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    (Original post by jamesl1)
    does greater lattice enthalpy mean the bonds are stronger or weaker ?
    If the lattice enthalpy is greater, it means more energy is required to break the bonds so the bonds are stronger

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    (Original post by rrelish)
    But that would not balance the hydrogens, you'll have 6 on the right and 2 on the left.
    The charges are also not balanced...
    oops i ment:

    2H2O + SO2 -->H2SO4 + 2H+ +2e-
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    Why does the emf, not change when the surface are of the platinum electrode is increased?

    I Bet its a simple answer :/
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    hi can anyone help me understand what the last part of the spefication means?
    'understand the chelate effect in terms of positive entropy change in these reactions'? (refereing to substitution of unidentate with bidentate and multidentate ligands forming more stable complexes)
    im panicking!!!!
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    still the same conc of ions and same temp so why would the position of equilibrium favour another way?
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    With the enthalpy change for a reaction is it:

    Bonds formed - bonds broken

    or

    Bonds broken - bonds formed
    ??
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    (Original post by boboyt)
    still the same conc of ions and same temp so why would the position of equilibrium favour another way?
    was that directed at me, if so then that kinda answers my questions
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    (Original post by Stirlo)
    With the enthalpy change for a reaction is it:

    Bonds formed - bonds broken

    or

    Bonds broken - bonds formed
    ??
    bonds broken - bonds formed
    ( only know cus i resat :P )
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    (Original post by sms21)
    hi can anyone help me understand what the last part of the spefication means?
    'understand the chelate effect in terms of positive entropy change in these reactions'? (refereing to substitution of unidentate with bidentate and multidentate ligands forming more stable complexes)
    im panicking!!!!
    When say [Cu(H20)6]2+ reacts with something that isnt unidentate (so multidentate) lets say EDTA4- you end up with more molecules in the products than the reactants. This means there has been an increase in entropy. As systems favour a higher more positive entropy the product formed is much more stable.

    [Cu(H2O)6]2+ + EDTA4- =====> CuEDTA2- + 6H2O
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    (Original post by sms21)
    hi can anyone help me understand what the last part of the spefication means?
    'understand the chelate effect in terms of positive entropy change in these reactions'? (refereing to substitution of unidentate with bidentate and multidentate ligands forming more stable complexes)
    im panicking!!!!
    The chelate effect is when a complex ion has unidente ligands replaced by bidente or multidente ligands

    Say you have Cu(H20)6 2+ and add EDTA 4-, the equation is

    Cu(H20)6^2+ + EDTA^4- ----> CuEDTA^2- + 6H20

    There is an increase in entropy as the number of particles increases from 2 to 7
    This is more stable because the reverse reaction would cause a decrease in entropy which is very hard to happen. Hope that helps
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    (Original post by sms21)
    hi can anyone help me understand what the last part of the spefication means?
    'understand the chelate effect in terms of positive entropy change in these reactions'? (refereing to substitution of unidentate with bidentate and multidentate ligands forming more stable complexes)
    im panicking!!!!
    i think it just means know that entropy increases when you substitute a ligand for a multidentate ligand, cus more molecules are formed on right hand side
    someone else might know a bit more!
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    Yay I know my colours now. Just periodicity to go over. :mad2:
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    Can anyone help with question 7c of Jan 11.

    2MnO4– + 6H+ + 5H2O2 2Mn2+ + 8H2O + 5O2

    Moles MnO4– = (24.35/1000) x 0.0187 = 4.55 x 10^–4
    Moles H2O2 = (4.55 x 10^–4) x 5/2 = 1.138 x 10^–3

    Moles H2O2 in 5 cm^3 original
    = (1.138 x 10^–3) x 10 = 0.01138
    Original [H2O2] = 0.01138 x (1000/5) = 2.28 mol dm^–3
    (allow 2.25-2.30)

    I had it correct uptil the bit in bold. But I don't get why you times by 10. Don't you only do that in order to get the moles in the 250cm3 solution? and not the moles in the 5cm3 solution. :confused:
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    (Original post by InItToWinItGetIt?)
    Can anyone help with question 7c of Jan 11.

    2MnO4– + 6H+ + 5H2O2 2Mn2+ + 8H2O + 5O2

    Moles MnO4– = (24.35/1000) x 0.0187 = 4.55 x 10^–4
    Moles H2O2 = (4.55 x 10^–4) x 5/2 = 1.138 x 10^–3

    Moles H2O2 in 5 cm^3 original
    = (1.138 x 10^–3) x 10 = 0.01138
    Original [H2O2] = 0.01138 x (1000/5) = 2.28 mol dm^–3
    (allow 2.25-2.30)

    I had it correct uptil the bit in bold. But I don't get why you times by 10. Don't you only do that in order to get the moles in the 250cm3 solution? and not the moles in the 5cm3 solution. :confused:
    You are working out the moles in 25cm, then you times it by 10 to get it in 250 and then use the 5cm to work out the conc.
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    (Original post by InItToWinItGetIt?)
    Can anyone help with question 7c of Jan 11.

    2MnO4– + 6H+ + 5H2O2 2Mn2+ + 8H2O + 5O2

    Moles MnO4– = (24.35/1000) x 0.0187 = 4.55 x 10^–4
    Moles H2O2 = (4.55 x 10^–4) x 5/2 = 1.138 x 10^–3

    Moles H2O2 in 5 cm^3 original
    = (1.138 x 10^–3) x 10 = 0.01138
    Original [H2O2] = 0.01138 x (1000/5) = 2.28 mol dm^–3
    (allow 2.25-2.30)

    I had it correct uptil the bit in bold. But I don't get why you times by 10. Don't you only do that in order to get the moles in the 250cm3 solution? and not the moles in the 5cm3 solution. :confused:
    It acts for the concentration of the original sample so of the 250cm3. That's why you times it by 10, I think
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    (Original post by azaking)
    You are working out the moles in 25cm, then you times it by 10 to get it in 250 and then use the 5cm to work out the conc.
    (Original post by Stirlo)
    It acts for the concentration of the original sample so of the 250cm3. That's why you times it by 10, I think
    But then why don't you divide by 50 to get moles in 5cm3 and then do c = n/v, instead of going straight to c = n/v?
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    (Original post by Sparkly-Star)
    :cry2: :hugs: Aww, I know how you feel. :cry2: :hugs: We'll get there.
    Thanks :hugs:
    I've finally finished that chapter :party:
    But as Jay Z say its "On to the next one" lol
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    When writing the conventional cell rep. e.g in JAN 10 question 3C
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF


    How do you know if to write the OH- before the H2O or H2O before the OH- ? or does it not matter

    Thank you
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    (Original post by GoodOl'CharlieB)
    Thanks :hugs:
    I've finally finished that chapter :party:
    But as Jay Z say its "On to the next one" lol
    Yay :jive: I finished 16 too, just doing some exam questions then 'On to the Next One'... :p:
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    (Original post by rrelish)
    When writing the conventional cell rep. e.g in JAN 10 question 3C
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF


    How do you know if to write the OH- before the H2O or H2O before the OH- ? or does it not matter

    Thank you
    I see you've posted this loads of times and I wish I knew the answer but I don't. Can anyone else PLEASE answer this thanks?
 
 
 
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