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    (Original post by goldlock)
    Yeah I agree with you, they put the haemoglobing stuff and platin stuff in january which was the kind of biology material - got a good feeling about this one
    yeah I agree


    I can also see an enthalpy question showing up and a question where you have to find the order of reactants
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    (Original post by arvin_infinity)
    Someone will be a life-saver and list all the use of transition metal we need to know! :rolleyes:

    Chromium---> stainless steel
    ...
    We need to know the uses aswell? oh my! i've asked for someone to post up the colours plenty of times and no one cares i have no idea what ones we need to know....
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    (Original post by 786girl)
    WHAT IS THE MOST DIFFICULT TOPIC? in your opinion. state one!!

    1. Redox equations...
    2. Redox Equations
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    I never understand Electrode potentials and half-cells.
    Entropy and Gibbs energy questions are difficult if you're not good at math.

    When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is
    –15 kJ mol–1.
    The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of –24 kJ mol–1.
    Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF

    can someone explain the answer to the question thanks?
    1. Potassium is a smaller ion than Rb, therefore has a greater charge density than it.
    2. It therefore has a greater attraction to the F ion.
    3. This means KF has a greater Lattice enthalpy than RbF
    4. Higher value for Lattice enthalpy means it is harder to break, therefore more energy needed to break bonds for enthalpy change of solution.

    I can't remember whether or not that is a 4 or 5 marker. Those should get you at least 4, maybe 5 marks.
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    (Original post by Pandit Bandit)
    how would you go about doing this type of question guys?
    I'll share a great recipe for Cooking up a balanced ionic redox equation

    FeO4^2- decomposes in the presence of H+ ions forming Fe (III) ions, oxygen and water. Construct the ionic equation for this reaction
    1. First, write the products and reactants down.

    FeO4^2- + H+ --> Fe3+ + O2 + H2O


    2. Find out what things change oxidation states, and exchange of electrons.

    Fe: +6 to +3, need 3e-
    O: 2- to 0 in O2 gives up 2e-, and unchanged for H2O


    3. Work out the ratios needed of oxidants and reductants.

    Need 2 Fe to balance with 3 O.


    4. Balance the equation concerning the ions involved in 3.


    2 Fe to balance with 3 O
    Therefore need at least 2 x FeO4^2-

    Oxygen is already in FeO4^2- , so we don't need to add any to the equation.

    Since there are 2Fe's, we'll need to use 3 O's to reduce it, so 3 O's will be oxidised to O2, therefore there will be 1.5 mol of O2 in equation (or double the whole lot if you like whole numbers).


    4FeO4^2- + H+ --> Fe3+ + 3O2 + H2O

    This is the part that requires thinking, and once you've done this, the rest is easy going


    5. Balance the rest, making sure the number of atoms on each side is equal.

    4FeO4^2- + H+ --> 4Fe3+ + 3O2 + H2O
    (4FeO4^2-, so will get 4Fe3+)

    4FeO4^2- + H+ --> 4Fe3+ + 3O2 + 10H2O
    (6O's are used to make 3O2 on RHS, so out of 16 O's on LHS, 10 are left for the H2O, so there'll be 10H2O)

    4FeO4^2- + 20H+ --> 4Fe3+ + 3O2 + 10H2O
    (10H2O on RHS, so we need 20H+ on RHS to enable that to happen)



    4FeO4^2- + 20H+ --> 4Fe3+ + 3O2 + 10H2O

    (6. Optional: draw a giraffe playing croquet with a pterodactyl when bored.)

    Et voila, one balanced redox equation, serves 2-4 (marks).
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    (Original post by goldlock)
    Yeah I agree with you, they put the haemoglobing stuff and platin stuff in january which was the kind of biology material - got a good feeling about this one
    Brief answer to it pls
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    (Original post by cws121)
    When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is
    –15 kJ mol–1.
    The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of –24 kJ mol–1.
    Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF

    can someone explain the answer to the question thanks?
    K+ is smaller than Rb+, so has a greater charge density than Rb+ which means the lattice enthalpy value for KF is more negative than RbF as the K+ ions can exert more attraction on the F- ions. a more negative lattice enthalpy value means a more positive negative lattice enthalpy change to get from KF(s) back to K+(g) and F-(g) so the lattice enthalpy affects the enthalpy change of solution more than the enthalpy change of hydration, even though the enthalpy change of hydration of K+ should be more negative than Rb+ as its smaller and attracts more to H2O molecules.
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    I think the hardest part is remembering the difference between rate constant, Kc, Ka, Kw and K stab. the titration equations are pretty demanding too - and buffer solutions. to many equations to remember all at once!
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    (Original post by threerose92)
    I always thought the enthalpy change of solution is endothermic and therefore always positive :confused:
    Enthalpy change of solution can be positive or negative, watch out!

    In the text book it only shows +ve, which is misleading.
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    Do we need to know any experiments for this paper?
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    (Original post by goldlock)
    What they've done is, they said "balance any atom that changes the oxidation number" - now, iodine changes oxidation number from minus 1 to 0, so they have to balance the iodine atoms (S changes oxidation number too but it's already balanced)
    ok cool, thank you so much.
    and so because you balance the HI because of the I, the H doesnt change oxidation number does it? so you dont need to balance that?
    ?

    thanks for your help! thank you soo so so so much!
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    (Original post by KnuckleheadNinja)
    ok cool, thank you so much.
    and so because you balance the HI because of the I, the H doesnt change oxidation number does it? so you dont need to balance that?
    ?

    thanks for your help! thank you soo so so so much!
    Nope as you can see, H stays as +1 throughout so you'd have to balance it later on, after you've accounted for all the oxidation number changes in step 2 of that example
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    (Original post by Twinkles)
    Do we need to know any experiments for this paper?
    In a previous paper a six mark question was on planning an experiment to calculate the enthalpy change of neutralisation. So some familiarity with experiments would help, it seems
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    (Original post by Twinkles)
    Do we need to know any experiments for this paper?
    I'd learn the whole mess regarding thiosulfate titrations and the colour changes there, I have a hunch it might come up
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    (Original post by SmartFool)
    for the love of god can anyone tell me the colours of compounds/transition metals we need to know please!
    On the addition of NaOH

    Ion Solution Precipitate

    Cu2+ pale blue pale blue
    Co2+ pink blue (turns beige in air)
    Fe2+ pale green green (turns brown in air)
    Fe3+ pale yellow rusty-brown

    Ligand Substitution


    Complex Ion Ligand added Observations

    [Cu(H2O)6]2+ NH3 Pale blue precipitate, excess, deep blue solution


    HCl(Cl) Pale blue -> green -> yellow solution

    [Co(H2O)6]2+ HCl (Cl- ligand) Pink -> blue solution
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    (Original post by goldlock)
    Nope as you can see, H stays as +1 throughout so you'd have to balance it later on, after you've accounted for all the oxidation number changes in step 2 of that example
    oh wow, you are my personal hero. Redox freaks me out and night before the exam? Not gunna lie, was panicking a little... thank you so much!
    what would i do without SR eh.
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    (Original post by sprazcrumbler)
    In a previous paper a six mark question was on planning an experiment to calculate the enthalpy change of neutralisation. So some familiarity with experiments would help, it seems
    That's why I asked.. I wasn't able to answer that Q because I just did not bother with any practicals I did in class :/
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    (Original post by SmartFool)
    for the love of god can anyone tell me the colours of compounds/transition metals we need to know please!
    This is good revision for me Hope this helps.


    Coloured compounds


    - Cu2+ are blue
    -Cr3+ are green
    -Cr2O7"2- are orange
    -MnO4- are purple
    -Co2+ are blue

    Ligand substitution

    - [Cu(H2O)6] + 4NH3 --><--- [Cu(NH3)4(H2O)2 + 4H2O
    pale blue solution deep blue solution
    -[Cu(H2O)6] + 4Cl- --><--- [CuCl4]2- + 6H2O
    pale blue solution yellow solution (not green which is a common mistake)
    -[Co(H2O)6] + 4Cl- ---><---- [CoCl4]2- + 6H2O
    pale pink solution deep blue solution

    Precipitation

    -Fe2+ + 2OH- ---> Fe(OH)2
    pale green solution to a pale green ppt (can be oxidised and turns rusty brown)
    -Fe3+ + 3OH- ---> Fe(OH)3
    yellow solution to a brown rusty ppt
    -Cu2+ + 2OH- ---> Cu(OH)2
    pale blue solution to pale blue ppt
    -Co2+ + 2OH- ----> Co(OH)2
    pink solution to a blue ppt
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    can any of you guys help me with this question....I am in need of desperado hellp, I'd be soooo grateful!!

    (Original post by Rogercbinboy)
    Having an understanding of the MnO4- and the sodium thiosulphate titrations is useful I think, not sure about any other.
    (Original post by Sawasdee)
    Revision all night for me perhaps :bumps: :coma:

    I HOPE the examiners don't give us an essay-based paper. They suppose to test us on our chemistry knowledge. I HATE those longer answer questions (worth a whole grade etc.)

    Anyway, good luck with your exam tomorrow and any upcoming ones
    (Original post by Guinea Pig Lover)
    here you go
    (Original post by darkiee)
    Yeah the specimen paper scary 20marks loool, Thats an hint to get prepared for essay question
    (Original post by darkiee)
    Yeah the specimen paper scary 20marks loool, Thats an hint to get prepared for essay question
    (Original post by Sawasdee)
    The most enjoyable question of the paper :sigh:
    Isn't it weird that what you hate the most always(or often) worth A LOT and can be a life-changing question --"
    (Original post by Sawasdee)
    The most enjoyable question of the paper :sigh:
    Isn't it weird that what you hate the most always(or often) worth A LOT and can be a life-changing question --"
    (Original post by Kalamari Dave)
    x



    CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION ITS MAKING ME GO CRAZY...IT'S FROM THE JANUARY 2011 PAPER, I'VE ALSO PROVIDED THE PAPERS FOR ANYONE WHO NEEDS IT...BUT PLEASE HELP.
    OK IT'S QUESTION 7Ai & 7bii

    FOR 7aI IT SAYS divide THE MOLES OF S2O32- by 4 to get moles of O2 in 25cm-3....WHY do you divide by 4....??????????

    and then why do you need the concentration of O2....surely they want sample of DOC in mg dm-3.....so doesn't that mean divide the mass of O2 in mg by the volume of river water

    OMG IM SO CONFUSED

    help!!!

    http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=30 6192&ResourceId=3842825 ----------PAPER

    http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=30 6192&ResourceId=3842826--------MArK Scheme
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    (Original post by DrDr)
    Enthalpy change of solution can be positive or negative, watch out!

    In the text book it only shows +ve, which is misleading.
    oh!!! thanks
 
 
 
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