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AQA CHEM4 and CHEM5 revision thread 2016

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Original post by Signorina
Yes ur correct in sense for the born haber cycle you would use -695 because you need two electron affinities

However The question is "what's the electron affinity of chlorine?

Think of it as:
2 x electron affinity of Cl = -695

You just want the electron affinity of Cl so

Electron affinity of Cl = -695/2

It's hard to explain and I understand why you're struggling to understand it but just remember the definition of electron affinity... The energy change when one mole of gaseous atoms gains one mole of electrons. What you found out is the energy change when TWO moles of gaseous atoms ......etc

Hence why you need to divide by two.


Thanks a lot your explanation has made it much clearer in my head
Original post by ahsan_ijaz
Thanks a lot your explanation has made it much clearer in my head


Is this exam for the old A Level spec??
(edited 7 years ago)
Original post by L1234567
Is this exam for the old A Level spec??


Yes
This is exam for the new a level spec hasn't even started


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Original post by Bloom77
Yes
This is exam for the new a level spec hasn't even started


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Ohh, I just wanted to know whether they will still make papers for the new A Level papers.
Reply 44
can someone help me with this question 5b(i) to 5(b)iv its structure determination

I got the answer right when asked to deduce the structure of P, BUT my structure was the other way around, (so the same as the mark scheme if read in the opposite direction) do you get penalised for this, as i didnt see anything that said you would in the additional comments. if you do, then can someone explain to me how you know which way to draw the structure. thank you!
past paper question 5biv: http://filestore.aqa.org.uk/subjects/AQA-CHEM4-QP-JAN13.PDF
mark scheme: http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-MS-JAN13.PDF
Original post by chzm
can someone help me with this question 5b(i) to 5(b)iv its structure determination

I got the answer right when asked to deduce the structure of P, BUT my structure was the other way around, (so the same as the mark scheme if read in the opposite direction) do you get penalised for this, as i didnt see anything that said you would in the additional comments. if you do, then can someone explain to me how you know which way to draw the structure. thank you!
past paper question 5biv: http://filestore.aqa.org.uk/subjects/AQA-CHEM4-QP-JAN13.PDF
mark scheme: http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-MS-JAN13.PDF


I did that as well but i think there is no problem.
Hi all, can you please post the hardest questions in unit 4 or 5 that you have experienced. Also, please state the source it is from if possible.Thanks a lot.
For the polymers topic, what exactly do we have to specificically have to memorise/regurgitate re: Kevlar, Nylon and terylene.

Do we need to know the monomers for each by heart? Or just how to form the polyester/polyamide once they give us the monomers or?

Thanks
Original post by hi-zen-berg
For the polymers topic, what exactly do we have to specificically have to memorise/regurgitate re: Kevlar, Nylon and terylene.

Do we need to know the monomers for each by heart? Or just how to form the polyester/polyamide once they give us the monomers or?

Thanks


I understand you have to know the polymers by heart, but if you memorise the monomers, then you can make the polymers easily :smile:


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Hi
I don't seem to understand the answer to this question
ImageUploadedByStudent Room1464851560.777058.jpg

Answer:
ImageUploadedByStudent Room1464851581.754970.jpg


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Original post by Bloom77
Hi
I don't seem to understand the answer to this question



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So enthalpy of solution is when the ionic compound is dissolved in enought water making sure the ions dont interact. When this happens, the equation will show 1 reactant producing 2 reactants. So the no. Of moles on RHS is higher than LHS. increase in moles=increase in entropy so the value of T¤S will always be bigger than H and so G will always be negative.
Original post by theboss1998
So enthalpy of solution is when the ionic compound is dissolved in enought water making sure the ions dont interact. When this happens, the equation will show 1 reactant producing 2 reactants. So the no. Of moles on RHS is higher than LHS. increase in moles=increase in entropy so the value of T¤S will always be bigger than H and so G will always be negative.


OMFG YOUR A STAR!
Thank you!


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Original post by Bloom77
OMFG YOUR A STAR!
Thank you!


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No problem. Happy to help!

Any body else who has a question either post it here or PM me and I will explain it.
I was wondering if someone could explain 8ei for me? I answered it so that it acted as a ligand on Aluminium but in the mark scheme it acts as a base- how can you distinguish whether it will act as a ligand or a base?
Original post by youngar
I was wondering if someone could explain 8ei for me? I answered it so that it acted as a ligand on Aluminium but in the mark scheme it acts as a base- how can you distinguish whether it will act as a ligand or a base?


Ok so as a ligand you get full substitution. As a base the diaminoethane acts as brownsted-lowry base, so it will accept a proton. It will react exactly like limited ammonia, so in limited ammonia you get NH4+, so with the diaminoethane you will get +H3N-CH2-CH2-NH3+ because it accepts the proton.
Reply 55
Original post by youngar
I was wondering if someone could explain 8ei for me? I answered it so that it acted as a ligand on Aluminium but in the mark scheme it acts as a base- how can you distinguish whether it will act as a ligand or a base?


It gave you a clue in the question. It says 1,2-diaminoethane (en) acts as a base and as a ligand. It then gives you 2 cases, one where it reacts with cobalt (II) ions which is a common question so you you should know that en acts a ligand with Co(II) compounds. The next situation, from the clue in the question you should suspect that en will acts as a base - as a Bronstead-Lowry base and not as a ligand.
Original post by youngar
I was wondering if someone could explain 8ei for me? I answered it so that it acted as a ligand on Aluminium but in the mark scheme it acts as a base- how can you distinguish whether it will act as a ligand or a base?


Check out the last page of chem revise transition metal notes



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Original post by B_9710
It gave you a clue in the question. It says 1,2-diaminoethane (en) acts as a base and as a ligand. It then gives you 2 cases, one where it reacts with cobalt (II) ions which is a common question so you you should know that en acts a ligand with Co(II) compounds. The next situation, from the clue in the question you should suspect that en will acts as a base - as a Bronstead-Lowry base and not as a ligand.


One thing I don't understand about this question is that the biddentat ligand does not replace all the waters in aluminium.

And sorry, I meant page 11 on chem revise


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Original post by rxns_00
Anyone got anything on organic synthesis? I can't deal with it man, I have no clue when it comes to answering a synthesis question lol


Aye I lav organic synthesis.
Original post by theboss1998
Ok so as a ligand you get full substitution. As a base the diaminoethane acts as brownsted-lowry base, so it will accept a proton. It will react exactly like limited ammonia, so in limited ammonia you get NH4+, so with the diaminoethane you will get +H3N-CH2-CH2-NH3+ because it accepts the proton.


Original post by B_9710
It gave you a clue in the question. It says 1,2-diaminoethane (en) acts as a base and as a ligand. It then gives you 2 cases, one where it reacts with cobalt (II) ions which is a common question so you you should know that en acts a ligand with Co(II) compounds. The next situation, from the clue in the question you should suspect that en will acts as a base - as a Bronstead-Lowry base and not as a ligand.


Original post by Bloom77
Check out the last page of chem revise transition metal notes



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Thank you all, but I think I'm still missing something... I understand that it can act both as a base and as a ligand but why does it act as a base in this case? en is in excess so I assumed it would be a ligand.


Edit: okay ignore me, aluminium doesn't undergo ligand exchange with ammonia so it wouldn't with this either... Silly me haha.
(edited 7 years ago)

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