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    (Original post by 16dan2life)
    depends on the compound. Let's say the question is to do with NaCl, we'd expect atomisation to be 1/2Cl2 -> Cl

    If they gave us, Cl2 -> 2Cl, then this is bond dissociation, so we need to half this value, to work out the atomisation.
    So if they gave us Cl2 -> Cl2 for a a compound like MgCl2, we would just use the bond enthalpy?
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    (Original post by lifeisgood2012)
    you blud... in the table it says lattice DISSOCIATION... YOU NEED LATTICE FORMATION SO YOU CHANGE THE + TO A NEGATIVE
    Thanks man, didn't realise that! i definitely won't make that mistake in the exam tomorrow now!!!
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    (Original post by 16dan2life)
    thats with NaOH, not diaminoethane
    Oh okay thanks very much


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    (Original post by problematic)
    Hey guys, quick question and sorry if this has already been answered.

    I;ve attached a table from june11 paper q 5 and its asking to do a cell representation using the Zn and Ag2o equestions.

    I just wanted to know why Pt boundaries are penalised, and what is the general rule when to put the Pt and when not to.

    thank you
    You have two solids on both sides therefore a platinum electrode isn't required.
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    (Original post by Atz23)
    hello, can someone explain to me question 3d on Jan 2013 please? it would be a big help....I do not get why you do 70-189.
    question is saying that gaseous water has condensed to liquid water. In other words, H2O(g)-->H2O(l)

    seeing as delta G is zero, and we're not given a value for delta H of liquid water, we're left with deltaH - T(deltaS). We need to work out delta H, so our equation is:

    delta H = deltaS(T)
    delta S = 70-189 = -119, divide it by 1000 to change the units of it into K J mol -1 = -0.119

    so, -0.119 times 373 = -44.4 Kjmol-1
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    (Original post by lewiszorz)
    In the electrochemical cell, where it's like Zn and Cu, how do you know that Zn has more tendency to oxidise to Zn2+ than Cu?
    If the emf of Zn is more negative than Cu it will be oxidised
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    (Original post by marleyxd)
    So if they gave us Cl2 -> Cl2 for a a compound like MgCl2, we would just use the bond enthalpy?
    Cl2 -> 2Cl *

    yes
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    (Original post by hawraaj313)
    thanks, it was on the mark scheme thats why
    which markscheme was this?
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    (Original post by popnit)
    No i know that, how do you figure out the equation? More specifically the 2 moles of each reactant except H2O2
    You know its yellow solution so you know its:

    2[Co(NH3)6]^(2+) + H2O2 ----> 2[Co(NH3)]^(3+) + 2OH-
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    (Original post by 16dan2life)
    Cl2 -> 2Cl *

    yes
    yeah thats what i meant haha, cheers!
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    (Original post by Atz23)
    hello, can someone explain to me question 3d on Jan 2013 please? it would be a big help....I do not get why you do 70-189.
    Your trying the work out delta S so you do:
    sum of delta S of products - sum of Delta S of reactants = delta S
    so use the Detla S values of H2O (g) -> H2O (l)
    So 70 (H2O (l)) - 189 (H2O (g))

    Hope this helped and you can understand what I've written
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    Could someone pls explain the Qs8b, the final bit that says Suggest a mechanism for the catalysed reaction....
    http://filestore.aqa.org.uk/subjects...5-QP-JUN12.PDF
    Thanks
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    do we need to know the CrO4 2- -> Cr2O7 2- equation?
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    (Original post by MrMeep2580)
    You know its yellow solution so you know its:

    2[Co(NH3)6]^(2+) + H2O2 ----> 2[Co(NH3)]^(3+) + 2OH-
    I know but I can't figure out why you put 2 moles!
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    (Original post by Melissa.S.)
    So confused, for lattice enthalpies , when the experimental values in born haber cycles is bigger than the value from theory, it means the bonding is stronger because the ions are polarised, so it contains partial covalent bonding. But then in the periodicity unit Al203 has a lower melting point because it contains covalent bonding??

    So does covalent character in ionic bonding make bonds stronger or weaker??
    I was also really confused about that!!! But i think the polarisation makes it stronger than the theoretical. I am not too sure but I have made myself just accept it and live with it.
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    (Original post by mcollin1)
    do we need to know the CrO4 2- -> Cr2O7 2- equation?
    Yes. It was asked in Jan 2013.

    Reagent- H2SO4

    2CrO42- + 2H+ --> [Cr2O7]2- + H2O

    CrO42- = yellow solution
    Cr2O7 2- = Orange solution
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    I know it's been posted before but i can't find the link to the January paper, could someone put up the link or quote the post it was in please! thanks

    edit: found it!
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    Can anyone explain how you calculate the temperature at which the reaction is NOT feasible?
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    (Original post by mcollin1)
    do we need to know the CrO4 2- -> Cr2O7 2- equation?
    yes its:

    2CrO4^(2-) + 2H+ ---> Cr2O7^(2-) + H2O
    Yellow soln. Orange soln.
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    (Original post by 16dan2life)
    Yes. It was asked in Jan 2013.

    Reagent- H2SO4

    2CrO42- + 2H+ --> [Cr2O7]2- + H2O

    CrO42- = yellow solution
    Cr2O7 2- = Orange solution
    thanks
 
 
 
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