OCR Chemistry A 2016 unofficial mark scheme 27/05/16

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    (Original post by Dynamic_Vicz)
    I think i got a B on the paper and not quite the A grade. What marks will i need in the next paper (Depth) to get an overall A grade for AS Chemistry.
    What are you basing the B off of? There are no grade boundaries for this spec. Nobody knows, is the short answer. I guess I could just say "get 100% and that should help get an A".
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    (Original post by Dynamic_Vicz)
    I think i got a B on the paper and not quite the A grade. What marks will i need in the next paper (Depth) to get an overall A grade for AS Chemistry.
    If you got a B that equates to 70 UMS if you got the exact mark for a B , so to get an A overall (160 UMS) you'd need to get 90 UMS. I predict that an A will be ~ 58 on this paper - so if we assume the next paper is similar (probably not) then you'd need to get 65-70/70 In my opinion. (For 90+ UMS)
    This is speculative, and there's no way of basing estimated grades. However, I feel these grade boundaries will be roughly correct;

    100 UMS (A) - 69+
    90 UMS (A) - 64+
    80 UMS (A) - 58
    70 UMS (B) - 50
    60 UMS (C) - 44
    50 UMS (D) - 37
    40 UMS (E) - 31
    0 UMS (U) - <= 30

    Remember everyone - a C can be turned into an A overall with 100 UMS on paper 2; get revising for Depth.
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    (Original post by jadeemma)
    I got -19.5 ... Didn't you have to times the number of moles by three?
    I got -19.5 too... I don't see what I did wrong
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    I'm glad that you also got loads of B answers in the multiple choice.
    I couldn't quite remember the reagents but as a whole I think the calculations were nice and solveable and as a whole the paper wasn't too difficult.
    For more it felt incredibly similar to the practice paper and specimen paper that they posted which I did as mocks.
    Grade boundaries will probably be high but hopefully paper 2 is also very similar to the practice paper
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    Was the last mechanism worth 2 marks?
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    (Original post by alkaline.)
    I thought it was tetrahedral bc that's just generally what it is in alkanes
    Ive realised now that in graphene each carbon is bonded to 3 other carbons and theres one free floating or something so idk if it's trigonal planar ... ://

    upon googling: http://virtuallaboratory.colorado.ed...ter3txt-3.html
    "This geometry is called trigonal planar and the C–C–C bond angle is 120°"

    fuc.
    Literally just read this from the same link you posted.

    "But in fact, carbon does form four bonds in graphite (carbon almost always forms four bonds - a central principle of organic chemistry.) The trick is that the four bonds are not always equivalent; in graphite, the fourth bond is not formed by the sp2 bonding orbitals, but rather involves an unhybridized 2p atomic orbital."

    So let's not over think it!


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    (Original post by MsDanderson)
    Literally just read this from the same link you posted.

    "But in fact, carbon does form four bonds in graphite (carbon almost always forms four bonds - a central principle of organic chemistry.) The trick is that the four bonds are not always equivalent; in graphite, the fourth bond is not formed by the sp2 bonding orbitals, but rather involves an unhybridized 2p atomic orbital."

    So let's not over think it!


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    CGP revision guide says its tetrahedral
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    (Original post by The gains kinggg)
    what calc do you have
    I have a Sharp EL-W531 and also a Casio FX9750-Gii
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    (Original post by laney1999)
    CGP revision guide says its tetrahedral
    For graphene, as each carbon has three bonding pairs of electrons, it's trigonal planar
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    (Original post by medicapplicant)
    Alright good and I swear that water of crystallisation question answer was CrCl3.6H20 like someone above
    How many marks was this question?
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    (Original post by Toffo132)
    If you got a B that equates to 70 UMS if you got the exact mark for a B , so to get an A overall (160 UMS) you'd need to get 90 UMS. I predict that an A will be ~ 58 on this paper - so if we assume the next paper is similar (probably not) then you'd need to get 65-70/70 In my opinion. (For 90+ UMS)
    This is speculative, and there's no way of basing estimated grades. However, I feel these grade boundaries will be roughly correct;

    100 UMS (A) - 69+
    90 UMS (A) - 64+
    80 UMS (A) - 58
    70 UMS (B) - 50
    60 UMS (C) - 44
    50 UMS (D) - 37
    40 UMS (E) - 31
    0 UMS (U) - <= 30

    Remember everyone - a C can be turned into an A overall with 100 UMS on paper 2; get revising for Depth.
    what is the weighting for each exam?
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    (Original post by laney1999)
    How many marks was this question?
    6H20 is correct and i think 3/4 marks
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    (Original post by alkaline.)
    I thought it was tetrahedral bc that's just generally what it is in alkanes
    Ive realised now that in graphene each carbon is bonded to 3 other carbons and theres one free floating or something so idk if it's trigonal planar ... ://

    upon googling: http://virtuallaboratory.colorado.ed...ter3txt-3.html
    "This geometry is called trigonal planar and the C–C–C bond angle is 120°"

    fuc.
    Revision guide said that diamond was tetrahedral
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    (Original post by SGHD26716)
    Revision guide said that diamond was tetrahedral
    Graphene and diamond are allotropes of carbon, they have different structures
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    (Original post by Toffo132)
    If you got a B that equates to 70 UMS if you got the exact mark for a B , so to get an A overall (160 UMS) you'd need to get 90 UMS. I predict that an A will be ~ 58 on this paper - so if we assume the next paper is similar (probably not) then you'd need to get 65-70/70 In my opinion. (For 90+ UMS)
    This is speculative, and there's no way of basing estimated grades. However, I feel these grade boundaries will be roughly correct;

    100 UMS (A) - 69+
    90 UMS (A) - 64+
    80 UMS (A) - 58
    70 UMS (B) - 50
    60 UMS (C) - 44
    50 UMS (D) - 37
    40 UMS (E) - 31
    0 UMS (U) - <= 30

    Remember everyone - a C can be turned into an A overall with 100 UMS on paper 2; get revising for Depth.
    An A is going to be less than 58, no matter how easy you think the paper was. 58 would be 83%. The highest it has ever been is 81% and that was on one of the old papers, back when only half the course was assessed in each test, add to that the fact that it is a new syllabus and the average A boundary over the past few years has been 75% I would assume 52 for an A.
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    I think the grade boundaries will be:

    A - 53/54
    B - 47
    C - 40
    D - 35
    E - 28

    *The paper was a bit challenging at the end. It won't be higher in my opinion because this is a new specification and this spec covers content from all four modues. The old specification had exams split into different modules and it didn't even exceed or got any closer to 80% for an A. Usually an A should be 56, however as this is a new exam being introduced, I do not believe A would be 56 or any higher.
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    (Original post by laney1999)
    How many marks was this question?
    The question was 3 marks.
    2 marks probably for finding the empirical formula.
    1 mark for giving your answer in terms of water of crystallisation.

    I got CrCl3. 6H20 as well.
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    any other questions people remember?
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    So what were the 2 steps for the last question with the 2 OH groups?
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    (Original post by oni176)
    I think the grade boundaries will be:

    A - 53/54
    B - 47
    C - 40
    D - 35
    E - 28

    *The paper was a bit challenging at the end. It won't be higher in my opinion because this is a new specification and this spec covers content from all four modues. The old specification had exams split into different modules and it didn't even exceed or got any closer to 80% for an A. Usually an A should be 56, however as this is a new exam being introduced, I do not believe A would be 56 or any higher.
    This seems realistic. Did anyone keep a record of their multiple choice answers? I rushed through it and think I got quite a few wrong but I got: BDDCCCBDCBCBACBADBAD
 
 
 
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