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    OK people tell me if this true or false.

    For buffer calculations: the conc. of the HA and A- are actually the reacting concentrations ( i.e. calculate the no. of moles of HA and A- that have fully reacted with each other then divide by the combined volume )
    For Kc calculations we use equilibrium concentrations.
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    (Original post by reneetaylor)
    I look at it like it is more proportional to I than to H, do you see what I mean? As H quadruples, so does the initial rate
    When I doubles in expt 1 and 2 so does the initial rate,
    And finally when I doubles in expt 2 and 3, so does the initial rate.

    Also the initial rate changed in expt 1 and 2 with respect to I, so therefore I is first order, you can therefore expect it to have an effect on each initial rate of reaction.

    I know it may seem a bit sketchy and risky, as because H SEEMS to have an effect on the initial rate, but I'm going with proportionality alone!

    Hopefully someone can give a better answer, I'm not fantastic at explaining things

    i think I understand what you mean

    so for instance experiments 1 and 2 for I- the conc is x2 and the initial is also x2 but for conc of H+ x1 and intitial is x 2 , comparing experiments 2 and 3 for I- conc is x 2 and initial is x2 and for conc of I- is 2 and initial is x 2, comparing 1 and 3 I- x4 and initial x 4, and for conc H+ is x2 and x4, but because in experiment 1 and 2 the H+ the conc was x2 and initial x1, this means H+ will be 0 order?
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    (Original post by DudeBoy)
    When is the exam, I got told it was afternoon, anyone?
    It is next month
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    (Original post by meg0001)
    How to you work out the rate determing step? the worked example in the book is really confusing
    well i think;
    looking at the full equation you can tell if it is the rds compared to the rate equation as orders and moles will match up,
    if you look at the rate equation (the order is the number of moles reacting in the rate determining step) so you can work out how many moles you will have and you just figure out the intermediate?
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    (Original post by JimmyA*)
    Kw= [H+] * 100[H+]
    This is because the concentration of OH- is 100 times more
    kw=100[H+]^2
    (1*10^-14)/100= [H+]^2
    1*10^-16 = [H+]^2
    1*10^-8= [H+]
    -log(1*10^-8)= 8
    why 2 H+ ? there should be OH- and H+ ?
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    (Original post by kevloui)
    nicee, my turn, but more energy related
    Define: Entropy (1)


    What does a negative entropy indicate (1)
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    do we have to show the dissociation equation for the salt of the weak acid or just the weak acid. e.g. CH3COOH ==> H+ + CH3COO- AND CH3COO-Na+ ==> Na+ + CH3COO-
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    (Original post by kuku2013)
    thanks, do we have to show the dissociation equation for the salt of the weak acid or just the weak acid. e.g. CH3COOH ==> H+ + CH3COO- AND CH3COO-Na+ ==> Na+ + CH3COO- e.
    hmm.. id say you do the dissociation of the acid, to be on the safe side
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    (Original post by DavidH20)
    You know the oxidation state of vanadium is +5. The oxidation state of oxygen is always -2 (in the examples we'll come across), so to balance out the oxidation numbers in the compound overall 2 vanadium atoms and five oxygen ones are required.
    was the (5) in bracket representing the oxidation number of vanadium?
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    (Original post by otrivine)
    why 2 H+ ? there should be OH- and H+ ?
    Because it say's the concentration of OH- is 100 times that of H+ so you can replace the OH- with 100*H+
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    (Original post by kevloui)
    hmm.. Id say you do the dissociation of the acid, to be on the safe side
    what about the salt of the acid
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    (Original post by JimmyA*)
    Because it say's the concentration of OH- is 100 times that of H+ so you can replace the OH- with 100*H+
    Okay , thanks

    I am stuck on rates, when you have a reactant you want to get the order but the other 2 have varying concentrations which experiments do u compare?
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    (Original post by otrivine)
    Define: Entropy (1)


    What does a negative entropy indicate (1)
    The level of disorder in a thermodynamic system
    Indicates that the reaction is feasible;

    Why does Sodium chloride have a less exothermic lattice enthalpy than magnesium oxide? (3)
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    (Original post by kuku2013)
    what about the salt of the acid
    Like this CH3COOH ==> H+ + CH3COO-
    But if given the other reactant then; CH3COOH + NaOH ==> H20+ + CH3COO-Na+
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    (Original post by Myocardium)
    Ah ok Would you say you feel more prepared this time round? Are you resitting f324 next week (I am - sat it in jan, didn't go well, haven't even started past papers for it yet got so many unit 5 exams ) What did you get in jan this year for f324? Sorry for all the qs - I find chem to be a bit of a hit and miss for me so I'm a little worried about these exams :/
    No worries, I got a C in that too. So I went down from a B in that too. I feel as though I'm more prepared, focusing solely on Chem helps too. Yep, its kind of the same thing for me - even the things i think went well, end up going so wrong at times. Yep resitting both Chem exams.
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    (Original post by kevloui)
    The level of disorder in a thermodynamic system
    Indicates that the reaction is feasible;

    Why does Sodium chloride have a less exothermic lattice enthalpy than magnesium oxide? (3)
    yes

    no=I was looking for the word , the system is more ordered



    This is because Mg2+ ion has a greater charge density than Na+ ions, Mg2+ has a smaller ionic radius than Na+ ion,

    The oxygen ion has a greater charge density, and has a smaller ionic radius than Cl-,

    and hence, there is a greater electrostatic attraction between the Mg2+ and O2- ions and hence more energy is needed to overcome the attractive forces between the ions
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    (Original post by kevloui)
    Like this CH3COOH ==> H+ + CH3COO-
    But if given the other reactant then; CH3COOH + NaOH ==> H20+ + CH3COO-Na+
    oh ok, thanks
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    (Original post by meg0001)
    How to you work out the rate determing step? the worked example in the book is really confusing
    To work out the rate determining step, of course you must know the rate equation.

    Taking the example in the book, we are conveniently given the rate equation

    rate= K [NO2]2.

    (NO2 + CO -> NO + CO2) Overall equation

    From this alone, we can figure out the number of moles present in the first/ slowest step.

    STEP 1

    Regarding to the rate equation, there should be two molecules of NO2

    NO2+ NO2
    Now, we must form a product and an intermediate - the product present in the overall equation

    Reacting NO2 with NO2 should give you NO and NO3.
    This way you can see that we have an intermediate, and one of the products present in the overall equation, as chemical reactions take place in a series of steps, and some intermediates are often formed (NO3 in this case)

    We can then compare the equation we'e devised so far with the overall equation.
    It makes sense to introduce CO into the next equation, also CO is a reactant in the overall equation.

    STEP 2

    We're not really happy with our intermediate, NO3 so we use it in the next step by reacting it with CO.

    NO3 + CO

    Reacting these guys together gives you CO2 and NO2.

    Comparing this to the overall equation

    NO2 + CO -> NO + CO2

    We have all of our products, and we reacted the necessary molecules (CO and NO2)
    Looking at the steps, you can see we've ticked all of the boxes.

    Hope this helped a little...
    Correct me if I've made any mistakes anyone!
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    Quick question: To work out Kc how do you know when to use moles or the concentration?
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    (Original post by otrivine)
    yes

    no=I was looking for the word , the system is more ordered



    This is because Mg2+ ion has a greater charge density than Na+ ions, Mg2+ has a smaller ionic radius than Na+ ion,

    The oxygen ion has a greater charge density, and has a smaller ionic radius than Cl-,

    and hence, there is a greater electrostatic attraction between the Mg2+ and O2- ions and hence more energy is needed to overcome the attractive forces between the ions
    I was gonna say that too, but I saw that it was 1 mark and didnt bother, sigh.
    And yes perfect!
    my turn!
 
 
 
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