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Author Topic: VCE Chemistry Question Thread  (Read 2312975 times)  Share 

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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8625 on: June 12, 2020, 10:33:00 am »
+5
Hey everyone, I just wanted to ask you all something and maybe for some advice. I’ve just had two SACs for Unit 1 Chem. I ended up getting a 68 and a 45... I don’t understand why though, like I did all the practice and everything, and it isn’t the “exam stress/nervousness” I do it the exact same way I do the practice but it’s still wrong. My teacher’s advice was that “you chose the hardest subject in this school, and it’s only going to get harder, I cannot do more to help other than teach it well, and I do.” What do I do? I do like the subject and I do study and everything but I do it right in the practice but completely wrong in the Tests/ exams? I still have an exam left this semester. What do I do? Should I...drop or persist and keep going ? Any advice on how to fix this problem or what I should do going forward ?

One of my favourite teachers used to always say to us, "practice doesn't perfect - perfect practice makes perfect". Now, he said this in regards to singing, but I feel it holds just as true with typical study.

You can't just do the practice questions and expect to do well. That's just practice, and all it will do is teach you how to answer those specific questions. So, perfect practice would be doing every single possible question ever? Of course not, you'd be sitting at your desk forever.

IMHO (and I encourage everyone reading this to comment as well, and say what works for you, or what you think "perfect practice" looks like), "perfect practice" is all about thinking about things physically and drawing an understanding of them. If a question asks you, "I mixed 50L of ethanol with 20L of ethanoic acid, how much ethyl ethanoate will I get?" - don't just start chugging through maths. Draw up the reaction scheme, make sure you know how ethyl ethanoate is even being made. Which part is the ethanol, and which part is the acid? What are the by-products, how are they forming? Next, what equations do you need? Can you combine them to make the calculation in one step? Is it beneficial to do it that way? Why those equations, why not other equations?

Don't just answer the question in front of you, try to answer every question that could be asked. And never start by just plugging numbers in - draw everything out. Something that I think gets lost on students a lot is that chemistry is a physical science. It's tangible, you can touch it with your own hands, and actually do this experiments. So those questions describing experiments ABSOLUTELY should be able to be drawn and understood, both on a macroscopic level (i.e., as if you were doing it), but also on the atomic level.

My understanding of chemistry fundamentally changed in university when I was forced to draw out what was happening on an atomic level. It never even mattered if I was right or wrong - by being forced to do it, I started thinking deeper, and my grades improved as a natural consequence.

So yeah, a lot there, but recommendations:
-Draw everything. Draw the atoms, draw the molecules, draw the cauldron bubbling - if you think it's relevant, draw it
-Try to understand everything as deeply as possible
-And of course, only take this advice when studying. If it helps you answer the question, draw it - but you also want to save time in an exam, so try to either draw quick, or learn to draw in your head

Ok. Thanks

For the phenol molecule how much is used annually

I couldn’t find how much is used only how much is produced

For this question
What is the purpose of using concentrated sulfuric acid in esterification reactions
A to dehydrate the system
B To catalyse the reaction
C  to be a reactant
D to hydrate the system

Why is it b

For the phenol, I got nfi, sorry man. For the sulfuric acid, the answer is honestly just - because it is. You might want to go over your esterification notes some more.

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Re: VCE Chemistry Question Thread
« Reply #8626 on: June 12, 2020, 10:42:06 am »
+5
Don't just answer the question in front of you, try to answer every question that could be asked. And never start by just plugging numbers in - draw everything out. Something that I think gets lost on students a lot is that chemistry is a physical science. It's tangible, you can touch it with your own hands, and actually do this experiments. So those questions describing experiments ABSOLUTELY should be able to be drawn and understood, both on a macroscopic level (i.e., as if you were doing it), but also on the atomic level.

This is so true! My approach to chem was very much “mmmkay so this is the formula we learnt in class and here are the numbers so let’s just chuck everything into an equation and hope for the best”. It worked well for me in maths but chem is definitely different.

My chem teacher once said that the difference between a low 40 and mid/high 40 in chem is that the former KNOWS the content, but the latter UNDERSTANDS it.

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Gogurt

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Re: VCE Chemistry Question Thread
« Reply #8627 on: June 13, 2020, 01:59:36 pm »
0
Can someone explain why the mole ratios between lead and the electrons is 1:1? I genuinely don't understand the logic behind it.
Thanks! (for d)

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8628 on: June 13, 2020, 05:14:57 pm »
0
Do you think it’s better to actually write your own set of notes or just use the atar notes/ other people’s notes.

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8629 on: June 14, 2020, 04:37:38 am »
+7
Can someone explain why the mole ratios between lead and the electrons is 1:1? I genuinely don't understand the logic behind it.
Thanks! (for d)

Might help if you think of it as 2:2 instead of 1:1. Mathematically, the ratio is the exact same, of course - but remember that the half-reaction that the lead goes through is:

Pb -> Pb2+ + 2e-

Except, there's not one lead atom in that equation - there's two of them. "But hang on", you might say. "There's two lead atoms, so don't both of them require 2 electrons???"

Well, you'd be exactly right - except, the reduction half reaction is this:

Pb4+ + 2e- -> Pb2+

So you see, the two electrons move from one lead molecule, to the OTHER lead molecule. So while both need 2 electrons, if those 2 electrons are the same, then you don't need 4 electrons for everything to happen! Hence, per equation, there are 2 lead molecules, but only 2 electrons moving around. This is more obvious if you go through the process of building the full equation yourself:

Pb -> Pb2+ + 2e-
Pb4+ + 2e- -> Pb2+

Next, add the sulfuric acid:
Pb + H2SO4 -> PbSO4 + 2e- + 2H+
Pb4+ + H2SO4 + 2e- -> PbSO4 + 2H+

Now, turn the Pb4+ into PbO2, predicting where the oxygens will likely move:
Pb + H2SO4 -> PbSO4 + 2e- + 2H+
PbO2 + H2SO4 + 2e- -> PbSO4 + 2H2O

Finally, balance out any discrepancies and add them both together:
Pb + H2SO4 -> PbSO4 + 2e- + 2H+
PbO2 + H2SO4 + 2H+ + 2e- -> PbSO4 + 2H2O
--------------------------------------------------------
Pb + PbO2 + H2SO4 -> 2PbSO4 + 2H2O

Which matches what they've given us. Notice that in the final step (which shows the full half-reactions), there are only two electrons that move, but also 2 lead atoms in the final reaction.

Do you think it’s better to actually write your own set of notes or just use the atar notes/ other people’s notes.

Depends entirely on the person - for me, I like to write my own notes, because the process of writing my notes helps me understand what's going on and commit it to memory.

thatdumbstudent

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Re: VCE Chemistry Question Thread
« Reply #8630 on: June 16, 2020, 09:59:49 pm »
0
how do you name this hydrocarbon?
would this be 2,3,4-trimethylbutamine? We haven't really gone through amines in class so i'm a little confused. I also don't know if it should be butan- or pentan- ? do you need a line connecting to an amine or does that count as carbon (?)

Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8631 on: June 17, 2020, 06:39:10 pm »
+4
how do you name this hydrocarbon?
would this be 2,3,4-trimethylbutamine? We haven't really gone through amines in class so i'm a little confused. I also don't know if it should be butan- or pentan- ? do you need a line connecting to an amine or does that count as carbon (?)
To answer your last question first, the line connecting to the amine group does not indicate an additional carbon (as can be seen in the drawing below).
When naming molecules its important we identify and name based off the longest carbon chain - in this case it would be 5 carbon's long. We then number the carbons along this longest chain prioritizing the most important functional groups; in this case we have alkyl groups and an amine group. The Amine is more important so we number a 5 carbon chain in such a way to give the amine group the lowest number possible (in this case this is carbon number 2). This makes the molecule 3,4-dimethyl pentan-2-amine.

Hopefully this clarifies any confusions but feel free to point out anything I might not have explained clearly enough!
« Last Edit: June 17, 2020, 07:11:08 pm by Erutepa »
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thatdumbstudent

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Re: VCE Chemistry Question Thread
« Reply #8632 on: June 17, 2020, 06:49:35 pm »
0
To answer your last question first, the line connecting to the amine group does not indicate an additional carbon (as can be seen in the drawing below).
When naming molecules its important we identify and name based off the longest carbon chain - in this case it would be 5 carbon's long. We then number the carbons along this longest chain prioritizing the most important functional groups; in this case we have alkyl groups and an amine group. The Amine is more important so we number a 5 carbon chain in such a way to give the amine group the lowest number possible (in this case this is carbon number 2). This makes the molecule 3,4-dimethyl pentan-2-amine.

Hopefully this clarifies any confusions but feel free to point out anything I might not have explained clearly enough!


hi yep i've recently found out the correct answer but your explanation did clarify more stuff so thank you! :-)
also, just another question, if i was given a name (4-ethyl-3-methylpent-2-ene) and i had i drew the structural formula out but lets say the double bond was located at the right. For the semi-structural formula, do i have to write the structure from left to right or does that not matter? if that makes any sense

so like CH3CH(CH2CH3)C(CH3)=CHCH3

btw i dont know if that's right in the first place haha but what i mean is you can see the double bond is towards the end however you'd still read it as 4-ethyl-3-methylpent-2-ene right? or do i need to draw it backwards to the double bond is located at the front?

whys

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Re: VCE Chemistry Question Thread
« Reply #8633 on: June 17, 2020, 07:06:58 pm »
+4
hi yep i've recently found out the correct answer but your explanation did clarify more stuff so thank you! :-)
also, just another question, if i was given a name (4-ethyl-3-methylpent-2-ene) and i had i drew the structural formula out but lets say the double bond was located at the right. For the semi-structural formula, do i have to write the structure from left to right or does that not matter? if that makes any sense

so like CH3CH(CH2CH3)C(CH3)=CHCH3

btw i dont know if that's right in the first place haha but what i mean is you can see the double bond is towards the end however you'd still read it as 4-ethyl-3-methylpent-2-ene right? or do i need to draw it backwards to the double bond is located at the front?
The structural formula is read left to right, but the double bond could be at the start or end. If you decide to put the double bond at the end of the structural formula, then all other functional groups must be numbered from right to left, and vice versa. With 4-ethyl-3-methylpent-2-ene, putting the double bond at the end means that the methyl will be on the carbon to the left of that, as you've shown, with the ethyl on the next carbon (as you've also shown). As long as it represents the molecule, it will be correct. You can always check this by drawing it out and seeing if it still matches the name. Personally, it is much easier to write the structural formula from left to right. It is also not a requirement to show double/triple bonds in the structural formula, but that is a personal choice and they won't mark you down if you decide to include it/not include it.
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Azila2004

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Re: VCE Chemistry Question Thread
« Reply #8634 on: June 17, 2020, 07:40:23 pm »
0
Hello,

I am a little confused about this question: 4.15 grams of tungsten was burned in chlorine and 8.95 grams of tungsten chloride (WCl6) was formed. Find the relative atomic mass of tungsten.

Isn’t the relative atomic mass just the number on the periodic table? I’m not really used to practising this exact type of question, so help would be appreciated :)
Just someone who likes to learn a lot of questions.

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IThinkIFailed

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Re: VCE Chemistry Question Thread
« Reply #8635 on: June 19, 2020, 04:30:47 pm »
0
Hey, I’m struggling to identify the variables in these experimental set ups:

Experiment 1:
Place two drops of cyclohexane on one watch glass and two drops of cyclohexene on another. Light the two liquids with a match and record the appearance of each flame

Experiment 2:
Using a fume cupboard, place cyclohexane to a depth of 1 cm in one test tube and cyclohexene to the same depth in another tube. Add two drops of a solution of iodine in hexane to each liquid. If there is no immediate reaction, stopper the tube and place it in sunlight for 5 minutes. Record your observations

Experiment 3:

Shake five drops of cyclohexane with 10 drops of water in a test tube. Note whether the saturated hydrocarbon is soluble in water. Repeat using the unsaturated hydrocarbon, cyclohexene

My attempt at identifying the variables is this:
Experiment 1:
Independent variable: type of hydrocarbon (saturated or unsaturated)
Dependent variable: Completion of combustion/ quality of flame

Experiment 2:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable:
Progression of a reaction with iodine (addition reaction)

Experiment 3:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable: solubility in water

Could someone please verify if I’m right, and if I’m wrong, point me in the right direction?
Thanks
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8636 on: June 19, 2020, 04:57:04 pm »
+5
Hello,

I am a little confused about this question: 4.15 grams of tungsten was burned in chlorine and 8.95 grams of tungsten chloride (WCl6) was formed. Find the relative atomic mass of tungsten.

Isn’t the relative atomic mass just the number on the periodic table? I’m not really used to practising this exact type of question, so help would be appreciated :)

I'm gonna be honest, I might need to think harder about this, but I'm not convinced you have enough information for this unless you use a periodic table to get the molecular weight of chlorine.

You are correct, of course - the relative atomic mass is the same as the molecular weight. Which begs the question, if you've got a periodic table to get the molecular weight of chlorine, why not use it to get the amount of chlorine? Is there more to the question you didn't give us? (Like previous parts)

Hey, I’m struggling to identify the variables in these experimental set ups:

Experiment 1:
Place two drops of cyclohexane on one watch glass and two drops of cyclohexene on another. Light the two liquids with a match and record the appearance of each flame

Experiment 2:
Using a fume cupboard, place cyclohexane to a depth of 1 cm in one test tube and cyclohexene to the same depth in another tube. Add two drops of a solution of iodine in hexane to each liquid. If there is no immediate reaction, stopper the tube and place it in sunlight for 5 minutes. Record your observations

Experiment 3:

Shake five drops of cyclohexane with 10 drops of water in a test tube. Note whether the saturated hydrocarbon is soluble in water. Repeat using the unsaturated hydrocarbon, cyclohexene

My attempt at identifying the variables is this:
Experiment 1:
Independent variable: type of hydrocarbon (saturated or unsaturated)
Dependent variable: Completion of combustion/ quality of flame

Experiment 2:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable:
Progression of a reaction with iodine (addition reaction)

Experiment 3:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable: solubility in water

Could someone please verify if I’m right, and if I’m wrong, point me in the right direction?
Thanks


Looks fine to me

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8637 on: June 29, 2020, 01:22:08 pm »
+1
Hi guys! I’m a year 12 revising U3 Chem and was wondering to what extent should we know about energy conversions? I came across a checkpoints question which was asking which is the most efficient. Does anyone know the pathway?

I.e thermal——> mechanical ——-> electrical?

Or is this more for the galvanic cells topic? (I’m going over fuels atm)
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Re: VCE Chemistry Question Thread
« Reply #8638 on: June 30, 2020, 04:45:51 pm »
0
Hi I was wondering if someone could explain to me in detail how does varying the surface area of electrodes affect the mass of metal formed deposited on the cathode in an electrolytic cell and consequently increase the rate of the chemical reaction? And why can the mass deposited deviate from the theoratical mass deposited using Faraday's Law?

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Re: VCE Chemistry Question Thread
« Reply #8639 on: June 30, 2020, 07:46:29 pm »
+10
Hi I was wondering if someone could explain to me in detail how does varying the surface area of electrodes affect the mass of metal formed deposited on the cathode in an electrolytic cell and consequently increase the rate of the chemical reaction? And why can the mass deposited deviate from the theoratical mass deposited using Faraday's Law?
When you say deviated from theoretical mass, are you speaking of possible errors?
And for the other first part increasing surface area increases number of collisions, hence faster rate of reaction.