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Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

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    (Original post by goldlock)
    But the mark scheme doesn't agree with that either - it says that the rate has to divided by four
    (Original post by Salliana23)
    Oh no sorry about that

    Edit - Maybe it's because the products are 2HCHO : 1/2 O2
    So the molar ratio is 4:1
    The initial rate is the initial rate of formation of HCHO not the rate of formation of all of the products (that's what I was getting confused about) so I think that's why the rate is divided by 4.

    I'll delete the other post - looked at the mark scheme for the first two parts of the question but I should have checked the rest!
    Just posted the answer guys ^

    (Original post by blush.ox)
    :gasp: i get that now, thanks
    You have already rated a post by this user recently!
    Hahaha no problemo!
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    in june 2010 (f325) paper, How did you manage to balance this ionic equation?

    BrO3- + ___Br- + ___H+ ---> ____Br2 + ____H2O

    (answers are; 5,6,3,3 in that order) Can someone explain to me how to figure this out thanks
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    (Original post by RyuHADOUKEN)
    Just posted the answer guys ^



    Hahaha no problemo!
    Oh dear I should have checked to see if anyone else had replied thanks for your help!
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    (Original post by Ecstacy.)
    in june 2010 (f325) paper, How did you manage to balance this ionic equation?

    BrO3- + ___Br- + ___H+ ---> ____Br2 + ____H2O

    (answers are; 5,6,3,3 in that order) Can someone explain to me how to figure this out thanks
    You need to use oxidation numbers to balance this question.

    BrO3- Each oxygen has an oxidation number of -2 so three make -6. As the charge is -1, Br must have an oxidation of +5 in this compound

    Bromine in BrO3 is reduced from +5 in BrO3- to 0 in Br2. This means that 5 electrons are needed in the half equation to reduce it
    BrO3 + 5e- + 6H+ -----> 1/2Br2 + 3H2O

    Br- has an oxidation number of -1 so it is oxidised
    Br- -----> 1/2Br2 + e-

    To balance the oxidation numbers, 5Br- are needed (so that there are 5 electrons on each side). This means that there will be 3Br2 altogether

    So the two half equations combined make

    BrO3- + 5Br- + 6H+ -------> 3Br2 + 3H2O

    Hope this helps
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    (Original post by Ecstacy.)
    in june 2010 (f325) paper, How did you manage to balance this ionic equation?

    BrO3- + ___Br- + ___H+ ---> ____Br2 + ____H2O

    (answers are; 5,6,3,3 in that order) Can someone explain to me how to figure this out thanks
    there are 3 Os in BrO3- so there must be 3 H2O, 3 H2O means 6H+
    you need to make the minus charges equal the positive charges
    6+
    BrO3-
    so you need another 5-
    so 5Br-
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    hi guys, in complex metal ions... does anyone know the difference between the tetrahedral and square planar shape because tetrahedral has come up on a few exam q's but only square planar in the ocr textbook? thanks
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    (Original post by ChubbyRain)
    hi guys, in complex metal ions... does anyone know the difference between the tetrahedral and square planar shape because tetrahedral has come up on a few exam q's but only square planar in the ocr textbook? thanks
    [CuCl4]2- is tetrahedral
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    Can somebody please help me with this question?
    f325 june 2010 paper Q7b)

    the markscheme says

    n(KMnO4) = 0.0200?23.451000 = 4.69 ? 10–4 (mol) ?
    n(H2O2) = 5/2 ? 4.69 ? 10–4 = 1.1725 ? 10–3 (mol) ?
    n(H2O2) in 250 cm3 solution
    = 10 ? 1.1725 ? 10–3 = 1.1725 x 10–2 (mol) ?

    concentration in g dm–3 of original H2O2

    = 40 ? 1.1725 ? 10–2 ? 34 = 15.9 (g dm–3) ?

    n(O2) = 5/2 ? 4.69 ? 10–4 = 1.1725 ? 10–3 (mol) ?
    volume O2 = 24.0 ? 1.1725 ? 10–3 = 0.0281 dm3

    but i don't understand where they got the "40" from

    can somebody please please help me
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    f325 june 2010 paper Q7b)

    http://www.ocr.org.uk/download/pp_10...n_gce_f325.pdf
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    Does anybody else feel that some module 3 based questions can be difficult? I have a really bad memory when it comes to all the colour changes during ligand substitution reactions and the colours of precipitates formed when NaOH is added to a transition metal ion! Any advice on what I should do?
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    (Original post by ChubbyRain)
    hi guys, in complex metal ions... does anyone know the difference between the tetrahedral and square planar shape because tetrahedral has come up on a few exam q's but only square planar in the ocr textbook? thanks
    Tetrahedral ones are 4 ligands spread all around the TM ion. Bond angles are 109 degrees.

    Square planar is like octahedral, but with the top and bottom ligands missing. There are 4 ligands, bond angles are 90 degrees.
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    when working out [OH-] of something like Ca(OH2) and im given the Ca(OH2) conc. why do i have to multiply the conc by x2? i assumed
    [Ca(OH2)=[OH-] but i know thats incorrect....
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    (Original post by Pandit Bandit)
    when working out [OH-] of something like Ca(OH2) and im given the Ca(OH2) conc. why do i have to multiply the conc by x2? i assumed
    [Ca(OH2)=[OH-] but i know thats incorrect....
    if you write out the dissociation of Ca(OH)2 it will be
    Ca(OH)2 -> Ca2+ + 2OH-
    so its 1 mole of Ca(OH)2 : 2 moles of OH-
    so you times the concentration by 2
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    ahh i see! thanks!
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    (Original post by maniRB)
    Does anybody else feel that some module 3 based questions can be difficult? I have a really bad memory when it comes to all the colour changes during ligand substitution reactions and the colours of precipitates formed when NaOH is added to a transition metal ion! Any advice on what I should do?
    Hey someone stated apparently we dont need to learn the colour changes??? I've spent ages learning it so I'm not really sure if we have to??? Can someone please clarify?
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    (Original post by hey-hey-hey)
    Hey someone stated apparently we dont need to learn the colour changes??? I've spent ages learning it so I'm not really sure if we have to??? Can someone please clarify?
    Yes we do have to learn them. For the ligand substituion reactions and the precipitation reactions
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    (Original post by hey-hey-hey)
    Hey someone stated apparently we dont need to learn the colour changes??? I've spent ages learning it so I'm not really sure if we have to??? Can someone please clarify?
    We need to know the colour changes that take place during the precipitation reactions with transition metals. We also need to know the colour changes of only a few ligand substitution reactions. I think they are Copper(II) ions with ammonia, Copper(II) ions with conc HCl, and Cobalt(II) ions with conc HCl.

    I think thats it.
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    (Original post by apo1324)
    We need to know the colour changes that take place during the precipitation reactions with transition metals. We also need to know the colour changes of only a few ligand substitution reactions. I think they are Copper(II) ions with ammonia, Copper(II) ions with conc HCl, and Cobalt(II) ions with conc HCl.

    I think thats it.
    Which transition metal ones, though? The spec. isn't hugely clear if memory serves, and the textbook has a whole load...my teacher thinks it'll just be the Fe 2+ and 3+, the manganese/manganate, and a couple of other common ones.
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    (Original post by Vinchenko)
    Which transition metal ones, though? The spec. isn't hugely clear if memory serves, and the textbook has a whole load...my teacher thinks it'll just be the Fe 2+ and 3+, the manganese/manganate, and a couple of other common ones.
    There are 4 precipitation reactions involving transtion metals which we have to know the colour changes for (and they are shown in the book), I think.
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    (Original post by mazam)
    Yes we do have to learn them. For the ligand substituion reactions and the precipitation reactions

    (Original post by apo1324)
    We need to know the colour changes that take place during the precipitation reactions with transition metals. We also need to know the colour changes of only a few ligand substitution reactions. I think they are Copper(II) ions with ammonia, Copper(II) ions with conc HCl, and Cobalt(II) ions with conc HCl.

    I think thats it.
    Thanks!! So all that revision isnt for nothing then...Thanks guys

    +rep
 
 
 
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