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    Using Ka for a diprotic acid e.g. H2A --> 2H+ + A2-
    Would the expression be
    Ka = [H+]^2 * [A2-] / [H2A]? very confused about this
    I know a similar thing is done with Kc
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    (Original post by Thisshouldbeapun)
    Using Ka for a diprotic acid e.g. H2A --> 2H+ + A2-
    Would the expression be
    Ka = [H+]^2 * [A2-] / [H2A]? very confused about this
    I know a similar thing is done with Kc
    No

    Kc I'd different
    With kc, the formula is products over reactants to the power of the number of moles

    With ka the formula is always - [H+] x [A-] / [HA]



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    Help please
    2bi

    Thank you

    I sort of understand the answer
    But I would like to know why

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    (Original post by High Stakes)
    Aye I lav organic synthesis.
    Mo u r an odd one
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    How would you name that?

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    (Original post by Glavien)
    How would you name that?

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    would it be N,N-ethylpropylbutylamine?
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    (Original post by theboss1998)
    would it be N,N-ethylpropylbutylamine?
    I got N-ethyl-N-propylbutan-1-amine but I'm not sure if it's correct.


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    (Original post by Glavien)
    I got N-ethyl-N-propylbutan-1-amine but I'm not sure if it's correct.


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    That seems right! How did you do it?
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    (Original post by theboss1998)
    That seems right! How did you do it?
    Hi,
    Since it's a tertiary amine it's missing a + charge on the N - how I know it's a tertiary amine is the 3 alkyl chains attached to the nitrogen -

    So you name it according to the rules of nomenclature - butylethylpropylamine or N-butylethylpropylamine





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    (Original post by Glavien)
    How would you name that?

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    Also personally I wouldn't have the N but as its not a N subsitututed amide


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    does anyone know whether a catalyst effects the rate constant, also is the rate expression and the rate equation the same thing?
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    (Original post by theboss1998)
    Only temperature affects the value of K. Also, yes they are.
    thank you!
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    (Original post by Superbubbles)
    does anyone know whether a catalyst effects the rate constant, also is the rate expression and the rate equation the same thing?
     \displaystyle k=Ae^{-E_a/(RT)} .
    The activation energy would be lower in presence of a catalyst, so the rate constant would increase.
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    (Original post by Superbubbles)
    thank you!
    Sorry my mistake I thought that was Kc, yes catalyst does affect it as B_9710 stated above. Sorry once more.
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    (Original post by theboss1998)
    Sorry my mistake I thought that was Kc, yes catalyst does affect it as B_9710 stated above. Sorry once more.
    no worries!
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    (Original post by B_9710)
     \displaystyle k=Ae^{-E_a/(RT)} .
    The activation energy would be lower in presence of a catalyst, so the rate constant would increase.
    thank you for clearing up the confusion! very much appreciated
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    does anyone know what pka is actually measuring, there doesnt seem to be a proper definition for it in the AQA text book :/
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    (Original post by Superbubbles)
    does anyone know what pka is actually measuring, there doesnt seem to be a proper definition for it in the AQA text book :/
    Acidity. Lower the pKa the higher the acidity.
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    (Original post by B_9710)
    Acidity. Lower the pKa the higher the acidity.
    I know at the half- neutralisation point A- = HA so pH= pKa but why is this only for weak acids or does it also apply for strong acids??, in the indicators section of the text book it says "the colour change of most indicators takes place over a pH range of around 2 units, centred around the value of pka for the indicator" but it doesnt talk about weak acids in this context
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    Chemistry unit 5
    January 2010
    Question 3h
    Is it okay yo say:
    Electrolysis of water to obtain hydrogen uses fossil fuels so not carbon neutral


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