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    (Original post by Ayman!)
    I've attached it a few posts ago, please take a look



    It's better to understand the trend rather than use the data booklet, but yours is very good exam technique
    Thanks!
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    Hey guys, can someone explain why the reagent for step C is Br2? Since when could you heat an alkane with Br2 and make a bromoalkane? I thought you needed UV light etc. for that process...?!

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    An experiment is carried out to check the oxidation number of chromium inchromium(II) ethanoate Cr2(CH3CO2)4(H2O)2.
    1.00 g (2.66 × 10−3 mol) of chromium(II) ethanoate is dissolved in 25 cm3 of1.00 mol dm−3 sulfuric acid.

    The solution is diluted with distilled water until the volume is 250 cm3. 25.0 cm3 portions of the diluted solution are titrated with 0.00750 mol dm−3potassium manganate(VII).

    Calculate the volume of potassium manganate(VII) needed to oxidize thechromium(II) ions present in each 25.0 cm3 portion to the +6 oxidation state.The manganese is reduced to the +2 oxidation state.

    Question 21d from the IAL Jan 2016 paper, I cant seem to figure out why the ratio of manganate to chromium +2 is 4:5

    Any ideas?
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    (Original post by gabby07)
    Hey guys, can someone explain why the reagent for step C is Br2? Since when could you heat an alkane with Br2 and make a bromoalkane? I thought you needed UV light etc. for that process...?!

    I think the UV light acts as the source of heat in this instance.
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    (Original post by Z.e.e)
    An experiment is carried out to check the oxidation number of chromium inchromium(II) ethanoate Cr2(CH3CO2)4(H2O)2.
    1.00 g (2.66 × 10−3 mol) of chromium(II) ethanoate is dissolved in 25 cm3 of1.00 mol dm−3 sulfuric acid.

    The solution is diluted with distilled water until the volume is 250 cm3. 25.0 cm3 portions of the diluted solution are titrated with 0.00750 mol dm−3potassium manganate(VII).

    Calculate the volume of potassium manganate(VII) needed to oxidize thechromium(II) ions present in each 25.0 cm3 portion to the +6 oxidation state.The manganese is reduced to the +2 oxidation state.

    Question 21d from the IAL Jan 2016 paper, I cant seem to figure out why the ratio of manganate to chromium +2 is 4:5

    Any ideas?
    MnO4- + 8H+ 5Cr2+ --> 5Cr3+ + Mn2+ + 4H2O.
    Combine half equations of cr3+ and MnO4- in acidic conditions. Should be 1:5 not 4:5
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    (Original post by Franckenstar)
    MnO4- + 8H+ 5Cr2+ --> 5Cr3+ + Mn2+ + 4H2O.
    Combine half equations of cr3+ and MnO4- in acidic conditions. Should be 1:5 not 4:5
    The mark scheme quotes "4 mol manganate(VII) react with 5 mol ofchromium (II)"

    The question says chromium oxidises from +2 to +6 though

    -------

    I got it now Cr goes from +2 to +6 so +2 ---> +6 + 4e-
    Mno4- goes from +7 to +2 so +7 + 5e- ----> +2

    I forgot you needed to cross multiply to get equate the electrons

    ratio is then 4 moles of manganate with 5 moles of chromium
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    How od you do part c) I hate questions like this!! Is there an easy method?? Don Pedro K. C0balt samb1234 anyone?!

    Name:  Screen Shot 2016-06-19 at 11.32.46.png
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    (Original post by gabby07)
    How od you do part c) I hate questions like this!! Is there an easy method?? Don Pedro K. C0balt samb1234 anyone?!

    Name:  Screen Shot 2016-06-19 at 11.32.46.png
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    Um what I did for that question isn't using the x and y thing but by saying that the contraction of 20cm^3 is due to water condensing. So 10 of the hydrocarbon produces 20 of water. Water has 2 H atoms so a mole of the hydrocarbon has 4 moles of H atoms. And it says with KOH volume contracts by 40, so that is volume of CO2. 10 of the hydrocarbon produces 40 of CO2 and CO2 contains one C atom, so a mole of the hydrocarbon has 4 moles of C atoms. Therefore C4H4, and I checked with the x and y thing whether it was right.
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    (Original post by C0balt)
    Um what I did for that question isn't using the x and y thing but by saying that the contraction of 20cm^3 is due to water condensing. So 10 of the hydrocarbon produces 20 of water. Water has 2 H atoms so a mole of the hydrocarbon has 4 moles of H atoms. And it says with KOH volume contracts by 40, so that is volume of CO2. 10 of the hydrocarbon produces 40 of CO2 and CO2 contains one C atom, so a mole of the hydrocarbon has 4 moles of C atoms. Therefore C4H4, and I checked with the x and y thing whether it was right.
    Thank you so much. Much clearer in my head now!
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    (Original post by gabby07)
    Thank you so much. Much clearer in my head now!
    Np ☺
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    (Original post by C0balt)
    Np ☺
    what's the definition of standard electrode potential? its seems to be different every time. Would ( the potential difference when the electrode is connected to a standard hydrogen electrode at 1 moldm-3, a atm and 298 k) be a good answer?
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    How are these two displayed complexes different?
    Attached Images
     
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    (Original post by ramadeen)
    what's the definition of standard electrode potential? its seems to be different every time. Would ( the potential difference when the electrode is connected to a standard hydrogen electrode at 1 moldm-3, a atm and 298 k) be a good answer?
    Yeah that would be OK
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    (Original post by gabby07)
    How are these two displayed complexes different?
    They aren't, one maps to the other upon rotation.
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    (Original post by gabby07)
    How od you do part c) I hate questions like this!! Is there an easy method?? Don Pedro K. C0balt samb1234 anyone?!

    Name:  Screen Shot 2016-06-19 at 11.32.46.png
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    I literally have no clue how you would do this haha XD This is from an IAL paper, I presume?
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    (Original post by Don Pedro K.)
    I literally have no clue how you would do this haha XD This is from an IAL paper, I presume?
    Yup
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    (Original post by gabby07)
    Yup
    Haha thought so xD I want to understand how to do it, feel like we could get one like this after that flipping unit 4...Which paper is it from?
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    Name:  ImageUploadedByStudent Room1466339075.296143.jpg
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    What would z be


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    gabby07 dw I got it
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    (Original post by Don Pedro K.)
    gabby07 dw I got it
    C0balt explained it well in the previous page
 
 
 
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