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Aqa chem 4/ chem 5 june 2016 thread

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Are specific temperatures and pressures needed for is simply saying "high temperatures" for example sufficient?
Reply 401
Original post by Windowswind123
Are specific temperatures and pressures needed for is simply saying "high temperatures" for example sufficient?


For which processes? At AS we needed to memorise a few temps e.g. for fermentation, but I can't think of any we need to know for A2

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Reply 402
Really hoping for a tough buffer calc to lower grade boundaries because I just spent all afternoon getting my head around them!! :wink:

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Original post by Fibsy
For which processes? At AS we needed to memorise a few temps e.g. for fermentation, but I can't think of any we need to know for A2

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I've learnt 50 degrees Celsius for the nitration of benzene but never seen it come up!
Original post by Fibsy
For which processes? At AS we needed to memorise a few temps e.g. for fermentation, but I can't think of any we need to know for A2

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For most processes I have some kind of temperature down - don't know how reliable they are. Like for the nitration of benzene I have 60 degrees written down.

EDIT:
Original post by Coolworm
I've learnt 50 degrees Celsius for the nitration of benzene but never seen it come up!

Haha, we chose the same process :')
(edited 7 years ago)
http://filestore.aqa.org.uk/subjects/AQA-CHM4-W-QP-JAN08.PDF

http://filestore.aqa.org.uk/subjects/AQA-CHM4-W-MS-JAN08.PDF

Can someone please explain to me why the answer to 4bi is 5??? Any help would be much appreciated :smile:
Reply 406
Original post by Lolaturface
nope only temperature affects the value of the rate constant, K. Increase temp and K increases, decrease temp and K decreases. Nothing else affects.


pressure affects it for sure...
if you increase the pressure in a reaction but you don't change the temperature or concentrations, does the rate go up? (i hope you're thinking yes)
so for rate = K[a]if rate is going up, then one of K, [a], or must be going up.which one is it?
Original post by 12284
How can any of the molecules be spectrum IV then because can't they all lose 1 hydrogen? The answer is meant to be molecule B for spectrum IV but can't it lose a hydrogen from "CH3"?


Yeah but the are talking about 'major ' peaks above an m/z of 100. Those probably could occur but its probably not as abundant. The reason the H breaks off from the CO Group is because the CO+ Ion (acylium ion) has a stability due to the delocalisation of the positive charge across the CO which increases its abundance and likelihood of formation.

Towards the end of the link below, it goes in a bit more detail:

http://www.chemguide.co.uk/analysis/masspec/fragment.html#top
how does elimination of 2-hydroxybutanoic acid with h2s04 work?
Original post by lahigueraxxx
http://filestore.aqa.org.uk/subjects/AQA-CHM4-W-QP-JAN08.PDF

http://filestore.aqa.org.uk/subjects/AQA-CHM4-W-MS-JAN08.PDF

Can someone please explain to me why the answer to 4bi is 5??? Any help would be much appreciated :smile:


This is because the methyl group on the left is in the same environment and the rest of the hydrogens are in the different environment.
Original post by TheEpsilion
This is because the methyl group on the left is in the same environment and the rest of the hydrogens are in the different environment.


Oh crap I was doing it for carbon nmr!! Haha thank you!
Reply 411
Original post by JammyW15
Yeah but the are talking about 'major ' peaks above an m/z of 100. Those probably could occur but its probably not as abundant. The reason the H breaks off from the CO Group is because the CO+ Ion (acylium ion) has a stability due to the delocalisation of the positive charge across the CO which increases its abundance and likelihood of formation.

Towards the end of the link below, it goes in a bit more detail:

http://www.chemguide.co.uk/analysis/masspec/fragment.html#top


Thanks! that was really helpfu!
Reply 413
Hiya can anyone help, why in question 2cii did the mark scheme not divide all the values to get the concentration?
is it a mistake in the mark scheme?
Reply 414
Original post by dingdon
how does elimination of 2-hydroxybutanoic acid with h2s04 work?


you dont need to know the mechanism, for elimination using h2so4 but you need to know where the double bond goes c-c=c-cooh in your case


dividing the moles by the volume 1.005dm-3 (since new total) is what you be doing, try it with this and without you'll get the same answer, but i always use volumes just in case
Reply 417
Original post by Sniperdon227
dividing the moles by the volume 1.005dm-3 is what you be doing, try it with this and without you'll get the same answer, but i always use volumes just in case


if you divide by the volumes you get the wrong ph


That question is a bit eh. 4.13 is pretty much 4.1 which can be used and not the O-H group. The question also states that Compund X has an ester and ketone group in it so no OH is present. The alkyl group in that case is CH2CH3
Original post by Sniperdon227
you dont need to know the mechanism, for elimination using h2so4 but you need to know where the double bond goes c-c=c-cooh in your case


how did you know where double bond goes?

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