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

    2
    ReputationRep:
    thankyouuuuuu
    Offline

    0
    ReputationRep:
    Could someone please answer a few questions I have about the specimen paper:
    1. 2c(ii) says that 'Four moles gaseous reactant forming' but isn't it three?
    2. 3b(i) - I don't understand how this question has been done
    3. 3b9ii0 - Cell e.m.f is 0.06 V but I got 0.60V (0.77-0.17) - I don't understand how it's 0.06V

    http://filestore.aqa.org.uk/subjects...5-W-SQP-07.PDF
    http://filestore.aqa.org.uk/subjects...5-W-SMS-07.PDF
    Offline

    1
    ReputationRep:
    do we need to know the ideal gas equation and when doing calculations, do you use the whole numbers to calculate, but write out your answers to 3 sig figures at each step?
    Offline

    0
    ReputationRep:
    Could someone please explain why the answer is I.E. = (+)1451 (kJ mol-1) to 1(b) on the June 2007 paper. I don't understand why and how it is this answer.

    Thanks

    http://www.a-levelchemistry.co.uk/AQ...02007%20ms.pdf
    http://www.a-levelchemistry.co.uk/AQ...une%202007.pdf
    Offline

    1
    ReputationRep:
    (Original post by starfish232)
    Could someone please answer a few questions I have about the specimen paper:
    1. 2c(ii) says that 'Four moles gaseous reactant forming' but isn't it three?
    2. 3b(i) - I don't understand how this question has been done
    3. 3b9ii0 - Cell e.m.f is 0.06 V but I got 0.60V (0.77-0.17) - I don't understand how it's 0.06V

    http://filestore.aqa.org.uk/subjects...5-W-SQP-07.PDF
    http://filestore.aqa.org.uk/subjects...5-W-SMS-07.PDF
    my guess is specimen papers are usually ridiculed with errors

    (Original post by starfish232)
    Could someone please explain why the answer is I.E. = (+)1451 (kJ mol-1) to 1(b) on the June 2007 paper. I don't understand why and how it is this answer.

    Thanks

    http://www.a-levelchemistry.co.uk/AQ...02007%20ms.pdf
    http://www.a-levelchemistry.co.uk/AQ...une%202007.pdf
    Think of it as an equation, to normally find enthalpy formation its equal to everything else added together.

    -642=150+242+736+x-728-2493 (x is the unkown)
    just rearrange
    -x=150+242+736+642-728-2493
    -x=-1451
    x=1451
    Offline

    0
    ReputationRep:
    if the surface area of the positive electrode increased what would be the effect on the emf?

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    in the january 2010 paper question 8a to do with electronegativity, in the mark scheme it states that the attraction of bond pair to nucleus increases? what bond pair is it on about?

    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    Leaving past papers till last few days lol

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by samfreak)
    if the surface area of the positive electrode increased what would be the effect on the emf?

    Posted from TSR Mobile
    Would there be no effect ... it's an equilibrium reaction, and increasing the surface area of a catalyst would have no effect on the emf, it would just mean that the reaction would reach equilibrium quicker...

    That's my opinion anyway, and my chemistry teacher said pretty much the same thing.
    Offline

    0
    ReputationRep:
    (Original post by idunnotbh)
    hey guys random question here but can anyone confirm this for me please...

    we do NOT have to learn the half equations and overall equations of re chargeable batteries and non re chargeable batteries, for example for lead acid batteries or zinc/carbon like in the text book?

    thank you
    i don't think so. I am just in case, but just need to be able to know how to derive them. I think...
    Offline

    0
    ReputationRep:
    (Original post by igobyursula)
    Would there be no effect ... it's an equilibrium reaction, and increasing the surface area of a catalyst would have no effect on the emf, it would just mean that the reaction would reach equilibrium quicker...

    That's my opinion anyway, and my chemistry teacher said pretty much the same thing.
    thanks :-)

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    In the book the spectrometry method is changing the vol of the ions, but everywhere else spectrometry is changing the frequency of visible light through a sample of a coloured complex (to see absorption). Is there a right one/preferred answer?
    Offline

    0
    ReputationRep:
    (Original post by samfreak)
    in the january 2010 paper question 8a to do with electronegativity, in the mark scheme it states that the attraction of bond pair to nucleus increases? what bond pair is it on about?

    Posted from TSR Mobile
    The pair of electrons shared in the bond between an atom of a period 3 element and any random atom. There isn't a specific bond in the question because you're supposed to realise that electronegativity only applies to bonds (at least at A-level it does)
    Offline

    0
    ReputationRep:
    (Original post by Crazy Crouton)
    Has anyone done the exam style paper in the back of the nelson thrones textbook? It seems quite weird and synoptic...
    Yeah, my teacher says the questions in the Nelson Thornes book aren't that great as some are taken from the old syllabus apparently.
    Offline

    0
    ReputationRep:
    (Original post by charch95)
    Yeah, my teacher says the questions in the Nelson Thornes book aren't that great as some are taken from the old syllabus apparently.
    The style of the old syllabus questions are a bit different I think. I'm doing them anyway just in case AQA decide to be a bit experimental this year. I suppose they are helpful in solidfying certain concepts...

    Though in some of the Unit 5 questions, there are some Unit 4 stuff,which I am ignoring
    Offline

    0
    ReputationRep:
    Chem 5 is on the day of my birthday, so I hope it won't be too bad....
    Offline

    0
    ReputationRep:
    Slightly panciking with the calcs on moles in this unit..... For the jan 2010 paper question 8c... how do we know it forms that??? and how do we know the equation???

    IT would be very helpful if someone can tell me a step by step method of approaching these calcs questions
    Offline

    0
    ReputationRep:
    (Original post by lifeisgood2012)
    Slightly panciking with the calcs on moles in this unit..... For the jan 2010 paper question 8c... how do we know it forms that??? and how do we know the equation???

    IT would be very helpful if someone can tell me a step by step method of approaching these calcs questions
    The issue with the calculations on Unit 5 is that there are different types of calculations that they may ask you. For example, they could ask you about redox titrations, or a back titration.

    In this example, and I have it in front of me now, there are two main equations.

    i) a periodicity equation. The reaction between P4O10 and 6H20 to form 4H3PO4

    ii)acid-base neutralisation between NaOH and Phosphoric Acid.

    You work out the moles of Sodium Hydroxide using the conc. and the volume given. You use the ratios in your Acid-base neutralisation equation between the sodium hydroxide and the phosphoric acid to work out the moles of phosphoric acid. From that you can work out the initial amount of P4O10 using the equation between P4O10 and 6H20.

    There isn't really any universal step by step method on these kind of questions as they may be asking for different things. I just keep practising them.
    Offline

    0
    ReputationRep:
    (Original post by JSN)
    + can some explain why the reactions with en will be different for 8ei and ii

    http://filestore.aqa.org.uk/subjects...5-QP-JAN12.PDF
    Hiya...

    Sorry if someone else has already answered this question for you.

    With 2+ hexa-aqua ions, basically (well my chemistry teacher told me because there is no distinction in the book(s))

    Its 3en which forms [M(en)3]2+ + 6H2O

    With 3+ hexa aqua ions it does something different.

    en is H2NCH2CH2NH2

    This is the following equation

    2[M(H2O)6 ]3+ + 3H2NCH2CH2NH2 ----> 2[M(H2O)3(OH)3] + 3H3NCH2CH2NH3

    This is because 3+ ions are slightly more acidic and can donate H+ more easily. Think of the reactions between the hexa aqua ions and carbonate ions.

    Hope this helps. I was completely muddled by it all initially. There is not actually any distinction in the book (I use Nelson Thornes) at all, but I think you are supposed to apply knowledge of the differences between the reactions betweeen carbonate with 2+ hexa aqua ions and 3+ hexa aqua ions. en is a bidentate ligand, as is the carbonate ion.
    Offline

    0
    ReputationRep:
    (Original post by idunnotbh)
    thank you! what kind of understanding? lol like basically do you we just need to know that when they're being re charged the equations go the other way?

    and to answer your question above, yeah sadly it is
    Yeah exactly! just the definitions of primary and secondary cells, what happens when discharging and recharging etc.... This is the first time i'm hearing it's synoptic... i know biology is synoptic but chemistry....ah mannnnnnn!!
 
 
 
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.