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OCR Chemistry A Exam Thread (Breadth - May 27 2016 and Depth - June 10 2016) Watch

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    (Original post by usmanahmed321)
    it was butanoic aCid
    you had TO times the EF C2H4O by two to get C4H802 OF WHICH THE MR WAS 88
    THERE WAS ALSO AN M/Z PEAK AT 88 SUGGESTING IT WAS BUTANOIC ACID
    IT COULDNT HAVE BEEN C2H40 SIMPLY BECUASE THAT ISNT A CARBOXYLIC ACID
    THE IR SPEC CLEARLY INDICATED IT WAS A C ACID
    THE MAS SPEC ALSO SAID THE MOST INTENSE PEAK WAS CAUSED BY A CARBOCATION WHICH WAS AT 43 OR 44
    SO THE MR WAS 88
    anyone agree?
    I think for sure it was an isomer of butanoic acid, but I thought the only way for it to make a secondary carbocation was for it to have a methyl group, so methylpropanoic acid?
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    (Original post by so_321)
    Wasn't it 67.5
    It was 67.43... which rounds to 67.4%
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    (Original post by 11234)
    67 percent alright to 2sf yea? Maybe?
    What did you get?
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    (Original post by 4nonymous)
    Yh it would. Also I think for yield it was 67.4%
    As for the last question 43 peak does work with butanoic acid because its (CH3CH2CH2)+ ? Although I aint sure
    I think I might have gotten 67.4% just can't remember I know I had sixty something . 4 or 5
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    (Original post by 4nonymous)
    It was 67.43... which rounds to 67.4%
    it was 67.45 which rounds to 67.5%
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    (Original post by 4nonymous)
    It was 67.43... which rounds to 67.4%
    Is 67 alright
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    No + o3 --> no2 o2
    no2 + o --> no + o2
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    thank god other people also put 2-methylpropanoic acid😅😅 what do you guys think grade boundaries will be like then? still 80% for an A as it was quite an easy paper ?
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    (Original post by 11234)
    Is 67 alright
    Probably
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    For the last question I put butanoic acid instead of 2 methyl propanoic acid :facepalm: Figured it out 30 seconds after the exam finished ffs

    How many marks will I lose? 2? 3?

    Posted from TSR Mobile
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    (Original post by Tyler213)
    it was 67.45 which rounds to 67.5%
    No it was 67.43. Did you use exact values or rounded values because I used exact.
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    (Original post by awsomedude)
    what did people write for the ozone depletion propagation steps?
    no• + o3 -> no2• + o2
    no2• + o -> no• + o2
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    (Original post by Tyler213)
    it was 67.45 which rounds to 67.5%
    Yeah I think that's what it was
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    Did anyone get near 5.7x10-4 for rate
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    (Original post by 4nonymous)
    No it was 67.43. Did you use exact values or rounded values because I used exact.
    They will probably allow 67.4 to 67.5
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    (Original post by asinghj)
    bo• + o3 -> no2• + o2
    no2• + o -> no• + o2
    Would we lose marks dor puting the dot on the wrong element I put it next to nitrogen
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    (Original post by TH3-FL45H)
    For the last question I put butanoic acid instead of 2 methyl propanoic acid :facepalm: Figured it out 30 seconds after the exam finished ffs

    How many marks will I lose? 2? 3?

    Posted from TSR Mobile
    1, if that. If a lot of people but butanoic acid they might accept it... because it's not wrong as such, it's just a different isomer. I did the same as you
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    (Original post by asinghj)
    They will probably allow 67.4 to 67.5
    Yh was thinking the same
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    nobody wanna help me do the unofficial mark scheme?
    http://www.thestudentroom.co.uk/show....php?t=4156327
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    in the enthalpy profile diagram did it say not to label the activation energy?
 
 
 
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