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Author Topic: HSC Chemistry Question Thread  (Read 1040589 times)  Share 

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jakesilove

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Re: Chemistry Question Thread
« Reply #510 on: July 27, 2016, 11:13:52 pm »
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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
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Maz

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Re: Chemistry Question Thread
« Reply #511 on: July 28, 2016, 12:58:04 pm »
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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
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jakesilove

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Re: Chemistry Question Thread
« Reply #512 on: July 28, 2016, 02:39:58 pm »
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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
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Re: Chemistry Question Thread
« Reply #513 on: July 28, 2016, 04:07:56 pm »
+1
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.

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anotherworld2b

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Re: Chemistry Question Thread
« Reply #514 on: July 29, 2016, 01:30:13 am »
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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

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Re: Chemistry Question Thread
« Reply #515 on: July 29, 2016, 01:52:14 pm »
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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 :)
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jakesilove

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Re: Chemistry Question Thread
« Reply #516 on: July 29, 2016, 02:54:49 pm »
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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



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
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jakesilove

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Re: Chemistry Question Thread
« Reply #517 on: July 29, 2016, 03:02:56 pm »
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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.



We can now convert this into moles.



We know that moles of gas occupy a fixed volume at a certain temperature: specifically, 24.79L per mole.



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
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Re: Chemistry Question Thread
« Reply #518 on: July 29, 2016, 03:36:19 pm »
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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 :)
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Re: Chemistry Question Thread
« Reply #519 on: July 29, 2016, 03:47:43 pm »
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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?

RuiAce

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Re: Chemistry Question Thread
« Reply #520 on: July 29, 2016, 03:59:05 pm »
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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.
« Last Edit: July 29, 2016, 04:04:14 pm by RuiAce »

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Re: Chemistry Question Thread
« Reply #521 on: July 29, 2016, 04:20:34 pm »
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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  :'(

RuiAce

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Re: Chemistry Question Thread
« Reply #522 on: July 29, 2016, 04:39:15 pm »
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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.
« Last Edit: July 29, 2016, 05:09:27 pm by RuiAce »

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Re: Chemistry Question Thread
« Reply #523 on: July 29, 2016, 05:27:25 pm »
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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?

RuiAce

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Re: Chemistry Question Thread
« Reply #524 on: July 29, 2016, 05:29:30 pm »
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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)