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    (Original post by redsoules)
    What would be REALLY helpful, is it someone could just remember and label like this:

    1.
    2.
    3.
    etc


    with the marks that question offered. Then I may be able to roughly calculate it. I just can't remember what each question was worth and estimating without that leave me in a mess.
    Where is the unofficial markscheme
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    (Original post by redsoules)
    Others did. It's not correct though unfortunately. The 2Fe2 signifies a covalent bond (the lower 2 on the right of Fe) and covalent bonds are not comprised of metals (non-metals only). So it really should be 4Fe
    I got 2FeO3 + 3C -> 2Fe + 3CO2
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    Yaasssss chem sucks so badddd
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    (Original post by Jamie Vardy)
    Realistically a C will be 38, a B 44, an A 51 and an A* most likely 56.
    Your'e having a laugh, at max A* will be 53/4
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    it was v easy last year swell and grade boundaries were not amazingly high either
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    (Original post by Jamie Vardy)
    ARE YOU AN IDIOT?

    You've forgotten about the grade boundaries...
    LMAO this actually made me laugh so hard !!
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    (Original post by Teenw123)
    Your'e having a laugh, at max A* will be 53/4
    Ikr at least A* would be 51/52 and A would be 41/42
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    For the question about Iron and Steel if i wrote it's an alloy so it's stronger what mark would i get out of 3 ?
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    (Original post by Elliepinhorne)
    I got 2FeO3 + 3C -> 2Fe + 3CO2
     \text{Fe}\text{O}_3 ?
    Iron(VI) oxide?
    That wouldn't be right.
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    Same with CIE
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    (Original post by Elliepinhorne)
    I got 2FeO3 + 3C -> 2Fe + 3CO2
    You can't have that because the question/words printed on the paper didn't have that my darling.

    Fe2O3 + C -> ....... + CO2

    Balance this out like so...

    2Fe2O3 + 3C ----> 4Fe + 3CO2
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    (Original post by Teenw123)
    Your'e having a laugh, at max A* will be 53/4
    A* will be 50 marks max
 
 
 
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