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    (Original post by otrivine)
    Not really, ok I will do a lot of examples for you!
    Thanks and are you sure the second one is only CO2 and not 2CO2 cause otherwise its not going to work at all in my eyes, even looked at my notes and other notes and nothing seems to help with that equation.


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    (Original post by otrivine)
    Not really, ok I will do a lot of examples for you!
    Okay I think I've completed it now, it all makes sense in my head so I hope it is right.




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    (Original post by MathsNerd1)
    Okay I think I've completed it now, it all makes sense in my head so I hope it is right.




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    Not quite right, nearly there and you forgot to put the charge on the cr2o7, it should be cr2o7 2-
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    how is everyone revising- I find it goes through one ear and out the other :/
    BLERGH so jel of those who just need a D in this for an A overall
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    (Original post by otrivine)
    Not quite right, nearly there and you forgot to put the charge on the cr2o7, it should be cr2o7 2-
    Grr, I'm so forgetful!!! Just let me finish this maths paper and then I'll fix it all, I definitely need some work on these apparently even though its just mainly Maths so I wouldn't have thought I'd struggle with it.


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    (Original post by MathsNerd1)
    Grr, I'm so forgetful!!! Just let me finish this maths paper and then I'll fix it all, I definitely need some work on these apparently even though its just mainly Maths so I wouldn't have thought I'd struggle with it.


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    I like you, so I will help you Tell me exactly which part you are stuck in, and I can write notes and attach for you!
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    (Original post by otrivine)
    I like you, so I will help you Tell me exactly which part you are stuck in, and I can write notes and attach for you!
    It's this isn't it?


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    (Original post by MathsNerd1)
    It's this isn't it?


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    No, the charges are not balanced!

    Firstly, show me what you got for your 2 half equations
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    (Original post by otrivine)
    No, the charges are not balanced!

    Firstly, show me what you got for your 2 half equations
    Oh, I think I've been looking at it all wrong then, here's my last attempt, otherwise I'll message my teacher to explain it all to me!



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    (Original post by MathsNerd1)
    Oh, I think I've been looking at it all wrong then, here's my last attempt, otherwise I'll message my teacher to explain it all to me!



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    No, you forgot to include the H+ in the second half equation!

    do you want me to post my solution?

    can you help me on orders please
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    (Original post by otrivine)
    No, you forgot to include the H+ in the second half equation!

    do you want me to post my solution?

    can you help me on orders please
    If you wouldn't mind that would be very useful to me and sure, what part are you getting stuck on?


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    (Original post by MathsNerd1)
    If you wouldn't mind that would be very useful to me and sure, what part are you getting stuck on?


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    of course

    its on page 160 question 1)b)i) I do not know why they would consider H2O2 as first order when the concentration remains the same and rate stays the same when comparing experiments 2 and 3. Also, [H+] should be first order not 0 order because the concentration increases but rate stays the same for expeiments 2 and 3?
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    (Original post by otrivine)
    of course

    its on page 160 question 1)b)i) I do not know why they would consider H2O2 as first order when the concentration remains the same and rate stays the same when comparing experiments 2 and 3. Also, [H+] should be first order not 0 order because the concentration increases but rate stays the same for expeiments 2 and 3?
    Okay well I'm not too sure how they got the H2O2 to be first order but maybe that's because they haven't shown us a second concentration for it to see how it's affected, you just have to take the information given to you as fact and proceed then. And I believe [H+] should be 0 because anything to the power 0 = 1 so it doesn't get affected by the concentration as the rate will remain constant.

    If a graph was drawn it would be a straight horizontal line and have you got the reaction orders for the others and created a rate equation with its units?


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    (Original post by MathsNerd1)
    Okay well I'm not too sure how they got the H2O2 to be first order but maybe that's because they haven't shown us a second concentration for it to see how it's affected, you just have to take the information given to you as fact and proceed then. And I believe [H+] should be 0 because anything to the power 0 = 1 so it doesn't get affected by the concentration as the rate will remain constant.

    If a graph was drawn it would be a straight horizontal line and have you got the reaction orders for the others and created a rate equation with its units?


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    Okay! thanks

    here is the attachment for the equation redox.
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    (Original post by otrivine)
    Okay! thanks

    here is the attachment for the equation redox.
    Thanks, I understand now as I forget about the Cr2 and just assumed it was for just a single Cr, my bad but I'm now doing more of these equations to build up my confidence in them, thanks again


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    (Original post by MathsNerd1)
    Thanks, I understand now as I forget about the Cr2 and just assumed it was for just a single Cr, my bad but I'm now doing more of these equations to build up my confidence in them, thanks again


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    No worries, want to revise again this evening ?
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    (Original post by otrivine)
    No worries, want to revise again this evening ?
    Sure thing, about the same time too? Cause that way I'll be able to plan around then to get other things done like Maths and all my other Chemistry units.


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    (Original post by MathsNerd1)
    Sure thing, about the same time too? Cause that way I'll be able to plan around then to get other things done like Maths and all my other Chemistry units.


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    sure, lets start at 5pm is that fine?
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    (Original post by otrivine)
    sure, lets start at 5pm is that fine?
    Erm, yeah that should be fine. Still only for 2 hours though.


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    (Original post by MathsNerd1)
    Erm, yeah that should be fine. Still only for 2 hours though.


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    ok
 
 
 
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