfactant molecules continue to adsorb into the grease, decreasing the surface tension between grease and water
4. Ion-dipole force, being a stronger intermolecular force than dispersion force, pulls grease off the surface and this forms spherical droplets known as micelles
5. The negatively charged heads on the soap repel each other, preventing grease and dirt from joining back together and this keeps micelles dispersed throughout the solution and therefore achieving the emulsifying action

Post by: MysteryMarker on August 15, 2016, 09:18:54 pm
Would the general equation for cellulose to glucose be

(n-1)H2O + n(-C6H10O6-) --> n(-C6H12O6)

or just the same reaction but instead of (n-1) H2O its just H2O

Just a bit confused ;P
Post by: RuiAce on August 15, 2016, 10:02:59 pm
Would the general equation for cellulose to glucose be

(n-1)H2O + n(-C6H10O6-) --> n(-C6H12O6)

or just the same reaction but instead of (n-1) H2O its just H2O

Just a bit confused ;P
(n-1) is correct.

Just remember: For your cellulose, the n is a subscript under the dash (a bit hard to demonstrate on forums):
-(C6H10O5)n-

And when it's glucose there's no need for a dash obviously
Post by: Maz on August 15, 2016, 11:40:13 pm
Hey,
I have a test coming up on soap and detergents, would you know any common questions that come up on a topic like that? or any things I should watch out for?

Thankyou :)
Post by: Happy Physics Land on August 16, 2016, 08:15:26 pm
Hey,
I have a test coming up on soap and detergents, would you know any common questions that come up on a topic like that? or any things I should watch out for?

Thankyou :)

Hey mq123!

They can ask you various forms of questions about soaps and detergents that can largely be summed up in six categories:

- Explain how soaps are produced through saponification
- Explain the cleaning action of soaps/detergents
- Explain why detergents have been developed to replace soaps
- Explain the uses of soaps and detergents, in relation to their properties
- Explain how soaps/detergents act as emulsifiers
- Assess/Discuss the environmental and social impacts of soaps and detergents

What I recommend, along with knowing how to verbally address these questions, is to know how to draw the chemical composition of soaps/detergents (anionic cationic and non-ionic), or at least, their skeletal structure. With soaps you need to ensure that you include your carboxylate head group. With anionic detergents, make sure you include the sulfonate head group. With cationic detergents, make sure you include the quarternary ammonium group. Finally with non-ionic detergents you may like to show the hydroxyl head group. These are what distinguishes the types of detergents and their different uses and properties would rely on the function of these head groups.
Post by: conic curve on August 17, 2016, 01:01:40 pm
Need help with the following

1. A student dissolves 5.3g of anhydrous sodium carbonate (Na2CO3) in water and then adds more water to make the volume up to 125 ml
a)why did the student first dissolve it on water and then make up the final volume by adding water
b)Why did the student use Anhydrous Na2CO3?

2. Calculate how many grams of solute are in 500mL of 0.5mol L^-1 sodium cholride (nacl) solution

Thanks Guys ;D
Post by: anotherworld2b on August 18, 2016, 06:24:37 pm
Hi I was wondering how do you write precipitation reactions?
I am confused on how cations and anions are involved in the reaction?
I am overall confused in general ???
Post by: RuiAce on August 18, 2016, 06:33:53 pm
Hi I was wondering how do you write precipitation reactions?
I am confused on how cations and anions are involved in the reaction?
I am overall confused in general ???
What do you mean by how to "write" precipitation reactions? You write them the same way you write everything else.

Also a cation is just an ion that has a positive charge (e.g. Na+) whereas an anion is an ion with a negative charge (e.g. Cl-)
They're not really important either unless you meant "ionic" equations.
Post by: conic curve on August 18, 2016, 08:12:07 pm
Hi I was wondering how do you write precipitation reactions?
I am confused on how cations and anions are involved in the reaction?
I am overall confused in general ???

I'm not exactly sure how cations and anions are involved in the reaction but all I know is that Cations and anions are formed when a metal loses electrons and a non metal gains those electrons (I don't know why that is the case). This then allows the electrostatic attraction between the positive charge and the negative charge to create an ionic compound

Moderation edit: Removed tutoring service website link
Post by: jyce on August 18, 2016, 08:30:41 pm
Hi I was wondering how do you write precipitation reactions?
I am confused on how cations and anions are involved in the reaction?
I am overall confused in general ???

Cations and anions are indeed involved in precipitation reactions. A precipitation reaction occurs when a positive ion (i.e., a cation) and a negative ion (i.e., an anion) combine to form an insoluble salt (the precipitate).

e.g., Ag+(aq) + Cl-(aq) -> AgCl(s)
Post by: RuiAce on August 18, 2016, 10:43:07 pm
Cations and anions are indeed involved in precipitation reactions. A precipitation reaction occurs when a positive ion (i.e., a cation) and a negative ion (i.e., an anion) combine to form an insoluble salt (the precipitate).

e.g., Ag+(aq) + Cl-(aq) -> AgCl(s)
That being said, this is an example of an ionic equation.
Post by: anotherworld2b on August 19, 2016, 01:05:18 am
That being said, this is an example of an ionic equation.

Then how would you do this question? Im confused on what anions and cations to join together?  :-\
Post by: RuiAce on August 19, 2016, 08:03:13 am
Then how would you do this question? Im confused on what anions and cations to join together?  :-\
This requires you to know what's called your solubility rules.

You need to identify which combination of ions will not dissolve in water. That is, those that are insoluble in water.

Take the first one. According to your solubility rules, Na+, NH4+ and K+ are always soluble. Leaving behind the option of Mg2+

CH3COO- and NO3- are both soluble always. This leaves Cl- and CO32-

But MgCl2 is soluble. In fact, CO32- is ALWAYS INSOLUBLE unless paired with a group 1 metal ion or the ammonium ion.

Hence the precipitate is MgCO3

So we can build our equation by nothing more than looking at what we have:
MgCl2(aq) + (NH4)2CO3(aq) -> 2 NH4Cl(aq) + MgCO3(s)
Post by: anotherworld2b on August 20, 2016, 02:08:08 am
Thank you very much rui ace  :D
This requires you to know what's called your solubility rules.

You need to identify which combination of ions will not dissolve in water. That is, those that are insoluble in water.

Take the first one. According to your solubility rules, Na+, NH4+ and K+ are always soluble. Leaving behind the option of Mg2+

CH3COO- and NO3- are both soluble always. This leaves Cl- and CO32-

But MgCl2 is soluble. In fact, CO32- is ALWAYS INSOLUBLE unless paired with a group 1 metal ion or the ammonium ion.

Hence the precipitate is MgCO3

So we can build our equation by nothing more than looking at what we have:
MgCl2(aq) + (NH4)2CO3(aq) -> 2 NH4Cl(aq) + MgCO3(s)
Post by: onepunchboy on August 20, 2016, 12:59:21 pm

What are the tests for the products?
Halp
Post by: RuiAce on August 20, 2016, 01:21:22 pm

What are the tests for the products?
Halp
In the first-hand investigation, the most convenient way to demonstrate electrolysis is by means of a Hoffman Voltameter.

We need to keep in mind the half equations that occur during the electrolysis.

For concentrated NaCl solution, the following half equations occur:
Anode: 2 Cl- -> Cl2(g) + 2e-
Cathode: 2 H2O(l) + 2e- -> H2(g) + 2 OH-(aq)

The presence of the chlorine gas at the anode can be tested using litmus paper. The chlorine gas will bleach the litmus, causing it to turn pale.
The presence of the hydrogen gas at the cathode can be tested using the pop test. Set the hydrogen on fire and anticipate a pop.
The presence of the hydroxide ions at the cathode can be tested using an appropriate indicator. Pour an appropriate indicator into a portion of the liquid produced at the cathode, and observe colour changes.

For dilute NaCl solution, we have the electrolysis of water. Whilst the cathode equation is the same, the anode equation is changed.
Anode: 2 H2O(l) -> O2(g) + 4 H+ + 4e-

The presence of the hydrogen ions at the anode can also be tested using an appropriate indicator.
The presence of the abundance of oxygen can be tested using the burning splint test. Raise a burning splint over the region with excess oxygen. The splint lies on fire itself.

Further info on the burning splint test can be researched on Google.
Post by: onepunchboy on August 20, 2016, 01:42:51 pm
Thanks rui !, just another one (http://uploads.tapatalk-cdn.com/20160820/0172b99b7a155eef262038829afb9f66.jpg)

I cant think of any other reasons for 6 marks other than conc. Nacl requires less voltage and produces more favourable products..
Halp
Post by: anotherworld2b on August 20, 2016, 02:48:07 pm
I am able to do simple precipiation reactions with 2 solutions but how you do this question?
Post by: anotherworld2b on August 20, 2016, 07:30:33 pm
I have a prac test that is about using precipiation reactions to identify 4 unknown solutions. We will be given the names of the solutions but we need to allocate the right name to the  correct solution. I was wondeing what would be the best way to approach doing this prac?
Post by: Happy Physics Land on August 20, 2016, 11:02:01 pm
Thanks rui !, just another one (http://uploads.tapatalk-cdn.com/20160820/0172b99b7a155eef262038829afb9f66.jpg)

I cant think of any other reasons for 6 marks other than conc. Nacl requires less voltage and produces more favourable products..
Halp

Hey Onepunchboy!

The chlor-alkali industry favours the use of concentrated NaCl(aq) over molten NaCl(l) because NaOH is a favourable product that can only be produced using concentrated brine and it has a major role in industries that involve the manufacturing of soaps, papers and synthetic fibres. In the electrolysis of Molten NaCl, the products are Na(l) and Cl(g). Whilst Na(l) does have some importance in the manufacture of indigo, pharmaceuticals and petrol additives, the use of NaOH in industry is much more diverse and dominant. In addition, like what you have correctly justified, it takes an energy input of 4.07V for the electrolysis of molten NaCl to take place (2Cl- ——> Cl2(g) + 2e-, Eox= -1.36V. Na+ + e- ---> Na, Ered = -2.71V) whilst it only takes an energy input of 2.19V for the electrolysis of aqueous NaCl. Consequently, chlor-alkali companies can effective reduce energy consumptions, reduce the cost involved and maximise their profits.
Post by: Happy Physics Land on August 20, 2016, 11:09:57 pm
I am able to do simple precipiation reactions with 2 solutions but how you do this question?

Hey Anotherworld2b!

This question requires some familiarity with the precipitation rules and observation of the compounds. If we look carefully, amongst all the compounds, potassium, sodium, ammonium and nitrate compounds would all be soluble according to our solubility rules. Sulphide is not included as a part of our solubility rules and in fact, sulphide is unable to form a precipitate compound with any of those cations. So this leaves us with lead and chloride. If you are experienced with these sorts of questions and familiar with solubility rules, you should immediately recognise that lead (II) chloride would produce a white coloured precipitate. So the ionic equation for the formation of this precipitate would be as follow:

Pb2+(aq) + 2Cl-(aq) -----> PbCl2(s)
Post by: Happy Physics Land on August 20, 2016, 11:27:38 pm
I have a prac test that is about using precipiation reactions to identify 4 unknown solutions. We will be given the names of the solutions but we need to allocate the right name to the  correct solution. I was wondeing what would be the best way to approach doing this prac?

Hey Anotherworld2b!

You definitely asked the right person, because I did the same prac exam as you in term 1 with 5 unknown solutions. I scored 25/25 in that exam and this one simple table trick I will show you below has really helped me! So essentially how it works is you construct two tables. You will be given on the day what all the four chemicals are, except you dont really know what each individual chemical is. So on one table, you will be constructing a table like what I have shown below, with your different unknown chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your experimental data.

(http://imgur.com/K9SkvMS.png)

On another table, you will be constructing a table like what I have also shown below, with your different KNOWN chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your theoretical data obtained through your solubility rules.

(http://imgur.com/Yb9LQIS.png)

So now what we do is we match the outcomes. For instance, whichever chemical in the first table that is able to form 2 precipitates should be matched to whichever chemical in the second table that is also able to form 2 precipitates. This way we can identify that the two chemicals are the same and hence identify the unknown chemical as I have shown below.

(http://imgur.com/h8aaklC.png)

But now, sometimes you can get situations where you have 2 chemicals from the first table that can form equal amounts of precipitates. Teachers shouldnt give you this sort of situations but if this situation does happen, normally it would not matter which one you would identify as the correct chemical. But, there can be special cases, like what I have shown below, that can help you to determine the correct chemical when two particular chemicals can form equal amounts of precipitates.

(http://imgur.com/X716dG7.png)

I hope my guide helps, and I certainly really enjoyed writing this explanation and I really hope you can achieve a high mark by following what I suggested above!

Best Regards
Happy Physics Land
Post by: anotherworld2b on August 21, 2016, 01:02:41 am
thank you very much for your guide Happy Physics Land link  ;D
I had no idea how to even start preparing for this practical but now I feel I have gathered a wealth of confidence :D
I have some questions to ask if that's okay to make sure I thoroughly understand

So basically for the first table you physically add the chemicals together and observe whether there is a precipitate or not.
eg. A+B = NR. Where A is one unknown and B is another unknown chemical. Then repeat for the other parts of the table?
Then for the parts of table where its AA, BB and ect do you leave that empty because your testing for example chemical A with chemical A? Which is done for the second table as well?
Would making a table of some kind to write down the appearance of the reaction such as the colour of the precipitate be necessary as well? How would you know if a precipitate is slightly soluble?
I am also a bit confused about the last part about identifying the chemical when there are two particular chemicals can form equal amounts of precipitates.
I was wondering what other things did you have to do in your prac exam apart from identifying the solutions?
I apologise for asking so many questions and would really like to say that I really appreciate your help

Hey Anotherworld2b!

You definitely asked the right person, because I did the same prac exam as you in term 1 with 5 unknown solutions. I scored 25/25 in that exam and this one simple table trick I will show you below has really helped me! So essentially how it works is you construct two tables. You will be given on the day what all the four chemicals are, except you dont really know what each individual chemical is. So on one table, you will be constructing a table like what I have shown below, with your different unknown chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your experimental data.

(http://imgur.com/K9SkvMS.png)

On another table, you will be constructing a table like what I have also shown below, with your different KNOWN chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your theoretical data obtained through your solubility rules.

(http://imgur.com/Yb9LQIS.png)

So now what we do is we match the outcomes. For instance, whichever chemical in the first table that is able to form 2 precipitates should be matched to whichever chemical in the second table that is also able to form 2 precipitates. This way we can identify that the two chemicals are the same and hence identify the unknown chemical as I have shown below.

(http://imgur.com/h8aaklC.png)

But now, sometimes you can get situations where you have 2 chemicals from the first table that can form equal amounts of precipitates. Teachers shouldnt give you this sort of situations but if this situation does happen, normally it would not matter which one you would identify as the correct chemical. But, there can be special cases, like what I have shown below, that can help you to determine the correct chemical when two particular chemicals can form equal amounts of precipitates.

(http://imgur.com/X716dG7.png)

I hope my guide helps, and I certainly really enjoyed writing this explanation and I really hope you can achieve a high mark by following what I suggested above!

Best Regards
Happy Physics Land
Post by: cajama on August 21, 2016, 09:33:07 pm
Hi I feel like these kinds of questions should be simple since its a multiple choice q. but I still struggle to answer them. Please help! (Question 19 - 2015 HSC Chem. Exam | Screenshots of the question attached)
Post by: jakesilove on August 21, 2016, 09:54:38 pm
Hi I feel like these kinds of questions should be simple since its a multiple choice q. but I still struggle to answer them. Please help! (Question 19 - 2015 HSC Chem. Exam | Screenshots of the question attached)

Hey!

Firstly, we know that the molar mass of Lead Chloride is 278.1g/mol. Therefore, we have
$n=\frac{13}{278.1}=0.0467 moles$

Since there is one mole of Lead in every mole of Lead Chloride, we have equal as many moles of Lead! We can therefore calculate the concentration in the original sample

$C=\frac{0.0467}{0.05}=0.0093 mol L^{-1}$

And that's that! Somewhere there my order of magnitude is a bit off, but that doesn't matter; clearly, the answer is C. Hope this helps!
Post by: cajama on August 21, 2016, 10:49:16 pm
Hey!

Firstly, we know that the molar mass of Lead Chloride is 278.1g/mol. Therefore, we have
$n=\frac{13}{278.1}=0.0467 moles$

Since there is one mole of Lead in every mole of Lead Chloride, we have equal as many moles of Lead! We can therefore calculate the concentration in the original sample

$C=\frac{0.0467}{0.05}=0.0093 mol L^{-1}$

And that's that! Somewhere there my order of magnitude is a bit off, but that doesn't matter; clearly, the answer is C. Hope this helps!

Ahh I see now but how do we know how many moles of lead there are for every lead chloride. Is it because in PbCl2 there is only one mole of lead in the molecule?
Where as, say if the precipitate was for whatever reason Pb3(PO4)2, there would be 1/3 moles for every mole of Pb3(PO4)2?
Post by: jakesilove on August 21, 2016, 10:54:36 pm
Ahh I see now but how do we know how many moles of lead there are for every lead chloride. Is it because in PbCl2 there is only one mole of lead in the molecule?
Where as, say if the precipitate was for whatever reason Pb3(PO4)2, there would be 1/3 moles for every mole of Pb3(PO4)2?

Almost! For every mole of Pb3(PO4)2, there are actually 3 mole-s of Pb! The little number is the number of 'moles' within the compound. So, for PbCl2, there is two moles of Cl and one mole of Pb. Makes sense?
Post by: cajama on August 21, 2016, 11:03:17 pm
Almost! For every mole of Pb3(PO4)2, there are actually 3 mole-s of Pb! The little number is the number of 'moles' within the compound. So, for PbCl2, there is two moles of Cl and one mole of Pb. Makes sense?

Oh yeah I get it completely now. Thank you so much!
Post by: conic curve on August 24, 2016, 05:25:00 am
Does the pressure a substance exerts on the container increase as it goes from solid to liquid or gas? Or is it the other way around?
Post by: RuiAce on August 24, 2016, 08:21:54 am
Does the pressure a substance exerts on the container increase as it goes from solid to liquid or gas? Or is it the other way around?
That way is correct.

Pressure only ever relates to the quantity of gaseous substances present in a (generally speaking, closed) system. Liquids and solids are not counted, so if gas is converted into liquid/solid we go the other way and decrease pressure.
Post by: massive on August 24, 2016, 04:46:31 pm
hey guys, how do you do this :S (answer is d btw)
Post by: RuiAce on August 24, 2016, 05:39:22 pm
hey guys, how do you do this :S (answer is d btw)
Because our system is constantly at 25oC, 100kPa, observe that VM = 24.79 L mol-1 is constant.

Since n = V/VM, because the molar volume is fixed the reaction that produces the greatest moles of gas will also produce the largest volume.

Write out the relevant chemical equation for each and every one of your reaction to determine the relevant moles. Keep in mind that as HCl is the excess reagent, it does not matter how many moles of it we use.

A) CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)
nCaCO3 = 50/100.9 = 0.49554... mol
nCO2 = nCaCO3 = 0.4955... mol

B) Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g)
nZn = 50/65.38 = 0.7647... mol
nH2 = nZn = 0.7647... mol

C) Na2CO3(s) + 2 HCl(aq) -> 2 NaCl(aq) + CO2(g) + H2O(l)
nNa2CO3 = 50/105.99 = 0.4717... mol
nCO2 = nNa2CO3 = 0.4717... mol

D) 2 Na(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2(g)
nNa = 50/22.99 = 2.1748... mol
nH2 = 1/2 nNa = 1.0874... mol

By comparison, the greatest moles of gas produced occurs in reaction D.

Note that the purpose of writing all the equations is to identify the mole ratio. The gas produced itself doesn't matter because we are talking about VOLUME, not mass.
Post by: massive on August 24, 2016, 05:46:37 pm
OOOHH i get it now thanks!

Theres also one other thing i don't get- its about equilibrium. I understand the whole lcp stuff about temperature and pressure but i don't get everything about concentration, i.e. when you add things to the system.

For q9 i know that the answer is b, i was just wondering if part c and d have any effect on the equilibrium.

Also for 10, this is what i was talking about the concentration stuff, how do you do it
Post by: massive on August 24, 2016, 06:21:20 pm

$C=\frac{0.0467}{0.05}=0.0093 mol L^{-1}$

Hey Jake why do you use 50ml for the volume? they also say that 25ml of sodium chloride was used, don't you add the volumes together :S
Post by: RuiAce on August 24, 2016, 07:03:13 pm
OOOHH i get it now thanks!

Theres also one other thing i don't get- its about equilibrium. I understand the whole lcp stuff about temperature and pressure but i don't get everything about concentration, i.e. when you add things to the system.

For q9 i know that the answer is b, i was just wondering if part c and d have any effect on the equilibrium.

Also for 10, this is what i was talking about the concentration stuff, how do you do it
With Q9, C and D would end up shifting the equilibrium to the left because you introduce more product, not reactant.

So.

If there is an excess of either reactant or product, then the equilibrium has been disturbed (or it was never at equilibrium). Therefore, according to LCP, the system wants to get rid of that EXCESS.

Conversely, if there is a shortage of either reactant or product, then LCP predicts that the system wants to bring in more of what's lacked to address the shortage. This is how equilibrium is (re-)established.

So for Q10, where the effect is not immediately obvious, you must analyse what happens.

You introduced hydrochloric acid. Obviously it's not going to react with water or the ammonium ion here. Does it react with ammonia?
No. The HCl will have a tendency to react with that hydroxide ion, which is clearly a base.

HCl + OH- -> Cl- + H2O

So since we take out OH- through the introduction of the acid, the equilibrium will shift to the right.

Now, immediately I can tell this question is beyond the HSC's scope, however it's not unreasonable. This is because of unnecessarily introducing option C: Equation is driven to completion.

A few drops of HCl is probably insufficient to do so. In practicality? Yes, this is possible if you had an unlimited amount of HCl. This is because you end up introducing so much acid that all the base ends up neutralised, and you have excess acid instead.

So for this question? It should be A.
Post by: massive on August 24, 2016, 07:15:32 pm

So for Q10, where the effect is not immediately obvious, you must analyse what happens.

You introduced hydrochloric acid. Obviously it's not going to react with water or the ammonium ion here. Does it react with ammonia?
No. The HCl will have a tendency to react with that hydroxide ion, which is clearly a base.

HCl + OH- -> Cl- + H2O

So since we take out OH- through the introduction of the acid, the equilibrium will shift to the right.

wait so are we just meant to learn this? is this part of module 2? and yeah the answer is A
Post by: RuiAce on August 24, 2016, 07:28:53 pm
wait so are we just meant to learn this? is this part of module 2? and yeah the answer is A
You were already taught it. It's just a process of deduction. And yes of course it counts under The Acidic Environment
Post by: massive on August 24, 2016, 11:15:30 pm
Guys how do you do part b for the question attached. I was just confused because there is no equation, so how do you use mole ratio comparison like you do if you want to figure out the percent of sulfate in fertilisers.
Post by: jakesilove on August 25, 2016, 10:57:20 am
Guys how do you do part b for the question attached. I was just confused because there is no equation, so how do you use mole ratio comparison like you do if you want to figure out the percent of sulfate in fertilisers.

You don't need any sort of equation; what you're given is plenty. First, using the molar mass of the substance and the mass you've been given, find the moles. Then, recall that for a substance

$X_nY_m$

that this means that for every one mole of the substance, there are n moles of X and m moles of Y. So, for one mole of the substance in this question, there will be 2 moles of phosphorus. I'll let you do the maths
Post by: anotherworld2b on August 25, 2016, 11:14:33 am
I was wondering would my answer to this question be correct?

PF5 abd PH3 have different shapes because of the structure and cimposition of molecukes each one contains. This determines which intermolecular forces are present,  and these forces determine the physical properties of the material

I also wanted to ask what is the difference between a dipole, polar bond and net dipole?
Post by: Happy Physics Land on August 26, 2016, 06:05:19 pm
I was wondering would my answer to this question be correct?

PF5 abd PH3 have different shapes because of the structure and cimposition of molecukes each one contains. This determines which intermolecular forces are present,  and these forces determine the physical properties of the material

I also wanted to ask what is the difference between a dipole, polar bond and net dipole?

Hey Anotherworld2b!

You are certainly right! PF5 and PH3 do have different shapes because of the structure and composition of the molecules! But this would be a little too general as an answer. When you are asked questions of a similar kind, what you need to include in your answer in "VSEPR principles" and "electronegativity". So according to VESPR rules, molecules are arranged in particular 3D shapes that aim to keep the valence electrons on elements as far apart as possible (since electrons repel each other) in order to maintain a stable structure. The main differences between PF5 and PH3 here are the number of elements involved and the contrast between fluorine's electronegativity and hydrogen's electronegativity. Because there are 5 fluorine elements, the molecule would take on a bitrigonal pyramidal structure with bond angles of 120 degrees and 90 degrees in order to fit all these fluorine atoms in a stable arrangement (more closely packed). In PH3, because there is only 3 hydrogen atoms, it takes on a trigonal pyramidal shape of 120 degrees between all hydrogen atoms (more loosely packed). Because fluorine has a high electronegativity, or a high affinity towards electrons, they must be kept considerably far away from each other to prevent the decomposition of the structure as fluorine tries to gain other electrons in the molecule. These factors all influence the shape and composition of PF5 and PH3.
Post by: Happy Physics Land on August 26, 2016, 06:16:47 pm
I was wondering would my answer to this question be correct?

PF5 abd PH3 have different shapes because of the structure and cimposition of molecukes each one contains. This determines which intermolecular forces are present,  and these forces determine the physical properties of the material

I also wanted to ask what is the difference between a dipole, polar bond and net dipole?

A dipole is where there is a pair of temporary opposite partial charges on opposite sides of a molecule.
A polar bond is the dipole-dipole interaction (for instance, the partial negative charge on one molecule and the partial positive charge on another molecule) between polar molecules.
A net dipole is essentially just a net partial charge. A substance without a net dipole would often be symmetrical in shape and bondings. A substance with a net dipole would have different electronegativities at the head and tail of the molecule.
Post by: massive on August 26, 2016, 11:46:36 pm
guys just a quick question, i don't really understand this working out:
For part iii of the q attached, why is the answer just 0.16g/L. Isn't that meant to be 0.16g/100ml and then it becomes 1.6g/L???
thnx
Post by: anotherworld2b on August 27, 2016, 12:58:27 am
Hi i was wondering if i could get my current answers checked and get help with the ones left blank.
Also whats the difference between ionisation and dissociation? They seem to be the same thing to me  :-\
Post by: anotherworld2b on August 27, 2016, 05:43:04 pm
The other part because the attachment was too large
Post by: Happy Physics Land on August 27, 2016, 10:20:00 pm
Hi i was wondering if i could get my current answers checked and get help with the ones left blank.
Also whats the difference between ionisation and dissociation? They seem to be the same thing to me  :-\

Hey anotherworld2b!

Your colour identification for CuSO4, CuCO3 and potassium manganate are incorrect. CuSO4 should have a blue colour and CuCO3 under normal conditions should have a green colour. KMnO4 is a weird one but it should have a purple colour.

In the context of chemistry, ionisation and dissociation can be used interchangeably, they both mean the splitting up of a compound into its component ions when its in an aqueous form.
Post by: Happy Physics Land on August 27, 2016, 10:33:01 pm
The other part because the attachment was too large

In your exams I'm sure you wont be asked about most of these chemicals listed in your second post - you are more likely to be asked about the colour of the common types of precipitates. Cr(NO3)3 (chromium nitrate) is usually dark-violet in colour. The mixture of CrCl3 and Mn(NO3)2 is a little more tricky. Basically all it asks you to do is to find the colour of CrCl3 solution and the colour of Mn(NO3)2 solution and find the resultant colour of the mixture of two colours. The colour of CrCl3 is violet and the colour of Mn(NO3)2 is white. So mixing the two colours together, the resulting mixture should be a light purple colour.
Post by: anotherworld2b on August 28, 2016, 01:10:00 pm
thank you very much for your help
I have another question if that's okay
I am really confused on how to know when to use which concentration unit and how to convert from one to the other?
- %v/v   %w/w and % v/w

Could I get advice before tomorrow? I have a chem test on this tomorrow  :-\
Post by: RuiAce on August 28, 2016, 01:11:53 pm
thank you very much for your help
I have another question if that's okay
I am really confused on how to know when to use which concentration unit and how to convert from one to the other?
- %v/v   %w/w and % v/w

Could I get advice before tomorrow? I have a chem test on this tomorrow  :-\
That's odd. You're going to have to provide questions that interrelate between %v/v and %w/w

Also dissolution is actually dissolving something. If you dissolve Na2CO3 you're not exactly "ionising" because the transfer of electrons has already happened. You're just splitting the molecule apart.

As opposed to reacting sodium with chlorine gas: 2 Na + Cl2 -> 2 NaCl. This is ionising
Post by: RuiAce on August 28, 2016, 01:17:44 pm
guys just a quick question, i don't really understand this working out:
For part iii of the q attached, why is the answer just 0.16g/L. Isn't that meant to be 0.16g/100ml and then it becomes 1.6g/L???
thnx
Yeah. They should've just considered concentration in g/L instead of be like mass per litre = ...

Post by: anotherworld2b on August 28, 2016, 02:53:25 pm
I believe b interrelates %v/v and %w/w but im not sure
How would you tell when to use a particular concentration unit?
Post by: jakesilove on August 28, 2016, 03:18:13 pm
I believe b interrelates %v/v and %w/w but im not sure
How would you tell when to use a particular concentration unit?

Parts per million is the equivalent of

$ppm= mg/L = mg/kg$

So, you can relate ppm and %w/w (as mg/kg is w/w, you may just need to convert units!)
Post by: massive on August 29, 2016, 03:31:11 pm
Guys how do you do this multiple choice question. And also in general how do you decide what electrolyte you want to use??
Post by: massive on August 29, 2016, 05:04:21 pm
Guys the answer is D for this question, but couldn't it be A as well???
Post by: jakesilove on August 29, 2016, 05:43:12 pm
Guys how do you do this multiple choice question. And also in general how do you decide what electrolyte you want to use??

Hey! For the first question, you just sort of need to know that when you use a platinum electrode (ie. an inert electrode), we generally use a substance like HCl that contains Hydrogen (sometimes, we just use water) which get reduced to form Hydrogen gas. I'm not sure of the best explanation here, except to say that none of the other answers make sense, and we don't have the required metal in the cathode itself.
Post by: jakesilove on August 29, 2016, 05:46:40 pm
Guys the answer is D for this question, but couldn't it be A as well???

As far as I can tell, the answer could also be A. The equilibrium should shift away from any increase in concentration. Not sure why the answer is D, and not A, although D is definitely correct
Post by: RuiAce on August 29, 2016, 06:00:18 pm
Guys the answer is D for this question, but couldn't it be A as well???
As far as I can tell, the answer could also be A. The equilibrium should shift away from any increase in concentration. Not sure why the answer is D, and not A, although D is definitely correct
Does the fact that the reactant is a solid play a role?
Post by: jyce on August 31, 2016, 12:26:32 am
Does the fact that the reactant is a solid play a role?

Yes, I believe it does. Adding more CaCO3(s) does not change its concentration, as it's a solid. Therefore, adding more CaCO3 does not cause a shift in the equilibrium position, leaving D as the correct option.
Post by: massive on August 31, 2016, 12:40:41 am
Yes, I believe it does. Adding more CaCO3(s) does not change its concentration, as it's a solid. Therefore, adding more CaCO3 does not cause a shift in the equilibrium position, leaving D as the correct option.

Wait what, isn't the concentration still increasing because technically there's more reactants and the system wants to counteract this by decreasing the reactants, thus it shifts right ? :S
Post by: Jakeybaby on August 31, 2016, 12:55:23 am
Wait what, isn't the concentration still increasing because technically there's more reactants and the system wants to counteract this by decreasing the reactants, thus it shifts right ? :S
The concentration of the solids are constant, only (aq) and (g) are in the equilibrium expressions. You can exclude pure solids and pure liquids from the equilibrium expressions.
Post by: RuiAce on August 31, 2016, 07:38:36 am
Yes, I believe it does. Adding more CaCO3(s) does not change its concentration, as it's a solid. Therefore, adding more CaCO3 does not cause a shift in the equilibrium position, leaving D as the correct option.
What I thought. I remember being taught this but it was a blur.
Wait what, isn't the concentration still increasing because technically there's more reactants and the system wants to counteract this by decreasing the reactants, thus it shifts right ? :S
When you add more moles of solid, you add the same amount of volume of the solid to counteract a potential increase in concentration. Given C=n/V, if n and V go up in the exact same proportion, the concentration is not affected.

Note that it is the CONCENTRATION of the substance which shifts the equilibrium, not the actual quantity (moles)
Post by: massive on September 01, 2016, 02:29:57 am
Hey guys, could someone explain how to do hydrolysis of salts to me, i still don't get how to do them :/ (like determining whether the salt is acid or base)
Thanks heaps!
Post by: conic curve on September 01, 2016, 10:43:26 am
Hey guys, could someone explain how to do hydrolysis of salts to me, i still don't get how to do them :/ (like determining whether the salt is acid or base)
Thanks heaps!

I don't know whether or not this is useful, but maybe you could take a look at this: http://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Equilibria/Solubilty/Hydrolysis

Hopes that helps  :D
Post by: RuiAce on September 01, 2016, 01:35:29 pm
Hey guys, could someone explain how to do hydrolysis of salts to me, i still don't get how to do them :/ (like determining whether the salt is acid or base)
Thanks heaps!
When something is dissolved in water, the ions that form the compound are dissociated (e.g. Na2CO3(s) -> 2 Na+(aq) + CO32-(aq)). To determine the nature of the dissolved substance as acidic or basic, we need to look at the ions that are formed.

Take this example first. CO32- is basic. However Na+ is a neutral cation. Therefore the resulting solution is basic and the original salt was also basic.

Take NH4Cl. NH4+ is acidic, however Cl- is a neutral cation. Therefore we have something acidic.

If we take NaCl, then both Na+ and Cl- are neutral so the result is neutral.

Simply put, analyse the cation and anion separately.
Post by: anotherworld2b on September 02, 2016, 01:29:33 am
How would you solve question b ii and c?
Post by: jakesilove on September 02, 2016, 10:36:48 am
How would you solve question b ii and c?

Hey! The formula we need for these questions is

$C_1V_1=C_2V_2$

For i)

$10(5)=200(C_2)$
$C_2=0.25mol/L$

For ii), we WANT a certain volume and concentration out, and we HAVE an initial concentration. The only thing we don't have is the initial volume, which is perfect, because that's exactly how the formula works!

$0.25(V_1)=(100)(0.1)$
$V_1=80mL$

For c), again it's just a simply application of the formula. Remember than C1 is your initial concentration, V1 is your initial volume, C2 is your final concentration, and V2 is your final volume.
The trick to this solution is that the TOTAL final volume is going to be 10ml+55mL=65mL

$C_1(10)=(0.48)(65)$
$C_1=3.12mol/L=3.1mol/L (2 sig figs)$
Post by: anotherworld2b on September 02, 2016, 07:05:09 pm
Thank you
Hey! The formula we need for these questions is

$C_1V_1=C_2V_2$

For i)

$10(5)=200(C_2)$
$C_2=0.25mol/L$

For ii), we WANT a certain volume and concentration out, and we HAVE an initial concentration. The only thing we don't have is the initial volume, which is perfect, because that's exactly how the formula works!

$0.25(V_1)=(100)(0.1)$
$V_1=80mL$

For c), again it's just a simply application of the formula. Remember than C1 is your initial concentration, V1 is your initial volume, C2 is your final concentration, and V2 is your final volume.
The trick to this solution is that the TOTAL final volume is going to be 10ml+55mL=65mL

$C_1(10)=(0.48)(65)$
$C_1=3.12mol/L=3.1mol/L (2 sig figs)$
Post by: onepunchboy on September 03, 2016, 07:33:10 pm

Yo can anyone help me out on q 16 and 17 ?
Thankss
Post by: jakesilove on September 03, 2016, 10:12:39 pm

Yo can anyone help me out on q 16 and 17 ?
Thankss

Not entirely sure for Q16; is the answer supposed to be A?

For Q17, let's think about what's actually going on. If we had chosen the correct indicator, the endpoint would have been somewhere in the 8-10 range (strong base, weak acid). Therefore, the pH should start on the lower end (ie. pH of 2-3, due to the weak acid) and strong base should be added until the pH is 8-10. However, if the indicator will change colors when the pH reaches 3-4, then much LESS of the base will be required! The greater the pH change, the more base would be required. Since we've used LESS base than expected to neutralise the acid, it will be expected that the acid is WEAKER. Therefore, I believe that the answer is B
Post by: jakesilove on September 03, 2016, 10:15:18 pm

Yo can anyone help me out on q 16 and 17 ?
Thankss

As far as I can tell, the Cathode should be positive as electrons flow TOWARDS it (remember, Reduction It Gains (RIG)). This won't come 'from' the electrolyte, but rather from the anode. If you consider the anode to be the external circuit, I imagine the answer is D
Post by: massive on September 04, 2016, 02:52:11 pm
Hey guys, i was doing a question for my option topic (industrial chem) and the question was to find the equilibrium constant (K). However the question didn't give the concentrations of the reactants and products in mol/L but rather gave it in kPa. When instructed to find K, the answers just substituted these kPa values into the formula. I was just confused, i thought you could only sub the concentration that are in mol/L. :S
Post by: jakesilove on September 04, 2016, 03:14:03 pm
Hey guys, i was doing a question for my option topic (industrial chem) and the question was to find the equilibrium constant (K). However the question didn't give the concentrations of the reactants and products in mol/L but rather gave it in kPa. When instructed to find K, the answers just substituted these kPa values into the formula. I was just confused, i thought you could only sub the concentration that are in mol/L. :S

Since the formula relies on ratios, if everything is in a gaseous form, the relative pressures will be equivalent to concentration. This assumes ideal gas etc etc. but we can assume all of that in a question like this. For instance, if I combined 1kPa of Nitrogen and 2kPa of Hydrogen, because the pressure is across a specific unit area (which is the same in both measurements), and because Gas will occupy a constant space per mole regardless of the chemical composition, we can assume that Hydrogen is 'twice as concentrated'. I didn't do Industrial, and I'm sure there are way more complicated reasons why this trick works, but basically if you ever see another question like this just use the kPa measurements! May not make 100% sense, just turns out to be what you need to do.
Post by: RuiAce on September 04, 2016, 03:19:45 pm
Since the formula relies on ratios, if everything is in a gaseous form, the relative pressures will be equivalent to concentration. This assumes ideal gas etc etc. but we can assume all of that in a question like this. For instance, if I combined 1kPa of Nitrogen and 2kPa of Hydrogen, because the pressure is across a specific unit area (which is the same in both measurements), and because Gas will occupy a constant space per mole regardless of the chemical composition, we can assume that Hydrogen is 'twice as concentrated'. I didn't do Industrial, and I'm sure there are way more complicated reasons why this trick works, but basically if you ever see another question like this just use the kPa measurements! May not make 100% sense, just turns out to be what you need to do.
Oh dear. Seems unfair to throw that on them. Yeah that's abnormal for the HSC
Post by: massive on September 04, 2016, 04:03:47 pm
Thanks Jake, just one more q, Why is it when pressure is increased the concentrations of everything increases instantaneously but then equilibrium shifts to side with fewer moles of gases. Also does this instant increase happen with any of the other variables (e.g. temperature or concentration). What i mean is, if the temperature is increased, does everything else increase suddenly then equilibrium shifts right or left (depending if exo/endo) and everything decreases or increases according to this new equilibrium position?
Post by: RuiAce on September 04, 2016, 04:09:27 pm
Thanks Jake, just one more q, Why is it when pressure is increased the concentrations of everything increases instantaneously but then equilibrium shifts to side with fewer moles of gases. Also does this instant increase happen with any of the other variables (e.g. temperature or concentration). What i mean is, if the temperature is increased, does everything else increase suddenly then equilibrium shifts right or left (depending if exo/endo) and everything decreases or increases according to this new equilibrium position?
The concentration of the solids are constant, only (aq) and (g) are in the equilibrium expressions. You can exclude pure solids and pure liquids from the equilibrium expressions.
What I thought. I remember being taught this but it was a blur.When you add more moles of solid, you add the same amount of volume of the solid to counteract a potential increase in concentration. Given C=n/V, if n and V go up in the exact same proportion, the concentration is not affected.

Note that it is the CONCENTRATION of the substance which shifts the equilibrium, not the actual quantity (moles)
Temperature is different. Temperature depends on your enthalpy change value, that is, your value for ΔH. The value for ΔH has nothing to do with the state your substances are in, be it solid, liquid or gaseous.

If you mean the actual concentrations, which are best depicted on a graph, then no, only changes in concentrations spike the graph up. This is because a change in temperature is nothing but a change in temperature - we are not introducing or extracting a substance relative to the volume of the vessel. Pressure or increases in concentration disturbs the ratio of substances (moles) to the volume of the vessel (volume, and note C=n/V), whereas temperature does not.

Note that the more gradual changes later are just a consequence of Le Chatelier's principle. The spikes are when we suddenly change concentrations in a matter of a second, whereas what comes next is the system trying to shift it's equilibrium to counter the change. (Once again - temperature does NOT forcibly disturb what's in there immediately, as opposed to e.g. taking out 2L of hydrogen.)
Post by: massive on September 04, 2016, 04:15:16 pm
We had this discussion on pressure already.Temperature is different. Temperature depends on your enthalpy change value, that is, your value for ΔH. The value for ΔH has nothing to do with the state your substances are in, be it solid, liquid or gaseous.

If you mean the actual concentrations, which are best depicted on a graph, then no, only changes in concentrations spike the graph up. This is because a change in temperature is nothing but a change in temperature - we are not introducing or extracting a substance relative to the volume of the vessel. Pressure or increases in concentration disturbs the ratio of substances (moles) to the volume of the vessel (volume, and note C=n/V), whereas temperature does not.

Note that the more gradual changes later are just a consequence of Le Chatelier's principle. The spikes are when we suddenly change concentrations in a matter of a second, whereas what comes next is the system trying to shift it's equilibrium to counter the change. (Once again - temperature does NOT forcibly disturb what's in there immediately, as opposed to e.g. taking out 2L of hydrogen.)
OHHH thanks man i get it now! btw, if the question gives the concentration of one substance we can't do mole ratio comparison using the balanced equation to get the concentration of the other species right?
Post by: RuiAce on September 04, 2016, 04:17:40 pm
OHHH thanks man i get it now! btw, if the question gives the concentration of one substance we can't do mole ratio comparison using the balanced equation to get the concentration of the other species right?
Well technically since n=CV, if you just leave V as something you don't know (like x in maths) you could carry from there.

But in general, no. Because such a method is asking too much for HSC chemistry students.
Post by: massive on September 04, 2016, 04:27:16 pm
ohk nws, thankyou. I just have one last question (hopefully)

For the equilibrium reaction below, would adding acid or base affect the equilibrium position at all?

N2O4 <---> 2NO2
Post by: RuiAce on September 04, 2016, 04:31:58 pm
ohk nws, thankyou. I just have one last question (hopefully)

For the equilibrium reaction below, would adding acid or base affect the equilibrium position at all?

N2O4 <---> 2NO2
That equilibrium is between two gases. Highly doubt an acid or base will do anything to it
Post by: jakesilove on September 04, 2016, 04:46:10 pm
That equilibrium is between two gases. Highly doubt an acid or base will do anything to it

Ditto. It would only have an effect if the substance (namely, Nitrogen Dioxide) was in aqueous form, which would have a different equilibrium anyway.
Post by: massive on September 04, 2016, 11:45:08 pm
GUYS, for dehydration of ethanol and hydration of ethylene what are the states for each species. There are some notes that say that water is gas and others that say it's liquid...so confused D:
Post by: Sine on September 04, 2016, 11:49:35 pm
GUYS, for dehydration of ethanol and hydration of ethylene what are the states for each species. There are some notes that say that water is gas and others that say it's liquid...so confused D:
water is a gas as the reaction requires such a high temperature. In VCE examiners still accept liquid though. i'm not aware of how it's marked in HSC.

EDIT: minor fix
Post by: jakesilove on September 05, 2016, 12:00:04 am
water is a gas as the reaction requires such a high temperature. In VCE examiners still accept liquid though. i'm not aware of how it's mark in HSC.

Same with the HSC. Basically, whether you put gas or liquid doesn't matter (although generally liquid is safer). Just as long as you include a state!
Post by: massive on September 05, 2016, 05:15:30 pm
Is HPO-4 and acid? Shouldn't it be a base, becuase H2SO4 is a strong acid and HPO-4 is its conjugate base ??
Post by: RuiAce on September 05, 2016, 05:38:25 pm
Is HPO-4 and acid? Shouldn't it be a base, becuase H2SO4 is a strong acid and HPO-4 is its conjugate base ??
Phosphoric acid and sulfuric acid are two different things. Sulfuric acid being acidic doesn't have anything to do with the nature of the hydrogen phosphate ion.

Presumably you meant HSO4 or something. Please clarify and then we'll get back to you
Post by: massive on September 05, 2016, 06:09:48 pm
Phosphoric acid and sulfuric acid are two different things. Sulfuric acid being acidic doesn't have anything to do with the nature of the hydrogen phosphate ion.

Presumably you meant HSO4 or something. Please clarify and then we'll get back to you

Yeah that's what i meant, my bad!
Post by: RuiAce on September 05, 2016, 08:56:04 pm
A conjugate base doesn't necessarily have to be basic in water in it's own right.

Consider the massively long sequence something like phosphoric acid goes through:

H3PO4 <-> H2PO4- + H+ <-> HPO42- + 2 H+ + PO43- + 3 H+

You can say that dihydrogen phosphate is the conjugate base of phosphoric acid. But at the same time, dihydrogen is the conjugate acid of hydrogen phosphate. So what do you reckon? Is dihydrogen phosphate an acid or a base?

The conclusion is that conjugate bases and acids are not enough to simply determine whether your substance is acidic or basic. Conjugate acids only relate B-L theory between a specific acid and a base, not on a generic scale. An amphiprotic substance can easily enough be either.

Your example is, however, unique. HSO4- exerts acidic properties in water. This is one of which you must know.
Post by: katherine123 on September 07, 2016, 04:55:41 pm
for questions involving the production of sulfuric acid process
do i need to mention the different temperature conditions and % conversion ie initially 1000 degrees, 550 degrees (97% conversion), 400 degrees (99.7% conversion)   or is there a specific compromise temperature
Post by: RuiAce on September 07, 2016, 06:16:57 pm
for questions involving the production of sulfuric acid process
do i need to mention the different temperature conditions and % conversion ie initially 1000 degrees, 550 degrees (97% conversion), 400 degrees (99.7% conversion)   or is there a specific compromise temperature
Well if you want to maximise your marks you would try to remember all three parts to as accurate as you can.

With temperature you'd want to remember that they're different for each of the three parts. Initially 1000 degrees yes but that's just for part 1.

Part 2 is where all the equilibrium principles come in - catalytic oxidation of sulfur dioxide into sulfur trioxide. I just memorised somewhere around 450 degrees C though I think. (But I did mention the first bed giving a 70% yield and the last giving a 99.7%)

It's not specific either I don't think. But it shouldn't be too far out from 450. 400 and 550 look acceptable.
Post by: zoeh on September 08, 2016, 04:07:16 pm
Hi,
Why are cyclohexene and cyclohexane used instead of just hexane and hexene when comparing reactivity of alkenes and alkanes?
Thankyouu :)
Post by: jakesilove on September 08, 2016, 04:18:47 pm
Hi,
Why are cyclohexene and cyclohexane used instead of just hexane and hexene when comparing reactivity of alkenes and alkanes?
Thankyouu :)

Hey!

This is a really good question. It's important to understand this, because I've seen multiple HSC questions asking about it!

Basically, we just need an Alkane and an Alkene to test, right? Unfortunately, most Alkenes have EXTREMELY low boiling points. Therefore, it is different to directly compare an Alkane and corresponding Alkene (because it's fairly difficult to test a gas. We want two liquids)! Cyclohexane/ene, due to its cyclical structure, 'sticks' together and remains a liquid much more easily. Additionally, it is non-reactive (won't explode upon testing) and fairly non-toxic.

Post by: zoeh on September 08, 2016, 05:24:02 pm
Hey just another question,
In the shipwrecks option, why do acidic environments accelerate corrosion?
Post by: RuiAce on September 08, 2016, 05:37:22 pm
Hi,
Why are cyclohexene and cyclohexane used instead of just hexane and hexene when comparing reactivity of alkenes and alkanes?
Thankyouu :)
May be of interest - I actually did the experiment with BOTH cyclohexene/cyclohexane and hex-1-ene/hexane. BOTH produced the same results you'd anticipate.

In fact, I don't know if Jake will disagree but I reckon BOTH are acceptable from my first-hand experience.

What we CANNOT use is something like ethane/ethylene. For reasons he pointed out.
Post by: jakesilove on September 08, 2016, 08:18:51 pm
May be of interest - I actually did the experiment with BOTH cyclohexene/cyclohexane and hex-1-ene/hexane. BOTH produced the same results you'd anticipate.

In fact, I don't know if Jake will disagree but I reckon BOTH are acceptable from my first-hand experience.

What we CANNOT use is something like ethane/ethylene. For reasons he pointed out.

I would generally stick to Cyclohexane/ene, because I've seen questions outlining the specific benefit of using that chemical.