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# Aqa chem 4/ chem 5 june 2016 thread

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1. (Original post by bat_man)
Can someone pls explain question 4b from the june 2010 paper of chem4. Thanks (:

Paper : http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

Mark scheme : http://filestore.aqa.org.uk/subjects...W-MS-JUN10.PDF
So the 5 peaks on the M/Z are
Carbon-12
Hydrogen-1
Oxygen-16
Chlorine-35
Chlorine-37

And then the for the most abundant peak
1) get an average abundance for Cl-35 and Cl-37

So (37x25+35x75)/100

Which gives you 35.5

2)most abundant peak is the molecular ion
So basically add up the Mr

12*12 + 1*4 + 2*16 + 4*35.5 = 322.0

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2. (Original post by Super199)
Yh if you don't mind, I've not seen a qs like this before

I find it helps to write out the actual chemical equations so you can see what's going on

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3. can someone explain when you times by two and when you divide by two for a diprotic and dibasic molecule
4. (Original post by Gezza_O'Brien)
So the 5 peaks on the M/Z are
Carbon-12
Hydrogen-1
Oxygen-16
Chlorine-35
Chlorine-37

And then the for the most abundant peak
1) get an average abundance for Cl-35 and Cl-37

So (37x25+35x75)/100

Which gives you 35.5

2)most abundant peak is the molecular ion
So basically add up the Mr

12*12 + 1*4 + 2*16 + 4*35.5 = 322.0

Posted from TSR Mobile
Right, so i get the 5 peaks now. That makes sense, but still with the MR, i got 322 as well but the answer is 320.. :/
5. (Original post by Gezza_O'Brien)
So the 5 peaks on the M/Z are
Carbon-12
Hydrogen-1
Oxygen-16
Chlorine-35
Chlorine-37

And then the for the most abundant peak
1) get an average abundance for Cl-35 and Cl-37

So (37x25+35x75)/100

Which gives you 35.5

2)most abundant peak is the molecular ion
So basically add up the Mr

12*12 + 1*4 + 2*16 + 4*35.5 = 322.0

Posted from TSR Mobile
Oh wait nvm, it says 322.0 as well, i didnt read the mark scheme fully. thanks!!
6. (Original post by bat_man)
Right, so i get the 5 peaks now. That makes sense, but still with the MR, i got 322 as well but the answer is 320.. :/
It says 320.0 or 322.0, the reason people get 320.0 is because they took the average of Cl as 35 because they rounded the decimal down. Don't worry 322.0 is the correct (and more accurate) answer

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7. (Original post by bat_man)
Can someone pls explain question 4b from the june 2010 paper of chem4. Thanks (:

Paper : http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

Mark scheme : http://filestore.aqa.org.uk/subjects...W-MS-JUN10.PDF
The molecular formula of TCDD is C12H4O2Cl4Chlorine exists as two isotopes 35Cl (75%) and 37Cl (25%).Deduce the number of molecular ion peaks in the mass spectrum of TCDD andcalculate the m/z value of the most abundant molecular ion peak.

So we're looking at Cl which has two Isotopes 35,(75%) and 37 (25%)

So it Could be the Mr of C12H4O2 + 35 + 35 + 35+35
Or 37 + 37 + 37 +37
Or 37+37+35+35
Or 37+35+35+35
Or 35+37+37+37

Could be any combination of isotopes, 5 peaks formed
M/z of most abundant ion peak (35 is 75% abundant so) you would add Mr of C12H14O2 to (4X35) which gives you 320.0

Hope that helps!
8. http://filestore.aqa.org.uk/subjects...4-QP-JUN13.PDF
http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

Q3e) stuck on the final bit worked out which was in excess, it was H2SO4, then worked out the conc of H+ (0.014928x1000/46) is this = to 2[H+] or [H]+ im confused is it equal to 2[H+] since weve already multiplied the moles by 2???
9. (Original post by RedDevil1997)
Attachment 548855 Can someone please explain how this has 11 peaks in its Carbon 13 nmr?!?
Hope this helps! By the way, number 8 is weirdly back in the benzene ring, I forgot to mark it so had to go back to it!
Attached Images

10. (Original post by Sniperdon227)
http://filestore.aqa.org.uk/subjects...4-QP-JUN13.PDF
http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

Q3e) stuck on the final bit worked out which was in excess, it was H2SO4, then worked out the conc of H+ (0.014928x1000/46) is this = to 2[H+] or [H]+ im confused is it equal to 2[H+] since weve already multiplied the moles by 2???
There's 0.02904 moles of H+ (2x the moles of H2SO4) and 0.014112 moles of OH-. Excess moles of H+ = 0.014928. Concentration of H+ = 0.014928/(46*10^-3). -log(10) of this = 0.49 to 2 dp.
11. (Original post by Gezza_O'Brien)

I find it helps to write out the actual chemical equations so you can see what's going on

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hmm. So HY decreases so you take away.
Y- increases so you add?
12. (Original post by Parallex)
There's 0.02904 moles of H+ (2x the moles of H2SO4) and 0.014112 moles of OH-. Excess moles of H+ = 0.014928. Concentration of H+ = 0.014928/(46*10^-3). -log(10) of this = 0.49 to 2 dp.
yeah but its diprotic so does multiplying the moles by two in previous steps make 2[H+]
13. (Original post by Sniperdon227)
yeah but its diprotic so does multiplying the moles by two in previous steps make 2[H+]
Yes.
14. http://filestore.aqa.org.uk/subjects...W-QP-JUN08.PDF

Can someone help me with q1di?
15. (Original post by Sniperdon227)
http://filestore.aqa.org.uk/subjects...4-QP-JUN13.PDF
http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

Q3e) stuck on the final bit worked out which was in excess, it was H2SO4, then worked out the conc of H+ (0.014928x1000/46) is this = to 2[H+] or [H]+ im confused is it equal to 2[H+] since weve already multiplied the moles by 2???
Nope it's equal to H+, you only multliply the concentration by 2 when working out moles of H2S04 because it's got 2H+ in it

So xs moles you just do over total volume to get new [H+], then do -log[H+]
16. (Original post by Super199)
hmm. So HY decreases so you take away.
Y- increases so you add?
Yeah exactly. So if you were to add, say, HCl instead of NaOH, then there is more H+ ions in the equilibrium that need to be used, so the equilibrium shifts to the left (to use up H+) so there is less Y- and more HY, so you take the moles of HCl from Y- and you add them to HY, then put the new moles into Kc blah blah blah, does that make sense? Tbf it's pretty confusing but you've just got to think about it in terms of what's happening to the equilibrium

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17. (Original post by hopingmedicinae)
Nope it's equal to H+, you only multliply the concentration by 2 when working out moles of H2S04 because it's got 2H+ in it

So xs moles you just do over total volume to get new [H+], then do -log[H+]
yeah but that 'new [H+]' must be equal to 2[H+] surely
18. (Original post by Sniperdon227)
yeah but that 'new [H+]' must be equal to 2[H+] surely
What do you mean? H2SO4 fully dissociates so the moles of H+ is 2x the moles of H2SO4. That's all it is.
19. (Original post by Parallex)
What do you mean? H2SO4 fully dissociates so the moles of H+ is 2x the moles of H2SO4. That's all it is.
In that question you multiply the moles of H2SO4 by two as you must since its diprotic, we find out that its in excess, by x moles, we divide xmoles by total volume this gives [H+] (or 2[H+]) -log the conc of H+ to give you your pH, whenever working out the pH of diprotic acids you must multiply by two?
20. (Original post by Sniperdon227)
In that question you multiply the moles of H2SO4 by two as you must since its diprotic, we find out that its in excess, by x moles, we divide xmoles by total volume this gives [H+] (or 2[H+]) -log the conc of H+ to give you your pH, whenever working out the pH of diprotic acids you must multiply by two?
Yeah. You're looking at how much excess there is after neutralisation. If it's a strong acid then you have to multiply by 2 because it will fully dissociate and 2H+ is available for neutralisation.

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