Advanced Higher Chemistry 2012 Thread

Reply 20

Sniggey

Original post by NMx11

I somehow got As in both prelims but I seem to have forgotten everything! :frown:

I should be able to get a B as long as the sqa aren't really horrible this year, think i'll cry if they are! :frown:

How are you feeling about tomorrow?

I'm sure you'll be fine! Yeah i'm a little bit anxious about the PPA questions as you might imagine, but fairly confident aside from that. I'm quite confident with units 1 & 2, I only hope there aren't many organic reactions in the exam tomorrow... :frown:

Reply 21

Babb_zz

6

omg HBSKCDBGCXKNCXHJVIHSZGCKJSAB is how im feeling just now, but apart from that, can aanyone tell mme hhow to work out question 40 in the multiple choice part of the 2011 paper, really appreciated

Reply 22

chemistrypupil

Hi, can anyone PLEASE help me with the following:

2007 q 5(c), 6(b), 13 (all)

2009 q 11 and 13

Reply 23

Hype en Ecosse

20

Original post by chemistrypupil

Hi, can anyone PLEASE help me with the following:

2007 q 5(c), 6(b), 13 (all)

2009 q 11 and 13

5c. Calculate the emf of the half-cell operating under standard conditions.

It's a zinc-copper half-cell, we've got the zinc being oxidised, and the electrons flowing to the copper electrode. Just look at the standard reduction potentials on page 11 of your databook. For Cu²⁺ we find that it's 0.34, and for Zinc, it's -0.76. Because the left side of the half-cell equation is being oxidised and not reduced, we take it away from the value of the right side.

0.34 - (-0.76) = 1.10V

Real easy one! Nice and quick. I'm currently revising reaction mechanisms and then the PPAs, so I don't want to spend loads of time doing the rest. Sorry. :redface:

Reply 24

Sniggey

Okay, I'm extremely confused about 2006 Q.2(a).

How would the end point of the titration be determined?

Correct me if i'm wrong, but I was sure the permanganate ion is purple in colour. In the question permanganate is reduced to Mn(II). Surely this would be a colour change of purple -> colourless?

Yet the marking scheme says it is the opposite? :s-smilie:

Reply 25

Sniggey

Original post by Babb_zz

omg HBSKCDBGCXKNCXHJVIHSZGCKJSAB is how im feeling just now, but apart from that, can aanyone tell mme hhow to work out question 40 in the multiple choice part of the 2011 paper, really appreciated

You can take out option C straight away as it doesn't have the formula C4H8O.
The molecule has 2 peaks in its proton NMR spectrum, and as proton NMR measures the different environments of hydrogen nuclei in a molecule, you're looking for a molecule with hydrogen atoms in 2* different environments from A, B or D.

*This is the bit i'm not sure about - I think the TMS must count as a peak, so you're looking for a molecule with only 1 environment of hydrogen? - D

Reply 26

NMx11

5

Original post by chemistrypupil

Hi, can anyone PLEASE help me with the following:

2007 q 5(c), 6(b), 13 (all)

2009 q 11 and 13

2007 5c.
Use the values from the data book p11 for the ion electron equations for zinc and copper. There are two different equations for copper but in the question it says Cu2+, Cu so you'll use the equation that has the E value 0.34.
To work out the emf i always use the saying "the top you turn, the lower you leave" so basically the equation that is furthest up in the table, reverse the E value (change from negative to positive value). In this example the zinc ostensibly higher up so:
E= 0.76+0.34, =1.1V

2007 qu 6b, I didn't get this one either so I'll let someone else explain! :L

Qu 13a (i)
To work out the mass of carbon:
you know that there is 12g of carbon in CO2 and in this example you have 3.52g of CO2 so,
Mass= 12/44 * 3.52 = 0.96g
For hydrogen there is 2g in H2O so,
Mass= 2/18 * 1.44 =0.16g
To work out the mass of oxygen, you know that there was 1.76g of X so,
Mass= 1.76 - mass of hydrogen - mass of carbon
= 1.76 - 0.16 - 0.96 = 0.64g

(ii) For carbon
You have 0.96g and so you divide this by its gfm, 0.96/12 = 0.08
Do this for hydrogen and oxygen and you should get:
C = 0.08 H = 0.16 O = 0.04
You then divide these numbers by the smallest of the numbers (0.04) and this should give you roughly:
C = 2 H= 4 O=1 and this is your formula! :biggrin:

b. Work out the molecular mass of the empirical formula (C2H4O) which will give you 44. Divide 88/44 = 2. Now double the empirical formula so you get C4H8O2 :smile:

c(i) You already know the formula. In the table where it says 'relative area under the peak' this tells you how many hydrogen atoms are attached to the carbon. Using p15 of your data book you can see that the shift at 10.8 must be COOH (acid group) because you already know the formula the rest is quite easy and you'll see that its butanoic acid :smile:

I'll do 2009 in a minute!

EDIT: Scott bet me to the first bit!

(edited 11 years ago)

Reply 27

Babb_zz

6

Original post by Sniggey

You can take out option C straight away as it doesn't have the formula C4H8O.
The molecule has 2 peaks in its proton NMR spectrum, and as proton NMR measures the different environments of hydrogen nuclei in a molecule, you're looking for a molecule with hydrogen atoms in 2* different environments from A, B or D.

*This is the bit i'm not sure about - I think the TMS must count as a peak, so you're looking for a molecule with only 1 environment of hydrogen? - D

Yeah man, you can score B out because it has three hydrogen environments which leaves A and C,

With A there are two hydrogen environments and fits the formula, but D only has one hydrogen environment, but like in the study guide they give and example and they don't include tthe TMS, im soon confused but thanks man

Reply 28

NMx11

5

2009 qu 11a.

From the mass spec you can see that the molecular mass is 88g (reading furthest to the right) Just like before you divide this by the mass of the empirical formula so:
88/44 = 2, so double the empirical formula so you get C4H8O2 :smile:

b (i) page 13 of the data book, absorption at 1745 is a C=O bond
(ii) For this i looked at the nmr spec. There are only 3 peaks so one of the carbon atoms has no hydrogen friends :frown:

So, the peaks at 1.4 and 3.6 are the same height so have the same numbers of hydrogen friends 1.4 is CH3 and 3.6 is CH3O. 2.4 is CH2, so put it all together (plus the carbon with no hydrogens) you can see that it is an ester (the carbon has an oxygen friend instead :smile:

) it is in fact methylpropanoate and we've also answered question d!
c. Basically using the above information just add up the masses till you get a fragment that equals 57 in this case it is COC2H5+, don't forget the + !

I hope that made sense! :L

13a. Basically we know that a superconductor has zero electrical resistance but in this example it stops working at 95K. Look at the graph in the answers and you'll see that the resistance increases after 95 K.

b (i) Just the same as before,
Divide % of each element by its molecular mass to give you:
Yittrium =0.1507 Ba=0.3 Cu=0.45 O=1.05
Divide all these by the smallest answer (0.1507) and this will give you:
Y Ba2 Cu3 O7 :biggrin:

(ii) Add 2,3 and 2 and divide by 3 = 2.33

(iii) I just kinda guessed this right! The fact that it says the metals don't change made me think that oxygen had to change. Since Cu was being reduced to 2 i just took one of the oxygen to make 6, YBa2Cu3O6! :dontknow:

Hope some of this was of help! I better finish off the past paper i was doing, haha!
Good luck tomorrow! :smile:

Reply 29

NMx11

5

Original post by Sniggey

I'm sure you'll be fine! Yeah i'm a little bit anxious about the PPA questions as you might imagine, but fairly confident aside from that. I'm quite confident with units 1 & 2, I only hope there aren't many organic reactions in the exam tomorrow... :frown:

Hopefully! Yeah units 1+2 are the best!
Good luck tomorrow :smile:

Reply 30

why-hello-there

Hey guys.

Might be a bit of a stupid question (and one I should really know the answer to after all these years :tongue:

), but is an HB2 pencil OK to use for the multi-choice? The pack I bought is labelled 'HB' only on the packaging, but on the pencil it says 'HB2'.

Just so I'm not paranoid that the machine won't read my answers properly :tongue:

Thanks!

Reply 31

username458263

9

Original post by Sniggey

Okay, I'm extremely confused about 2006 Q.2(a).

How would the end point of the titration be determined?

Correct me if i'm wrong, but I was sure the permanganate ion is purple in colour. In the question permanganate is reduced to Mn(II). Surely this would be a colour change of purple -> colourless?

Yet the marking scheme says it is the opposite? :s-smilie:

Because it's a titration you're adding in the permanganate dropwise and it immediately reacts, so the solution remains colourless. When the end-point has been reached, the permanganate ions will no longer react, so their purple colour stays. Hope that explains it :smile:

It feels really weird that I'm not doing English any more and Chemistry is my first exam... not feeling completely ready for this. Shall need to do some last minute revision of units 1 and 2 and the ********************* PPAs.

Reply 32

Sniggey

Original post by NMx11

Hopefully! Yeah units 1+2 are the best!
Good luck tomorrow :smile:

Thanks, good luck to you too!

Original post by derangedyoshi

Because it's a titration you're adding in the permanganate dropwise and it immediately reacts, so the solution remains colourless. When the end-point has been reached, the permanganate ions will no longer react, so their purple colour stays. Hope that explains it :smile:

It feels really weird that I'm not doing English any more and Chemistry is my first exam... not feeling completely ready for this. Shall need to do some last minute revision of units 1 and 2 and the ********************* PPAs.

Ah, thanks a lot. You're more familiar with unit 3 than the other 2? That's the complete opposite of most people! And I think we're all agreed that the PPAs are a major pain in the ass. Good luck!

Reply 33

Sniggey

Original post by Babb_zz

Yeah man, you can score B out because it has three hydrogen environments which leaves A and C,

With A there are two hydrogen environments and fits the formula, but D only has one hydrogen environment, but like in the study guide they give and example and they don't include tthe TMS, im soon confused but thanks man

Yeah, I double checked as well, TMS doesnt count as a peak. It must be that the groups surrounding any given methyl group actually affect the environment of the hydrogen, so in D there are actually 2 - Carbons bonded to the oxygen and carbons bonded to only other carbons - I can't see how any other way there would be 2 different environments... :frown:

Reply 34

username458263

9

Original post by Sniggey

Ah, thanks a lot. You're more familiar with unit 3 than the other 2? That's the complete opposite of most people! And I think we're all agreed that the PPAs are a major pain in the ass. Good luck!

I hated unit 3 at first but I've studied it much more than the other two. And there are so many easy marks eg explain what SN1 means, you say nucleophilic substitution, 1st order, and there's 2 marks with zero effort or thought. I knew Units 1 and 2 really really well for the prelim but I've been doing past papers and keep coming up with things I've slightly forgotten eg buffer solutions. :/

Reply 35

Luke_Storm

I'm feeling fairly confident about tomorrow, its my best subject and I got really high marks in the prelims.. Done a tonne of past paper Q's and PPA Q's too, so I think I should be alright.. Ahh well, good luck everyone!

Reply 36

Fifixx

Ha, just found this thread :wink:

Good luck today everyone!!!! :smile:

Reply 37

Marcus4004

2

Original post by Babb_zz

Yeah man, you can score B out because it has three hydrogen environments which leaves A and C,

With A there are two hydrogen environments and fits the formula, but D only has one hydrogen environment, but like in the study guide they give and example and they don't include tthe TMS, im soon confused but thanks man

Okay guys i will try to clear this question up... You always discount the TMS! Okay by looking at A there are 3 different hydrogen environments, i know if you look in your data book two of the environments come under the same group (the adjacent hydrogens to the carbonyl group). This does NOT mean they will be the same peak, it just means that there will be two peaks within the range given in the data book. So when you add the RCH3, you have 3 groups so A can be eliminated.

As guys have said B is incorrect and so is C.

On inspecting D there are only 2 environments hence the answer is D.

Hope that helps! Good Luck GUYS!!