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    Welcome Squad
    (Original post by abdullahharis)
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    machan hows prep in your place?
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    (Original post by RizK)
    machan hows prep in your place?
    Ahhh machan what u doing here?? Its Unit 6 lol

    Profile pic on point too!!!
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    (Original post by abdullahharis)
    Ahhh machan what u doing here?? Its Unit 6 lol

    Profile pic on point too!!!
    xD thanks!
    I'm gonna fail ._. you got any notes on organic synthesis for unit 6?
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    does anyone have a summary of all reactions (transition metals, organic etc) we need to know for unit 6?
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    (Original post by Quirky01)
    this is oxidation of alcohol..so cr2o7(2-) is reduced to cr(3+)
    Oh yeaaah!! Thaanks!!
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    (Original post by demotivated)
    Never heard of the fusion tube, just simple distillation to determind bp (despite its inaccuracy).

    Is this something that we need to know then?
    I don't know lol! That's what my teacher drew me.. (And he's the most reliable person in the world tbh)
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    (Original post by Adorable98)
    I don't get this question, so basically you add ethanol (which has an OH-so)
    Cr2O72- + OH- --> CrO42-+ H+

    So how come the answer's not CrO42- ??

    Well don't really need to think so much, just see orange to green and you definately know it's Cr2O7 to Cr+3
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    I have done the past years from 2011-2016, I dont even know what to do right now and my paper is like tomorrow. Any tips guys, notes please?
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    (Original post by RizK)
    xD thanks!
    I'm gonna fail ._. you got any notes on organic synthesis for unit 6?
    Here you go!
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    (Original post by syahmishaarani)
    I have done the past years from 2011-2016, I dont even know what to do right now and my paper is like tomorrow. Any tips guys, notes please?
    Revise the user guide and go through all your past papers! x)
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    (Original post by Aimen.)
    I don't know lol! That's what my teacher drew me.. (And he's the most reliable person in the world tbh)
    Okay. Does anyone know how to write the formulas of compounds with water of crystallisation?

    Like K2SO4.Cr(SO4)3.zH2O where z=12 or something, how do you figure these out, provided you know all the anions and cations?
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    (Original post by Aimen.)
    Well don't really need to think so much, just see orange to green and you definately know it's Cr2O7 to Cr+3
    :yy::yes:
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    (Original post by samb1234)
    A reaction is in general feasible if ecell is >0. We know that the reduction potentials are for the reducing direction, i.e X+Ae- ----> X^-A. What would it be for a species being oxidised, how would this affect the value given in the data booklet and how can we then use that to answer the question
    Hey, I am stuck on a point in this part

    Okay, so what I thought is that Sn is a reducing agent and it will be oxidized to Sn2+ , there E for Sn ias to be +0.14V, since it will be oxidised, it will be on LHS of cell, for the Vanadium, it has to be on RHS of cell and it has to be reduced, we have 4 potential equations that will have -1.00, +0.34.-0.26 and -1.18, we have to pick the species that has the most positive E that is +0.34 for the VO2+ to V3+

    Our Ecell is 0.34-(+0.14) = 0.2

    Can you tell me please where I got wrong, I am so confused :dontknow:
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    (Original post by PlayerBB)
    Hey, I am stuck on a point in this part

    Okay, so what I thought is that Sn is a reducing agent and it will be oxidized to Sn2+ , there E for Sn ias to be +0.14V, since it will be oxidised, it will be on LHS of cell, for the Vanadium, it has to be on RHS of cell and it has to be reduced, we have 4 potential equations that will have -1.00, +0.34.-0.26 and -1.18, we have to pick the species that has the most positive E that is +0.34 for the VO2+ to V3+

    Our Ecell is 0.34-(+0.14) = 0.2

    Can you tell me please where I got wrong, I am so confused :dontknow:
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    Posted from TSR Mobile
    Sn is a reducing agent, so we know that it itself will be oxidised, and therefore the E for Sn is +0.14 as you have mentioned. We know that the vanadium species are being reduced, i.e. the E values quoted in the question are the values we need to use. If we let X be the e value for the reduction of the vanadium, then we know that for the reaction to be feasible X +0.14>0. Can you take it from there?
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    (Original post by samb1234)
    Sn is a reducing agent, so we know that it itself will be oxidised, and therefore the E for Sn is +0.14 as you have mentioned. We know that the vanadium species are being reduced, i.e. the E values quoted in the question are the values we need to use. If we let X be the e value for the reduction of the vanadium, then we know that for the reaction to be feasible X +0.14>0. Can you take it from there?
    Yeah thank you!

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    (Original post by PlayerBB)
    Yeah thank you!

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    The mistake you made was the - sign. If you are doing it that way, i.e. reduced species -oxidised species, you do NOT change the sign of the ecell values, i.e. in this case you would have X--0.14
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    (Original post by samb1234)
    The mistake you made was the - sign. If you are doing it that way, i.e. reduced species -oxidised species, you do NOT change the sign of the ecell values, i.e. in this case you would have X--0.14
    Do we just simply input the values of the electrode potential i.e in this case
    Ecell= Ereduction-Eoxidation which is 0.34-(-0.14) correct?
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    (Original post by sabahshahed294)
    Do we just simply input the values of the electrode potential i.e in this case
    Ecell= Ereduction-Eoxidation which is 0.34-(-0.14) correct?
    yes
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    (Original post by samb1234)
    yes
    Alright. Thank you!
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    http://qualifications.pearson.com/co...e_20110119.pdf

    Q2(ii)

    The answer is:
    The concentration of iodine is proportional to the titre

    Why's that?
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