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    (Original post by Aerosmith)
    Hilarious how people just cannot accept that the question required you to have an appreciation of the fact that 1 mol is formed when standard 😂😂
    Standard just means standard conditions tho??
    Like pressure and temp.
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    (Original post by Parallex)
    No, it definitely didn't. It gave you Hf values in a table but it asked you to work out the standard enthalpy change for for the reaction shown.
    it did in the question guess we will just see when the paper comes out its not a big deal 2 or 3 marks calm down
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    (Original post by Mattk1997)
    Do you divide by 2 or not this is doing my head in now
    You don't.
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    Can someone please explain why increasing the pressure of oxygen would change that half cell equilibrium at all? I thought it was only if they were all gaseous?

    So I put that there was no effect
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    (Original post by Suits101)
    Substitution of unidentate ligands (water in this question) by bidentate (ethane-1,2-diamine in this question) or multidentate ligands leads to a more stable complex because of the increase in the number of particles so disorder increases, entropy increases, delta H is approximately 0 and delta G <= 0
    Ahhh i see, i understand the effect but our teacher never referred to it as that and aIways just said "remember it and spew it out when the time comes".

    cheers
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    (Original post by gh2)
    For the dividing by 2, I was looking at the question in the exam trying to figure out whether it was necessary or not! I think that if answers are given as for example: -189 kj for the enthalpy change for the reaction then that's fine, and if given as -94.5 kj mol-1 that's also fine. Both are the same it's just the units that change and I think if you changed units depending on your answer you should get the marks. Agree?
    Just to point out that the enthalpy change of -198 kj mol-1 still needs the 'mol-1' section to it, because that's the units for enthalpy change. Using 'mol-1' as an argument to justify dividing by 2 is a fallacy because mol-1 is not referring to 'per mole of SO3', it purely denoting 'per such reaction' which is in this case forms 2 moles of SO3
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    (Original post by Aerosmith)
    I think you're just a stubborn nerd who can't accept that it was perfectly fine to divide by 2 and work it out
    Lol what a way to throw a tantrum when proved wrong. 'stubborn nerd', pretty sure you were the one who mocked us earlier for 'not being able to read'. Also, how is your answer perfectly fine if all your values are half of what they should be? I'm pretty sure you weren't saying our answers would be perfectly fine if they were double the required value.

    Saying that, I'm not the one to usually try to make someone feel bad for getting an answer wrong on an exam (I'll leave that to you, it seems to be your forte) but the way you're acting like you couldn't possibly be wrong and insulting others in the process when in fact you are incorrect makes me not so sympathetic towards you.

    In the end of the day, it's two marks on one paper, it's not the end of the world so just move on
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    (Original post by FireBLue97)
    Lets get this straight
    1 mark for equation 2:5 ratio
    2nd mark for no of moles of Mno4- 0.02x(26.4/1000)
    3rd mark for finding moles of C2o4- using the ratio 0.00132
    4th mark for x10 to find in original 250 cm3
    5th mark for x by 88 Mr of C2O4 to get 1.16 as this is specifically for the total C2o4 present before titre so up to this point definitely right
    6th mark for the Mr of C2o4 which is 88
    7th mark for dividing by 2.29 or something then times 100% to get the percentage as 50.7
    Here are the official 7 marks for this question as both the 50.7 and 94.4% answers concern C2o4 pressent in the original sample 94.4 gives 8 marks whilst this is a 7 marker so why isnt this paper out of 101 dalmations
    You're forgetting that the question asked to write the equation between MnO4- and C2O42- (normally 2 marks)

    2 marks for equation
    1 mark for moles of MnO4-
    1 mark for moles of C2O42-
    1 mark for x 10
    1 mark for putting into equation
    1 mark for final answer
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    (Original post by emma_1111)
    how many marks do you lose for not converting degrees to kelvin?
    Bumping this because I also forgot to convert to k I think, if you guys give me the answer for using as degrees and answer for kelvin then I can see if I forgot to convert
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    (Original post by Spectral)
    Just to point out that the enthalpy change of -198 kj mol-1 still needs the 'mol-1' section to it, because that's the units for enthalpy change. Using 'mol-1' as an argument to justify dividing by 2 is a fallacy because mol-1 is not referring to 'per mole of SO3', it purely denoting 'per such reaction' which is in this case forms 2 moles of SO3
    Was just going to reply w/ the exact same thing. There's only one answer and that's the undivided value.
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    (Original post by Suits101)
    For the gold jewellery one did anyone say:

    E (Au+/Au) > E (O2/H2O) so Au+ ions oxidise water to oxygen and Au+ ions are reduced to Au
    I seid something like that and also I worked out the Emf and it was negative so reaction was no feasible.
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    (Original post by TheSpanishGuy)
    I seid something like that and also I worked out the Emf and it was negative so reaction was no feasible.
    Seems good
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    (Original post by RefusedAccess)
    Solubility question stated that solubility was mentioned as the amount of present MgCl2(s) available to dissolve in aq solution, and as the reaction went backwards, more MgCl2(s) was produced, and hence the solubility was increased, not descreased

    Either way increase or decrease, as long as your explanation was correct, then I believe you will get a minimum of 2/3
    If more mgcl2 is produced its less soluble - solubility would be it forming aq ions
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    unofficial MS made yet?
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    (Original post by emsieMC)
    Bumping this because I also forgot to convert to k I think, if you guys give me the answer for using as degrees and answer for kelvin then I can see if I forgot to convert
    50 degrees to 323 kelvin
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    it is 2 K for each of the born haber cycle like 2k(s) then 2K(g) and the 1/2 o2 to o(g) for atomisationand the for the Co3+question the product was definitely o2 not 1/2 o2 and what was the tick was HC2o4h H2so4 hcl hno3 please ?
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    Any idea on UMS for 90 marks?
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    (Original post by Spectral)
    Just to point out that the enthalpy change of -198 kj mol-1 still needs the 'mol-1' section to it, because that's the units for enthalpy change. Using 'mol-1' as an argument to justify dividing by 2 is a fallacy because mol-1 is not referring to 'per mole of SO3', it purely denoting 'per such reaction' which is in this case forms 2 moles of SO3
    Ahh fair enough i do see :/. However for the reaction you had 2 moles of SO2/SO3, therefore when you do the eqtn, in terms of units, a 'KJ mol-1' is multiplied by 'mol' (the 2 as there are 2 moles of SO3/SO2) therefore delta H is in KJ but for one mole you divid by 2 again to get KJ mol-1? sorry for complex stuff
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    (Original post by Mattk1997)
    50 degrees to 323 kelvin
    Sorry I meant final answer calculation the whole thing for if you used degrees and for if you used kelvin, just to see if I recognise the number
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    Would I lose marks for not having my coordinate bonds as arrows? I drew them like normal covalent bonds and now I'm kicking myself!
 
 
 
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