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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread watch

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    (Original post by PlayerBB)
    You need to know all organic reactions, tests, flame test, brief information about redox and Intermolecular forces, electron configuration, I don't know if there's anything else. Same I haven't revised for As yet it's just so annoying, there is so much to revise for each subject

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    Pretty sure you it is unnesasary to be going over flame tests lol.
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    (Original post by usman.zubair)
    Thats true but i think boundaries will be very high. As this is the last exam session for this series & many would be repeating OR giving complete chem A levels. So have to take it srsly :\
    so you think chem4 and 5 boundaries will be extremely high this year?
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    (Original post by Tomasio)
    so you think chem4 and 5 boundaries will be extremely high this year?
    No.im talking about unit 1 there.

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    (Original post by Tomasio)
    Pretty sure you it is unnesasary to be going over flame tests lol.
    Lol, that depends on which exam board are you, for Edexcel you definitely have to revise flame tests
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    Can someone help me with this, I thought It is Phenol because it is slightly soluble in water but very soluble in HCl but that complex with Cu2+ confuses me
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    (Original post by PlayerBB)
    Can someone help me with this, I thought It is Phenol because it is slightly soluble in water but very soluble in HCl but that complex with Cu2+ confuses me
    The amine because amines form coloured complex with Cu2+ ions. Adding aqueous copper sulphate is a functional group test iirc. And I am guessing phenylamine is more soluble in HCl than it is in water because it is protonated and gains charge, more easily hydrated. But not sure if I'm right.
    I remember doing that question though
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    (Original post by C0balt)
    The amine because amines form coloured complex with Cu2+ ions. Adding aqueous copper sulphate is a functional group test iirc. And I am guessing phenylamine is more soluble in HCl than it is in water because it is protonated and gains charge, more easily hydrated. But not sure if I'm right.
    I remember doing that question though
    Can you remind me of this functional group test with copper sulphate, I can't seem to remember it!!

    And yeah, that seems like a reasonable explanation, thank you!! And lol you have a good memory I guess
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    (Original post by PlayerBB)
    Can you remind me of this functional group test with copper sulphate, I can't seem to remember it!!

    And yeah, that seems like a reasonable explanation, thank you!! And lol you have a good memory I guess
    Well if adding [Cu(H2O)6]2+ to an organic compound gives blue precipitate (green if phenylamine) then it's got an amine group
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    Can someone add me to chemistry unit 1 thread
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    (Original post by C0balt)
    Well if adding [Cu(H2O)6]2+ to an organic compound gives blue precipitate (green if phenylamine) then it's got an amine group
    Ahh thank you Cobalt

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    (Original post by PlayerBB)
    Lol, that depends on which exam board are you, for Edexcel you definitely have to revise flame tests
    huh? I do edexcel and have never seen it u 5 past papers.
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    (Original post by Tomasio)
    huh? I do edexcel and have never seen it u 5 past papers.
    Okay but Edexcel are unpredictable so it's better to know them just in case......

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    (Original post by PlayerBB)
    Okay but Edexcel are unpredictable so it's better to know them just in case......
    That they definitely are and very good at it too as seen in the Unit 6 paper!
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    http://qualifications.pearson.com/co...e_20150119.pdf

    Could someone please explain 24 aiii please
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    (Original post by ST_123)
    http://qualifications.pearson.com/co...e_20150119.pdf

    Could someone please explain 24 aiii please
    What's the problem here? Like you are unable to solve the sum?
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    (Original post by ST_123)
    http://qualifications.pearson.com/co...e_20150119.pdf

    Could someone please explain 24 aiii please
    Could you link me to the markscheme for this question please. I'd like to check my answer
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    (Original post by HaydenMoussa)
    Could you link me to the markscheme for this question please. I'd like to check my answer
    https://c838cff4741acb48ae1ed62e5992...0Chemistry.pdf
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    (Original post by sabahshahed294)
    What's the problem here? Like you are unable to solve the sum?
    I just don't really know what to do and how to use all the data ive been given (and why ive been given those data).
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    (Original post by ST_123)
    I just don't really know what to do and how to use all the data ive been given (and why ive been given those data).
    First of all, you have the volume of the titre and the volume of HCl, find the number of moles for the base and the HCl using n=cv. This is done so as to get the moles of ethanoic acid remaining after reacting with NaOH as it's stated that NaOH reacts with both the acids with the word "both" being in bold. Now, once you get the moles, you take the difference of the moles of HCl and NaOH. Now that you have the moles remaining(for ethanoic acid), this is the no of moles at equilibrium for ethanoic acid. Then check the difference of the initial moles and the moles at equilibrium as well. This is to get the moles that have reacted and hence you'll get the moles of the alcohol and ester. For water, you already have the moles and now just you got to add up simply. Substitute the moles in your Kc expression as volume cancels out!
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    (Original post by sabahshahed294)
    First of all, you have the volume of the titre and the volume of HCl, find the number of moles for the base and the HCl using n=cv. This is done so as to get the moles of ethanoic acid remaining after reacting with NaOH as it's stated that NaOH reacts with both the acids with the word "both" being in bold. Now, once you get the moles, you take the difference of the moles of HCl and NaOH. Now that you have the moles remaining(for ethanoic acid), this is the no of moles at equilibrium for ethanoic acid. Then check the difference of the initial moles and the moles at equilibrium as well. This is to get the moles that have reacted and hence you'll get the moles of the alcohol and ester. For water, you already have the moles and now just you got to add up simply. Substitute the moles in your Kc expression as volume cancels out!
    Thanks i did the question now. Although i still don't quite understand how finding the mol of hcl and naoh and subtracting the two gives me the mol left in equilibrium of ethanoic acid.
 
 
 
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