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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD watch

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    (Original post by GetOverHere)
    No, because the whole point of an exam is to be a distinctly-unique entity from the one preceding it. The only influence last year has is in how OCR choose to write their paper, etc.

    Everyone's grade is marked raw, so out of 60 for F324. If we plot everyone's results, we get a bell curve, where some people do horribly, some people do amazingly, and most do decently.

    Imagine we have 100000 students who sat F324. We take the top 10000 scores (top 10%) and we find out what the lowest member of this group got. Say they got 56. That means 56 is the A* boundary. Obviously candidate 10001 may also have 56, so they'll also get an A*. Then we look at the next 10000. Let's say the lowest member gets 50 - an A become an 50. Candidates 20001-20054 may all also have 50, so they'll all get A's. We repeat this until we account for all grade boundaries and all students.

    ^ This is how grade boundaries are decided. No outside influences, just cold hard data. That way, we have a consistent rubric to match marks against (by taking 10% samples) and usually little variation in the percentage of people attaining a certain grade; 8.39% of students last year got an A* compared to 7.67% in 2014. Everybody is happy c:


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    Partly true, but also partly misinformed. Lets say one cohort of students is a lot more intelligent than a previous year, wouldn't that just skew the results and make it unfair? Grade boundaries are decided by many factors and this is one of them...
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    I did A2 chemistry last year, hence did F324 and F325

    How did you guys find this years paper?
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    (Original post by Lucy985)
    Would anyone mind helping me with question 3 (c) (ii) on the June 2012 paper? I seem to be getting the wrong number of moles for butanoic acid and I don't understand how or why the concentrations are calculated
    Ok, so the butanoic acid is mixed with NaOH and an acid-base reaction would obviously occur.
    CH3(CH2)2COOH + NaOH --> CH3(CH2)2COO-Na+ + H2O

    You work out the number of moles of butanoic acid: n= c x v/1000
    n(butanoic acid) = 0.0125mol
    n(NaOH) = 0.0025mol

    NaOH is the limiting reactant, so all of it is used up and forms 0.0025mol of the salt, CH3(CH2)2COO-Na+

    Because butanoic acid is in excess, only some of it is used up: 0.0125 - 0.0025= 0.01mol
    You then work out the new concentrations using c=n/v
    c(butanoic acid)= 0.01/(100/1000) = 0.1 mol dm^-3
    c(CH3(CH2)2COO-)= 0.0025/(100/1000) = 0.025 mol dm^-3

    Then you plug these concentrations into the buffer equation and you get a concentration of H+ of 6.04x10^-5.
    Then -log that answer and you get a pH of 4.22
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    (Original post by abdulmilad)
    I did A2 chemistry last year, hence did F324 and F325

    How did you guys find this years paper?
    F324 easier than last year's. A few questions were quite wordy, and there were a few traps that might have caught people out. No gigantic proton NMR at the end.
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    (Original post by AqsaMx)
    Attachment 554037

    Could some explain why the cell potential increases when water is added?
    Anyone??
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    http://us20.chatzy.com/11324377761338
    testing each other for f325 tomorrow
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    when water is added it causes the concentration of whatever ions to decrease, this causes the equilibrium to shift to the side where the ions are, in the direction of ion that have dcreased in conc. This results in a more negative / more negative electrode potential value, greater difference between electrode potential values, larger overall se potential value
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    (Original post by AqsaMx)
    Anyone??
    The addition of water means that Cu2+ concentration is decreased. The equilibrium will then shift to the left to restore the balance, leading to more Cu being oxidised to Cu2+. This releases more electrons, resulting in a bigger difference in electrode potential.
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    Anyone got a link to the 2015 F325 and MS ? Thanks
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    (Original post by Walkerz7)
    Hey guys could you help me with this question?
    "You are providednwi I three alcohols that are structural isomers
    CH3CH2CH2CH2OH
    CH3CH2CHOHCH3
    (CH3)3OH
    You have access to normal lab apparatus and an infrared spectrometer
    You should provide structural formula for any reaction and a description of how you will identify the three alcohols from any observations" it's a 6 marker plz help
    Firstly, identify the alcohols

    The fist alcohol is a primary alcohol

    The second alcohol is a secondary alcohol

    The third alcohol is a tertiary alcohol

    Testing the compounds ----->

    So you know that a primary alcohol can be oxidised to a carboxylic acid and a secondary alcohol can only be oxidised to a ketone.

    We use (H2SO4/K2Cr2O7) as reagents to carry out the oxidation of alcohol under reflux

    The third alcohol will have no change as it is a tertiary alcohol so it can't be oxidised.

    So we will see solution change from orange to green for the first 2 alcohols

    From the first alcohol we get a carboxylic acid, and for the second alcohol we get a ketone.

    Using a infrared spectrometer to analyse the compounds

    For the product(carboxylic acid) of the first alcohol you will find:

    Very broad absorption at 2500-3300 cm^-1 for O-H bond
    Absorption at 1640-1750 cm^-1 for C=O bond

    For the product(ketone) of the second alcohol you will only find:

    Absorption at 1640-1750 cm^-1 for the C=O bond
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    (Original post by M.Branson98)
    Ok, so the butanoic acid is mixed with NaOH and an acid-base reaction would obviously occur.
    CH3(CH2)2COOH + NaOH --> CH3(CH2)2COO-Na+ + H2O

    You work out the number of moles of butanoic acid: n= c x v/1000
    n(butanoic acid) = 0.0125mol
    n(NaOH) = 0.0025mol

    NaOH is the limiting reactant, so all of it is used up and forms 0.0025mol of the salt, CH3(CH2)2COO-Na+

    Because butanoic acid is in excess, only some of it is used up: 0.0125 - 0.0025= 0.01mol
    You then work out the new concentrations using c=n/v
    c(butanoic acid)= 0.01/(100/1000) = 0.1 mol dm^-3
    c(CH3(CH2)2COO-)= 0.0025/(100/1000) = 0.025 mol dm^-3

    Then you plug these concentrations into the buffer equation and you get a concentration of H+ of 6.04x10^-5.
    Then -log that answer and you get a pH of 4.22
    Thanks very much!
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    (Original post by Walkerz7)
    Hey guys could you help me with this question?
    "You are providednwi I three alcohols that are structural isomers
    CH3CH2CH2CH2OH
    CH3CH2CHOHCH3
    (CH3)3OH
    You have access to normal lab apparatus and an infrared spectrometer
    You should provide structural formula for any reaction and a description of how you will identify the three alcohols from any observations" it's a 6 marker plz help
    Is this F325? Looks like an F324 question.

    1. Primary alcohol
    2. Secondary alcohol
    3. I think you mean (CH3)3COH. Tertiary alcohol.

    1. Distil it to give carbonyl (aldehyde) (orange to green = oxidised). Positive test with Tollens therefore primary
    2. Distil/reflux it to give carbonyl (ketone) (orange to green = oxidised). Negative test with Tollens therefore secondary
    3. Distil/reflux it. No oxidation (no orange to green) therefore tertiary
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    I've seen the predictions have been posted on twitter and I'm confused what exactly is meant by "Effect of equilibrium shift on electrode potential". Please can someone explain, possibly with an example? Thanks!
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    You seem like an A/A* candidate. Are there any areas or types of questions that you struggle with in particular? Any tips you might have. Thanks

    Aiming for A/A*
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    (Original post by sophay_)
    I've seen the predictions have been posted on twitter and I'm confused what exactly is meant by "Effect of equilibrium shift on electrode potential". Please can someone explain, possibly with an example? Thanks!
    Got to so with electrode potentials when the standard conditions has been changed
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    (Original post by Rust Cohle)
    You seem like an A/A* candidate. Are there any areas or types of questions that you struggle with in particular? Any tips you might have. Thanks

    Aiming for A/A*
    No areas in particular. The application of knowledge questions are the hardest. Tips would be to ensure that you know all of the little details in the specification so that you can pick up as many of the 'easy' marks as possible.
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    (Original post by KB_97)
    Well f***. after the aqa core 3 disaster my only hope for a* is chemistry. It's never gonna happen.
    Gonna join you on that boat after the ocr core 3 disaster... Literally our entire uni offer depends on tomorrows paper wow
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    is the cathode the positive electrode?
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    2SO2 + O2 > 2SO3

    Explain what would happen to the pressure as the system was allowed to reach equilibrium

    I wrote the pressure would decrease as wquilibrium would shift to the left, but the MS says pressure decreases and fewer moles? Could anyone explain what Is meant by this?
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    Why is it difficult to use Born-Haber cycles to find the lattice enthalpy of sodium carbonate? I know it's something to do with the formation of the carbonate ions right?
 
 
 
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