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    (Original post by zef1995)
    Thank you! It's good that you feel better about it now
    Exactly, one more exam down, that's gotta be a good thing! Good luck for F324, some of my friends are taking that too so fingers crossed for a good paper!
    Have you got to meet any uni offers?
    I'm offered an ABB Dx ...you?
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    (Original post by Sarangtaec)
    I'm offered an ABB Dx ...you?
    I'm sure you'll be able to get that! Which uni is it if you don't mind me asking?
    I am applying to uni next year, but the unis I'm thinking of require in the region of AAB/AAA, fingers crossed!
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    (Original post by kuku2013)
    hey for the ph of the buffer question did we have to take away the ph of HCOOH from NaOH or visa versa ( I cant remember which way round it is)?
    The method I was taught was to find the moles, and then the concentration of A- and HA at equilibrium. As NaOH reacts with HCOOH, HCOOH is in excess, and all the NaOH reacts with it to form the salt (you work out moles NaOH and that equals the moles of salt formed). The moles of HCOOH at equilibrium is the moles of it to start with - moles NaOH reacted. Then, you work out concentration (in our paper, it was 1 dm^3) of the HCOOH and the NaOH, and plug it into the buffer calculation formula to work out [H+]. -log[H+] = pH. That's how I did it and I got 3.99 for the calculation - what did you get?
    The method you described may well work, I just haven't tried it personally xD
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    Can anyone explain to me how they worked out the last question, it's really bothering me LOL


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    (Original post by zef1995)
    I'm sure you'll be able to get that! Which uni is it if you don't mind me asking?
    I am applying to uni next year, but the unis I'm thinking of require in the region of AAB/AAA, fingers crossed!
    Oooh good luck! I've applied for Manchester =) I'm hoping to get AAB cause I don't think I'll get an A in Chemistry. Lowest I can possibly get is BBC so I really hope I don't get that LOL Dx
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    (Original post by Sarangtaec)
    Oooh good luck! I've applied for Manchester =) I'm hoping to get AAB cause I don't think I'll get an A in Chemistry. Lowest I can possibly get is BBC so I really hope I don't get that LOL Dx
    Thanks very much
    Manchester is great, I hope you can get in there! You never know, you may have done better in this exam than you think, and therefore be able to achieve your A
    I'm sure you won't! Either way, BBC sounds like a good insurance to have
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    Could someone please tell me what page the unofficial mark scheme is one on this thread please.

    Thank you
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    (Original post by zef1995)
    The method I was taught was to find the moles, and then the concentration of A- and HA at equilibrium. As NaOH reacts with HCOOH, HCOOH is in excess, and all the NaOH reacts with it to form the salt (you work out moles NaOH and that equals the moles of salt formed). The moles of HCOOH at equilibrium is the moles of it to start with - moles NaOH reacted. Then, you work out concentration (in our paper, it was 1 dm^3) of the HCOOH and the NaOH, and plug it into the buffer calculation formula to work out [H+]. -log[H+] = pH. That's how I did it and I got 3.99 for the calculation - what did you get?
    The method you described may well work, I just haven't tried it personally xD
    I think I got a similar answer!! its just that I remember doing a similar question in a past paper and thought it would be the same method. if you look on page 132 there is a long method by _HabibaH_, I did a the same method do you think its correct?? !!!
    Thanks anyway
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    (Original post by Holz888)
    It's OK, it's not bothering me
    Well, the entire mechanism must show the overall equation. In mine, the first step was the rate determining step.
    So we started with 2NO2 (it looks like two molecules need to bang into each other for the reaction to proceed, rather than one just falling apart). I decided to do
    2NO2 --> 2NO + O2
    then a fast (NOT rate determining) step could be O2 +2CO --> 2CO2

    Now, if you take the entire mechanism and condense it, you get:
    2NO2 + O2 + 2CO --> O2 + 2CO2
    O2s on both sides cancel
    2NO2 + 2CO --> 2CO2
    Simplify
    NO2 + CO --> CO2
    and that's your initial equation!

    LOL this didnot even come up but thanks somuch for your help
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    (Original post by kuku2013)
    I think I got a similar answer!! its just that I remember doing a similar question in a past paper and thought it would be the same method. if you look on page 132 there is a long method by _HabibaH_, I did a the same method do you think its correct?? !!!
    Thanks anyway
    Yay!
    Yeah me too, I saw a very similar question in June 12, is that where you saw the similar question?
    I've looked at it and it looks correct to me! If you did something similar then that may be a good sign
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    (Original post by zef1995)
    Yay!
    Yeah me too, I saw a very similar question in June 12, is that where you saw the similar question?
    I've looked at it and it looks correct to me! If you did something similar then that may be a good sign
    Did you get

    greater,smaller,greater for the box filling?

    and what did you get for aqueous acid and base

    I did the two equations,

    and talked about oxidisng agent, equilibrium shifting ?
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    has any one got the paper
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    (Original post by kattyratty)
    Can anyone explain to me how they worked out the last question, it's really bothering me LOL


    Posted from TSR Mobile
    Last Question

    Mol of Ni(DMG)2 = 2.57/288.8 = 8.90x10-3 = mol of hydrated Ni compound in 2.5g

    Mr(hydrated Ni compound)=2.5/8.9x10-3= 280.8g/mol

    Mass of H2O=2.5-1.38=1.12g
    Mol H2O in 2.5g = 1.12/18 = 0.062

    Mol H2O/mol Ni compound = 0.062/8.9x10-3 = 7

    Therefore NiX.7H2O

    Mr(X) = Mr(hydrated Ni compound) - 7xMr(H2O) - Mr(Ni)
    = 280.8-126-58.7
    = 96.1

    Therefore X= SO4

    So NiSO4.7H2O
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    (Original post by zef1995)
    Yay!
    Yeah me too, I saw a very similar question in June 12, is that where you saw the similar question?
    I've looked at it and it looks correct to me! If you did something similar then that may be a good sign
    yep that where I saw the question June 2012!! luckily I was going through the paper just before the exam n remembered the calculation. Had to ask on tsr cause there was only 5 of us doing the exam n the other 4 did a different method which got me worried!!!! thanks!!
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    (Original post by otrivine)
    Did you get

    greater,smaller,greater for the box filling?

    and what did you get for aqueous acid and base

    I did the two equations,

    and talked about oxidisng agent, equilibrium shifting ?
    No I didn't, I completely guessed that one because I was very unsure :/
    I'm sorry, which question was that, with the acid and base?
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    (Original post by kuku2013)
    yep that where I saw the question June 2012!! luckily I was going through the paper just before the exam n remembered the calculation. Had to ask on tsr cause there was only 5 of us doing the exam n the other 4 did a different method which got me worried!!!! thanks!!
    Ah, thank goodness we saw it beforehand
    At least you know you've probably done it right, from what I'm gathering! No worries
    Good luck with any other exams you may have!
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    (Original post by zef1995)
    Ah, thank goodness we saw it beforehand
    At least you know you've probably done it right, from what I'm gathering! No worries
    Good luck with any other exams you may have!
    good luck to you too!!!! are you retaking f324?
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    Sensible predictions for the grade boundaries???
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    (Original post by kuku2013)
    good luck to you too!!!! are you retaking f324?
    Thank you very much! Nope I'm not, only got Maths C4 left, that should be interesting :P
    What about you?
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    (Original post by captain anonymous)
    sensible predictions for the grade boundaries???
    a- 73
    b- 65
    c- 57
    d- 49
 
 
 
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