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    (Original post by _HabibaH_)
    Which paper is this from please? I can't really see what you're saying fully from that equation alone.
    Jan 2011, 4bii
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    (Original post by Tendael)
    Jan 2011, 4bii
    I understand what you're saying. It seems like a chance thing.

    But it does say on the paragraph before that the overall equation is the combustion equation so I'm thinking you work backwards.
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    Why does an increase in rate not lead to an increase in the rate constant?
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    Does anyone know where the comment is that the person put on with all the legacy past papers? I can't find it anywhere.
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    (Original post by ofudge)
    Why does an increase in rate not lead to an increase in the rate constant?
    The rate constant increases with temperature, because of particles exceeding activation energy and hence, more frequency of collision increases
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    (Original post by ofudge)
    Why does an increase in rate not lead to an increase in the rate constant?
    because it's a constant. K literally means the constant, so remains constant whatever you do to the reactants as you cannot change it. It's the combination of the value of K and the molecules in the rate equation and to what power they're raised that determines the rate. The only thing that changes it is increasing temperature.
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    (Original post by ofudge)
    Why does an increase in rate not lead to an increase in the rate constant?
    It does increase k but only if temp is increased. Otherwise k remains constant.
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    (Original post by bmpink)
    because it's a constant. K literally means the constant, so remains constant whatever you do to the reactants as you cannot change it. It's the combination of the value of K and the molecules in the rate equation and to what power they're raised that determines the rate.
    Thank you very much. So even with increased/decreased temperature it remains the same?

    Are there any other "constants" in this exam that we need to know stay the same?

    Thanks a lot for your help.
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    does anyone want to revise asking each other some hard questions as pracTISE?
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    (Original post by ofudge)
    Thank you very much. So even with increased/decreased temperature it remains the same?

    Are there any other "constants" in this exam that we need to know stay the same?

    Thanks a lot for your help.
    That's ok! The rate constant only depends on the temperature. That's the only thing that can change it, because of increasing number of collisions per second. YES! There's Kc, equilibrium constant, which only changes with temperature and not pressure or anything else, Ka the acid dissociation constant, and Kstab the stability constant. I'm pretty sure that's all of them. Oh and Kw the ionic product of water, which increases with increasing temperature(I think! double check, there is a question on Kw in Jan 2011 I think)
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    During the analysis of brass, 1.65 g of the alloy was reacted with concentrated nitric acid. The resulting solution was neutralised, transferred to a volumetric flask and made up to 250 cm3 using distilled water.
    An excess of aqueous potassium iodide was added to a 25.0 cm3 portion of the solution from the volumetric flask and the liberated iodine was titrated with 0.100 mol dm–3 sodium thiosulphate. 20.0 cm3 of aqueous sodium thiosulphate were required to remove the iodine.


    The reactions taking place in this titration may be summarised as follows.
    2Cu2+ + 4I–  2CuI + I2
    I2 + 2S2O32–  2I– + S4O62–
    (i) Calculate the amount, in moles, of sodium thiosulphate in 20.0 cm3 of solution.



    for this I get 0.0025 moles and mark scheme gets 0.002 ?
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    (Original post by otrivine)
    During the analysis of brass, 1.65 g of the alloy was reacted with concentrated nitric acid. The resulting solution was neutralised, transferred to a volumetric flask and made up to 250 cm3 using distilled water.
    An excess of aqueous potassium iodide was added to a 25.0 cm3 portion of the solution from the volumetric flask and the liberated iodine was titrated with 0.100 mol dm–3 sodium thiosulphate. 20.0 cm3 of aqueous sodium thiosulphate were required to remove the iodine.


    The reactions taking place in this titration may be summarised as follows.
    2Cu2+ + 4I–  2CuI + I2
    I2 + 2S2O32–  2I– + S4O62–
    (i) Calculate the amount, in moles, of sodium thiosulphate in 20.0 cm3 of solution.



    for this I get 0.0025 moles and mark scheme gets 0.002 ?
    The MS is right as (20 x 0.10)/1000 = 0.002

    You used 25cm3 instead of 20cm3
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    (Original post by bmpink)
    That's ok! The rate constant only depends on the temperature. That's the only thing that can change it, because of increasing number of collisions per second. YES! There's Kc, equilibrium constant, which only changes with temperature and not pressure or anything else, Ka the acid dissociation constant, and Kstab the stability constant. I'm pretty sure that's all of them. Oh and Kw the ionic product of water, which increases with increasing temperature(I think! double check, there is a question on Kw in Jan 2011 I think)
    Ah that's great thanks I knew what they were but didnt realise they were constants! oops :L
    Need 115/150 in this exam to get my A... pressures on! good luck to you.
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    (Original post by _HabibaH_)
    The MS is right as (20 x 0.10)/1000 = 0.002

    You used 25cm3 instead of 20cm3
    opps sorry made , what about these two

    For every one mole of Cu2+ ions present in solution, deduce the amount, in moles, of S2O32– ions needed for the titration.


    answer ......................... mol
    [1]
    (iii) What is the amount, in moles, of Cu2+ ions present in 25.00 cm3 of solution?
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    (Original post by otrivine)
    During the analysis of brass, 1.65 g of the alloy was reacted with concentrated nitric acid. The resulting solution was neutralised, transferred to a volumetric flask and made up to 250 cm3 using distilled water.
    An excess of aqueous potassium iodide was added to a 25.0 cm3 portion of the solution from the volumetric flask and the liberated iodine was titrated with 0.100 mol dm–3 sodium thiosulphate. 20.0 cm3 of aqueous sodium thiosulphate were required to remove the iodine.


    The reactions taking place in this titration may be summarised as follows.
    2Cu2+ + 4I–  2CuI + I2
    I2 + 2S2O32–  2I– + S4O62–
    (i) Calculate the amount, in moles, of sodium thiosulphate in 20.0 cm3 of solution.



    for this I get 0.0025 moles and mark scheme gets 0.002 ?
    where did you get this paper?
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    the colour between

    Fe2+ and Mno4 -- > Mn2+ + Fe3+ e.t.c

    does the colour change from purple, to almost pink(colourless)
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    (Original post by otrivine)
    opps sorry made , what about these two

    For every one mole of Cu2+ ions present in solution, deduce the amount, in moles, of S2O32– ions needed for the titration.


    answer ......................... mol
    [1]
    (iii) What is the amount, in moles, of Cu2+ ions present in 25.00 cm3 of solution?
    0.002 moles?
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    (Original post by _HabibaH_)
    0.002 moles?
    they put 1 ?
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    (Original post by ofudge)
    Ah that's great thanks I knew what they were but didnt realise they were constants! oops :L
    Need 115/150 in this exam to get my A... pressures on! good luck to you.
    115 isn't bad! I need about 83% if I didn't go up in my F321 retake, but I genuinely have no idea how it went because it's so much easier than A2 I feel like I have nothing to gage it on! Good luck! If you need any more help ask me because I need the practice!!
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    (Original post by ofudge)
    Why does an increase in rate not lead to an increase in the rate constant?
    An increase in 'rate'? What does it mean to increase rate? You increase/decrease particular factors to influence the rate equation, like temp, concentration and pressure.
 
 
 
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