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    2Marks definitions
    1Mark reacting 1 and 2
    1 Mark for reason
    1Mark reacting 6 and 7
    1 Mark saying why

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    (Original post by Z1228)
    It is lattice enthalphy isn't it?
    no dH solution
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    (Original post by smileez_101)
    so Cu(OH)2 wont react with methanoic acid to form the Cu(HCOO)2 salt?
    Well it's a weak acid reacting with a precipitate that's formed by a reaction with an acid, so I don't think so. I might be totally wrong btw just letting you know cause I almost wrote that too!
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    (Original post by Z1228)
    It is lattice enthalphy isn't it?
    Oh, did you get detla G as -1700ish?
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    Is this the full answer? I don't get it 🤔
    (Original post by lai812matthew)
    there are no protons and carbons.
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    (Original post by HFancy1997)
    2Marks definitions
    1Mark reacting 1 and 2
    1 Mark for reason
    1Mark reacting 6 and 7
    1 Mark saying why

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    how many marks will i lost if i mistaken write 5 and 6 instead of 6 and 7. but i write a correct equation and explanation at the end.....
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    (Original post by smileez_101)
    so Cu(OH)2 wont react with methanoic acid to form the Cu(HCOO)2 salt?
    An acid can react with a solid to produce the salt. I think it's fine. They've confused parts of the question.
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    Damn, for the buffer solution i got the PH of 6.5 but i rounded it up to 7 and i stated it was neutral.

    And for Cr acting both oxidising and reducing agent, i assumed that it was asking for Cr in general and not Cr3+ and that they have to happen simultaneously so i did 4 and 6 i think? The two reactions that have Cr in them since one of those reaction goes oxidation way and the other go towards the reducing way.
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    acid can react with insoluble base.
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    (Original post by Cal 1)
    Well it's a weak acid reacting with a precipitate that's formed by a reaction with an acid, so I don't think so. I might be totally wrong btw just letting you know cause I almost wrote that too!
    if that is true then the exam board made a damn good trick
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    (Original post by lai812matthew)
    acid can react with insoluble base.
    But that was a solid that was already formed by a reaction with an acid, most likely a strong one. Why would it suddenly react with methanoic acid?
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    That paper made me cry.
    Have no idea if I did well but I feel my chances of uni are blown
    Hahahaha asif I thought I'd aim for A* at the beginning of A2
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    What was the ph of buffer solution and how did you calculate it?
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    (Original post by tcameron)
    That paper made me cry.
    Have no idea if I did well but I feel my chances of uni are blown
    Hahahaha asif I thought I'd aim for A* at the beginning of A2
    same. but i have already prepared my tears for this paper before it starts.....
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    What was the answet to this question
    (Original post by smileez_101)
    That question on oxidising and reducing agent was 6 marks and I def got that reducing equation wrong because I said it would react with 3 - 6 redox systems instead of just one.
    How many marks do people think thats worth?
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    is anyone working on an unofficial markscheme ?
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    (Original post by nimoooooooo)
    What was the ph of buffer solution and how did you calculate it?
    3.43 I think, and you just worked out the moles used, then divide each by 1000. No need to deal with reacting masses because conjugate base was given. Plug into Kaacidoversalt, then -log c:


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    What was the buffer pH? Really confused me, I don't think the acid was in excess?
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    (Original post by k.russell)
    I went for the ol' CuCO3, which is also insoluble but I think it would get the mark no? You want to end up with the Cu(HCOO-), because the carbonate and the acid would definitley react, even if it was solid before
    That's true, and overall it may well still work, and so might Cu(OH)2, it's just a dodgey bet considering it technically wanted the Cu2+ ions to be the thing reacting.
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    (Original post by GetOverHere)
    3.43 I think, and you just worked out the moles used, then divide each by 1000. No need to deal with reacting masses because conjugate base was given. Plug into Kaacidoversalt, then -log c:


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    do you think that working out conc of both acid and the salt would be 2/4 marks or 1?
 
 
 
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