ATAR Notes: Forum
Archived Discussion => Mathematics and Science => 2013 => Exam Discussion => Victoria => Chemistry => Topic started by: saedf on November 12, 2013, 01:22:08 pm
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If you do please post them here
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I had A,B,A,C,C,A,A,C,B,B,C,A,D,D,D,C,B,D,C,A,D,C,C,A,B,D,B,D,B,C with one or two guesses
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I had ABACCAACCBCDBDDCBDCADCCABDBDBDD So 4 of ours are different, although I did rush them when I put them down so I could've possibly copied some incorrectly
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The first MC answer was 5 sig figs
Damn :(. I mustn't have read it properly. Was the 2 sig fig value unrelated to what it wanted you to calculate?
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The first MC answer was 5 sig figs
We have four answers saying it was two. Are you sure?
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I said A, because it was the lowest number of sig figs, but I remember there was some weird rules for multiplying and dividing (which you had to do to get the pressure) that I never bothered to learn.
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Don't judge me, I actually struggled with the first question, sig figs are like the worst for me I put A which was 2 but I actually stared at it for like 4 minutes!! hahaha omg but I went well with the rest of the MC, that first question was like woah simple stuff.
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I said A, because it was the lowest number of sig figs, but I remember there was some weird rules for multiplying and dividing (which you had to do to get the pressure) that I never bothered to learn.
I thought those rules were for margin of error? Maybe I'm wrong.
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abadcaaccb
cabddcbdcb
accaddbcbc
but I did copy them down pretty hastily so there's bound to be errors
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LOL I love how you can't detect sarcasm on the internet, sorry guys lol
Yeah it was 2 sig figs obviously hahaha.
I didn't get that one was an Fe stick (I think?) there were three separate beakers with different solutions. No idea what to do
Use the electrochemical series to see if a spontaneous redox reaction occurs between the solution and fe stick
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Had to check if they could spontaneously react am I right? Just look at your chem series. It was Beaker 1
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1 & 2** don't know why half my comment was deleted
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What did you guys get for the back titration question? I can't remember it exactly... sth like which of the following 3 cases are best for back titration. I chose ii and iii (not the volatile one), but now I think the answer should be all 3 :(
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All 3, but I was unsure about the 2nd one.
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I think the 2nd should be fine if we think it this way:
Putting excess diluted acid & solid, let them react. Back titration for the excess amount of acid...
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All 3, but I was unsure about the 2nd one.
Yeah I'm sure of all 3, at least I hope.. It's a shame no one appears to have been able to grab a copy of the exam :( I don't really know, do you ask your VCAA people at the end if they have any spares you can take??
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When do the suggested solutions come out?
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Yeah I'm sure of all 3, at least I hope.. It's a shame no one appears to have been able to grab a copy of the exam :( I don't really know, do you ask your VCAA people at the end if they have any spares you can take??
Imagine how much easier it would be if VCAA just put their exams up on the site right after they were finished.
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Wrote my answers down in there
ABADCAACCB
CABADCBDCA
ACCDCDBDBC
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Wrote my answers down in there
ABADCAACCB
CABADCBDCA
ACCDCDBDBC
What do people mean by writing down MC answers???
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What do people mean by writing down MC answers???
Just copied them onto the data booklet.
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Just copied them onto the data booklet.
ohhhhhhhhh makes sense!
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wasn't the last one d?
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what did you guys write for the hydrogen peroxide MC about why no reaction was evident?
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I'm pretty sure it would react with the water?
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I'm pretty sure it would react with the water?
In the series hydrogen peroxide is both a stronger oxidant and reductant than water; so I don't think that comes into the equation
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what did you guys write for the hydrogen peroxide MC about why no reaction was evident?
The reaction rate was too slow to observe a reaction
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Shit, I had D and then changed it
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for the question which asks when back titration should be used, i wrote ii) and iii) because i dont see how something being volatile makes back titration better than normal titration. anyone else?
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i dont see how something being volatile makes back titration better than normal titration.
According to my Chemistry teachers and notes on chemistry, it is important to use a back titration for volatile substances, because excess solution can capture gas that might otherwise escape.
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for the question which asks when back titration should be used, i wrote ii) and iii) because i dont see how something being volatile makes back titration better than normal titration. anyone else?
100% sure that it is used for volatile substances.
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So from looking at adam_s95, saedf, Chazef, and randomposter's chosen answers, I read through the exam and I think the answers for the multi-choice are:
(http://i.imgur.com/Ds4DjHP.png)
Though I'm not 100% sure.. Esp. with Q20 & 25:
(http://i.imgur.com/Rsw5H3W.png)
(http://i.imgur.com/GMwTW4C.png)
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So from looking at adam_s95, saedf, Chazef, and randomposter's chosen answers, I read through the exam and I think the answers for the multi-choice are:
(http://i.imgur.com/Rsw5H3W.png)
(http://i.imgur.com/GMwTW4C.png)
WHERE DID YOU GET THOSE PICS FROM? :D
can you please post the full thing? :)
nvm, found it... can't believe got q2 wrong :(
Oh, and mine were:
A C A C C A A C C B C A B D C B D C D D C C A B D B D B C B D
lol... I think I messed up a few :P
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I'm pretty sure #30 was C. Because Fe has a charge of two whereas the others have a charge of one, and so only half the number of moles of Fe will be deposited compared with K and Ag.
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I'm pretty sure #30 was C. Because Fe has a charge of two whereas the others have a charge of one, and so only half the number of moles of Fe will be deposited compared with K and Ag.
Fairly sure it was D, Potassium doesn't form a solid as water reduces in preference to it.
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Fairly sure it was D, Potassium doesn't form a solid as water reduces in preference to it.
ahhh... tricky. :P
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I'm pretty sure #30 was C. Because Fe has a charge of two whereas the others have a charge of one, and so only half the number of moles of Fe will be deposited compared with K and Ag.
No potassium formed since it is below water on the electro chem series. And it was asking for mass iirc, so you had to times it by the molar mass too... don't think it made much of a difference tho
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So from looking at adam_s95, saedf, Chazef, and randomposter's chosen answers, I read through the exam and I think the answers for the multi-choice are:
(http://i.imgur.com/Ds4DjHP.png)
Though I'm not 100% sure.. Esp. with Q20 & 25:
(http://i.imgur.com/Rsw5H3W.png)
(http://i.imgur.com/GMwTW4C.png)
Fairly sure 25 is D. It doesn't oxidise water in preference to itself, and they are standard half cells.
Also are you sure 28 is D, I think it's C, but might be wrong. Also not sure about 20, I put D but again could be wrong and not confident.
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28 seems like a tricky question, I though it was B because all the salt bridge has to do is to provide a steady stream of anions/cations, hence the main reason why KNO3 is used is because it is unreactive.
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pretty sure answer for question 20 is d.
1. obviously we can eliminate b and c as they different volumes by same conc. as volume does not affect %ionisation
2. comparing a and d, since the ka values are the same for both we can set up equilibrium constant expressions
for a) ka=[H3O+]2/0.01
therefore [H3O+]=0.1ka
% ionisation=0.1ka/0.01 x 100=1000ka
for d) ka=[H3O+]2/1
therefore [H3O+]=ka
% ionisation=ka/1 x 100=100ka
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but for q 28 dont the cations of the salt bridge flow to the cathode side? and the anion of the salt bridge flow to the anode? cause since anode is the site of oxidation, anion will need to be released to balance the positive charge formed due to the oxidation. vice versa So i thought it couldnt be D so i chose C as the best answer.
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I thought being soluble in water isn't the main reason why it is a good salt bridge, the more important thing is that K+ is a weak reductant so it does not react in preference of say iron or something.
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but for q 28 dont the cations of the salt bridge flow to the cathode side? and the anion of the salt bridge flow to the anode? cause since anode is the site of oxidation, anion will need to be released to balance the positive charge formed due to the oxidation. vice versa So i thought it couldnt be D so i chose C as the best answer.
I chose C but didn't even realise this bit, you're right NO3- flows into the anode half cells, as positive chargers are 'formed' there.
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I'm pretty sure 30 is C. I know water is oxidised over potassium normally, but I thought at a high molar concentration such as 1M, potassium was favoured over water? I'm not 100% sure on this though, so I don't know...
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I'm pretty sure 30 is C. I know water is oxidised over potassium normally, but I thought at a high molar concentration such as 1M, potassium was favoured over water? I'm not 100% sure on this though, so I don't know...
That's true for Cl-, where there is a small difference in half cell voltages (0.13V), but there is a big difference between water and potassium (over 2v)
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I thought being soluble in water isn't the main reason why it is a good salt bridge, the more important thing is that K+ is a weak reductant so it does not react in preference of say iron or something.
But that option also says that nitrate is a strong oxidant, which would make it a poor choice for an electrolyte.
28 is C, as all the other options are incorrect. D is incorrect because i) oxidation occurs at the anode. Oxidation is a loss of electrons, which means a positive charge builds up at the anode. Thus, anions, which are negatively charged, not K+ (which is a cation) like option D suggests, must flow to the anode to balance this positive charge ii) reduction occurs at the cathode. Reduction is a gain of electrons, which means a negative charge builds up at the cathode. Thus, cations, which are positively charged, not NO3- (which is a cation) like option D suggests, must flow to the cathode to balance this negative charge.
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For MC I've got:
1. A
2. B
3. A
4. D
5. C
6. A
7. A
8. C
9. C
10. B
11. C
12. A
13. B
14. D
15. D
16. C
17. B
18. D
19. C
20. D
21. A
22. C
23. C
24. A
25. D
26. D
27. B
28. C
29. B
30. D
Hopefully I've transcribed that right from my scrap paper! Short answer coming soon as soon as I finish typing them up (omg so tedious).
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For question 4, still really confused...
II said being 'insoluble in water but soluble in dilute acid', then shouldn't it be something like a gravimetric analysis because something would be precipitated out etc., instead of being a volumetric one?
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For question 4, still really confused...
II said being 'insoluble in water but soluble in dilute acid', then shouldn't it be something like a gravimetric analysis because something would be precipitated out etc., instead of being a volumetric one?
This question is saying that the sample itself is insoluble in water. So I need to dissolve it using an acidic or basic solvent, and then figure out how much of the solvent was required.
With gravimetric analyses, the sample itself is soluble in water, e.g. fertiliser. So I'd dissolve my fertiliser and then maybe add barium chloride. Barium ions would react with the sulfate ions in fertiliser to form barium sulfate, which is insoluble in water. Then I'd be able to measure how much sulfate was in the fertiliser. But the sample itself when it contained sulfate was soluble in water!
On a simple level, you can't precipitate a sample that isn't soluble in water.
So if I had a limestone sample and I dunked it in water, it would just be a big ol' rock in water. Hence what I'd need to do is dissolve it in acid instead. The reason it dissolves in acid is because acids are more corrosive than water, and more importantly calcium carbonate is a base. So I'd add a known excess of acid to dissolve the limestone, then titrate that excess with a base to figure out how much acid was left over after dissolve the sample. I'd then be able to calculate how much was actually required for the sample, thus how much calcium and/or carbonate was in the limestone (save for contaminations).
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Could you explain why 20 is D please? I resorted to just going with the highest conc of ethanoic acid but that's obviously wrong haha
and what do you reckon is a good score on the multiple choice judging by the standard of the exam for an A+?
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Could you explain why 20 is D please? I resorted to just going with the highest conc of ethanoic acid but that's obviously wrong haha
and what do you reckon is a good score on the multiple choice judging by the standard of the exam for an A+?
Since ethanoic acid is a weak acid, you could use the Ka values in the data booklet to work out that it's D, figuring out the [H3O+] for each one, then dividing by the initial concentration of acid and multiplying by one hundred.
EDIT: Omg my logic for the second bit is so poorly explained it doesn't sound right at all. Trying to think of a better way to explain. But the answer is definitely D!
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Yeah I just resorted to the highest concentration favoring a net forward reaction for Q. 20, although I didn't specifically do calculations with Ka values to confirm
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So if I had a limestone sample and I dunked it in water, it would just be a big ol' rock in water. Hence what I'd need to do is dissolve it in acid instead. The reason it dissolves in acid is because acids are more corrosive than water, and more importantly calcium carbonate is a base. So I'd add a known excess of acid to dissolve the limestone, then titrate that excess with a base to figure out how much acid was left over after dissolve the sample. I'd then be able to calculate how much was actually required for the sample, thus how much calcium and/or carbonate was in the limestone (save for contaminations).
Oh thanks that makes so much sense... got so messed up with all the conceptions :-[
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Could you please explain Question 13 of multiple choice? I must've stuffed up my oxidation numbers because I got that they were all reduction reactions. XD
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Could you please explain Question 13 of multiple choice? I must've stuffed up my oxidation numbers because I got that they were all reduction reactions. XD
Thank god I'm not the only one who got that haha, I thought I was going crazy
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By the sounds of things, I think we got a lot of the same answers. XD
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Could you please explain Question 13 of multiple choice? I must've stuffed up my oxidation numbers because I got that they were all reduction reactions. XD
Which ON were you using?
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In NaNO3, I got an oxidation of +5 for N; in NO2, I got an oxidation of +4; in NH2, I got an oxidation number of -2. I think I made an error here though because in NH the oxidation number is -1... my logic is probably wrong.
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By the sounds of things, I think we got a lot of the same answers. XD
TWINS!
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Which ON were you using?
I used the N oxidation number for step 1, reducing from an ON of +5 to +4.. then in step 2, again looking at N, reducing from an ON of +4 to +2... then the only MC option that had both step 1 and 2 was D.
I feel like that isn't right... but someone have a check?
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Umm I just knew that addition of H2 was reduction..? And addition of electronegative elements like nitrogen or oxygen, cause oxidation.. :-\
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Maybe looking at the oxidation number of nitrogen wasn't right...
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Maybe looking at the oxidation number of nitrogen wasn't right...
Oxygen found its way into step I and was formed in the produce, that was oxidation, I crossed out all the options with I and was left with II which I evidenced with addition of H2 is reduction.. I think?
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This one required some intuition as well as just the ONs. Step 1 and step 3 both involve the addition of oxygen! And they also involve the removal of hydrogen in the participating 'functional groups'. Step 2 involves the removal of oxygen and the addition of hydrogen, thus reduction.
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Thushan's solutions have D!
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Thushan's solutions have D!
I'm not sure whether to cry :'( where are the solutions?
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wow such confuse
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Hmm will have to wait for the reasoning - even using ON I'm not convinced, because of the influence the carbon attached to NO2, NH2 and NH should have on the ON of Nitrogen.
Fair enough. I was tossing up between the two answers, but if oxidation occurs doesn't reduction have to occur too? Or am I making up things in my head haha
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Fair enough. I was tossing up between the two answers, but if oxidation occurs doesn't reduction have to occur too? Or am I making up things in my head haha
Yep but there are two partners in redox! If reduction happens to one, oxidation happens to the other.
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Yep but there are two partners in redox! If reduction happens to one, oxidation happens to the other.
So then part I and III would involve reduction reaction as well because oxidation occurs?
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So then part I and III would involve reduction reaction as well because oxidation occurs?
Mmm but I assume it's referring to oxidation or reduction of the organic molecule, not the other reactants.
Working it out 'mathematically' using the VCE redox laws, question D is correct. Intuitively, question B makes more sense. I'm just not certain on the maths because it involves ON of organic molecules which is perhaps tangential to the VCE course, hence why I'm backing option B. I didn't even try to do the maths the first time I did it haha.
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Can someone explain question 21?
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Can someone explain question 21?
It says to calculate the 'approximate pH'. If you look at the Ka values for phosphoric acid, the second and third ionisations are so weak that they barely have any effect on the hydronium ion concentration of the solution. Thus when calculating pH to 1 significant figure (as it is 0.1M), you only need to use the first Ka value as the effect of the second and third ionisations is so negligible it won't change the pH by very much (probably somewhere in the realm of 0.0001).
If you try to calculate pH using only the first Ka value and then try calculating it using all three Ka values, you'll find that your answers are approximately the same! They'll be the same for quite a few sig figs, at least.
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Mmm but I assume it's referring to oxidation or reduction of the organic molecule, not the other reactants.
Working it out 'mathematically' using the VCE redox laws, question D is correct. Intuitively, question B makes more sense. I'm just not certain on the maths because it involves ON of organic molecules which is perhaps tangential to the VCE course, hence why I'm backing option B. I didn't even try to do the maths the first time I did it haha.
Fair enough. Oh well, I guess I'm trying to convince myself I got it right although B does seem more feasible haha
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Fair enough. Oh well, I guess I'm trying to convince myself I got it right although B does seem more feasible haha
We'll see! I hope for your sake it is D, since you're the one who did the exam!
EDIT: Thushan has changed his answer to B. Will wait for his reasoning.
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Mmm but I assume it's referring to oxidation or reduction of the organic molecule, not the other reactants.
Working it out 'mathematically' using the VCE redox laws, question D is correct. Intuitively, question B makes more sense. I'm just not certain on the maths because it involves ON of organic molecules which is perhaps tangential to the VCE course, hence why I'm backing option B. I didn't even try to do the maths the first time I did it haha.
In the question is says with steps of the synthesis involve a reduction reaction, it doesn't explicitly say 'the molecule undergoes reduction in steps...'
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In the question is says with steps of the synthesis involve a reduction reaction, it doesn't explicitly say 'the molecule undergoes reduction in steps...'
It does to an extent actually. It says that it's an 'organic pathway for synthesis of paracetamol', so what we're doing is tracking how we go from phenol to paracetamol. The other reactants only have fleeting participation and they serve only to evolve that phenol molecule, so we're most interested in what happens to that.