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    (Original post by Benjy100)
    Are marks lost if I represented aluminium ethanoate like this ... Al(CH3COO-)3 with the minus sign there? Or do they overlook such things?
    Doubt it matters.
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    I think I understand the confusion over this rate equation business now.

    What we were supposed to do is draw a tangent to the graph at the point where t=40 and use dy/dx to get the gradient of the tangent, which is equal to the rate of reaction at that point. What I did was take the values of x and y that the tangent went through, which was about 4.5 on the y-axis and 220 on the x-axis, giving me a value of 0.02moldm^-3s^-1. I think the people that got 0.04moldm^-3s^-1 took the x and y values at the point at which t=40, which is where y = 3.5 and x = 75 or something like that.

    I think the first method is right but I'm not sure.
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    (Original post by MathsNerd1)
    Get off this thread cause you're clearly too rude to talk to others and jealous just cause not everyone sucks at Chemistry as I'm assuming you do!
    I dont count people such as yourself credible human beings, more... machines. maybe perhaps a subspecies of human that is more akin to an albino rat
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    (Original post by needtosucceed=))
    the question told you it had an oxidation state of 6 (VI) therefore it has to be FeO4 2-
    Sorry thats what I meant, accidentally typed the 2!
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    The question on rate of reaction, I put a point at where t=40 and drew a line tangent, and got coordinates (40,concentration to where 40 was) and ( 120, conc to whatever that was)

    then did rate of reaction = change in conc of reactant or product divided by time taken?
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    Is there any unofficial ms?
    i saw the google doc earlier... but it gone missing??!
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    (Original post by NerdMeister)
    I think I understand the confusion over this rate equation business now.

    What we were supposed to do is draw a tangent to the graph at the point where t=40 and use dy/dx to get the gradient of the tangent, which is equal to the rate of reaction at that point. What I did was take the values of x and y that the tangent went through, which was about 4.5 on the y-axis and 220 on the x-axis, giving me a value of 0.02moldm^-3s^-1. I think the people that got 0.04moldm^-3s^-1 took the x and y values at the point at which t=40, which is where y = 3.5 and x = 75 or something like that.

    I think the first method is right but I'm not sure.
    I took values that the tangent went through and got about 0.04.
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    (Original post by NerdMeister)
    I think I understand the confusion over this rate equation business now.

    What we were supposed to do is draw a tangent to the graph at the point where t=40 and use dy/dx to get the gradient of the tangent, which is equal to the rate of reaction at that point. What I did was take the values of x and y that the tangent went through, which was about 4.5 on the y-axis and 220 on the x-axis, giving me a value of 0.02moldm^-3s^-1. I think the people that got 0.04moldm^-3s^-1 took the x and y values at the point at which t=40, which is where y = 3.5 and x = 75 or something like that.

    I think the first method is right but I'm not sure.
    I drew a tangent and worked out the gradient of the tangent and got 0.04ish

    Did you get the scales wrong because I'm pretty sure it wouldn't have been 220s...
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    (Original post by otrivine)
    The question on rate of reaction, I put a point at where t=40 and drew a line tangent, and got coordinates (40,concentration to where 40 was) and ( 120, conc to whatever that was)

    then did rate of reaction = change in conc of reactant or product divided by time taken?
    yep I think thats right
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    (Original post by alow)
    I drew a tangent and worked out the gradient of the tangent and got 0.04ish

    Did you get the scales wrong because I'm pretty sure it wouldn't have been 220s...
    I don't know, I'm very confused now
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    (Original post by Sweatynerd)
    I dont count people such as yourself credible human beings, more... machines. maybe perhaps a subspecies of human that is more akin to an albino rat
    I'm far from a machine, I'm only looking at 85+ most likely which tells you I still make mistakes and I don't only just revise I've got other things to put to my name so I'm just going to ignore you from now on as you clearly talk a load of crap!
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    (Original post by NerdMeister)
    I think I understand the confusion over this rate equation business now.

    What we were supposed to do is draw a tangent to the graph at the point where t=40 and use dy/dx to get the gradient of the tangent, which is equal to the rate of reaction at that point. What I did was take the values of x and y that the tangent went through, which was about 4.5 on the y-axis and 220 on the x-axis, giving me a value of 0.02moldm^-3s^-1. I think the people that got 0.04moldm^-3s^-1 took the x and y values at the point at which t=40, which is where y = 3.5 and x = 75 or something like that.

    I think the first method is right but I'm not sure.
    No. Did what you did, got around 0.04...
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    (Original post by alow)
    I think mine were 3.25/75 or something close to that, got an answer of 0.04ish i think
    I drew the tangent at 40 seconds and got approximately 0.04 as well! Let's hope we were right!
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    (Original post by needtosucceed=))
    yep I think thats right
    I got 0.02 something , in the June 2011 they accepted a range of answers so pretty sure its fine,


    what did u put for the optical isomer
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    Does anyone think this paper was really hard? Harder than Jan2013? I hope so because Jan 2013 had grade boundary of 70/100 for an A.
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    If anyone has an unofficial mark scheme could you send it to me please!
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    (Original post by otrivine)
    I got 0.02 something , in the June 2011 they accepted a range of answers so pretty sure its fine,


    what did u put for the optical isomer
    I'm pretty sure they wouldnt accept a gradient that is half as steep as it should be....

    probably more like a range of 0.35<dy/dx<0.45
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    (Original post by otrivine)
    I got 0.02 something , in the June 2011 they accepted a range of answers so pretty sure its fine,


    what did u put for the optical isomer
    I had x3 of the c2o4 ligands
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    (Original post by alow)
    I'm pretty sure they wouldnt accept a gradient that is half as steep as it should be....

    probably more like a range of 0.35<dy/dx<0.45
    I think between 0.02 to 0.04 something! in june 2011 how much difference did they accept , it was a huge variety right?
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    (Original post by needtosucceed=))
    I had x3 of the c2o4 ligands
    I put 2 x C2O42- and 2 H2O

    cause they put (C2O4)2 , I thought only 2 x C2O4 2-
 
 
 
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