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    (Original post by Schoolio93)
    I don't think you need to know any of the above. Because there is a very high possibility that they will give you unfamiliar equations and tell you to construct half equations. I really was struggiling with redox equations so I ended up doing random equations for like 3 hours with my teacher at school. Complex Redox equations and ligand substitution are quite tricky.
    My teacher told me that the application redox equations would be given to us and they might ask to recall these and then we have to write balanced equation(s) by cancelling out the electrons on either side
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    (Original post by Schoolio93)
    Lattice enthalpy is not more important than enthalpy change of solution! It is just these two enthalpies are two different things. Lattice enthalpy measures the strength of the ionic bond whereas enthalpy of solution is the enthalpy which takes place when one mole of a compound is dissolved in water.

    I guess for this unit Lattice enthalpy is more important, because you effectively need to work it out using the born-haber cycle. But usually no enthalpy is more important than the other. you cannot have a reaction happening if any of the enthalpies will be missing.
    No I completely agree and realised I made a small typo in my Q but in the Jan 2011 paper question d) gives a mark to say:

    Idea that ?H(solution) is affected more by lattice enthalpy than by hydration enthalpy

    didn't really understand that and the textbook doesn't really clear anything up
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    Can anyone explain Qu. 7B on JUN 2010 paper please!!!!

    ive tried looking at the mark scheme, and theres a 40 in there thats come from nowhere
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    (Original post by potentialmedicalstudent)
    Can anyone explain Qu. 7B on JUN 2010 paper please!!!!

    ive tried looking at the mark scheme, and theres a 40 in there thats come from nowhere
    1st part: 0.02x0.02345=0.000469mol of KMnO4
    2KMnO4:5H2O2 so 0.000469x2.5=0.0011725=mols of H2O2 in 25cm3
    25x40 is putting it into dm3 so 25x40 = 1dm3 or 1000cm3
    so 0.0011725x40 or 0.0011725/0.025 = 0.0469mol dm3 of H2O2
    then do 0.0469x (16x2+1x2) = 1.5946 g dm3

    2nd part: 0.000469 KMnO4:O2= 2:5
    so 0.000469x2.5 = 0.0011725mol of O2
    n=v/24 so 0.0011725x24 = 0.02814dm3

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    (Original post by intellectual1)


    That book was written by the examiners, they highly recommended their textbook at a Chemistry conference last December :rolleyes: The Stretch and Challenge questions within them and Extension exercises are great for preparing for the A* Stretch and Challenge The teacher answer guide with it is well worth it, not sure if I should be posting it on here though...

    F324 is only 1 hour and the final exam for most people...it won't be anything like the past papers much, also the examiners are VERY SPECIFIC with the curly arrows and mechanisms:eek:
    Ah ok thankyou

    Yep they are very specific. I did F324 in January and asked for my paper back. I lost a mark in reduction of carbonly bonds where my arrow wasnt touching the double bond. It was slightly above bad times. I remember telling everyone that was the best question in the paper.. and i lost a stupid mark there so yep, curly arrows!!!!

    I dont think i have the time to order that book and stuff now but i would really appreciate it if you could give some tips or what sort of calculations or topics to focus alot on.. if you understand? Pleaseee thanks!!!
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    Will someone please explain the 'Fuel Cell' to me? I really don't understand the textbook or revision guide?!
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    (Original post by wilsea05)
    25x40 is putting it into dm3 so 25x40 = 1dm3 or 1000cm3
    thanks for the help, but im really going to display my stupidity

    but is that always the case to work out dm3, you have to times the cm3 by 40?

    because i get everything else, except the 40 LOL
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    Guys,

    What is the half equation for H202-------->H20 (quite tricky)

    And also CH3CH20H--------> CH3C00H(again quite tricky)
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    (Original post by J DOT A)
    Guys,

    What is the half equation for H202-------->H20 (quite tricky)
    I'm gonna have a go and say

    2e- + 2H+ + H2O2 ---> 2H2O
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    Anyone have the F325 Jan 2011 mark scheme please?
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    (Original post by Rickesh)
    Anyone have the F325 Jan 2011 mark scheme please?
    http://www.thestudentroom.co.uk/show....php?t=1595938

    its on this page
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    (Original post by wilsea05)
    1st part: 0.02x0.02345=0.000469mol of KMnO4
    2KMnO4:5H2O2 so 0.000469x2.5=0.0011725=mols of H2O2 in 25cm3
    25x40 is putting it into dm3 so 25x40 = 1dm3 or 1000cm3
    so 0.0011725x40 or 0.0011725/0.025 = 0.0469mol dm3 of H2O2
    then do 0.0469x (16x2+1x2) = 1.5946 g dm3

    2nd part: 0.000469 KMnO4:O2= 2:5
    so 0.000469x2.5 = 0.0011725mol of O2
    n=v/24 so 0.0011725x24 = 0.02814dm3

    But for part one, you haven't taken dilutions into account, g of H202 is going to be 10x more,

    Not trying to be pedantic!
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    Hey mate
    how do you determine whether an electrode potential becomes more negative or positive by changing the conc of the ion (half cell)

    my rule was if eq shifts to left- becomes more negative
    if eq shifts to the right becomes more postive
    However that theory was undermine when it didn't work for a particular reacvtion.
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    Hello

    can anyone explain to me Question 7 on the jan 2011 paper please??
    Also what are the grade boundaries for this paper

    Thanks in advance
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    Guys what does it mean if an electrode potential is negative compared to that of the standard hydrogen electrode. I know a negative electrode means it goes to the left hand side. However when setting up a half cell, do the electrons flow from Hydrogen to the other thing you are trying to work out? But then Hydrogen produces now electrons as its voltage is O.Ov so then wth is going on LOL.
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    (Original post by J DOT A)
    Guys what does it mean if an electrode potential is negative compared to that of the standard hydrogen electrode. I know a negative electrode means it goes to the left hand side. However when setting up a half cell, do the electrons flow from Hydrogen to the other thing you are trying to work out? But then Hydrogen produces now electrons as its voltage is O.Ov so then wth is going on LOL.
    If it's negative compared to the Hydrogen half-cell, it means that it is the negative electrode and it releases the electrons which to travel to the H+ half cell:
    2H+ + 2e- --> H2
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    (Original post by voices1)
    Hey mate
    how do you determine whether an electrode potential becomes more negative or positive by changing the conc of the ion (half cell)

    my rule was if eq shifts to left- becomes more negative
    if eq shifts to the right becomes more postive
    However that theory was undermine when it didn't work for a particular reacvtion.
    Because the half-cells are at equilibrium the it will work on Le Chatlier's principle so it depends on the reaction on hand.

    But strictly speaking a half-cell is measured under standard conditions of 1 mol dm-3
    so therefore you shouldn't really need to change concentration.

    I hope this has helped, or if you want to ask something specific, I'll be happy to help
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    (Original post by haydyb123)
    But for part one, you haven't taken dilutions into account, g of H202 is going to be 10x more,

    Not trying to be pedantic!
    i did do that when i did the paste paper and was writing it all down, but was just doing it quickly to show where the 40 came from
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    (Original post by Rickesh)
    Hello

    can anyone explain to me Question 7 on the jan 2011 paper please??
    Also what are the grade boundaries for this paper

    Thanks in advance
    Do you want the calculation specifically or?
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    Just did the January 2011 Paper. It wasn't too bad except the odd question about planning an experiment for working the enthalpy change of neutralisation and I didn't get the Q7 calculation AT ALL, well the first part was ok by working out the moles but thats about where I got to

    Can anyone help me with that, at the same time explaining whats going on? Thanks
 
 
 
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