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    (Original post by azz92)
    June 2010 f325
    Q7 b)
    The student diluted 25cm3 of a solution of hydrogen peroxide with water and made the solution up to 250cm3. The student Titrated 25cm3 of this solution with 0.02 moldm-3 KMno4 under acidic conditions. The volume of KMno4(aq) required to reach the end point was 23.45cm3.

    • calculate the concentration in gdm3 , of the undiluted hydrogen peroxide solution
    • What volume of oxygen gas, measured at RTP, would be during this Tirtration



    THE mark scheme is rubbish HELP!!

    oh i think u may need this 2MnO4- + 6H+ + 5H2O2 -> 2Mn2+ + 8H2O + 5O2 this is the equation for 7a)

    I got 0.399 gdm and 280 cm of o2
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    (Original post by CoventryCity)
    Yes you need to know the hydrogen fuel cell one they are

    1/2 O2 + H20 +2e--------> 2OH- +0.4V
    2H2O +2e- ----------> H2 + 2OH- -0.83V

    These are the half equations
    Do we need to know the half equations? Or just the overall equation??
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    raj16 you never finished the qeustion what about the volume of oxygen gas can someone tell me why they got 0.0281 dm3
    June 2010 qeustion 7b)
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    (Original post by Josh0806)
    Could somebody please explain to me how to balance this equation...

    FeO42(–) + H(+) ? Fe(3+)+ O2+ H2O

    ...to give this...

    4FeO42(–) + 20H(+) ? 4Fe(3+)+ 3O2+ 10H2O
    Fe has oxidation state 6+ in FeO4 (2-) and oxidation state 3+ in Fe3+, so has changed by -3

    O has oxidation state 2- in FeO4(2-) and oxidation state 0 in O2, so the change is +3, however, 2 oxygen atoms have been used to form O2, so the overall change is +4

    to balance the lowest common factor of -3 and +4 is 12, so you multiplu FeO4(2-) by 4 to get:

    4FeO4(2-) + n[H+] --> 4Fe(3+) + H2O + 02

    now look at the overall molecular charges, as they have to balance on both sides:

    we have (4x-2) + n(1) --> (4x3+) + 0 + 0

    which is -8 + n1+ --> 12+
    we need to get 12+ on both sides, so we need to add 20H+ ions to the left to get:

    4FeO4(2-) + 20[H+] --> 4Fe(3+) + H2O + 02

    simple balancing from here, 20H+ on right, so you need 10H2O:

    4FeO4(2-) + 20[H+] --> 4Fe(3+) + 10H2O + 02

    16 oxygens on left, 10 on the right, so we need 6 more, so times O2 by 3

    4FeO4(2-) + 20[H+] --> 4Fe(3+) + H2O + 302

    there's your answer
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    (Original post by percy93)
    I got 0.399 gdm and 280 cm of o2
    anyone agree
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    Has anyone realised that Born Haber cycles have not come in Jan 2011, so they'll defintely come up tommorow?
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    (Original post by azz92)
    raj16 you never finished the qeustion what about the volume of oxygen gas can someone tell me why they got 0.0281 dm3
    June 2010 qeustion 7b)
    Use the formula n = v/24

    Rearrange it, v = n x 24

    As you got the n = (1.1725 x 10^–3) x 24 = 0.0281 dm3
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    Okay, I have read and done the past paper and somethings you people come up with in this forum JUST SCARE ME COMPLETELY , makes me feel like I DONT KNOW ANYTHING
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    (Original post by azz92)
    raj16 you never finished the qeustion what about the volume of oxygen gas can someone tell me why they got 0.0281 dm3
    June 2010 qeustion 7b)
    sorry! i forgot.

    .
    You knw mols of H202 1.1725x10-3.
    ratio 5:5 so no of mols of 02 is the same.
    using mols=vol/24
    1.1725x10-3 X 24=0.0281
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    (Original post by SmartFool)
    ANYONE PLEASE! How do you work out where the equivalence line is on titration curves as Ive seen on a past paper where the line is at 33cm3 and not 25cm3
    It isn't always 25
    It is in the middle of the straight section of the titration curve!
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    Do we have to remember any equations, such as the redox titrations concerning MnO4 and Fe?
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    Someone tell me what we need to know about "H for the future"
    and "Storage and fuel cell" ! My teacher left it for us to read
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    (Original post by ViralRiver)
    Do we have to remember any equations, such as the redox titrations concerning MnO4 and Fe?
    As far as I remember u can deduce them from the data given! But let me know if there is any :rolleyes:
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    (Original post by arvin_infinity)
    Someone tell me what we need to know about "H for the future"
    and "Storage and fuel cell" ! My teacher left it for us to read
    umm pretty much the advantages and disadvantages of hydrogen fuel cells and storage of hydrogen...i think anyway, well that's all i've came across so far
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    Okay, so in the textbook it gives that table showing what happens when you add NaOH to different solutions.

    For example, it says when you add NaOH to Co(H20)6 you get a pink solution and a blue precipitate.

    Does the pink solution form from the NaOH as well or is that already there and the NaOH just makes a precipitate?
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    I really need to do well on this tomorrow, theres no hope in bio. PLZ BE A DECENT PAPER PLZ :sigh:.
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    (Original post by percy93)
    anyone agree
    Surely you need to divide your 0.399 by 25/1000 because you've figured out the grams in 25cm3 but you need in dm3 so I get 15.9g dm3. which seems big. Hmm maybe I'm wrong, someone correct me.
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    (Original post by Vidja)
    Okay, so in the textbook it gives that table showing what happens when you add NaOH to different solutions.

    For example, it says when you add NaOH to Co(H20)6 you get a pink solution and a blue precipitate.

    Does the pink solution form from the NaOH as well or is that already there and the NaOH just makes a precipitate?
    nooooo, NaOH just acts as the source of OH- ions, as its when complexes react with OH- ions that precipitation occurs

    The pink solution is the Co2+ ions as they are pink in aqueous solutions, so the fact that there is still some pink left suggests that not all of the Co2+ ions have reacted with the OH-
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    (Original post by arvin_infinity)
    Someone tell me what we need to know about "H for the future"
    and "Storage and fuel cell" ! My teacher left it for us to read
    You need to know the hydrogen-oxygen fuel cell which is
    1/O2 + H2O + 2e- --------> 2OH- +0.4V
    2H2O + 2e- --------> H2 + 2OH- -0.83V

    These are the half equations for it

    Fuel cells are good as more efficient and less pollution
    Storage of Hydrogen at ridiculous conditions in liquid form. Or stored on adsorbers or absorbers.
    -ves are difficult to transport, adsorbers and absorbers have limited life
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    (Original post by Rocks:))
    Surely you need to divide your 0.399 by 25/1000 because you've figured out the grams in 25cm3 but you need in dm3 so I get 15.9g dm3. which seems big. Hmm maybe I'm wrong, someone correct me.
    15.9 is correct
 
 
 
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