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    (Original post by mina.)
    What? apart from completely messing that up i also forgot a question? Typical me..
    Talking about missing a question. The last question was about the enthalpy change diagram right? because I had no time and was rushing the paper, hopefully I didnt miss anything..especially at the very end
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    (Original post by bala123)
    Talking about missing a question. The last question was about the enthalpy change diagram right? because I had no time and was rushing the paper, hopefully I didnt miss anything..especially at the very end
    I didnt have a clue for that one. Nice how you go into the exam confident and walk out ***
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    so does anyone have the unofficial markscheme ?
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    For the 4 mark question about the mass of salt towards the end i got 7.44g. they gave you the Ph 7.2 and the concentration of the acid, 0.1 and they gave ka. so you worked out the hydrogen ion concention. then you used the formula:
    ka = [H+] x salt/acid to work out the concentration of the salt which was 0.06something. n=c x v and the volume was 1dm-3 so the concentration of the salt was the number of moles and you multiplied it by the Mr120 which gave 7.44g.
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    I had absolutely no clue how to answer the 'explain why the student is wrong' question.
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    (Original post by mina.)
    Agree with it all.
    Couldn't find the original post so I'll agree with your agreeing of it =P.
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    (Original post by limetang)
    I had absolutely no clue how to answer the 'explain why the student is wrong' question.
    I put that it was an equilibrium and that as the H+ was used up more would be made from the undisscoiated bit and eventually the whole of the acid will be used up.
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    (Original post by limetang)
    I had absolutely no clue how to answer the 'explain why the student is wrong' question.
    I wrote weak acids dont fully dissociate so the concentration of H+ will be lower
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    (Original post by limetang)
    I had absolutely no clue how to answer the 'explain why the student is wrong' question.
    Ha! It's one of those questions that you look at and are like..
    "He's wrong? Let me read this again.. His conclusion definitely seems right to me. **** I'm just as stupid as he is! I can feel OCR laughing at me right now.."
    Though on a more serious note, I said that the concentration of the acid or alkali is likely to affect the volume required?
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    Wasnt the bond angle on the SF6 109 degrees?
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    (Original post by limetang)
    I had absolutely no clue how to answer the 'explain why the student is wrong' question.
    I had on idea either i just made it up saying it partially dissociated because it was a weak acid so all of 7.5cm3 of NaOH is not used up so its more than enough. lol
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    (Original post by tom108)
    Wasnt it just 120 x 0.1 ?? (in short)
    (Original post by Mathelete)
    I got that too! And it is right!! My chemistry teacher checked it!!
    I think we're forgeting its a 4 mark question... if its four marks, it wouldnt be that easy. I did that at first and realised the calculation was too short and is probably worth only 2 marks. I did some other thing using Ka and got like 19.4 or summin like that
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    (Original post by arsenalman)
    Wasnt the bond angle on the SF6 109 degrees?
    It was 90 as it had 6 flourines around it. It is 109 when there are 4 groups around the molecule
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    (Original post by tkoki1993)
    I think we're forgeting its a 4 mark question... if its four marks, it wouldnt be that easy. I did that at first and realised the calculation was too short and is probably worth only 2 marks. I did some other thing using Ka and got like 19.4 or summin like that
    It was 7.44g
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    (Original post by 04ayasin)
    It was 90 as it had 6 flourines around it. It is 109 when there are 4 groups around the molecule
    Argh and i realised i drew the 3 d shape wrong lol.
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    (Original post by arsenalman)
    I had on idea either i just made it up saying it partially dissociated because it was a weak acid so all of 7.5cm3 of NaOH is not used up so its more than enough. lol
    It partially dissociates but when all the H+ reacts with the NaOH, more dissociates and it keeps happening until all the H+ have been used up. So, the fact that is a weak acid is irrelevant.
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    I got 19.2g, but from rounding to something like 0.16moldm-3 at one point, but if I used the complete value in the calculator I would get 19.35-something. I don't know if this is right, there seems to be 12g, 7.4g and 19g appearing quite often. At least we'll get working out marks though?
    There'll probably be one for finding the conc. of H+ ions at least..
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    (Original post by arsenalman)
    It was 7.44g
    are you completely sureeeE????? lool
    I'm trying to hang on to as many marks as possible to maintain my sanity. When I realise my answer is wrong, I still probably delude myself its right just to remain happy lol
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    (Original post by arsenalman)
    I had on idea either i just made it up saying it partially dissociated because it was a weak acid so all of 7.5cm3 of NaOH is not used up so its more than enough. lol
    Oh I think I just made chemistry up for it. I said something along the lines of. "That equation uses a salt therefore it doesn't act as a weak acid"
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    (Original post by brendan.)
    I got 19.2g, but from rounding to something like 0.16moldm-3 at one point, but if I used the complete value in the calculator I would get 19.35-something. I don't know if this is right, there seems to be 12g, 7.4g and 19g appearing quite often. At least we'll get working out marks though?
    There'll probably be one for finding the conc. of H+ ions at least..
    woohoo!!! someone else with 19.4!!!
    the happiness has returned

 
 
 
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