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    I have a number of multiple choice questions which i got wrong. Since the markscheme only says the letter of the answer i dont know where i went wrong.

    There are 14 - sorry

    https://drive.google.com/open?id=0Bx...FVVUTM5Tzkwa28
    https://drive.google.com/open?id=0Bx...HFtSEROZlpTWnM
    https://drive.google.com/open?id=0Bx...VB0c2VzQmZ4Rms
    https://drive.google.com/open?id=0Bx...U82eGRWRFBwZjg
    https://drive.google.com/open?id=0Bx...W9uMUNhd2VxeTA
    https://drive.google.com/open?id=0Bx...V9SSVdjdTNLM3M
    https://drive.google.com/open?id=0Bx...TFkYUs5bDQwRW8
    https://drive.google.com/open?id=0Bx...051MzhHaV9WZkU
    https://drive.google.com/open?id=0Bx...TFGczhmLThYYkU
    https://drive.google.com/open?id=0Bx...Vk3QXNUUFZNVDQ
    https://drive.google.com/open?id=0Bx...WNxSDRzOFVDLTQ
    https://drive.google.com/open?id=0Bx...W5VSkZveWNyRW8
    https://drive.google.com/open?id=0Bx...kl4VzU1VWRXcWc
    https://drive.google.com/open?id=0Bx...W1TLU5zNGVDZlE
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    Done

    I dont expect all of them to be answered but i appreciate any help given
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    11. the key word is added, not the total volume.
    14. write out the electronic configuration (1s2 etc), if you don't now see, come back to me/us.
    15. small, highly charged cation + big anion = lots of covalent character
    21. F doesn't contain 2 mol of ester... 0 + 0.5 =/= 2.5
    23. C (0) -> CO2 (+4)
    27. [Mg2+] = [OH-]/2 (since 1mol Mg(OH)2 -> Mg2+ + 2OH-) therefore k = 0.5x10-3 x (1.0x10-3)2 = D
    30. if it weren't a lone pair, it would be 120o, but it is a lp, so the bonds are pushed slightly closer together than 120o i.e. A
    32. you drew a Lewis Dot structure for NH2- showing a lone pair! it actually has 2x lp
    35. what did you put? I think it is B
    41. what do you think X could be? (white ppt with AgNO3 which dissolves in NH3(aq)
    47. only temperature can change Kc. You showed the equation, the reverse reaction is basically A
    48. the combined DfH of H2O and D2O = 2x DfH of HDO i.e. DrH = 0 therefore B
    51. B is disproportionation
    52. 20 of C2H6 needs 70 of O2, producing 40 of CO2 and 60 of H2O, but note 30 of O2 didn't react. D
    55. bpt is determined by IMF strength, the intramolecular bond i.e. covalent bond doesn't break. The only IMF is a London dispersion, which is determined by the # of e- i.e the surface area.

    Did I miss any?

    Where are these from? CIE?
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    (Original post by Pigster)
    11. the key word is added, not the total volume.
    14. write out the electronic configuration (1s2 etc), if you don't now see, come back to me/us.
    15. small, highly charged cation + big anion = lots of covalent character
    21. F doesn't contain 2 mol of ester... 0 + 0.5 =/= 2.5
    23. C (0) -> CO2 (+4)
    27. [Mg2+] = [OH-]/2 (since 1mol Mg(OH)2 -> Mg2+ + 2OH-) therefore k = 0.5x10-3 x (1.0x10-3)2 = D
    30. if it weren't a lone pair, it would be 120o, but it is a lp, so the bonds are pushed slightly closer together than 120o i.e. A
    32. you drew a Lewis Dot structure for NH2- showing a lone pair! it actually has 2x lp
    35. what did you put? I think it is B
    41. what do you think X could be? (white ppt with AgNO3 which dissolves in NH3(aq)
    47. only temperature can change Kc. You showed the equation, the reverse reaction is basically A
    48. the combined DfH of H2O and D2O = 2x DfH of HDO i.e. DrH = 0 therefore B
    51. B is disproportionation
    52. 20 of C2H6 needs 70 of O2, producing 40 of CO2 and 60 of H2O, but note 30 of O2 didn't react. D
    55. bpt is determined by IMF strength, the intramolecular bond i.e. covalent bond doesn't break. The only IMF is a London dispersion, which is determined by the # of e- i.e the surface area.

    Did I miss any?

    Where are these from? CIE?
    Thank you soo much
    I really appreciate it.

    I havent had a chance to look at them today - ill go through them tomorrow and let you know if i dont understand anything.

    Im not too sure - im guessing AQA (my exam board) but they could also be from OCR:dontknow:
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    (Original post by Pigster)
    11. the key word is added, not the total volume.
    14. write out the electronic configuration (1s2 etc), if you don't now see, come back to me/us.
    15. small, highly charged cation + big anion = lots of covalent character
    21. F doesn't contain 2 mol of ester... 0 + 0.5 =/= 2.5
    23. C (0) -> CO2 (+4)
    27. [Mg2+] = [OH-]/2 (since 1mol Mg(OH)2 -> Mg2+ + 2OH-) therefore k = 0.5x10-3 x (1.0x10-3)2 = D
    30. if it weren't a lone pair, it would be 120o, but it is a lp, so the bonds are pushed slightly closer together than 120o i.e. A
    32. you drew a Lewis Dot structure for NH2- showing a lone pair! it actually has 2x lp
    35. what did you put? I think it is B
    41. what do you think X could be? (white ppt with AgNO3 which dissolves in NH3(aq)
    47. only temperature can change Kc. You showed the equation, the reverse reaction is basically A
    48. the combined DfH of H2O and D2O = 2x DfH of HDO i.e. DrH = 0 therefore B
    51. B is disproportionation
    52. 20 of C2H6 needs 70 of O2, producing 40 of CO2 and 60 of H2O, but note 30 of O2 didn't react. D
    55. bpt is determined by IMF strength, the intramolecular bond i.e. covalent bond doesn't break. The only IMF is a London dispersion, which is determined by the # of e- i.e the surface area.

    Did I miss any?

    Where are these from? CIE?
    15. AlBr3 ?
    21. Because there is 0 moles of ester at the start?
    30. Name:  Screenshot 2017-01-17 21.00.36.png
Views: 61
Size:  11.3 KB (from my revision guide :3)
    32. I did draw 2 lps but its faint - is the name of the shape 'bent'?
    35. I put B but ive written C as the right answer - should i double check with someone?

    Ill look through the rest tomorrow. Thanks for helping me
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    (Original post by kiiten)
    15. AlBr3 ?
    Agreed
    21. Because there is 0 moles of ester at the start?
    Agreed
    30. (from my revision guide :3)
    Shows how good (my) VSEPR is
    32. I did draw 2 lps but its faint - is the name of the shape 'bent'?
    Agreed
    35. I put B but ive written C as the right answer - should i double check with someone?
    I still think it is B.
    Ill look through the rest tomorrow. Thanks for helping me
    Agreed
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    (Original post by Pigster)
    Agreed
    21. Because there is 0 moles of ester at the start?
    Agreed
    30. (from my revision guide :3)
    Shows how good (my) VSEPR is
    32. I did draw 2 lps but its faint - is the name of the shape 'bent'?
    Agreed
    35. I put B but ive written C as the right answer - should i double check with someone?
    I still think it is B.
    Ill look through the rest tomorrow. Thanks for helping me
    Agreed
    30. Does that mean the revision guide is wrong? Ill check on google just to be sure.
    35. Yes it is B, i checked with someone
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    (Original post by kiiten)
    30. Does that mean the revision guide is wrong? Ill check on google just to be sure.
    35. Yes it is B, i checked with someone
    Yes, your revision guide appear to be correct.

    BUT as VSEPR is taught:

    The shape is based on a trigonal bipyramid and the equatorial angle should be 120 degrees, attenuated by the lone pair, which, as its acting from 120 degrees should have only a very small effect. This would predict a bond angle of approximately 118 degrees.

    Another piece of theory supporting a minimal effect is the reduced effect of VSEPR in the third period and above.

    I can only explain the answer by offering the possibility orbital hybridisation, i.e. reorganisation into a shape more corresponding to a square pyramid than a distorted trigonal bipyramid.

    I am going to have a browse through Cotton and Wilkinson tonight ...
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    (Original post by kiiten)
    30. Does that mean the revision guide is wrong? Ill check on google just to be sure.
    Two things:
    SF4's bond angles might simply be required knowledge for the spec that the questions are taken from.
    charco is right, there could be more to it than can be predicted by VSEPR... you'd think that with 2lp and 2bp H2S would be like water, i.e. 104.5, but it is 92.
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    (Original post by Pigster)
    11. the key word is added, not the total volume.
    14. write out the electronic configuration (1s2 etc), if you don't now see, come back to me/us.
    15. small, highly charged cation + big anion = lots of covalent character
    21. F doesn't contain 2 mol of ester... 0 + 0.5 =/= 2.5
    23. C (0) -> CO2 (+4)
    27. [Mg2+] = [OH-]/2 (since 1mol Mg(OH)2 -> Mg2+ + 2OH-) therefore k = 0.5x10-3 x (1.0x10-3)2 = D
    30. if it weren't a lone pair, it would be 120o, but it is a lp, so the bonds are pushed slightly closer together than 120o i.e. A
    32. you drew a Lewis Dot structure for NH2- showing a lone pair! it actually has 2x lp
    35. what did you put? I think it is B
    41. what do you think X could be? (white ppt with AgNO3 which dissolves in NH3(aq)
    47. only temperature can change Kc. You showed the equation, the reverse reaction is basically A
    48. the combined DfH of H2O and D2O = 2x DfH of HDO i.e. DrH = 0 therefore B
    51. B is disproportionation
    52. 20 of C2H6 needs 70 of O2, producing 40 of CO2 and 60 of H2O, but note 30 of O2 didn't react. D
    55. bpt is determined by IMF strength, the intramolecular bond i.e. covalent bond doesn't break. The only IMF is a London dispersion, which is determined by the # of e- i.e the surface area.

    Did I miss any?

    Where are these from? CIE?
    32. So my revision guide is right but for that particular example the angle is different?

    41. I put x=Mg Y=MgCl ?
    48. Whats DfH? I dont know what you mean
    51. Why is B redox? The only oxidation state that changes is chlorine
    52. I know what you mean but how do you know only 70cm3 reacts i.e. it doesnt say oxygen is in excess.
    55. What do you mean by # of e-? Are London dispersion forces Van der Waals?
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    32. Your guide is correct for SF4. My guess was based on VSEPR, which would predict that SF4 would be about 118. My earlier comment was that VSEPR isn't too quantitative. Sometimes VSEPR gets it wrong, and I used H2S as an example of where it gets it wrong.
    41. White ppt that dissolves in NH3(aq) must contain Cl-
    48. delta H formation
    51. Redox reactions are those in which oxidation states change! (also where there is oxidation there MUST be reduction)
    52. The balanced equation shows that 1 mol of ethane reacts with 3.5 mol of O2, 10 cm3 of ethane contains as many mol as 10 cm3 of O2, therefore 20 cm3 of ethane reacts with 70 cm3 of O2
    55. VdW is the name for all IMF including LDF. OCR A used to refer to LDF as VdW (incorrectly), i.e. they had hydrogen bonds, permanent dipole-dipole and VdW as the three you were expected to know about. Nowadays, they expect you to know about H-bonds, pd-d and LDF. BUT they actually were/are referring to the same things. They used to refer to VdW as id-id, not they call them LDF or id-d.
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    (Original post by Pigster)
    32. Your guide is correct for SF4. My guess was based on VSEPR, which would predict that SF4 would be about 118. My earlier comment was that VSEPR isn't too quantitative. Sometimes VSEPR gets it wrong, and I used H2S as an example of where it gets it wrong.
    41. White ppt that dissolves in NH3(aq) must contain Cl-
    48. delta H formation
    51. Redox reactions are those in which oxidation states change! (also where there is oxidation there MUST be reduction)
    52. The balanced equation shows that 1 mol of ethane reacts with 3.5 mol of O2, 10 cm3 of ethane contains as many mol as 10 cm3 of O2, therefore 20 cm3 of ethane reacts with 70 cm3 of O2
    55. VdW is the name for all IMF including LDF. OCR A used to refer to LDF as VdW (incorrectly), i.e. they had hydrogen bonds, permanent dipole-dipole and VdW as the three you were expected to know about. Nowadays, they expect you to know about H-bonds, pd-d and LDF. BUT they actually were/are referring to the same things. They used to refer to VdW as id-id, not they call them LDF or id-d.
    41. MgCl? Not sure how i answered this tbh, it confused me
    48. Whats DrH?
    51. So is the Cl being oxidised and reduced? (0 to +5 and 0 to -1 in different products)
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    41. Mg reacts with an element that makes a white ppt with Ag+, which dissolves in ammonia, i.e. it is a chloride, X = Cl2. Then do the molar calc to make MgCl2.
    48. DrH = enthalpy of reaction
    51. Agreed.

    Have you found out where they're from, yet? Pah, here's me helping you out and you won't return the favour!
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    which past paper is this from?
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    (Original post by Pigster)
    41. Mg reacts with an element that makes a white ppt with Ag+, which dissolves in ammonia, i.e. it is a chloride, X = Cl2. Then do the molar calc to make MgCl2.
    48. DrH = enthalpy of reaction
    51. Agreed.

    Have you found out where they're from, yet? Pah, here's me helping you out and you won't return the favour!
    41. Yep i got it now - i was using x as Cl- instead of Cl2 thats why my answer was wrong.
    48. Sorry could you explain again, im not sure what you mean

    You wanted me to find out? Sorry i didnt know. But like I said earlier im not sure, I think its a mixture from AQA past papers
    Thank you for helping me though! I really appreciate it

    (Original post by shohaib712)
    which past paper is this from?
    I think its from more than one
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    (Original post by kiiten)
    48. Sorry could you explain again, im not sure what you mean
    Starting with H2, D2 and O2, you can make H2O and D2O, the energy change is -286 + -294 = -580
    OR you can make 2HDO, the energy change is 2x-290 = 580
    THEREFORE the energy change for H2O + D2O -> 2HDO must be -(-286+ -294) + (2x-290) = 0
    i.e. H2O + D2O <-> 2HDO involves no energy change and hence, according to Le Chetalier, changing T will have no effect on the position of the equilibrium (in the same way as changing pressure has no effect on the position of an equilibrium involving the same number of gas molecules on the LHS and RHS, e.g. H2 + I2 <-> 2HI)

    Since yield is independent of temperature, B is your answer.
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    (Original post by Pigster)
    Starting with H2, D2 and O2, you can make H2O and D2O, the energy change is -286 + -294 = -580
    OR you can make 2HDO, the energy change is 2x-290 = 580
    THEREFORE the energy change for H2O + D2O -> 2HDO must be -(-286+ -294) + (2x-290) = 0
    i.e. H2O + D2O <-> 2HDO involves no energy change and hence, according to Le Chetalier, changing T will have no effect on the position of the equilibrium (in the same way as changing pressure has no effect on the position of an equilibrium involving the same number of gas molecules on the LHS and RHS, e.g. H2 + I2 <-> 2HI)

    Since yield is independent of temperature, B is your answer.
    Hey i have another multiple choice question if youre not too busy



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    (Original post by kiiten)
    Hey i have another multiple choice question if youre not too busy
    Ethanoic is a weak acid, so [H+] =/= [acid]
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    (Original post by Pigster)
    Ethanoic is a weak acid, so [H+] =/= [acid]
    Oh yeah of course.

    Wait but i still get the same answer (unless something has gone wrong ):

    Ka = [H+]^2 / [HA]

    10^-1 x 0.1 = [H+]^2
    [H+] = 0.1
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    (Original post by kiiten)
    Oh yeah of course.

    Wait but i still get the same answer (unless something has gone wrong ):

    Ka = [H+]^2 / [HA]

    10^-1 x 0.1 = [H+]^2
    [H+] = 0.1
    0.1M ethanoic acid does not have a pH of 1

    ka of ethanoic acid is 1.78 x 10-5
 
 
 
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