Join TSR now and get all your revision questions answeredSign up now

OCR Salters Chemistry F335 12th June 2013 Exam revision thread Watch

    Offline

    1
    ReputationRep:
    (Original post by 11hokj1)
    can someone PLEASE help me with this question 5(f)(ii)

    http://www.ocr.org.uk/Images/79341-q...-by-design.pdf

    thanks in advance!


    Posted from TSR Mobile
    Admitted I had to reverse engineering from the markscheme to get this (i'm not sure if this is a proper explanation) but here we go.

    We're told that the solution is mixed with 1/3 of the volume needed to neutralise; therefore the ratio of [salt] to [acid] is 1/3:2/3

    Using Ka = [H+] * [salt]/[acid], substituting values into [salt]/[acid] leaves this to equal 1/2.

    Then doing -log(2*Ka) = 2.73 which is your answer (I don't think this is entirely right as I think the answer is something to do with the extent of neutralisation between the mixtures too). Hope this helps
    Offline

    0
    ReputationRep:
    (Original post by martynsteel)
    Admitted I had to reverse engineering from the markscheme to get this (i'm not sure if this is a proper explanation) but here we go.

    We're told that the solution is mixed with 1/3 of the volume needed to neutralise; therefore the ratio of [salt] to [acid] is 1/3:2/3

    Using Ka = [H+] * [salt]/[acid], substituting values into [salt]/[acid] leaves this to equal 1/2.

    Then doing -log(2*Ka) = 2.73 which is your answer (I don't think this is entirely right as I think the answer is something to do with the extent of neutralisation between the mixtures too). Hope this helps
    Do we just ignore the given concentration values for a question like this?
    I can't figure out how to get from the concentration values given --> 1/2
    Offline

    0
    ReputationRep:
    (Original post by 11hokj1)
    can someone PLEASE help me with this question 5(f)(ii)

    http://www.ocr.org.uk/Images/79341-q...-by-design.pdf

    thanks in advance!


    Posted from TSR Mobile
    this questions been confusing me for like half an hour now and it's only one mark which just rubs it in

    I've been just expecting it to be standard Ka / (salt/acid) = [H+]
    then minus log of that but it's never the right answer but whenever I do Ka / HA/salt I get the right answer and I have no idea why waahhhh
    Offline

    1
    ReputationRep:
    (Original post by Kreayshawn)
    Do we just ignore the given concentration values for a question like this?
    I can't figure out how to get from the concentration values given --> 1/2
    Right I think I've eventually got it, I reckons its due to the fact that we're told 1/3 of the solution is diluted, and the ratio of [salt]/[acid] HAS to be 2 (from reverse engineering the answer).

    So my final calculating came to be:

    (0.0045*0.1)/((2/3)*0.05*0.027) which works out to be 1/2 exactly. I think the trick of this question is that the concentration of the acid (HA) has been partially diluted.
    Offline

    11
    ReputationRep:
    Can someone maybe come up with a document that contains recurring questions that come up every now and again?
    Offline

    0
    ReputationRep:
    (Original post by martynsteel)
    Right I think I've eventually got it, I reckons its due to the fact that we're told 1/3 of the solution is diluted, and the ratio of [salt]/[acid] HAS to be 2 (from reverse engineering the answer).

    So my final calculating came to be:

    (0.0045*0.1)/((2/3)*0.05*0.027) which works out to be 1/2 exactly. I think the trick of this question is that the concentration of the acid (HA) has been partially diluted.
    aaaaaaahhh you're the best man thanks very much - I was missing out the 2/3

    can't believe it was only 1 mark!!
    Offline

    1
    ReputationRep:
    (Original post by Tikara)
    aaaaaaahhh you're the best man thanks very much - I was missing out the 2/3

    can't believe it was only 1 mark!!
    For all the stress it's caused me it's so not worth one mark! But thank you!!
    Offline

    0
    ReputationRep:
    (Original post by martynsteel)
    Right I think I've eventually got it, I reckons its due to the fact that we're told 1/3 of the solution is diluted, and the ratio of [salt]/[acid] HAS to be 2 (from reverse engineering the answer).

    So my final calculating came to be:

    (0.0045*0.1)/((2/3)*0.05*0.027) which works out to be 1/2 exactly. I think the trick of this question is that the concentration of the acid (HA) has been partially diluted.
    thank you!
    Offline

    1
    ReputationRep:
    (Original post by Kreayshawn)
    thank you!
    You're welcome!
    Offline

    0
    ReputationRep:
    Can someone please help me with Q2d on June 2010?
    Why id-id bonds but not hydrogen bonds since both can form hydrogen bonds?
    Thanks
    Offline

    1
    ReputationRep:
    (Original post by super121)
    Can someone please help me with Q2d on June 2010?
    Why id-id bonds but not hydrogen bonds since both can form hydrogen bonds?
    Thanks
    They both have the ability to form two hydrogen bonds (two lones pairs on each oxygen) per molecule so talking about hydrogen bond strength seems redundant in this case. All compounds exhibit ID-ID (some experience additional bonding on top of this) and one compound is more branched so it makes more sense to talk about ID-ID forces here.
    Offline

    0
    ReputationRep:
    (Original post by AmirHabeeb)
    They both have the ability to form two hydrogen bonds (two lones pairs on each oxygen) per molecule so talking about hydrogen bond strength seems redundant in this case. All compounds exhibit ID-ID (some experience additional bonding on top of this) and one compound is more branched so it makes more sense to talk about ID-ID forces here.
    But one's more branched than the other, so surely you can say that it forms weaker intermolecular bonds because the chains cannot get as close together?
    Offline

    0
    ReputationRep:
    Do we need to know about partial pressure calculations? (Kp)
    It's in my CGP revision guide but don't think I've seen it show up in any past papers and I can't see it on the specification
    Offline

    0
    ReputationRep:
    (Original post by Kreayshawn)
    Do we need to know about partial pressure calculations? (Kp)
    It's in my CGP revision guide but don't think I've seen it show up in any past papers and I can't see it on the specification
    Page 174 in chemical ideas says 'In this course, we do not deal quantitively with Kp'
    Offline

    2
    ReputationRep:
    Can someone please explain to me how to work out how many hydrogen environments there are in a compound in NMR?

    When explaining, examples would be most appreciated.

    Thank you 😃


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by super121)
    Page 174 in chemical ideas says 'In this course, we do not deal quantitively with Kp'
    thanks!
    hope there isn't too much else in this CGP guide that I might end up learning unnecessarily..

    Is an ion-dipole bond the same thing as an ion-permanent dipole bond?
    Offline

    11
    ReputationRep:
    (Original post by Kreayshawn)
    thanks!
    hope there isn't too much else in this CGP guide that I might end up learning unnecessarily..

    Is an ion-dipole bond the same thing as an ion-permanent dipole bond?
    Yeah it is
    Offline

    0
    ReputationRep:
    (Original post by nukethemaly)
    Yeah it is
    Thanks



    Can anyone demonstrate what this would look like? Jan 11 2D, pasted the mark scheme on the screenshot but I don't fully know what kind of diagram they are after
    Name:  jan112d.png
Views: 607
Size:  47.2 KB
    Offline

    0
    ReputationRep:
    (Original post by Kreayshawn)
    Thanks



    Can anyone demonstrate what this would look like? Jan 11 2D, pasted the mark scheme on the screenshot but I don't fully know what kind of diagram they are after
    Name:  jan112d.png
Views: 607
Size:  47.2 KB
    first diagram on page 96 in the cgp revision guide
    • PS Reviewer
    Offline

    16
    ReputationRep:
    (Original post by calmpeach)
    Hey guys, hope revision is going well for you.
    I've got something that I can't get my head around and it'll be great if could help me

    On p55 in new edition salters revision guide by OCR it says on the top right corner (for visible absorption spectroscopy):
    absorption is more intense and the wavelength of lamda-max increases for organic molecules with large delocalised systems.

    correct me if I'm wrong, but I thought as the delocalisation increases, the excitation energy of electrons would decrease?
    help!
    You are correct, except that you have got confused between wavelength and frequency - a longer wavelength = lower frequency = less energy.
 
 
 
Poll
Which Fantasy Franchise is the best?

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

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