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Advanced Higher Chemistry 2012 Thread

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  • View Poll Results: How did you find the 2012 exam, compared to the last few years' papers?
    A lot easier than previous years
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    Easier than previous years
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    Pretty much the same
    17.24%
    Harder than previous years
    51.72%
    Much harder than previous years
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    • Thread Starter
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    Thought i would make this thread early to discuss the build up and any after exam comments

    The revision is going okay so far - unit 3 is difficult though trying to remember all the different reagents used.
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    Brilliant, I was wondering when someone was going to make this thread! I find U3 fun as its quite logical, U2 however is so borrring!!
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    Haha, i thought it would be a good idea

    I'm the other way round - i quite enjoy units 1 and 2, but just can't seem to get my head around 60% of unit 3 :0

    We should do a joint attempt at the exam :P haha
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    haha!
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    Hi guys, do any of you know how to work out the multiple choice question 16 in the 2010 paper and multiple choice question 19 and 20 in the 2011 paper, really appreciate it
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    (Original post by Babb_zz)
    Hi guys, do any of you know how to work out the multiple choice question 16 in the 2010 paper and multiple choice question 19 and 20 in the 2011 paper, really appreciate it
    2010 Q16

    So they tell you the standard enthalpy of combustion of hydrogen is -286 kJmol^-1

    If you write out the standard enthalpy of combustion of hydrogen you get:

    H2(g) + 1/2 O2 (g) --> H2O (l)

    Now if you look at the standard enthalpy of formation, you get the exact same equation (enthalpy change when 1 mole of a compound is formed from its elements, in their normal states, under standard conditions).

    So the enthalpy change is exactly the same, -286 kJmol^-1, which is answer A.

    2011 Q19

    The enthalpy change that they've asked you to find is from the top line to the bottom line of the thermochemical cycle, so to work it out you need to know all the values in the cycle. Therefore we need to determine the enthalpy change for the line which represents:

    H2(g) + 1/2 O2 (g) --> H2O (l)

    This represents the standard enthalpy of combustion of hydrogen, because it is the combustion of one mole of it in excess oxygen under standard conditions (cos all the chemicals are in their standard states).

    If you check in the data book, you'll see that the standard enthalpy of combustion of hydrogen is -286 kJmol^-1.

    Now you just need to plug all the numbers into your calculator, going round the cycle (arrows pointing up are positive, arrows pointing down are negative).

    So you end up with:

    -286 + 432 + 248 – 916 – 46 = -568 which is answer A.


    Hope that helped

    And to join in with the general chat, I find Unit 1 very boring, quite like Unit 2 and hate Unit 3 (although I am gradually coming to terms with it, which is a good job seeing as the exam is so very soon!).
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    (Original post by Chiringuito)
    2010 Q16

    So they tell you the standard enthalpy of combustion of hydrogen is -286 kJmol^-1

    If you write out the standard enthalpy of combustion of hydrogen you get:

    H2(g) + 1/2 O2 (g) --> H2O (l)

    Now if you look at the standard enthalpy of formation, you get the exact same equation (enthalpy change when 1 mole of a compound is formed from its elements, in their normal states, under standard conditions).

    So the enthalpy change is exactly the same, -286 kJmol^-1, which is answer A.

    2011 Q19

    The enthalpy change that they've asked you to find is from the top line to the bottom line of the thermochemical cycle, so to work it out you need to know all the values in the cycle. Therefore we need to determine the enthalpy change for the line which represents:

    H2(g) + 1/2 O2 (g) --> H2O (l)

    This represents the standard enthalpy of combustion of hydrogen, because it is the combustion of one mole of it in excess oxygen under standard conditions (cos all the chemicals are in their standard states).

    If you check in the data book, you'll see that the standard enthalpy of combustion of hydrogen is -286 kJmol^-1.

    Now you just need to plug all the numbers into your calculator, going round the cycle (arrows pointing up are positive, arrows pointing down are negative).

    So you end up with:

    -286 + 432 + 248 – 916 – 46 = -568 which is answer A.


    Hope that helped

    And to join in with the general chat, I find Unit 1 very boring, quite like Unit 2 and hate Unit 3 (although I am gradually coming to terms with it, which is a good job seeing as the exam is so very soon!).
    THANKYOU SOO MUCH, can you please help me out in unit 2 because I actually have no cclue really appreciate it mate
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    (Original post by Babb_zz)
    THANKYOU SOO MUCH, can you please help me out in unit 2 because I actually have no cclue really appreciate it mate
    Haha sure, if you have any specific questions I'm happy to try to explain, if I can
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    Absolutely dreading this exam on Monday! I did so well in the prelim but i seem to have forgotten everything I know nothing about PPAs!
    Would anybody be able to explain 2008 qu 5b, 6, 8b. Would be greatly appreciated
    Good luck with the studying!
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    Anyone got a marking scheme for 2001?
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    (Original post by NMx11)
    Absolutely dreading this exam on Monday! I did so well in the prelim but i seem to have forgotten everything I know nothing about PPAs!
    Would anybody be able to explain 2008 qu 5b, 6, 8b. Would be greatly appreciated
    Good luck with the studying!
    5b

    The enthalpy of combustion of a substance refers to the enthalpy change when one mole of a substance reacts completely with oxygen.
    The question says that chromium reacts to produce chromium (III) oxide, and gives us the formula for this in the table: Cr2O3.

    Writing out the balanced equation for chromium's combustion:
    2Cr + 1.5O2 -> Cr2O3.

    This is the same as the enthalpy of formation of Cr2O3, the value for which is given in the table as -1140. However, it is not the correct equation for the enthalpy of combustion of chromium, since 2 moles are involved instead of 1. So, to get the correct equation for chromium's enthalpy of combustion, we divide that equation by 2, and therefore also divide its enthalpy change (-1140) by 2, giving us the answer of -570 kJ/mol.

    6

    a) 2 moles thiosulphate react with 1 mole iodine
    n = cv = 0.1 * 0.0205 = 0.00205 moles thiosulphate reacted
    divide by 2 = 0.001025 moles iodine reacted

    b) 1 mole ClO- produces 1 mole iodine so 0.001025 moles ClO- present in the 25cm3 of solution.
    So 0.01025 moles present in the 250cm3 of solution.

    A mistake here might be to calculate the concentration of ClO- in the solution, which would lead to 0.01025/0.25 = 0.041.
    However, the question asks for the concentration of the ClO- in the bleach, of which there is 10cm3 in the solution, so the correct equation is 0.01025/0.01 = 1.025 moles per litre.

    8b
    From question 6 we know that 2 moles thiosulphate react with 1 mole iodine.

    i) 0.025 * 0.0188 = 0.00047 moles thiosulphate reacted
    so divide by 2 to get 0.000235 moles iodine
    0.000235/0.01 = 0.0235 moles per litre

    Same procedure for ii with different numbers!
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    (Original post by Armaros)
    5b

    The enthalpy of combustion of a substance refers to the enthalpy change when one mole of a substance reacts completely with oxygen.
    The question says that chromium reacts to produce chromium (III) oxide, and gives us the formula for this in the table: Cr2O3.

    Writing out the balanced equation for chromium's combustion:
    2Cr + 1.5O2 -> Cr2O3.

    This is the same as the enthalpy of formation of Cr2O3, the value for which is given in the table as -1140. However, it is not the correct equation for the enthalpy of combustion of chromium, since 2 moles are involved instead of 1. So, to get the correct equation for chromium's enthalpy of combustion, we divide that equation by 2, and therefore also divide its enthalpy change (-1140) by 2, giving us the answer of -570 kJ/mol.

    6

    a) 2 moles thiosulphate react with 1 mole iodine
    n = cv = 0.1 * 0.0205 = 0.00205 moles thiosulphate reacted
    divide by 2 = 0.001025 moles iodine reacted

    b) 1 mole ClO- produces 1 mole iodine so 0.001025 moles ClO- present in the 25cm3 of solution.
    So 0.01025 moles present in the 250cm3 of solution.

    A mistake here might be to calculate the concentration of ClO- in the solution, which would lead to 0.01025/0.25 = 0.041.
    However, the question asks for the concentration of the ClO- in the bleach, of which there is 10cm3 in the solution, so the correct equation is 0.01025/0.01 = 1.025 moles per litre.

    8b
    From question 6 we know that 2 moles thiosulphate react with 1 mole iodine.

    i) 0.025 * 0.0188 = 0.00047 moles thiosulphate reacted
    so divide by 2 to get 0.000235 moles iodine
    0.000235/0.01 = 0.0235 moles per litre

    Same procedure for ii with different numbers!
    Thank you sooo much! It seems so simple now!
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    Oh dear - 24 hours to go, and i'm struggling a bit :0

    Been doing lots of multiple choice questions though, and been improving each time
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    Well, this is my overview of the course...

    Unit 1 - Good
    Unit 2 - Good
    Unit 3 - The most horrible topic known to mankind.
    PPAs - Just depends on which ones appear in the paper, but overall, OK.

    I think I am well enough prepared for the exam, but can anyone recomend a site where I can get older exam papers? I feel like I need a bit more practice....
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    (Original post by Steven W)
    Well, this is my overview of the course...

    Unit 1 - Good
    Unit 2 - Good
    Unit 3 - The most horrible topic known to mankind.
    PPAs - Just depends on which ones appear in the paper, but overall, OK.

    I think I am well enough prepared for the exam, but can anyone recomend a site where I can get older exam papers? I feel like I need a bit more practice....
    Behold:
    http://web.me.com/ginneswatsonkelso/...st_Papers.html
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    Thank-You
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    (Original post by NMx11)
    Absolutely dreading this exam on Monday! I did so well in the prelim but i seem to have forgotten everything I know nothing about PPAs!
    Same here, I need to revise about three quarters of the PPAs from scratch, I cannot remember a single thing from the ones we did at the start of the year

    Don't even get me started on all the organic reactions in Unit 3...

    And to boot, the SQA chemistry past paper service seems to be overloaded with traffic today, I keep getting an 'internet explorer is not working' message when I try to download the PDFs

    Edit - I can download them no problem on google chrome
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    (Original post by Sniggey)
    Same here, I need to revise about three quarters of the PPAs from scratch, I cannot remember a single thing from the ones we did at the start of the year

    Don't even get me started on all the organic reactions in Unit 3...

    And to boot, the SQA chemistry past paper service seems to be overloaded with traffic today, I keep getting an 'internet explorer is not working' message when I try to download the PDFs

    Edit - I can download them no problem on google chrome
    I attempted to learn to the PPAs earlier, wasn't very successful!
    Haha, yeah unit 3...
    I'm stressing so much, i need a B to get into uni!
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    (Original post by NMx11)
    I attempted to learn to the PPAs earlier, wasn't very successful!
    Haha, yeah unit 3...
    I'm stressing so much, i need a B to get into uni!
    I'm doing some last-minute past papers, trying to put the PPAs out of my ming for now...

    Are you confident? What did you get in your prelims etc?
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    I somehow got As in both prelims but I seem to have forgotten everything!
    I should be able to get a B as long as the sqa aren't really horrible this year, think i'll cry if they are!
    How are you feeling about tomorrow?

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