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    (Original post by LewisEdwards1234)
    Of course there is! Judging the rest of the paper went well you could drop a minimum of 25 marks and still get an A, the calculation questions were worth a total of 9
    Let's say I dropped 30 marks on this exam
    I got 43/45 on my coursework
    Can I still get an A...do I need to do really well on f335?
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    (Original post by Marli-Ruth)
    Let's say I dropped 30 marks on this exam
    I got 43/45 on my coursework
    Can I still get an A...do I need to do really well on f335?
    43/45 on coursework will gain you around 86 ums, and 60/90 possibly around 65 ums so thats 151, so you're still in a strong position to get an A depending how AS went. You need 240 ums for an A so a minimum of 89 ums and going by 2015 grade boundaries 80 marks (which is an A grade) is worth 96 ums so I imagine you'd be fine!
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    (Original post by charliexox)
    What did people put for the vapours thing about high temperatures??
    Luckily for me I happened to do the bromate/bromide reaction (with acid) for my coursework! So I knew the rate equation without having to work it out.

    For the vapours question, I'm fairly sure that bromine is a volatile liquid (thinking back to the halogens last year?), and I'm also fairly sure that bromate and bromide in solution aren't volatile as I had to heat those up in my investigation. My logic was that bromine is the only volatile liquid, therefore the measured amount of bromine (or absorbance if you're doing colorimetry I guess) would be lower than the actual amount of bromine produced due to evaporation. I then went on to say that the measured rate would be lower than the actual rate because of the loss of bromine from solution.

    What I wrote is probably completely wrong, but it was only 2 or 3 marks so...
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    (Original post by LewisEdwards1234)
    43/45 on coursework will gain you around 86 ums, and 60/90 possibly around 65 ums so thats 151, so you're still in a strong position to get an A depending how AS went. You need 240 ums for an A so a minimum of 89 ums and going by 2015 grade boundaries 80 marks (which is an A grade) is worth 96 ums so I imagine you'd be fine!
    I only got 228 UMS at AS
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    (Original post by lyricalvibe)
    Yeah I'll upload it soon-ish unless it's urgent for you rn :P?

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    yes please, quite urgent thanks
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    (Original post by lyricalvibe)
    Yeah I'll upload it soon-ish unless it's urgent for you rn :P?

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    yes please, quite urgent thanks
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    (Original post by phagocyte1998)
    Luckily for me I happened to do the bromate/bromide reaction (with acid) for my coursework! So I knew the rate equation without having to work it out.

    For the vapours question, I'm fairly sure that bromine is a volatile liquid (thinking back to the halogens last year?), and I'm also fairly sure that bromate and bromide in solution aren't volatile as I had to heat those up in my investigation. My logic was that bromine is the only volatile liquid, therefore the measured amount of bromine (or absorbance if you're doing colorimetry I guess) would be lower than the actual amount of bromine produced due to evaporation. I then went on to say that the measured rate would be lower than the actual rate because of the loss of bromine from solution.

    What I wrote is probably completely wrong, but it was only 2 or 3 marks so...
    That's exactly what I wrote! I couldn't think of anything else that was volatile in that reaction.
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    (Original post by Chloestar)
    That's exactly what I wrote! I couldn't think of anything else that was volatile in that reaction.
    What was the question exactly, would the temperature have effected the rate constant also?
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    (Original post by phagocyte1998)
    I think I agree with most of those answers. The only thing that I remember doing differently is I put aldehyde for the functional groups in Vanillin or whatever rather than ketone.
    Now I think about it, I'm 99% sure I put aldehyde too! It was ketone in the second molecule. Thanks, I'll change that now.
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    Can I measure the bromine reaction using titration?
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    (Original post by HecTic_H)
    Can I measure the bromine reaction using titration?
    I think that is acceptable not 100% but I see no reason why it wouldn't work.
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    (Original post by HecTic_H)
    Can I measure the bromine reaction using titration?

    I think you can
    I put colourimetry.. Is that wrong?
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    (Original post by Marli-Ruth)
    I think you can
    I put colourimetry.. Is that wrong?
    Pretty sure colorimetry is correct, because it came up before.
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    (Original post by SamuelN98)
    What was the question exactly, would the temperature have effected the rate constant also?
    I think the question was something along the lines of "The student thinks that some volatile substances may be lost if the experiment is carried out at a higher temperature. Explain how the calculated rate of reaction would be affected."

    In other words I think the gist is to do with how the measured rate would differ from the actual rate. I may be completely wrong, but that's how I interpreted the question. In any case, the actual question was probably nothing like what I just said!
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    (Original post by WittyGonStein)
    For the standard electrode potential question, it was +1.8 V, not -0.26 as it said the sign was the same as for the half-cell with iron. Hence they both had to be positive meaning that as for iron it was +0.77, to get +1.03, it had to be +1.8 for the other one
    I think by sign being the same it meant that iron would still be the positive electrode therefore the electrode potential for iron would be more positive than the electrode potential for vanadium. Therefore the electrode potential would be -0.26 (which is the actual electrode potential for the half equation).
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    (Original post by Lozzie97)
    I think by sign being the same it meant that iron would still be the positive electrode therefore the electrode potential for iron would be more positive than the electrode potential for vanadium. Therefore the electrode potential would be -0.26 (which is the actual electrode potential for the half equation).
    It's +1.80V they swapped the chromium for vanadium. They said the potential difference was +1.03V, and that the new electrode(vanadium) had the same sign as the opposing electrode (iron, +0.77)
    Potential difference is most positive minus least positive .
    So least positive plus potential difference is most positive
    (+0.77 + +1.03 = 1.80V)





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    Can anyone remember the diagram of the tanks and then which solution was which and what the answer was to that q? And if everyone posts what they think they roughlyu scored we might get an idea of the grade boundaries
    Think I got around 62 worst case
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    (Original post by marylee0000)
    I said organic polymer for coating the cell. Is that allowed too?
    I would think so
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    (Original post by Lozzie97)
    I think by sign being the same it meant that iron would still be the positive electrode therefore the electrode potential for iron would be more positive than the electrode potential for vanadium. Therefore the electrode potential would be -0.26 (which is the actual electrode potential for the half equation).
    Yes that was my reasoning, I also plugged the numbers back into the E cell equation to check and it seemed to work out ok, the purpose of the electrode was to see if it would work in place of the other electrode so it'd make sense if it was less than the iron one.
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    Name:  Vanadium.PNG
Views: 136
Size:  25.8 KB

    http://www.docbrown.info/page07/transition03V.htm

    Look at the table for vanadium 2+/3+, it's -0.26 exactly so I am fairly certain that those saying -0.26 in the exam were correct
 
 
 
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