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    (Original post by Funtry)
    I chose lactic acid for by removal of the other three:-
    Benzoic acid is carcinogenic (Unit 4, benzene based compounds tend to be carcinogenic), pyruvic acid is a ***** to get at (unless we want whole factories full of metabolising cells!) and if we used acetic acid then we'd have yummy vinegar sweets. That was my thought process anyway.
    I still find this so bizarre. I'm sure there was a systematic way of choosing, but hopefully I can replicate what you got on June 2012.

    January 2012 was slightly easier but I was still only around 80. The same questions on the Redox titrations got me.

    Will come back on Sunday or something, logging out of Studentroom for a few days.

    Good lucK!!!
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    (Original post by otrivine)
    did you get the right answer by any chance, can you show me your working.
    I just looked at the mark scheme which was
    rate = 1.0 × 10–12 /4 = 2.5 × 10–13 (mol dm–3 s–1), and worked backwards. I was really puzzled at first, but after re-reading it, it says the initial rate data in the table is for the formation of methanol, so you have to use that value

    Experiment 1's initial rate of formation of HCHO is 1.0 × 10–12, and the stoichiometry of HCHO : O2 is 2 : 0.5 from the equation:

    O3(g) + C2H4(g) <-> 2HCHO(g) + ½ O2(g)

    -you have to use the initial rate of formation of methanol that's given, and divide it by 4 - ( 2 : 0.5 = 4 : 1 ), to get the initial rate of formation of oxygen.
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    Revision for this is not going well. However I try to revise some sections it just does not go in. I scraped a B on my mock (which was Jan 2013) but I need an A to get into uni so I'm a bit worried

    I started making an idiot's guide to F325 for myself, going through the whole section in the textbook and explaining it in really easy ways with silly pictures, hopefully that'll help.
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    (Original post by LRJ)
    I just looked at the mark scheme which was
    rate = 1.0 × 10–12 /4 = 2.5 × 10–13 (mol dm–3 s–1), and worked backwards. I was really puzzled at first, but after re-reading it, it says the initial rate data in the table is for the formation of methanol, so you have to use that value

    Experiment 1's initial rate of formation of HCHO is 1.0 × 10–12, and the stoichiometry of HCHO : O2 is 2 : 0.5 from the equation:

    O3(g) + C2H4(g) <-> 2HCHO(g) + ½ O2(g)

    -you have to use the initial rate of formation of methanol that's given, and divide it by 4 - ( 2 : 0.5 = 4 : 1 ), to get the initial rate of formation of oxygen.
    Oh I see , trick question that , thank you
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    (Original post by otrivine)
    Oh I see , trick question that , thank you
    If its a 1 mark make sure you've read the question properly. You'll pretty much always just have to divide or multiply by something.
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    Is anyone going over the AS / F324 stuff, because F325 supposed to be synoptic?
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    (Original post by shiggle)
    Is anyone going over the AS / F324 stuff, because F325 supposed to be synoptic?
    Lucky enough for me I don't have to as I still know it. Unlucky for me is I'm sitting the other exams too, that's the reason I know it all -.-
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    (Original post by Phil90974)
    Could somebody help me June 2012 question 3cii please?
    I'm looking at the markscheme and it still isnt making much sense to me

    Many thanks
    You can work out the number of moles of butanoic acid (50/1000 * 0.25); and the no. of moles of NaOH (50/1000 * 0.05). However, the NaOH neutralised some of the acid. So... the acid after ome has been neutrolised to water is no moles of acid - no moles of NaOH.

    So now you have
    n(NaOH) = 50/1000*0.05 = 2.5x10-3mol
    n(Acid ) = 50/1000*0.25 - 50/1000*0.05 = 0.01mol


    Now you can work out the concentrations of both
    as C * V = N ; C = N / V.

    So c(NaOH) = 2.5x10-3 / (100 /1000) = 0.025 M
    and c(acid) = 0.01 / (100/1000) = 0.1 M
    *100 as both together (50cm3 + 50cm3)

    But, as:

    HA + NaOH --> H20 + A-NA+

    the NaOH == the COO- (
    CH3(CH2)2COO– )

    So now just put the concentrations of NaOH ( as ...COO-) and the acid (as ...COOH) and Ka, then rearrange to get H+ conc.

    Don't forget to -log10[H+] to get your pH.
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    Can someone help me with june 2011 question 8? I'm unsure as to how the copper precipitate is formed in the 2nd ionic equation as well as the equation for step 4. Thanks
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    (Original post by whitedaisy)
    Can someone help me with june 2011 question 8? I'm unsure as to how the copper precipitate is formed in the 2nd ionic equation as well as the equation for step 4. Thanks
    Step 2 :
    Because it says aq sodium carbonate, is added to neutralise any acid, you know that the CO32- must have reacted with H+ so:
    2H+ + CO32- --> CO2 + H2O (This is the gas that is formed)

    and the rest must've reacted with the Cu2+


    Cu2+ + CO3 2- ->> CuCO3 (This is the ppte that is formed)

    For Step 4 you know that CuI and I2 must be formed, so what I did was I wrote that as my products :

    ____________ --> CuI + I2

    (I think this reaction is in the spec)

    Then I put the most likely reactants :

    Cu2+ + I- --> CuI + I2

    and then balanced it (not that it's not 3I-, as then the charges on each sides of the equation don't match.

    2Cu2+ + 4I- --> 2CuI + I2
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    (Original post by shiggle)
    Step 2 :
    Because it says aq sodium carbonate, is added to neutralise any acid, you know that the CO32- must have reacted with H+ so:
    2H+ + CO32- --> CO2 + H2O (This is the gas that is formed)

    and the rest must've reacted with the Cu2+


    Cu2+ + CO3 2- ->> CuCO3 (This is the ppte that is formed)

    For Step 4 you know that CuI and I2 must be formed, so what I did was I wrote that as my products :

    ____________ --> CuI + I2

    (I think this reaction is in the spec)

    Then I put the most likely reactants :

    Cu2+ + I- --> CuI + I2

    and then balanced it (not that it's not 3I-, as then the charges on each sides of the equation don't match.

    2Cu2+ + 4I- --> 2CuI + I2
    Ahh thanks! What do you mean by the charges balancing out? I originally thought it was 3I- before looking at the mark scheme.
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    (Original post by whitedaisy)
    Ahh thanks! What do you mean by the charges balancing out? I originally thought it was 3I- before looking at the mark scheme.
    In reactions the total charge on the left hand side has to match the right hand side.

    For example the total charge on the left in this reaction is -3; and on the right it's 0, so it can't be right:

    Cu2+ + 3I- --> CuI + I2
    0 + (3*-1) =/= 0 +0

    But on this the total charges on both sided balance to 0:

    2Cu2+ + 4I- --> 2CuI + I2

    2*2 + 4*-1 = 0 + 0
    4 -4 0 + 0
    =0 =0


    EDIT: though of a better example :

    In balancing the oxidation of I- to I2 you first write out the two species and balance the atoms:

    2I- --> I2

    because there's 2 (one) negative Iodide ions on the left and I2 has no charge, you need to make the equation balance in terms of charge -- you need to either add 2+ on the the left hand side to make both sides equal 0, or add 2- on the right to make both sides equal -2; you do the latter:

    2I- --> I2 + 2e-
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    Anybody worried about timing? I've done all past papers but didn't do them under exam conditions apart from one as I've already done bits of each paper as I've been going along, so saw no point in it really.

    In the one I did I had 10 minutes to spare but I'd already done 3 of the 8 questions so had to think about them less, so probably would have run out of time in reality...

    So any tips for saving time? I seem to be really over the top on being neat with my drawing of molecules/measuring electrode or cell potentials, so gonna try and be a bit less over the top I think.
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    (Original post by shiggle)
    In reactions the total charge on the left hand side has to match the right hand side.

    For example the total charge on the left in this reaction is -3; and on the right it's 0, so it can't be right:

    Cu2+ + 3I- --> CuI + I2
    0 + (3*-1) =/= 0 +0

    But on this the total charges on both sided balance to 0:

    2Cu2+ + 4I- --> 2CuI + I2

    2*2 + 4*-1 = 0 + 0
    4 -4 0 + 0
    =0 =0


    EDIT: though of a better example :

    In balancing the oxidation of I- to I2 you first write out the two species and balance the atoms:

    2I- --> I2

    because there's 2 (one) negative Iodide ions on the left and I2 has no charge, you need to make the equation balance in terms of charge -- you need to either add 2+ on the the left hand side to make both sides equal 0, or add 2- on the right to make both sides equal -2; you do the latter:

    2I- --> I2 + 2e-
    Thank you! I can't believe we haven't been taught that for ionic equations, I know how to balance charges for half equations though thankfully.
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    Guys now I know its late but can I ask what everyone got for the practical exams out of 40? Do you think the grade boundaries will be high again this year?
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    (Original post by Gulzar)
    Guys now I know its late but can I ask what everyone got for the practical exams out of 40? Do you think the grade boundaries will be high again this year?
    31 unfortunately, I always crack during practicals, not sure why. I hoping that's at least 40 UMS.

    Grade boundaries are likely to be high, they are every year, make sure to do your best on the paper.

    It's funny how 80% on coursework equates to ~70% UMS and 70% on this paper equates to 80% UMS. They really need to sort out the papers lol.
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    (Original post by Gulzar)
    Guys now I know its late but can I ask what everyone got for the practical exams out of 40? Do you think the grade boundaries will be high again this year?
    I got 37/40 so I'm hoping that'll be high enough to get me the A grade, then all I have to do is smash to last two units!


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    (Original post by Gulzar)
    Guys now I know its late but can I ask what everyone got for the practical exams out of 40? Do you think the grade boundaries will be high again this year?
    38, that'll probably be a mid-low A. It's annoying seeming as when 80% of the schools do these practical's properly, the other 20%, who blatantly cheat get full marks and raise the grade boundaries for those who deserve a higher mark.
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    (Original post by FailedMyExams)
    31 unfortunately, I always crack during practicals, not sure why. I hoping that's at least 40 UMS.

    Grade boundaries are likely to be high, they are every year, make sure to do your best on the paper.

    It's funny how 80% on coursework equates to ~70% UMS and 70% on this paper equates to 80% UMS. They really need to sort out the papers lol.
    As quoted by the last person...its because some schools are so **** they have to cheat to get their numbers up and ultimately this makes the grade boundaries shoot up. so annoying!
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    Anybody want to do some questions?
 
 
 
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