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    Just only found this thread! Got 280/300 AQA AS, need an A for uni offer


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    I need help for these two questions. For the ∆H sol, I calculated it this way. -(-2526)-1890+641-801= +476 kJmol^-1. As for the second one, 476-2526=-2050kJmol^-1. However, the answers are (i) ∆Hosol = 641 – 801 = –160 kJ mol–1 [1] (ii) ∆Hohyd = (1890 – 2526 – 160)/2 = –398 kJ mol–1

    Can anyone explain the mark scheme to me? I can't quite get it.
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    (Original post by RMNDK)
    There are some mistakes in your equation of Co2+ with NH3.
    You also might want to give the colour of [Co(H2O)6]2+ before it is oxidised by oxygen.

    How come you haven't included the equation of Fe3+ with excess OH- ?
    The precipitate will dissolve in concentrated NaOH.
    Thankyou for your comments, I will edit the document. Plus I have just cross referenced Fe3+ with excess OH- in my CPG revision guide and it states it as 'no visible change' so I really don't know what to put to be honest.
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    (Original post by jyyl)
    Name:  Screen Shot 2016-04-23 at 5.17.55 PM.png
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    I need help for these two questions. For the ∆H sol, I calculated it this way. -(-2526)-1890+641-801= +476 kJmol^-1. As for the second one, 476-2526=-2050kJmol^-1. However, the answers are (i) ∆Hosol = 641 – 801 = –160 kJ mol–1 [1] (ii) ∆Hohyd = (1890 – 2526 – 160)/2 = –398 kJ mol–1

    Can anyone explain the mark scheme to me? I can't quite get it.
    Here you go.
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    anyone doing edexcel ial ?
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    (Original post by djmans)
    anyone doing edexcel ial ?
    Yeah

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    (Original post by RMNDK)
    There's a REALLY LONG way, and there's a much shorter way.

    Here's the short way, because **** the long way.
    Lol I say short, but my post is long. That's only because I just really really want you to understand it sorry. The working out when you come to write it is short.

    You agree that you have some moles of HA.
    The question is saying that the student adds some NaOH. This amount is equal to a third of the amount needed to neutralise it.

    In other words, a third of the HA has reacted with the NaOH. If it said, half the amount was added, then half of the HA would have been neutralised, and so on.

    When we react the HA and NaOH, this forms our A-.
    Spoiler:
    Show
    Cos remember, NaOH + HA ---> NaA + H2O
    and the NaA ---> Na+ + A-
    So as a result, the amount of A- we formed is equal to one third of the HA we started with.

    In other words...

     \mathrm{A}^- = \dfrac{1}{3}\times \mathrm{HA}.

    If you want to represent this as concentration, by all means divide it by the volume. But the volume will cancel out, which allows me to just use the moles.

    In fact I don't need to do that because...

     \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\mathrm{A}^-} \ \ \ \ \ becomes  \ \ \ \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\dfrac{1}{3} \times \mathrm{HA}}

    And now the HAs cancel out, which just leaves...

     \mathrm{H}^+ = \mathrm{K}_c \times 3

    This is now solvable for pH
    It's just like that first question of where the HA is half the number of A
    Spoiler:
    Show
    The long **** way would be to calculate the moles of each reagent, and then the concentration, then dividing one by the other, and you get funny numbers.
    Thank you for this.. I'm still quite confused as this is only a 1 mark question?! Surely I'm missing something 🙈 So annoying that the MS doesn't explain further...
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    Has anyone got OCR A-level june 2015 past papers F321, F322 , F324 and F325??
    Exams are almost here and I really need those past papers to revise as they are the only ones I haven't done yet!
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    (Original post by RMNDK)
    There's a REALLY LONG way, and there's a much shorter way.

    Here's the short way, because **** the long way.
    Lol I say short, but my post is long. That's only because I just really really want you to understand it sorry. The working out when you come to write it is short.

    You agree that you have some moles of HA.
    The question is saying that the student adds some NaOH. This amount is equal to a third of the amount needed to neutralise it.

    In other words, a third of the HA has reacted with the NaOH. If it said, half the amount was added, then half of the HA would have been neutralised, and so on.

    When we react the HA and NaOH, this forms our A-.
    Spoiler:
    Show
    Cos remember, NaOH + HA ---> NaA + H2O
    and the NaA ---> Na+ + A-
    So as a result, the amount of A- we formed is equal to one third of the HA we started with.

    In other words...

     \mathrm{A}^- = \dfrac{1}{3}\times \mathrm{HA}.

    If you want to represent this as concentration, by all means divide it by the volume. But the volume will cancel out, which allows me to just use the moles.

    In fact I don't need to do that because...

     \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\mathrm{A}^-} \ \ \ \ \ becomes  \ \ \ \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\dfrac{1}{3} \times \mathrm{HA}}

    And now the HAs cancel out, which just leaves...

     \mathrm{H}^+ = \mathrm{K}_c \times 3

    This is now solvable for pH
    It's just like that first question of where the HA is half the number of A
    Spoiler:
    Show
    The long **** way would be to calculate the moles of each reagent, and then the concentration, then dividing one by the other, and you get funny numbers.
    You made a mistake. Moles of HA after partial neutralisation would be 2/3 of original value.
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    (Original post by seph_muriel)
    Has anyone got OCR A-level june 2015 past papers F321, F322 , F324 and F325??
    Exams are almost here and I really need those past papers to revise as they are the only ones I haven't done yet!
    Ask your teacher. If s/he wants you to have them you'll get them.
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    (Original post by shiney101)
    So what would you call this reaction?
    Personally i think your original statement was perfectly correct!!!
    In the oxidation of alcohols.... acidified potassium dichormate ions are used. In the oxidation of a primary alcohol the cr is reduced to 3+ thus the solution turns orange to green (for aldehydes only).........
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    do you find chemistry hard or easy?
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    (Original post by TeachChemistry)
    You made a mistake. Moles of HA after partial neutralisation would be 2/3 of original value.
    Thank you for the correction
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    TeachChemistry

    Hey, i'm stuck on Q6.2 on the new specimen paper:

    The students collected a 20cm^3 sample of liquid and weighted it. The mass of the sample was 16g. The density of ethanol is 0.79gcm^-3 and that of water is 1.00gcm^-3. Use this data to work out the mass of ethanol in the sample collected. Assume volume of the sample = volume of water+ethanol.

    I attempted by setting volume as x and for ethanol x-20 and rearranged for x to get 19.05g which is wrong... How would you come across doing this?
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    (Original post by Dysprosium)
    TeachChemistry

    Hey, i'm stuck on Q6.2 on the new specimen paper:

    The students collected a 20cm^3 sample of liquid and weighted it. The mass of the sample was 16g. The density of ethanol is 0.79gcm^-3 and that of water is 1.00gcm^-3. Use this data to work out the mass of ethanol in the sample collected. Assume volume of the sample = volume of water+ethanol.

    I attempted by setting volume as x and for ethanol x-20 and rearranged for x to get 19.05g which is wrong... How would you come across doing this?
    Which exam board?
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    (Original post by TeachChemistry)
    Which exam board?
    Aqa
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    (Original post by TeachChemistry)
    Here you go.
    Thank you so much
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    (Original post by Dysprosium)
    TeachChemistry

    Hey, i'm stuck on Q6.2 on the new specimen paper:

    The students collected a 20cm^3 sample of liquid and weighted it. The mass of the sample was 16g. The density of ethanol is 0.79gcm^-3 and that of water is 1.00gcm^-3. Use this data to work out the mass of ethanol in the sample collected. Assume volume of the sample = volume of water+ethanol.

    I attempted by setting volume as x and for ethanol x-20 and rearranged for x to get 19.05g which is wrong... How would you come across doing this?
    What you have done so far gives you 19.05 cm3. So multiply by 0.79 to get mass.
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    Hi guys Hope revision is going well for you all I have one question that I'm stuck on (from the AQA June 2013 paper)- I've managed to solve pretty much all the redox calculation questions except this one
    I would appreciate any help- thank you!

    Compound Z is a complex that contains only cobalt, nitrogen, hydrogen and chlorine.A solid sample of Z was prepared by reaction of 50 cm3 of 0.203 mol dm–3 aqueouscobalt(II) chloride with ammonia and an oxidising agent followed by hydrochloric acid.When this sample of Z was reacted with an excess of silver nitrate, 4.22 g ofsilver chloride were obtained.Use this information to calculate the mole ratio of chloride ions to cobalt ions in Z.Give the formula of the complex cobalt compound Z that you would expect to beformed in the preparation described above.Suggest one reason why the mole ratio of chloride ions to cobalt ions that you havecalculated is different from the expected value.
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    (Original post by haj101)
    Hi guys Hope revision is going well for you all I have one question that I'm stuck on (from the AQA June 2013 paper)- I've managed to solve pretty much all the redox calculation questions except this one
    I would appreciate any help- thank you!

    Compound Z is a complex that contains only cobalt, nitrogen, hydrogen and chlorine.A solid sample of Z was prepared by reaction of 50 cm3 of 0.203 mol dm–3 aqueouscobalt(II) chloride with ammonia and an oxidising agent followed by hydrochloric acid.When this sample of Z was reacted with an excess of silver nitrate, 4.22 g ofsilver chloride were obtained.Use this information to calculate the mole ratio of chloride ions to cobalt ions in Z.Give the formula of the complex cobalt compound Z that you would expect to beformed in the preparation described above.Suggest one reason why the mole ratio of chloride ions to cobalt ions that you havecalculated is different from the expected value.
    So what don't you follow in the mark scheme?

    Moles of cobalt = (50 × 0.203)/1000 = 0.01015 mol
    Moles of AgCl = 4.22/143.4 = 0.0294
    Ratio = Cl- to Co = 2.9 : 1
    [Co(NH3)6]Cl3 (square brackets not essential)
    Difference due to incomplete oxidation in the preparation
 
 
 
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