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    For writing ionic equations of say Ca(OH)2 + 2Pyruvic Acid ----> (Pyruvate)2Ca2+ + 2H2O

    The ionic equation is apparently just H+ and OH- ----> H2O

    Can someone explain how, I don't get why the Pyruvic Acid to Pyruvate isn't also in the ionic equation if it lost a Hydrogen?
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    Any examples of dibasic calculations in past papers?
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    (Original post by VMD100)
    That is pretty convoluted but when you phrase it like that it does make sense.
    In answer to your original question I am 99% that is not in the spec, I remember my chem teacher really briefly mentioning it but I'm fairly certain he didn't even go into detail as it was not in the course.
    Oh okay great thanks!
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    Anyone got the mark scheme to the 2015 paper?
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    (Original post by Dinasaurus)
    For writing ionic equations of say Ca(OH)2 + 2Pyruvic Acid ----> (Pyruvate)2Ca2+ + 2H2O

    The ionic equation is apparently just H+ and OH- ----> H2O

    Can someone explain how, I don't get why the Pyruvic Acid to Pyruvate isn't also in the ionic equation if it lost a Hydrogen?
    cause it hasn't changed oxidation or physical state
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    (Original post by mechanism)
    cause it hasn't changed oxidation or physical state
    surely by losing a Hydrogen it has changed oxidation state.
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    I still don't really understand the June 2015 42iii question on the adding of the Mg to the buffer :/
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    In electrode potential
    does reduction happen on the positive or negative terminal?
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    (Original post by M0n1)
    In electrode potential
    does reduction happen on the positive or negative terminal?
    Oxidation produces electrons (which are negative) so oxidation occurs on the negative / reduction on the positive
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    (Original post by Rust Cohle)
    https://i.gyazo.com/49069f767bf5ea95...c6a979a5e2.png
    There is question. Since we know that the Chromium gains mass its going from aqeous ---> solid. So write that half equation:

    Cr -----> Cr3+ + 3e-

    The X electrode loses mass so it goes from solid ----> aqeous. In order to to help you form the half equation, we're told that the aqeous is XSO4 meaning oxidation state of X = +2.

    X2+ + 2e- -----> X

    -----------------------------------

    Now simply create a full equation with the two half equations:

    2Cr3+ + 3X ----> 3X2+ + 2Cr

    -------------------------------

    Moles of Cr we can find since we know its mass (1.456g) and the mr 52. n = m/Mr = 0.028

    Apply this to the full equation and you'll find that the moles of X = 0.042 using stoichiometry (molar ratios)

    We can know work Mr of X since we have the mass of 1.021g. Mr = m/n = 24.3 = Magnesium
    just a q how come u divided mole by cr only and nit by cr2so4


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    (Original post by ranz)
    just a q how come u divided mole by cr only and nit by cr2so4


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    Cr is formed as a solid thats why

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    (Original post by tcameron)
    I still don't really understand the June 2015 42iii question on the adding of the Mg to the buffer :/
    i remember that q, so hard.
    baically its apparently prettty simple my friends helped me.
    just calc how many moles of mg added(0.25), minus 1mol of the acid by that, thats whats left kf the acid. 0.25+1 mole of conjucate base is produced as orevious page wrote that conjucate base had 1moldm3 so that + mole of mg 0.25 then sub values into kc eq and find H+. its confusing hope this helped


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    (Original post by Dinasaurus)
    surely by losing a Hydrogen it has changed oxidation state.
    I don't think so
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    (Original post by Dinasaurus)
    surely by losing a Hydrogen it has changed oxidation state.
    It's becuase, in the prescence of Ca(OH)2 the Pyruvic acid is completely dissociated. Therefore it can be considered to be completely dissociated BEFORE the reaction takes place and therefore is a spectator ion for the reaction:

    2C3H3O3- + 2H+ + Ca2+ +2OH- --> Ca2+ +2C3H3O3- + 2H20

    The ions cancel on each side to give:

    2H+ +2OH- --> 2H2O
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    (Original post by ScarletRose)
    Anyone got the mark scheme to the 2015 paper?
    http://www.chemhume.co.uk/A2CHEM/Exa..._MS_June15.pdf
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    Can anyone please explain Jan 2013 Q8)C)ii ? I've worked it out before but I really don't remember how it's done.
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    What ratio of acid to acidoate should we use in a buffer?

    Am I supposed to always say 2:1?
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    Name:  chemistry.png
Views: 59
Size:  111.4 KB

    Hey can anyone quickly explain why the concentration of HCOOH is 0.24 and not 0.64 (3.2x0.2) in this buffer question please?
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    (Original post by smartalan73)
    Name:  chemistry.png
Views: 59
Size:  111.4 KB

    Hey can anyone quickly explain why the concentration of HCOOH is 0.24 and not 0.64 (3.2x0.2) in this buffer question please?
    Acid is in excess. So it'll react with all of the NaOH and you'll have some left over. You've got 0.64 moles of HCOOH and 0.4 moles of NaOH so 0.64-0.4=0.24 moles HCOOH left.
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    (Original post by smartalan73)
    Name:  chemistry.png
Views: 59
Size:  111.4 KB

    Hey can anyone quickly explain why the concentration of HCOOH is 0.24 and not 0.64 (3.2x0.2) in this buffer question please?
    0.64-0.4 quick
 
 
 
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