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    :woo: Just want it over with now
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    Could someone explain to me why the second electron affinity is +'ve. I think it was something to do with removing a negative charge?
    It'd be nice to have someone clarify though
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    (Original post by hiyarearl)
    Just done Jan 2011, got 72/100 which was a comfortable A it seems.

    Does anyone know what the raw mark was for 100% UMS for Jan 2011, out of interest?
    73 raw marks was for A* and 65 for a A, so 100% UMS should be 81 or above.

    Are you retake this paper again?
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    How would you do q7 B(ii) last question in jan 2011 paper?

    arrrghh
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    (Original post by Flux_Pav)
    does the ionic radius decrease as you move from left to right of periodic table?
    yes. Decreases as you go left to right and increases as you go up to down.
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    (Original post by hiyarearl)
    Just done Jan 2011, got 72/100 which was a comfortable A it seems.

    Does anyone know what the raw mark was for 100% UMS for Jan 2011, out of interest?
    I got 78 :P
    I think it was around 81/82 to get 150 UMS, but these were just my calculations
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    (Original post by asem93)
    When delta G is less than 0 !!
    yhh thats what I said...did u understand that!?!?
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    (Original post by TsunKit)
    Could someone explain to me why the second electron affinity is +'ve. I think it was something to do with removing a negative charge?
    It'd be nice to have someone clarify though
    It's the repulsion between the e and the negative ion

    More energy is required to overcome this repulsion.
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    (Original post by SteveScott)
    You need to know at least one of octohedral, tetrahedral and the flat one with four ligands, and at least one optical and stereoisomer, but I'd have thought two would be preferable.
    The flat one is a square planar right? cis-platin one?
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    (Original post by susan23)
    yhh thats what I said...did u understand that!?!?
    I am sorry I tried to quote the person that posed the question...ended up accidetly quoting you

    But yeah your answer hits the nail on the head...:cool:
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    (Original post by TsunKit)
    Could someone explain to me why the second electron affinity is +'ve. I think it was something to do with removing a negative charge?
    It'd be nice to have someone clarify though
    No the 2nd is usually exothermicThe 2nd is endothermic as you are adding a negative electron to an already negative ion, hence a lot of energy is needed to overcome the repulsion.
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    so which one is it that is like a born haber cycle but uses hydration and solution instead of atomisation and ionisation??
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    (Original post by gregory14db)
    No the 2nd is usually exothermic, as you are adding a negative electron to an already negative ion, hence a lot of energy is needed to overcome the repulsion.
    It is endothermic... You are contradicting yourself.
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    do we need to know what EDTA is used for and how?

    Pg 213
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    (Original post by gregory14db)
    No the 2nd is usually exothermic, as you are adding a negative electron to an already negative ion, hence a lot of energy is needed to overcome the repulsion.
    A lot of energy needed means it's endothermic.
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    Do we have to draw the lone pairs on ligands when drawing complexes????
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    (Original post by blush.ox)
    do we need to know what EDTA is used for and how?

    Pg 213
    It can't hurt, just rememberit's used in detergents, it's a pretty general answer to 'what is this chemical used for' in chemistry exams
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    (Original post by SteveScott)
    Haha

    I know pKa = -logKa , so 10^-pKa = Ka

    Ka = [H+][A-]/[HA]

    but i thought i saw something about pka being equal to pH for weak acids.....I dunno.
    At half neutralisation [A-]=[HA] so Ka=[H+] so pKa=pH BUT ONLY FOR HALF NEUTRALISATION
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    (Original post by mandn)
    At half neutralisation [A-]=[HA] so Ka=[H+] so pKa=pH BUT ONLY FOR HALF NEUTRALISATION
    Okay thanks, lol.
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    (Original post by mandn)
    At half neutralisation [A-]=[HA] so Ka=[H+] so pKa=pH BUT ONLY FOR HALF NEUTRALISATION
    wtf is half neutralisation :lolwut:
 
 
 
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