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

    16
    ReputationRep:
    (Original post by Sherlockedd)
    Yeah, but chem 5 is not synoptic to chem 4, only to chem 1 and 2 :L
    So can they not ask us pH calculations at all then?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by brittanna)
    So can they not ask us pH calculations at all then?
    nope, nothing from chem 4
    Offline

    0
    ReputationRep:
    You know in those 2 part questions, where the second part asks you to use the result from the first part...it says "if you have been unable to answer part *blah blah*, use the answer 1.2345 (this is not the correct answer)"

    Even if you did get the answer for part 1, would they penalise you if you used the value that they give you?
    Offline

    2
    ReputationRep:
    (Original post by Sherlockedd)
    Well the SHE has a pd of 0 by definition and it shows on the table they give you in the paper that Fe3+ has a positive pd
    What table? :confused:
    Offline

    1
    ReputationRep:
    (Original post by fowlerbean)
    I believe the support medium is just the surface for which the reaction may occur on, the catalyst is spread over it and then adsorption can take place.
    (Original post by Dirtybit)
    2. It forms Al(OH)4
    Thank you
    Offline

    2
    ReputationRep:
    (Original post by Poopy)
    You know in those 2 part questions, where the second part asks you to use the result from the first part...it says "if you have been unable to answer part *blah blah*, use the answer 1.2345 (this is not the correct answer)"

    Even if you did get the answer for part 1, would they penalise you if you used the value that they give you?
    Not at all, my chemistry teacher actually encourages us to use those values in that type of question because it's safe!
    Offline

    0
    ReputationRep:
    (Original post by Chris-69)
    I'm confused? What's with the 2:5 thing? Is it do do with carry over marks? I.E you only lose the accuracy mark? Because I don't see where the ****-up is
    Well maybe cockup is the wrong word. I just know that there is no mention in the specification of having to combine three half equations. It wasn't especially hard but obviously a lot of people got really confused so the mark scheme allowed a wide range of answers and you only lose a mark for getting the ratio itself wrong
    Offline

    0
    ReputationRep:
    Does anybody have a table for reactions of chromium, cobalt and manganate? I've made a table for the reactions of cu to al but I think another one with them in would really help. Thanks
    Offline

    0
    ReputationRep:
    Guys do we need to know all the equations for batteries and non rechargeable batteries?


    Posted from TSR Mobile
    Offline

    19
    ReputationRep:
    (Original post by pjanoo)
    Not at all, my chemistry teacher actually encourages us to use those values in that type of question because it's safe!

    Yes but any value that you use aslong as you have the correct working out you can still get the marks.
    Offline

    19
    ReputationRep:
    (Original post by kurdabora)
    Guys do we need to know all the equations for batteries and non rechargeable batteries?


    Posted from TSR Mobile
    Nah just alkaline and acidic Fuel cells.
    Offline

    11
    ReputationRep:
    (Original post by gingerandice)
    like in unit 2, do you get full marks just for the right answer in calculations or it is step by step marking? lol poor chemistry, been neglected by me over the weekend till bio5 has been castrated.
    Same here!
    Offline

    0
    ReputationRep:
    (Original post by Poopy)
    You know in those 2 part questions, where the second part asks you to use the result from the first part...it says "if you have been unable to answer part *blah blah*, use the answer 1.2345 (this is not the correct answer)"

    Even if you did get the answer for part 1, would they penalise you if you used the value that they give you?
    Interesting question! I'd say probably not, but I don't know why you would do that...
    Offline

    0
    ReputationRep:
    Can somebody please help explain this problem to me:

    In the specimen paper Q3cii) the markscheme says that increasing the surface area of an electrode has no effect on the emf. But in the Jan 10 paper Q3e) the mark scheme says that increasing the surface area of the electrode increases the rate of reaction.

    Can someone explain this to me because they seem to be contradicting eachother?
    Offline

    1
    ReputationRep:
    (Original post by jimmy_95)
    Can somebody please help explain this problem to me:

    In the specimen paper Q3cii) the markscheme says that increasing the surface area of an electrode has no effect on the emf. But in the Jan 10 paper Q3e) the mark scheme says that increasing the surface area of the electrode increases the rate of reaction.

    Can someone explain this to me because they seem to be contradicting eachother?
    I'm not 100% sure but... It has no effect on emf because all the ions are the same? Even if the reaction is faster, same amount of ions react.

    Could someone else confirm?
    Offline

    11
    ReputationRep:
    http://www.a-levelchemistry.co.uk/aq...W-QP-Jun02.pdf
    Could someone help me with Qs 2b and c. For part c, I keep getting +0.84, how is the answer -0.84?
    Offline

    2
    ReputationRep:
    (Original post by Dirtybit)
    Yes but any value that you use aslong as you have the correct working out you can still get the marks.
    Oh okay I suppose it would be quicker for the examiner though as the answer for the 'given value' is also on the mark scheme ready for them to see and mark, rather than follow your working out.
    Offline

    2
    ReputationRep:
    (Original post by popnit)
    http://www.a-levelchemistry.co.uk/aq...W-QP-Jun02.pdf
    Could someone help me with Qs 2b and c. For part c, I keep getting +0.84, how is the answer -0.84?
    for 2b I'm getting +1.93V, what are you getting?

    for 2c, I'm doing this:

    emf = (right hand side) - (left hand side)
    = (-0.44) - (0.40)
    = -0.84V

    You may have been taught to do (positive) - (negative) but doing the (RHS) - (LHS) is better
    Offline

    11
    ReputationRep:
    (Original post by pjanoo)
    for 2b I'm getting +1.93V, what are you getting?

    for 2c, I'm doing this:

    emf = (right hand side) - (left hand side)
    = (-0.44) - (0.40)
    = -0.84V

    You may have been taught to do (positive) - (negative) but doing the (RHS) - (LHS) is better
    Yep I got 1.93 but I don't understand the explanation part.

    For part c, I have been doing the RHS-LHS rule. How do you know which one is positive and which one is negative?

    Thanks btw!
    Offline

    0
    ReputationRep:
    (Original post by lantern)
    I'm not 100% sure but... It has no effect on emf because all the ions are the same? Even if the reaction is faster, same amount of ions react.

    Could someone else confirm?
    Thats the only explanation I could think of. But if the rate of reaction increases, then the number of electrons transferring from the negative to positive electron should increase, which should surely increase the emf.
 
 
 
Poll
If you won £30,000, which of these would you spend it on?

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.