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    (Original post by MathsNerd1)
    Well that paper was absolutely vile! It wasn't too bad until I got to the last few questions then it all just fell apart! :-/


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    How did you find the curve drawing for the titration and what did you get for n the last question
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    (Original post by otrivine)
    How did you find the curve drawing for the titration and what did you get for n the last question
    Pretty simple as the first was a strong acid base one and the second was a weak acid strong base one. I got n = 2 but not through many calculations, more from the knowledge of the oxidation states and colours of Vanadium. I probably won't get much of anything for that though :-/


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    (Original post by MathsNerd1)
    Pretty simple as the first was a strong acid base one and the second was a weak acid strong base one. I got n = 2 but not through many calculations, more from the knowledge of the oxidation states and colours of Vanadium. I probably won't get much of anything for that though :-/


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    Yes that is correct, and did you get magnesium for the 5 mark question with the chromium
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    (Original post by otrivine)
    Yes that is correct, and did you get magnesium for the 5 mark question with the chromium
    I'm sure it was only 4 marks and I couldn't see how I could do it, so I guessed with copper :-/


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    (Original post by MathsNerd1)
    I'm sure it was only 4 marks and I couldn't see how I could do it, so I guessed with copper :-/


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    oh , that was a tough one, what do you think you got, I think the Jan 2013 was the more difficult paper out of all the years especially it makes it difficult under exam conditions
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    (Original post by otrivine)
    oh , that was a tough one, what do you think you got, I think the Jan 2013 was the more difficult paper out of all the years especially it makes it difficult under exam conditions
    Like probably not even half? Just an appalling attempt really :-/


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    (Original post by MathsNerd1)
    Like probably not even half? Just an appalling attempt really :-/


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    (Original post by otrivine)
    oh , that was a tough one, what do you think you got, I think the Jan 2013 was the more difficult paper out of all the years especially it makes it difficult under exam conditions

    (Original post by MathsNerd1)
    I'm sure it was only 4 marks and I couldn't see how I could do it, so I guessed with copper :-/


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    (Original post by otrivine)
    Yes that is correct, and did you get magnesium for the 5 mark question with the chromium

    (Original post by MathsNerd1)
    Pretty simple as the first was a strong acid base one and the second was a weak acid strong base one. I got n = 2 but not through many calculations, more from the knowledge of the oxidation states and colours of Vanadium. I probably won't get much of anything for that though :-/


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    Hi guys, just wondering if you could help me on a question. On the june 2011 paper question 4dii, and 4e (how do you know it's lactic acid?)

    Thank you
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    (Original post by master y)
    Hi guys, just wondering if you could help me on a question. On the june 2011 paper question 4dii, and 4e (how do you know it's lactic acid?)

    Thank you
    I presume this is the magic tang question this is because Pka value is equivalent to the PH value
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    (Original post by otrivine)
    I presume this is the magic tang question this is because Pka value is equivalent to the PH value
    yeah... how do you that pka = pH
    Also, you can still get marks for saying like acetic acid for example?... I think i got 4/6 marks... not really sure, so would have got 5/6 if i said lactic
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    (Original post by master y)
    yeah... how do you that pka = pH
    Also, you can still get marks for saying like acetic acid for example?... I think i got 4/6 marks... not really sure, so would have got 5/6 if i said lactic
    Because both are involved in the anti log
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    If Your revising now, YOU WILL FAIL , IT IS TOO LATE!
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    Who dislike, it is the harsh truth!

    Viva La India
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    (Original post by H0ls)
    Question 8(c)(ii)
    Moles of KMnO4 = 13.2/1000 x (2.25 x 10^-2) = 2.97 x 10^-4
    Moles of Vn (50cm) = 0.126/50.9 = 2.47 x 10^-3
    Moles of Vn (10cm sample) = 2.47 x 10^-3 / 5 = 4.95 x 10^-4

    Therefore we have VN (4.95) : KMnO4 (2.95) (both with x 10^-4) so we roughly have a 5 : 3 ratio


    We know in MnO4 that the electron transfer is + 5e-, we have 3 moles. So we have 15e- being transferred.
    Therfore in 5 moles of Vn2+ we have 15e-. 15/5 = 3. Therefore we must have 3e- being transferred.

    VO3- has a V state of +5, so if we take away these 3 electrons then we have an oxidation state of 2+. This is also clarified by the violet colour of the solution which is given in the question.

    Half equation: MnO4- + 8H+ + 5e- ----> Mn2+ + 4H20
    Half Equation: Vn2+ 3H20 -----> V03- + 6H+ + 3e-
    Overall eq: 3MN04- + 5Vn2+ + 3H20 -----> 6H+ + 5V03- + 3Mn2+
    Oh God
    But how could you tell the ratio was 5:3???
    I mean I wouldnt have realised the ratio was 5:3 from looking at ration 1.67.
    Is there another way of finding that out?


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    (Original post by H0ls)
    6e)
    Firstly we know SO4 ions have a charge of -2, this means we are looking for a metal with a +2 charge.

    Chromium half cell: Cr3+(aq) + 3e- --> Cr(s) (as electrode gained mass)
    Therefore X Half cell: X(s) --> X2+(aq) + 2e- (as electrode loses mass)

    From balancing these two to form an equation we can deduce:
    3X(s) + 2Cr3+(aq) ------> 3X2+(aq) + 2Cr(s)

    So we know we have a 3:2 (X:CR) ratio.

    Number of moles of CR: 1.456/52 = 0.028 mol
    Number of moles of X: 0.028 x 1.5 = 0.0420 mol
    Therefore: mass/moles = MR so 1.021/0.0420 = 24.3 therefore must be MG!

    I struggled with this too, but I was gobsmacked with how it easy it was once my friend explained!
    Ah thank you, it seems so simple when you explain it like that I look back at my mock and there are so many simple mistakes made
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    (Original post by Gulzar)
    Oh God
    But how could you tell the ratio was 5:3???
    I mean I wouldnt have realised the ratio was 5:3 from looking at ration 1.67.
    Is there another way of finding that out?


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    Same here, I wouldn't have thought to put 1.67 into a fraction to get 5/3 so 5:3
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    (Original post by Gulzar)
    Oh God
    But how could you tell the ratio was 5:3???
    I mean I wouldnt have realised the ratio was 5:3 from looking at ration 1.67.
    Is there another way of finding that out?


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    It's just the ratio of moles. You round up the two numbers of moles to the nearest whole number and compare them. I may have made it seem more complicated than it is.
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    I do a great deal
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    (Original post by H0ls)
    6e)
    Firstly we know SO4 ions have a charge of -2, this means we are looking for a metal with a +2 charge.

    Chromium half cell: Cr3+(aq) + 3e- --> Cr(s) (as electrode gained mass)
    Therefore X Half cell: X(s) --> X2+(aq) + 2e- (as electrode loses mass)

    From balancing these two to form an equation we can deduce:
    3X(s) + 2Cr3+(aq) ------> 3X2+(aq) + 2Cr(s)

    So we know we have a 3:2 (X:CR) ratio.

    Number of moles of CR: 1.456/52 = 0.028 mol
    Number of moles of X: 0.028 x 1.5 = 0.0420 mol
    Therefore: mass/moles = MR so 1.021/0.0420 = 24.3 therefore must be MG!

    I struggled with this too, but I was gobsmacked with how it easy it was once my friend explained!

    (Original post by master y)
    Hi guys, just wondering if you could help me on a question. On the june 2011 paper question 4dii, and 4e (how do you know it's lactic acid?)

    Thank you
    ^^ Can i ask you to help me on my question please? Thank you!!
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    (Original post by master y)
    Hi guys, just wondering if you could help me on a question. On the june 2011 paper question 4dii, and 4e (how do you know it's lactic acid?)

    Thank you
    4d(ii) Well, we have two carboxylic acid groups on the molecule, in acid dissociation we have two possibility: Either one group dissociates, or both dissociates.

    Pka= 1.23 (this shows the strongest acid ph, therefore we can conclude that this is the molecule where we keep one carboxylic acid group.
    C2H2O4 ---> C2HO4- + H+

    Pka+ 4.19 (well this has a less acidic ph, so we know that is the molecule where we have lost both carboxylic acid groups)
    C2HO4- ----> C2O42- + H+

    (I was told to do the second equation as if the first carboxylic acid group had already been lost, i'm sure you can do it straight from the C2H2O4)


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    hi,can Anyone send me the jan 2013 paper?
    please and thank you
 
 
 
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