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March 29, 2024, 09:08:05 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313728 times)  Share 

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Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8850 on: October 24, 2020, 10:48:23 am »
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No there are no units for kc

p0kem0n21

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Re: VCE Chemistry Question Thread
« Reply #8851 on: October 24, 2020, 11:32:10 am »
+6
Are we expected to have units on a Kc calculation? Do we get penalised if we don't?


It somewhat depends on what you're answering. If you're just writing the expression for the equilibrium constant, then it probably isn't super necessary. However, to be safe, I would write:

[insert Kc expression here], where the units are [whatever your units are based on your expression]

Of course, if you're substituting values into the expression, then units would be expected. Besides, it's not too difficult to calculate the units if you already have the basic equilibrium constant expression!

No there are no units for kc

There can be units for the equilibrium constant. I'll show how to calculate them below.

Let's start off with a general equation from the textbook: aW + bX <-> cY + dZ   (lowercase letters are coefficients, uppercase letters are your chemical substances)

We know that Kc = ([W]a * [X]b)/([Y]c * [Z]d)

If you want to calculate your units, you just substitute your uppercase letters (your chemical substances) with M, your molarity. This gives you:

([M]a * [M]b)/([M]c * [M]d)  which can be expressed as Ma+b-c-d through index laws.

Of course, these letters don't make any sense if they're not put into the context of an example. So let's take the chemical equation 2NO + 2H2 <-> N2 + 2H2O

Then, the expression for our equilibrium constant expression is:

([NO]2 * [H2]2)/([N2]1 * [H2O]2)

As we did for the general equation, substitute the symbols of the chemical substances for M.

([M]2 * [M]2)/([M]1 * [M]2), simplified to M2+2-1-2,

which, when fully simplified, gives you the units of M.

Sometimes, you'll get crazy units like M3 which make no theoretical sense (for the scope of VCE at least). Sometimes, you'll get no units at all (M0). However, don't fret. If you've done the above calculation correct, then it's the right answer.

Sine

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Re: VCE Chemistry Question Thread
« Reply #8852 on: October 24, 2020, 11:57:45 am »
+9
Are we expected to have units on a Kc calculation - such as the M-1 etc ? Do we get penalised if we don't - my teacher has been and I'm just curious if the same thing applies with VCAA.
Yeah I would be including the units.
Looking at the 2017 exam 1 mark was given for the answer AND units. Units for Kc were also referenced in the 2018 examiners report.


Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8853 on: October 24, 2020, 09:33:27 pm »
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Why does low voltage decrease the chance of water getting electrolysed?

I’m also confused about during titration? Sometimes the mood of the diluted thing is = to the mood of the undiluted thing. But sometimes dilution factor is needed.


And sometimes you have a solution and you don’t dilute it but to find the tire you need to use ratios.

I’m literally in tears what is correct?

EDIT:

3. Are ethers in the VCE course? It’s in a STAV exam. But I didn’t recognise that ethyoxyethane has an ether functional group.
« Last Edit: October 24, 2020, 10:29:31 pm by Coolgalbornin03Lo »
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Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8854 on: October 25, 2020, 10:54:48 am »
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Why does low voltage decrease the chance of water getting electrolysed?

I’m also confused about during titration? Sometimes the mood of the diluted thing is = to the mood of the undiluted thing. But sometimes dilution factor is needed.


And sometimes you have a solution and you don’t dilute it but to find the tire you need to use ratios.

I’m literally in tears what is correct?

EDIT:

3. Are ethers in the VCE course? It’s in a STAV exam. But I didn’t recognise that ethyoxyethane has an ether functional group.

I’m adding onto this:

4. In electrolysis, if let’s say iron solid is connected to the negatively charged electrode which is The cathode, does that mean it can’t act as a reductant at the other graphite electrodes, so as a result water is the strongest reductant available in this aqueous solution?
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8855 on: October 25, 2020, 02:46:59 pm »
+5
Why does low voltage decrease the chance of water getting electrolysed?

So, chemistry isn't as simple as, "you've reached activation energy, reaction happens now". Energy levels are wobbly. For a given chemical system, you might measure that the activation energy is 135 kJ. This doesn't mean that for every pair of molecules, you need 135 kJ for them to react. One pair might need 140, or another might need 120, or there might be two that just refuse to react no matter how much energy you seem to pump into the system. And then if you wait another 5 minutes, those two molecules that refused to react might now react when you give them 1 kJ of energy. These energy levels are constantly moving, because the system is a quantum system, and quantum physics is confusing and makes no sense but that's how it works. The reason we can measure a discrete activation energy is because, on average, the whole system will react when you put in exactly 135 kJ - if you put in more, it will still all react, but if you put in less, there won't always be a reaction.

This is the same thing with the electrochemical signal. Those energy levels are wobbly and move around, so if you lower the voltage, you lower the chance of reacting those molecules that are of slightly higher energy.

Also note - if you didn't understand that, it's WAY beyond VCE. It just turns out that the result is simple, but the explanation is hard.

I’m also confused about during titration? Sometimes the mood of the diluted thing is = to the mood of the undiluted thing. But sometimes dilution factor is needed.


And sometimes you have a solution and you don’t dilute it but to find the tire you need to use ratios.

I’m literally in tears what is correct?

Think about the underlying chemistry. Draw a diagram to figure out when the unknown was diluted and when it wasn't. Figure out if you care about the dilution or not. For example (and I encourage you to draw along the actual glassware used and what the experiment would actually look like), I take 10 mL of 0.50 M sodium hydroxide and dilute it to 20 mL. I then titrate in 22.4mL of HCl. If I now want to know the concentration of HCl, do I need to include a dilution factor? No - I never diluted the HCl. I DID dilute the amount of NaOH, but that doesn't change the amount of mole, and that's what I'm going to use in the calculations.

Another example - I take a 20 mL solution of NaOH. I then take 1 mL of that, and dilute it to 50 mL. I then titrate this against 15.65 mL of 1.0 M HCl. If I want to find the concentration of the original 20mL solution, do I need to use a dilution calculation? Yes - a titration will only ever tell me the concentration of the solution in the titration. So, if I want to find the concentration of a different solution, I need to find the concentration of the 50 mL solution, then take that to the 20mL solution.

Similar example. I take a 20 mL solution of 5.0 M NaOH. I then take 1 mL of that, and dilute it to 50 mL. I then titrate this against 15.65 mL of HCl. If I want to find the concentration of HCl, do I need to use a dilution factor? No, I can I just use the 1 mL of NaOH in the calculation, because the dilution after that doesn't change the amount of mol.

Hopefully this all makes sense. I can't say it's as simple as, "if it's a dilution of the known compound, it doesn't matter", which it seems like all this is, because I'm not certain that's all it is, but I can't think of an example where you need to consider the dilution of the known compound. But if you think about what you care about, and draw the situation as you go along, hopefully then you'll be able to see when it does and doesn't matter.

You're also welcome to share some examples of these questions that confused you - potentially one where the dilution calc was used, and one where it wasn't, and we'll be able to help you out.

EDIT:

3. Are ethers in the VCE course? It’s in a STAV exam. But I didn’t recognise that ethyoxyethane has an ether functional group.

I did a ctrl+f for ether and found nothing, so I'mma say no

I’m adding onto this:

4. In electrolysis, if let’s say iron solid is connected to the negatively charged electrode which is The cathode, does that mean it can’t act as a reductant at the other graphite electrodes, so as a result water is the strongest reductant available in this aqueous solution?

I mean, electrochem isn't my best suit - but my understanding is that it doesn't matter, and the iron can still be oxidised

EDIT: So a quick google tells me that it doesn't matter what part of the battery the iron solid is attached to. It will become the anode if the iron is getting oxidised, because anode and cathode are designated based on where oxidation and reduction occurs, not by what part of the battery you connect it to. Having said all that, I don't think VCAA would ask something this weird, so I don't think you have to worry about it - and if you do find a VCAA question like this, please show it to me, because the examiner report will answer all.

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8856 on: October 25, 2020, 10:13:52 pm »
0

You're also welcome to share some examples of these questions that confused you - potentially one where the dilution calc was used, and one where it wasn't, and we'll be able to help you out.


Ooh thanks! I always feel bad sharing questions but I’m beyond stumped on this!

In question 5bii. of 2017 VCAA sample exam the mols of the diluted are assumed to be the same amount of moles in the original solution (at least according to solutions supplied by my school). But I used dilution factor twice while they used it once so my answer was a tiny bit too small.

However in Question 8c of the 2014 exam they do use dilution factor for the 4.95 mL aliquot?
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sazabo_c

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Re: VCE Chemistry Question Thread
« Reply #8857 on: October 25, 2020, 10:22:41 pm »
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Hello! I was wondering if anyone has any tips on remembering the conditions of reactions for organic chemistry. There are a lot of reactions and I have trouble remembering.

Thanks a million to whoever replies :)

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8858 on: October 25, 2020, 11:46:10 pm »
+5
Ooh thanks! I always feel bad sharing questions but I’m beyond stumped on this!

In question 5bii. of 2017 VCAA sample exam the mols of the diluted are assumed to be the same amount of moles in the original solution (at least according to solutions supplied by my school). But I used dilution factor twice while they used it once so my answer was a tiny bit too small.

However in Question 8c of the 2014 exam they do use dilution factor for the 4.95 mL aliquot?

So, I can't talk about the first question, because 5bii in 2017 was a mass spec question. As for 2014 - you never HAD to use a dilution factor. For example, the way I'd do this question is first figure out the concentration, then figure out the amount of mol in that aliquot:



And notice that I got the exact same answer? The dilution factor cuts out the middle man, it means there's one less thing to calculate, but I still got the right answer using my longer calculation that makes sense to me. And this is something I tell to all my maths students, because it's more relevant there, but it's still relevant here - the correct method is the one that makes sense to you, but still gets the right answer. So if you're confused by the dilution factor being used, just ignore it - and instead focus on doing this extra step.

If you're still confused by why I even had to take this extra step, again, it comes down to thinking about the physical situation. The amount of iodine (or the mol of iodine, or the number of molecules of iodine) that is in the standard solution is different to the amount in the aliquot. The concentration is the same, but the amount is different. So, you have to do an intermediate calculation to convert from one to the other. Whether that's a dilution calculation, or my extra calculation, you have to pick one.

Hello! I was wondering if anyone has any tips on remembering the conditions of reactions for organic chemistry. There are a lot of reactions and I have trouble remembering.

Thanks a million to whoever replies :)

If what you're struggling to remember is conditions such as exact temperatures, stirring, extraction, etc., those aren't required. Otherwise, I like to do it functional group and molecule by functional group and molecule. Ignore that giant flowchart that everyone says, "memorise this". That's stupid. Rote learning is stupid. Understanding, however, is amazing and will always lead to better learning. Focus on what functional groups can react in what ways, and things will often make sense. For example, it doesn't make sense that an alkene could just react with Br2, right? Both look fairly stable, there's no reason for the bromine to spontaneously go from a 0 charge state to a -1 charge state - you'd need to put in a bunch of energy - like UV light. So, the alkene bromination reaction must be UV catalysed.

Chocolatemilkshake

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Re: VCE Chemistry Question Thread
« Reply #8859 on: October 26, 2020, 06:25:50 am »
+5
Hello! I was wondering if anyone has any tips on remembering the conditions of reactions for organic chemistry. There are a lot of reactions and I have trouble remembering.

Thanks a million to whoever replies :)
I definitely agree that rote learning is not the answer although for me personally, it found it really beneficial to categorise the reactions and just write a list (previously all the reactions felt muddled in my brain and it was difficult to reach to certain information when I needed it. After having written it out I found it much easier to organise the content in my brain and thus gain a better understanding of it. This won't work for everyone but maybe it'll work for you). Although I've written out the products of the reactions in this list, as keltingmeith indicated, there's no need to rote learn this as it should just come naturally with the understanding of each reaction. Anyway see below...

Reaction pathways
SUBSTITUTION reactions (no C=C bonds)
- alkane + halogen: UV light/heat --> haloalkane
- haloalkane + NaOH --> alcohol
- haloalkane + ammonia --> amine

ADDITION reactions (C=C involved)
- alkene + hydrogen gas: metal catalyst--> alkane
- alkene + halogen --> haloalkane
- alkene + water: H3PO4/Heat --> alcohol
- alkene/alkyne molecules can form a polymer in presence of a catalyst

OXIDATION reactions --> alcohols oxidised in presence of H+/Cr2O72- or H+/MnO4-
- Primary alcohols --> aldehyde --> carboxylic acid
- Secondary alcohols --> ketone
- Tertiary alcohols --> don't oxidise

CONDENSATION reaction
- carboxylic acid + alcohol: H2SO4 (l) --> ester (esterification)

HYDROLYSIS reaction (pretty much opposite of above)
- ester + water: catalyst --> carboxylic acid + alcohol

As keltingmeith, no use rote memorising this but writing it out like this helps me (and it also shows that there actually aren't that many reactions you need to know). Another tip is when given a question figure out your reactant and what is the desired product. Generally once you've sorted these out in your brain, the required reaction comes naturally.
« Last Edit: October 26, 2020, 12:24:40 pm by Chocolatemilkshake »
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Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8860 on: October 26, 2020, 08:20:44 am »
+3
Ohhh thanks keltingmeith! I think I was confused because my teacher said the mols of diluted and undiluted are the same. And the other question which I was talking about was in the 2017 sample exam (5biii)

Also chocolatemilkshakes diagram was great for reaction pathways but (not to be picky) the Sulfuric acid in esterification either has to have concentrated next to it or be in a liquid state  :)
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dylan.kumar21

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Re: VCE Chemistry Question Thread
« Reply #8861 on: October 26, 2020, 09:36:38 pm »
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Is it true that you can only lose 1 mark due to significant figures in the exam??

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8862 on: October 27, 2020, 03:28:01 pm »
+2
Ohhh thanks keltingmeith! I think I was confused because my teacher said the mols of diluted and undiluted are the same. And the other question which I was talking about was in the 2017 sample exam (5biii)

Ahhhh. So first, what your teacher said is correct - the amount of compound, in mol, of a diluted and undiluted sample are the same. The only difference is how much water makes up the solution. Think of a real world example - say you have dissolved 5g of copper in 50mL of water. If you add ANOTHER 50mL of water, that copper isn't going anywhere. It hasn't decided to just yeet itself out of existence - it's still there.

The difference in this question (2014s), however, isn't that we have a diluted and undiluted sample. It's that not ALL of the solution was used. Back to the copper example - after diluting to 100 mL, I then take the beaker and pour 50 mL of that solution into another beaker. Where is the copper now? Well, it should be split evenly between the two beakers. So if I started out with 5g of copper, the beakers IN TOTAL should have 5g, which would mean 2.5g of copper in each beaker. Or maybe, I poured 20mL in the second beaker, so one has 80mL of solution, and the other has 20mL - then there'd be 4g of copper in one beaker, and 1g in the other. It's the same thing with the iodine - sure, I started with 15g of it, but not all of that 15g is going to be in the 4.95 mL that I used for the titration - only some of it is. It's not that the amount of mol changed in the dilution, because it won't - if the mol changes, then that means some of the material must have reacted with something else. What's actually happened is that when I take an aliquot, or part of a solution, away from the full solution, the some of the chemical will come with the new aliquot, and some will stay behind in the full solution.

As for the 2017 question, the reason you won't use a dilution factor is because in 5bii, you don't care about the full standard yet. You only care about what's in the aliquot. The aliquot hasn't been diluted. You had to dilute the stock solution to MAKE the aliquot, but the aliquot itself was never touched. That's all that answering these questions takes - thinking about the actual chemistry that's going on, and not just plugging in the numbers you find into equations.

Is it true that you can only lose 1 mark due to significant figures in the exam??

Short answer: most likely

Long answer: everyone says something to this effect, but it's not cut and dry. Firstly, VCAA has no official stance on this - having said this, there 100% is information that teachers get that we don't, so we need to trust them a bit here. Unfortunately, every teacher says something different. Some have said that you need to use sig figs for every question, but you can only lose up to 1 mark if you get things wrong. Eg, let's say question 2 and 5 both require sig figs. You get the sig figs wrong on both of these questions. You will only lose 1 mark. Let's say you got 2 right, but 5 wrong. You will only lose 1 mark.

Another thing I've heard is that there's only one question in the whole exam where you can lose the mark. Let's go back to the previous example, but let's say that question 2 is the question where you can lose the sig fig mark. If you get the sig figs wrong in both questions, again, you will only lose 1 mark. But, if you get the sig figs right in question 2, but wrong in question 5, then you will not lose any marks for sig figs.

The problem is - if the second scenario is the true one, we have NO way of knowing what question has the sig fig marks attached to it. It's also not clear if this sig fig mark is attached to the question (eg, question is out of 3 - 1 mark for working out, 1 mark for answer, 1 mark for sig figs), or if it's just a penalty if you get the sig figs wrong (eg, question is out of 2 - 1 mark for working out, 1 mark for answer. -1 if you get sig figs wrong), so we can't even try and guess it from the mark distribution. And again, this is us trusting our teachers, because VCAA don't give this information to us (that I can find, at least. Would be very happy for someone to prove me wrong here, tbh)

So, you can rest easy knowing the yes - at most, you'll likely only lose 1 mark for sig figs. You are, of course, better off just trying to get sig figs right (they're honestly not that hard to learn if you can convert to scientific notation, because then there's just one trick for you to use). But just in case you're worried about making a mistake in the heat of the moment (because yeah, turns out, exams can be stressful - and when stressed, you often make silly mistakes you normally wouldn't. Who knew????), don't worry - you'll probably only lose 1 mark at the most.

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8863 on: October 27, 2020, 10:13:50 pm »
0
how would you draw the structure for 3-ethyl-2-methyl-5-hexenoic acid

sweetiepi

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Re: VCE Chemistry Question Thread
« Reply #8864 on: October 27, 2020, 10:22:16 pm »
+4
how would you draw the structure for 3-ethyl-2-methyl-5-hexenoic acid
Have you had a shot at this yourself first? :)

Tips
- First: identify your parent chain. That would be 5-hexene. The double-bond is on the 5-6 carbon. Add the carboxylic acid to the 1 end.
- Second: On the third carbon of the hexene, you add an ethyl group
- Third: On the second carbon of the hexene, you add a methyl group
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