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    (Original post by Bruhh)
    Well its symmetrical so thats where the opposite sides comes from. The image is from a UCLA pdf on NMR.

    not sure symmetry has much to do with whether they can interact or not though. lol ok well not the same molecule but who knows
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    Guys if you have further exams, I highly suggest you not to waste time over this thread. Go and revise.

    Majority did bad so the grade boundaries will compensate for it.

    Thank me later.
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    (Original post by Bruhh)
    Hmm okay, but I thought if you had a group on the benzene the environments were slightly different like in this image:

    By the OCR data sheet that's 2 environments not 6. The attached group wasn't an alkyl group anyways like your picture shows. One environment on the left benzene-nitrogen thing, another on the benzene ring on the right attached to the amide. One one the amide.

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    (Original post by YeSand)
    Guys if you have further exams, I highly suggest you not to waste time over this thread. Go and revise.

    Majority did bad so the grade boundaries will compensate for it.

    Thank me later.
    But.. It's too fun
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    (Original post by Zuzuvela)
    By the OCR data sheet that's 2 environments not 6. The attached group wasn't an alkyl group anyways like your picture shows. One environment on the left benzene-nitrogen thing, another on the benzene ring on the right attached to the amide. One one the amide.

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    Well I think both are reasonably valid answers so we'll just have to see
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    freezing temp: -7
    n = 4
    y = 2

    anyone else got these?
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    what was the pressure?
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    (Original post by Zuzuvela)
    6 hydrogen environments is wildly wrong. NMR with more than 4 peaks just doesn't happen at A2.
    You may be correct -1.9 was an answer that a majority got. It wasn't clear what was meant by particles dissolved (already dissolved or to be dissolved) whether it was talking about the ions or the solute overall. Some of those answers were suggested btw not necessary what I put.

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    Wildly wrong?? lol sure
    I got 6 too
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    (Original post by Abbie_1511)
    Wildly wrong?? lol sure
    I got 6 too
    So did I


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    (Original post by Abbie_1511)
    Wildly wrong?? lol sure
    I got 6 too
    There was 6 hydrogen environments apparently according to my teacher but only 3 different ones if you know what I mean. So you used the numbers 1,2,3 and labeled each environment accordingly. The number 3 was used only once I just got the email from my teacher. He's not willing to publish the paper because he's not meant to have it

    Labels 1 and 2 were already done for you you just added a couple of ones and twos and a single 3. Basically if it was a NMR graph it would be 3 peaks total area representing 6 Hs
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    (Original post by Zuzuvela)
    By the OCR data sheet that's 2 environments not 6. The attached group wasn't an alkyl group anyways like your picture shows. One environment on the left benzene-nitrogen thing, another on the benzene ring on the right attached to the amide. One one the amide.

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    Thats 4 environments
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    (Original post by GetMeToUni99)
    what was the pressure?
    I got 0.036 but Im awful at the maths questions so someone else needs to verify
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    Has anyone done the chemistry A ocr paper on 27 May and 10 June 2016??


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    (Original post by WittyGonStein)
    I got 0.036 but Im awful at the maths questions so someone else needs to verify
    I also got 0.036 if I recall correctly.
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    (Original post by Bruhh)
    I also got 0.036 if I recall correctly.
    Would you mind telling me how if you remember. I am interested as it may well be something I need to know for uni if I get there. Thanks

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    (Original post by Zuzuvela)
    Would you mind telling me how if you remember. I am interested as it may well be something I need to know for uni if I get there. Thanks

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    Umm I can't remember the numbers exactly, but 1mol of gas takes up 24dm^3 in 1atm. I think it was asking about what the pressure of 1.5x10-3 moles of gas would be in a 24dm3 volume. So I just did 1.5x10-3 x 24 to get that answer, it'd be helpful if I had the question to refer to as I realise thats an awful explanation.
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    (Original post by Zuzuvela)
    Would you mind telling me how if you remember. I am interested as it may well be something I need to know for uni if I get there. Thanks

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    Just looked online and it says pressure and volume are inversely proportional and so I think you work out the volume and then divide 1 by ths value so I think I got it wrong
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    (Original post by WittyGonStein)
    Just looked online and it says pressure and volume are inversely proportional and so I think you work out the volume and then divide 1 by ths value so I think I got it wrong
    Yeah I got 666 Atmospheres I am still confused

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    hopefully we all did well
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    I think f334 boundaries will be

    A* - 70/90
    A - 64/90
    B - 58/90
    C - 53/90
    D - 47/90
    E - 42/90

    What does everyone else think
 
 
 
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