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Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

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    (Original post by viksta1000)
    your wrong


    if you work out the concentration of H2O2 in the 250, you know the diluted concentration of H2O2

    the amount of H2O2 in the dilute sample is 25cm3 in 250cm3, which is the same as 1/10th of the original sample, so if you x10 you get the amount in the original sample
    you know the answer to that question is 40 x n[H202] x 10. do you know why we are multiplying by 40?

    Thanks
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    (Original post by SmartFool)
    Thank you! but do you have the mark scheme to these questions anywhere? i need to go over quite a lot on that past paper especially the buffers bit! i can't seem to find it on the ocr website!
    Someone will be a life-saver and list all the use of transition metal we need to know! :rolleyes:

    Chromium---> stainless steel
    ...
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    (Original post by touran22)
    any one got any ideas its driving me nuts!
    First you divide the % composition of each compound by its Mr, so you get:

    11.11/14 : 3.17/1 : 41.27/52 : 44.45/16

    This gave you 0.79 : 3.17 : 0.79 : 2.78

    Then divide all ratios by the smallest number, so 0.79. This gives you:

    1 : 4 : 1 : 3.5

    Double everything to make 3.5 up to 7.

    2 : 8 : 2 : 7

    Which gives you N2H8Cr2O7
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    (Original post by Princess_perfect786)
    i got the answer as -54kJ but the actual answer is -54kJ/mol. what am i supposed to be dividing -54 by?
    To get to -54, you will have divided the value q (in kJ) by a molar quantity (moles), right, so your units are kilojoules/moles
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    I am pretty sure I am going to fail this exam tomorrow.
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    (Original post by threerose92)
    you know the answer to that question is 40 x n[H202] x 10. do you know why we are multiplying by 40?

    Thanks
    yeah number of c = n x 1000 / v

    v = 25

    so c = n x 1000/25 = n x 40
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    When you draw isomers using the ligand 'en' do you have to make sure that the carbons in CH2 are joined to the nitrogen, I mean does this have to be made clear. Could you lose marks?
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    REDOXplease help. Me. !

    so it says

    "Hydrogen Iodide, HI, is oxidised to Iodine, I2, by conc. H2SO4 which is reduced to H2S"

    so why is the equation (before balancing ox numbers)

    2HI + H2SO4 -> H2S + I2

    i get the equation but where the hell did that 2 infront of the HI come from? i thought you were supposed to balance AFTER doing oxidation numbers.

    (question is from page 183, worked example)
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    (Original post by KnuckleheadNinja)
    wow. that must be such a relief. what did you get on the AS mods for maths? thats exactly what i was trying to do. work REALLY hard on C1 and C2 and then can afford to lose a few for C3 and C4. but i dont think i could go in there and get a U and come out with an A!
    C1: 99
    C2: 100
    S1: 98
    Total for AS: 297/300

    C3: 79
    C4: Next week
    M1: 82
    Total needed for A = 480 - 297 - 79 - 82
    = 22 !!! :-)
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    for the love of god can anyone tell me the colours of compounds/transition metals we need to know please!
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    jan 11 mark scheme please?
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    (Original post by KnuckleheadNinja)
    I've got another question
    i SUCK at redox.
    like REALLY badly.

    so i was going over the examples on page 183 to try and help and .... no. the worked example on page 182 is fine. but
    when it got to step 2 on page 183, where RANDOMLY they've got 2HI and i dont see where that came from. at all. like, its not even balanced? why did they do that! seriously!?

    Help much appreciated.

    What they've done is, they said "balance any atom that changes the oxidation number" - now, iodine changes oxidation number from minus 1 to 0, so they have to balance the iodine atoms (S changes oxidation number too but it's already balanced)
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    (Original post by SmartFool)
    Thank you! but do you have the mark scheme to these questions anywhere? i need to go over quite a lot on that past paper especially the buffers bit! i can't seem to find it on the ocr website!
    Here they are in the same folder you downloaded.

    Page 63

    First question is about kP - partial pressure, do we need to learn?
    Attached Images
  1. File Type: pdf ocr_40998_ms_09_l_gce_jun[1].pdf (353.6 KB, 910 views)
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    I have a feeling the carbonic acid and hydrogencarbonate buffer system in the blood will be on the paper
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    (Original post by viksta1000)
    yeah number of c = n x 1000 / v

    v = 25

    so c = n x 1000/25 = n x 40
    thanks
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    (Original post by CoventryCity)
    I have a feeling the carbonic acid and hydrogencarbonate buffer system in the blood will be on the paper
    Yeah I agree with you, they put the haemoglobing stuff and platin stuff in january which was the kind of biology material - got a good feeling about this one
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    (Original post by Steven)
    First you divide the % composition of each compound by its Mr, so you get:

    11.11/14 : 3.17/1 : 41.27/52 : 44.45/16

    This gave you 0.79 : 3.17 : 0.79 : 2.78

    Then divide all ratios by the smallest number, so 0.79. This gives you:

    1 : 4 : 1 : 3.5

    Double everything to make 3.5 up to 7.

    2 : 8 : 2 : 7

    Which gives you N2H8Cr2O7
    thanks for your help...i see how they got the 07 but no were in the question does it hint that its ment to be doubled? is that just something we have to apply knowledge to?
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    (Original post by cws121)
    When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is
    –15 kJ mol–1.
    The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of –24 kJ mol–1.
    Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF

    can someone explain the answer to the question thanks?
    I always thought the enthalpy change of solution is endothermic and therefore always positive :confused:
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    2HI + H2SO4 -> H2S + I2

    2HI balances the I2

    In HI I will be -1 and I2 it will be 0


    on a side note...where have the oxygens gone?
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    WHAT IS THE MOST DIFFICULT TOPIC? in your opinion. state one!!

    1. Redox equations...
 
 
 
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