Hey there! Sign in to join this conversationNew here? Join for free

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

    Offline

    3
    ReputationRep:
    (Original post by mechanism)
    Can someone please explain how to construct the redox equation June 2014 Q8? Strugglingggggg
    Have you managed this now?
    Offline

    5
    ReputationRep:
    (Original post by ranz)
    but theyre both formed by condesation reaction isnt it?


    Posted from TSR Mobile
    Yes, H2O is released
    Offline

    5
    ReputationRep:
    (Original post by TeachChemistry)
    Have you managed this now?
    Yes, thank you. Was not on my game yesterday haha.
    Offline

    16
    ReputationRep:
    (Original post by rory58824)
    If you are unsure about the group, try writing all the information about every other peak, and then see what is remaining. It's a process of elimination in some way. See if that works out for you
    Thank you!

    Posted from TSR Mobile
    Offline

    10
    ReputationRep:
    (Original post by Hunnybeebee)
    I don't have a data sheet with me rn but say at 7ppm there's three different types of potential environments, which one do you select? Especially if the question hasn't hinted anything like it's an ester and you only have the spec to work from?

    Posted from TSR Mobile
    I struggle with it myself, but you would usually get some kind of an incling, e.g. a molecular formula (which may indicate no nitrogens present or a benzene ring at 7ppm)
    Offline

    2
    ReputationRep:
    does anyone have the june 2015 papers? need them for revision
    Offline

    16
    ReputationRep:
    (Original post by itsConnor_)
    I struggle with it myself, but you would usually get some kind of an incling, e.g. a molecular formula (which may indicate no nitrogens present or a benzene ring at 7ppm)
    ah okay thank you!
    Offline

    16
    ReputationRep:
    not sure if it has been discussed yet but with the f321 and f322 retakes, do you think that the grade boundaries will be a lot higher? Due to everyone doing this exam has done it before and most of them in A2 therefore have a better understanding of the AS?
    Offline

    2
    ReputationRep:
    Name:  image.jpeg
Views: 230
Size:  136.0 KB
    Guys please help with question (ii)
    The mark scheme doesn't make sense been trying to understand it but doesn't make sense
    Offline

    5
    ReputationRep:
    (Original post by Miminfl)
    Name:  image.jpeg
Views: 230
Size:  136.0 KB
    Guys please help with question (ii)
    The mark scheme doesn't make sense been trying to understand it but doesn't make sense
    I haven't seen the mark scheme for this answer but I'll try to tell you how'd I go about doing it:

    With solution A, there is 5x10^-3 moles of acid reacting with 2.5x10^-3 moles of NaOH therefore after the reaction has occured some acid will still remain as it is in excess. The acid will be in equilibrium with its conjugate base (ethanoate ions).

    But with solution B, there is 5x10^-3 moles of acid reacting with the same amount of NaOH, therefore there would be full neutralisation due to the 1:1 quantities. There would be no acid remaining as it all would have reacted, therefore no equilibrium will exist.
    Offline

    16
    ReputationRep:
    (Original post by Miminfl)
    Name:  image.jpeg
Views: 230
Size:  136.0 KB
    Guys please help with question (ii)
    The mark scheme doesn't make sense been trying to understand it but doesn't make sense
    In solution A, there is excess acid as there's 5x10^-4 of ethanoic acid, but there's 2.5x10^-4 of NaOH. As there is excess, there will be an equilibrium of CH3COOH, and it's conjugate base of CH3COO- (which was formed by CH3COONa).

    In solution B, these is equal amount of mol of each solution, therefore they will completely react:

    CH3COOH + NaOh --> CH3COONa + H2O. No equilibrium, therefore cannot be a buffer solution as it isn't a weak acid and its conjugate base.
    Offline

    2
    ReputationRep:
    (Original post by ReeceM1)
    I haven't seen the mark scheme for this answer but I'll try to tell you how'd I go about doing it:

    With solution A, there is 5x10^-3 moles of acid reacting with 2.5x10^-3 moles of NaOH therefore after the reaction has occured some acid will still remain as it is in excess. The acid will be in equilibrium with its conjugate base (ethanoate ions).

    But with solution B, there is 5x10^-3 moles of acid reacting with the same amount of NaOH, therefore there would be full neutralisation due to the 1:1 quantities. There would be no acid remaining as it all would have reacted, therefore no equilibrium will exist.
    Thank you, that really helped , I didn't know I had to find out the number of moles.
    Offline

    2
    ReputationRep:
    [QUOTE=Hunnybeebee;64748169]In solution A, there is excess acid as there's 5x10^-4 of ethanoic acid, but there's 2.5x10^-4 of NaOH. As there is excess, there will be an equilibrium of CH3COOH, and it's conjugate base of CH3COO- (which was formed by CH3COONa).

    In solution B, these is equal amount of mol of each solution, therefore they will completely react:

    CH3COOH + NaOh --> CH3COONa + H2O. No equilibrium, therefore cannot be a buffer solution as it isn't a weak acid and its conjugate base.[/QUOTaE]
    Thank you it makes sense now
    Offline

    16
    ReputationRep:
    A chemist checked the concentration of aqueous calcium hydroxide, Ca(OH)2, inthe sewage water by titration with 5.00 × 10–3 mol dm–3 hydrochloric acid. Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l) The chemist titrated 25.0 cm3 of the sewage water with 21.35 cm3 of HCl toreach the endpoint of the titration. Calculate the concentration, in mol dm–3, of the calcium hydroxide in the sewagewater.

    MS

    moles of HCl =10005 10 21.35 3× ×− = 1.067 × 10–4 mol (1)
    moles of Ca(OH)2 =2.1 067 104× = 5.34 × 10–5 mol (1)
    concentration of Ca(OH)2 = 40 × 5.34 × 10–5 = 2.136 × 10–3 mol dm–3 (1)

    Where did 40 in the 3rd point come from??
    Offline

    5
    ReputationRep:
    (Original post by Hunnybeebee)
    A chemist checked the concentration of aqueous calcium hydroxide, Ca(OH)2, inthe sewage water by titration with 5.00 × 10–3 mol dm–3 hydrochloric acid. Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l) The chemist titrated 25.0 cm3 of the sewage water with 21.35 cm3 of HCl toreach the endpoint of the titration. Calculate the concentration, in mol dm–3, of the calcium hydroxide in the sewagewater.

    MS

    moles of HCl =10005 10 21.35 3× ×− = 1.067 × 10–4 mol (1)
    moles of Ca(OH)2 =2.1 067 104× = 5.34 × 10–5 mol (1)
    concentration of Ca(OH)2 = 40 × 5.34 × 10–5 = 2.136 × 10–3 mol dm–3 (1)

    Where did 40 in the 3rd point come from??
    Does it mention somewhere further up in the stem of the question anything about diluting something to 1dm3 / or using 1dm3 of sewage water?

    If it does then it would be x40 because only 25cm3 is used in the titration, so this would be scaled up 40x to get the values for 1dm3.

    If it doesn't mention that then I'm not sure.
    Offline

    6
    ReputationRep:
    (Original post by Hunnybeebee)
    A chemist checked the concentration of aqueous calcium hydroxide, Ca(OH)2, inthe sewage water by titration with 5.00 × 10–3 mol dm–3 hydrochloric acid. Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l) The chemist titrated 25.0 cm3 of the sewage water with 21.35 cm3 of HCl toreach the endpoint of the titration. Calculate the concentration, in mol dm–3, of the calcium hydroxide in the sewagewater.

    MS

    moles of HCl =10005 10 21.35 3× ×− = 1.067 × 10–4 mol (1)
    moles of Ca(OH)2 =2.1 067 104× = 5.34 × 10–5 mol (1)
    concentration of Ca(OH)2 = 40 × 5.34 × 10–5 = 2.136 × 10–3 mol dm–3 (1)

    Where did 40 in the 3rd point come from??
    Ah i've seen somehting similar before in a MS, "Where did this number come from?". Funny enough it was also the number 40. Here, on this video, skip to 1 : 38 : 45 https://www.youtube.com/watch?v=VaJB...LxLvLnv9Jq3TO1

    or here's another thread that answered your q:
    http://www.thestudentroom.co.uk/show....php?t=1665699
    Offline

    16
    ReputationRep:
    (Original post by bakedbeans247)
    Ah i've seen somehting similar before in a MS, "Where did this number come from?". Funny enough it was also the number 40. Here, on this video, skip to 1 : 38 : 45 https://www.youtube.com/watch?v=VaJB...LxLvLnv9Jq3TO1

    or here's another thread that answered your q:
    http://www.thestudentroom.co.uk/show....php?t=1665699
    I somehow still got the same answer doing 25/1000 instead of multiplying by 50? I'll have a look in a bit but thank you for all the links! Nice when you get a informative response :P

    Anyone know where else to get acid and basis worksheets? Just did them all on physicsandmathstutor.com and got 79% (will obvs keep going until 100% :P) but anyone?
    Offline

    2
    ReputationRep:
    Give this a go, hundreds of questions here:

    http://www.a-levelchemistry.co.uk/OC...0A%20home.html
    Offline

    3
    ReputationRep:
    Anyone able to explain the ozone lysis of the second one? Thank you!

    Name:  image.png
Views: 253
Size:  123.9 KB
    Offline

    3
    ReputationRep:
    (Original post by AqsaMx)
    Anyone able to explain the ozone lysis of the second one? Thank you!

    Name:  image.png
Views: 253
Size:  123.9 KB
    Sorry but I haven't even heard of ozone lysis, which board is this for??

    Posted from TSR Mobile
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brexit voters: Do you stand by your vote?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.