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    (Original post by maisym00)
    For the equilibrium question I put: I- reacts with Ag to form agI (yellow ppt), therefore the position of equilibrium will shift to the left to restore I- conc. As a result, more aqueous iodine was formed so the position of equilibrium of 1 will also shift to the left, therefore more grey/black solid iodine is formed. For the electrochemistry question I needed up with two equations. The oxidising agent one where Cr+3 reacts with AL and the reducing agent one where Cr+3 reacts with feO4- or something. Anyone else?
    That's pretty much what I did, so yay
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    Anyone going to make an unofficial markscheme?
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    (Original post by Jsey)
    You find the work backwards through the titration to find the moles of Cu2+. Then you multiply the moles of Cu2+ by the molecular mass of Cu(HCOO)2 to find the mass in grams. Then you subtract this value from the grams they gave you in the question. Then you use the remaining grams to find the moles of water. Using moles equals mass over molar mass. Then lastly you divide the moles of water by the moles of the Cu2+ to get the molar ratio.
    Damn, I think after finding the moles of Cu2+ I worked out the molar mass of the wholecompund by dividing the mass they gave by moles of Cu2+. Then, I took away molar mass of Cu(HCOO)2 to from that to get the molar mass of water. Then divide by 18 to get 17 for X.
    So basically, I used the method on page 220 in the textbook.
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    Could we draw the diagrams in pencil (the harber cycle and the cell potential)? I'm afraid it wouldnt show up
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    (Original post by milemed11)
    the ph was given as 7 so i worked out the conc of h+ then divided kw by that to give oh-
    oh- > h+ so i put alkaline

    anyone else?
    yess yess
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    What I got for the 4-step mechanism:

    Overall equation: H2O2 + 2I- + 2H+ forms I2 + 2H2O

    H2O2 + I- forms IO- + H2O
    IO- + H+ forms HIO
    HIO + I- forms I2 + OH- (given in question)
    OH- + H+ forms H2O
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    (Original post by Fmathslad)
    i did it by finding out the concentration of (h+) and the concentration of (oh-) and since the concentration of (oh-) was most than (h+), this shows that it was in fact alkaline, but few others did it your way too so i assume your method might be correct
    Yeah same, the OH- concentration was higher than the H+ concentration so it seemed logical to say that the the solution was Alkaline.
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    (Original post by Fmathslad)
    the answer was 4.0002 exactly so i don't know how you got 3.5? lucky assumption i guess
    Looooool idk 😭😭
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    (Original post by cr7alwayz)
    Can anyone explain how x=4 for the last question if you remember your working out.
    1)Calculate moles of sodium thiosulfate ion thing?
    2) Ratio of that was 2:1 respect to I2, so times the moles by 2
    3) In question it said 2 moles of Cu3+ reacts with 1 mol of I2, so times the moles of I2 by 2 (ie sodium thiosulfate moles and Cu3+ moles were the same)
    4) Since the moles for Cu3+ were in 25cm3 FROM 250cm3, times it by 10 to get moles of Cu3+ in 250cm3
    4) work out molar mass of whole thing, they gave u grams of 2.226? it was like 225g
    5) Work out molar mass of of hydrated compounds, Cu(HCOO)2 (dot) xH20 so 153.5 (dot) x18
    6) equate that to molar mass of whole thing and solve for x

    Not sure about mole numbers and grams but that was my method
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    (Original post by cr7alwayz)
    Damn, I think after finding the moles of Cu2+ I worked out the molar mass of the wholecompund by dividing the mass they gave by moles of Cu2+. Then, I took away molar mass of Cu(HCOO)2 to from that to get the molar mass of water. Then divide by 18 to get 17 for X.
    So basically, I used the method on page 220 in the textbook.
    Thats correct method but x=4

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    (Original post by HFancy1997)
    Yes for thr 6 Marker Half cells q

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    Could you not react cr3+ with any thing below it when it came to showing it was reducing agent?

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    For the question about acidic or alkaline at 60'C with a pH of 7, did no one use 10^-7 = [H+]? Because that statement has to true by definition doesn't it? So I then divided the Kw at 60'C, (which was 9.4 blah blah x 10^-14) by 10^-7, giving 9.4 blah blah x 10^-7 for [OH-] then I said that [H+] > [OH-] so the solution is acidic. Sorry to go against the flow, but my method makes complete sense to me and I think it's right, if anyone can show me why that's wrong, please go ahead (or if you think I'm talking sense please back me up lol)
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    (Original post by ImNervous)
    Could you not react cr3+ with any thing below it when it came to showing it was reducing agent?

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    lol yeah, i wrote like 6 equations for that one :/
    I also reacted the Cr2O7 with the Cr3+ woops
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    (Original post by _NMcC_)
    Yeah same, the OH- concentration was higher than the H+ concentration so it seemed logical to say that the the solution was Alkaline.
    Isn't it neutral as pH is a measure of h+ conc which stays the same?
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    (Original post by maisym00)
    For the equilibrium question I put: I- reacts with Ag to form agI (yellow ppt), therefore the position of equilibrium will shift to the left to restore I- conc. As a result, more aqueous iodine was formed so the position of equilibrium of 1 will also shift to the left, therefore more grey/black solid iodine is formed. For the electrochemistry question I needed up with two equations. The oxidising agent one where Cr+3 reacts with AL and the reducing agent one where Cr+3 reacts with feO4- or something. Anyone else?
    I got the same except i left out the observations, did it explicitly ask you to mention what you observe in the question?
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    (Original post by k.russell)
    For the question about acidic or alkaline at 60'C with a pH of 7, did no one use 10^-7 = [H+]? Because that statement has to true by definition doesn't it? So I then divided the Kw at 60'C, (which was 9.4 blah blah x 10^-14) by 10^-7, giving 9.4 blah blah x 10^-7 for [OH-] then I said that [H+] > [OH-] so the solution is acidic. Sorry to go against the flow, but my method makes complete sense to me and I think it's right, if anyone can show me why that's wrong, please go ahead (or if you think I'm talking sense please back me up lol)
    10^-7<9.4*10^-7
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    (Original post by ImNervous)
    Could you not react cr3+ with any thing below it when it came to showing it was reducing agent?

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    No because the Cr2+ would be oxidised and you were strictly looking at cr3+

    There were two reactions

    1 and 2 Where cr3+ is reduced to 2+
    6 and 7(Bottom two) Where Cr3+ is oxidised to Cr6+

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    For the 6 marker on redox,
    Would it be 2 marks for definitions
    2 marks for equations and 2 marks for explaining?

    For some reason i did overall equations with each redox system and mentioned 2 where Cr is being oxidised and reduced- didnt mention anything positive/negative eletrode potential of the cells.
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    (Original post by MrZebraCookie)
    10^-7<9.4*10^-7
    yep, just realised that myself lol thanks anyway
    ffs oh well probs only -1
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    (Original post by HFancy1997)
    It was at 60 degrees C not 25, you dont use 25 degrees C Kw which is 10^-14, you use the one it gave in the question as it said 60 degrees.Think the answer was 10.76

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    Think I missed the 60 degrees for this part! Where did it say this happened at 60 degrees? Also, how many marks was this question?
 
 
 
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