2014 H2 Chemistry Paper 2, Paper 3 and Paper 1 discussions
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Originally posted by Angelofv3rdun:
How to draw the P2N2Cl6 ??
Cl3P=N-N=PCl3
or
(Cl)2PN(Cl)N(Cl)P(Cl)2 -
Originally posted by Helloheyho:
Question 3cii why not alkyne (triple bond)? It fits the molecula formula and will be reduced to pentane by H2 Ni too.
The best answer (expected by Cambridge) is indeed an alkyne. Just that Cambridge will probably (for A level purposes) accept a diene as well. -
Originally posted by BUTTanalALDEHYDE:
Hi!! For question 1c, is it possible to argue for H2O2 to be more acidic because the negative charge on conjugate base can be delocalised among the two electronegative oxygen atoms?
Acceptable. Though the word "delocalized" should be used in the context of resonance, while "dispersed" should be used in the context of induction. Here, only induction applies, not resonance (since both O atoms are sp3 hybridized). -
Stephen and Matthew, you guys better not disappoint me. And you guys damn well better not be complacent, be sure you continue to work hard and prepare well the next couple of days to get a perfect 40/40 score for your Paper 1!
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Originally posted by Adelineeeee (::
Does anyone have any idea on the mark range for B? oh and how did u guys find paper2? Many ppl said it was easy but i find the questions asked pretty weird ):
B cut-off score should be approx 62%.
Don't listen to these "many pple". Many Singapore JC / school taught answers are actually not fully correct (according to my standards anyway, maybe Cambridge might be merciful and meet them halfway).
You're right that this year's P2 and P3 have 'weird' questions, following the trend of the recent years 'weird' papers since 2010.
In my opinion, the difficulty of the 2010, 2011, 2012, 2013 and 2014 papers are approximately the same. Hence the cutoff score for A grade should remain approximately 75% this year. -
Originally posted by Adelineeeee (::
Hi, may i ask if anyone knows whats the percentage of our practical exam and how much can it help in pushing a overall B to A?
Just be sure you score a perfect or near perfect 40/40 score for your P1. There's your best chance to still get that A grade.
As for the % weightage of practical exam or SPA, check out page 4 of the following pdf document :
https://www.seab.gov.sg/content/syllabus/alevel/2015Syllabus/9647_2015.pdf -
A level chemistry woes since 2010
Whether it's good or bad, all the best for the rest of your papers ppl ! -
M isnt chiral cos got internal plane of symmetry what so its image is superimposable
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For p2n2cl6
If i drew cl3p=n-n=pcl3 correct?
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Originally posted by Alohamadda:
For p2n2cl6
If i drew cl3p=n-n=pcl3 correct?
Yes, that's correct too. -
but isnt that valency 4 for p and valency 2 for nitrogen? im abit confused
haha
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Originally posted by Alohamadda:
but isnt that valency 4 for p and valency 2 for nitrogen? im abit confused
haha
Valency is about the number of bonds. Double bond is considered 2 bonds. So for a triply bonded N2 molecule, each N atom still has a valency of 3. -
Here's a quick test for every single one of you taking Paper 1 next Mon. Time yourself to see whether you can complete this MCQ within 1 min. In fact, post your timing here. See who has the fastest timing for this MCQ!
Ready... Get Set.... Go !!!
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Sorry sir one more question!
As for the qst on whether L or M is chiral, can i say M is not chiral because it does not have any net rotation? ><
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Originally posted by Adelineeeee (::
Sorry sir one more question!
As for the qst on whether L or M is chiral, can i say M is not chiral because it does not have any net rotation? ><
No, you cannot say it doesn't rotate plane polarized light, because how would you know that? To know that, you need a sample of the species, and you would need to carry out the experiment using a polarimeter to confirm this.
Hence, the required answer is something that you can observe by looking at the structure of your drawn out M vs L, which is : only L has a sp3 C atom bonded to 4 different groups, while M does not. -
Hi sir, just want to ask you a few qns if you don't mind...
For qn3(a), what do u think are the marking pts?
and is this eqn correct, H3PO4 + NaI --> HI + NaH2PO4?
For 3c(ii), if i drew the compound Q as 2 alkene grp, what are the angles involved?
For 3d(i), if i got the empirical forumula and molecular formula correct, but the drawing of structure is wrong, how many marks will i get /4m?
For 1d(ii), what's the explanation for difference betwn theoretical and actual val?
Sorry for asking so many qns, and thanks in advance for ur help!!
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Sir
For qn 5) could u show the steps to calculate the actual mass of the 3 metals?
Thannks
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Originally posted by moonwalkers:
Hi sir, just want to ask you a few qns if you don't mind...
For qn3(a), what do u think are the marking pts?
and is this eqn correct, H3PO4 + NaI --> HI + NaH2PO4?
For 3c(ii), if i drew the compound Q as 2 alkene grp, what are the angles involved?
For 3d(i), if i got the empirical forumula and molecular formula correct, but the drawing of structure is wrong, how many marks will i get /4m?
For 1d(ii), what's the explanation for difference betwn theoretical and actual val?
Sorry for asking so many qns, and thanks in advance for ur help!!
3a) Yes, that's correct. Remaining 2 marks is for the balanced redox reaction equation between H2SO4 and HI, and 1 mark for stating (further explanation not required) that H2SO4 is more strongly oxidizing than H3PO4.
3cii) if you drew a C=C=C, the bond angles are 180 deg around central C atoms, and 120 deg around the other 2 C atoms.
3di) 2 out of 4 marks.
1dii) As a consequence of ring strain due to angle strain (ie. bond pair - bond pair inter-electron repulsions at such low angles weaken the bonds), the bond dissociation enthalpies in epoxides are significantly less endothermic, compared to in other molecules without such strain. -
Originally posted by Alohamadda:
Sir
For qn 5) could u show the steps to calculate the actual mass of the 3 metals?
Thannks
I won't provide detailed calculations (some JC/school or other tutor should make the full worked solutions online soon), just the concept behind the working.
Using Q=IT and Faraday's constant, you can determine the moles of electrons transferred, and hence moles and therefore mass of Cu present.
From sample mass of red complex, you can find moles of red complex = moles of Ni, hence you can find moles and therefore mass of Ni present.
Mass of Ag = 0.05g (given in the question). -
Thanks!
For 5(e)(i), i used
Mass of Cu = Q/2(96500) x 63.5
Is that correct?
then (e)(ii),
the actual Ag removed isn't it just 0.05g?
then to find mass of Cu removed, i use 0.95 - 0.05 - mass of Ni removed
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Originally posted by moonwalkers:
Thanks!
For 5(e)(i), i used
Mass of Cu = Q/2(96500) x 63.5
Is that correct?
then (e)(ii),
the actual Ag removed isn't it just 0.05g?
then to find mass of Cu removed, i use 0.95 - 0.05 - mass of Ni removed
Yes, for both Cu and Ag (Lol). -
Ok. Can i check with u answers for 1(e)(i)?
I think I'm wrong somewhere but can u verify if there is?
http://postimg.org/image/x72bx1tyl/
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Originally posted by moonwalkers:
Ok. Can i check with u answers for 1(e)(i)?
I think I'm wrong somewhere but can u verify if there is?
http://postimg.org/image/x72bx1tyl/
Your error is that you need to branch your straight chain as follows :
C is CH3CH(OH)CH(CH3)2
E and F are 2-methylbut-2-ene and 3-methyl-but-1-ene
A is C6H5CH2COCH(CH3)CH(CH3)2
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Hey sirrr. If i drew like skeletal formula and displayed formula for a organic compound like for the last structure, i juz drew a ring to show the hexane without the c and h but drew the full displayed formula for side chains, is there any issue?
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Originally posted by Alohamadda:
Hey sirrr. If i drew like skeletal formula and displayed formula for a organic compound like for the last structure, i juz drew a ring to show the hexane without the c and h but drew the full displayed formula for side chains, is there any issue?
No problem at all, since the question didn't specify any particular type (eg. skeletal, condensed or displayed) of structural formula, so Cambridge will accept any type, including a hybrid of all 3 types within your answer (exactly as you described).