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

    0
    ReputationRep:
    (Original post by hey_its_nay)
    does anyone have the jan 11 paper and markscheme?
    Attached Images
  1. File Type: pdf AQA-CHEM5-W-QP-JAN11.pdf (377.9 KB, 55 views)
  2. File Type: pdf AQA-CHEM5-W-MS-JAN11.pdf (202.6 KB, 51 views)
    Offline

    0
    ReputationRep:
    (Original post by BethBeth)
    1) Enthalpy of atomisation of Na
    2) Enthalpy of atomisation of Cl
    would it also be the same as:
    1)enthalpy of sublimation of Na
    2)bond dissociation of Cl
    Offline

    0
    ReputationRep:
    (Original post by strawberry_cake)
    Were there no chem5 papers for jan06? I can't find any
    Here you go..
    Attached Images
  3. File Type: pdf AQA-CHM5-W-MS-JUN06.pdf (192.4 KB, 616 views)
  4. File Type: pdf AQA-CHM5-W-QP-JUN06.pdf (139.2 KB, 282 views)
    Offline

    0
    ReputationRep:
    (Original post by sam_xo)
    Here you go..
    Oh I just realised you said Jan 06. Lol sorry, not got them!
    Offline

    0
    ReputationRep:
    (Original post by jimmy303)
    The question tells you hydrogen peroxide is converted into O2 so that gives you the half equation:

    H2O2 --> O2 +2H+ + 2e-
    (remember that the oxidation state of oxygen in H2O2 is only -1)
    Oxygen's oxidation state changes from -1 to 0 and there are two so two electrons are given off.

    You are supposed to be able to remember the half reaction for the reduction of MnO4- (and Cr2O7(2-) in other questions), which is:

    MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O
    Manganese's oxidation state changes from +7 to +2 so five electrons are added.

    Make the number of electrons on both reactions the same so the oxidation of H2O2 is multiplied by 5 and the reduction of MnO4- is multiplied by 2 (making 10 electrons each).

    5H2O2 --> 5O2 +10H+ + 10e-
    2MnO4- + 16H+ + 10e- --> 2Mn2+ + 8H2O

    Cancel out things that appear on both sides of the equation (in this case H+ and e-) and merge.

    5H202 + 2MnO4- + 6H+ ---> 5O2 + 2Mn2+ + 8H2O
    Thanks for that
    Offline

    1
    ReputationRep:
    (Original post by BethBeth)
    Both CGP and Nelson Thornes says pale pink, just consulted them Best to stick with that methinks :cool:

    No idea why someone negged you for this? :confused:
    Ah ok, if another book says it as well then it should be alright. Thanks

    lol meh, rep is very low down my priority list atm
    Offline

    0
    ReputationRep:
    (Original post by strawberry_cake)
    Were there no chem5 papers for jan06? I can't find any
    Oh I just realised you said Jan 06. Lol sorry, not got them!
    Offline

    1
    ReputationRep:
    Can someone please write out the equation of Co2+ with HCl?
    Offline

    0
    ReputationRep:
    What colour is [Cr(NH3)6)3+ I am pretty sure the NT book has it wrong?
    Offline

    0
    ReputationRep:
    (Original post by Wish I Could Change This)
    Can someone please write out the equation of Co2+ with HCl?
    [Co(H20)6)2+ +4Cl- = [CoCl4]2- + 6H20
    Offline

    0
    ReputationRep:
    (Original post by Wish I Could Change This)
    Can someone please write out the equation of Co2+ with HCl?
    Co(H2O)6 + 4Cl- ----> Co(Cl4)2- + 6H2O
    Offline

    0
    ReputationRep:
    (Original post by NRican)
    What colour is [Cr(NH3)6)3+ I am pretty sure the NT book has it wrong?
    I thought it was purple. What does the NT book say?
    Offline

    0
    ReputationRep:
    Don't know if anyone's asked this already (toooo many pages) but in the specification it says - "be able to explain why values from mean bond enthalpy calculations differ from those determined from enthalpy cycles".

    But don't enthalpy cycles use mean bond enthalpies? Or do they mean Born-Haber cycles which use enthalpies of electron affinity and things like that rather than bonds broken vs bonds made??
    Offline

    0
    ReputationRep:
    (Original post by starburst92)
    anyone?
    yeah we do:

    Reduction: O2(g) + 2H2O(l)+ 4e- > 4OH-(aq)

    Oxidation: 2H2(g) + 4OH-(aq) > 4H2O(l) + 4e-

    Overall: 2H2(g) + O2(g) > 2H2O(l)
    Offline

    0
    ReputationRep:
    (Original post by teddyWS)
    would it also be the same as:
    1)enthalpy of sublimation of Na
    2)bond dissociation of Cl
    I don't like enthalpy of sublimation - never heard of it referred to as that, stick with enthalpy of atomisation and you will definitely get the marks..

    I suppose it is bond dissociation in the case of 1/2Cl2(g) ---> Cl(g)
    But if you got this in a Born-Haber Cycle of MgCl2 for instance you would have to times enthalpy of atomisation by two.

    Enthalpy of atomisation = Enthalpy change attending the formation of one mole of gaseous atoms from the element in it's standard state.
    Best be careful and stick to one definition.
    Offline

    0
    ReputationRep:
    (Original post by InItToWinItGetIt?)
    Ah ok, if another book says it as well then it should be alright. Thanks

    lol meh, rep is very low down my priority list atm
    True, true! I don't get it sometimes though Might lose sleep on it after the exam
    Offline

    0
    ReputationRep:
    What synoptic stuff do we need to know from the other units?
    Offline

    0
    ReputationRep:
    (Original post by NRican)
    What colour is [Cr(NH3)6)3+ I am pretty sure the NT book has it wrong?
    Its purple i think
    Offline

    0
    ReputationRep:
    hey could anyone help me with redox equations im just not sure when you include only one reagent and product for example in the specimen paper,

    3 (a)An aqueous solution of sulfur dioxide was reacted in separate experiments as follows.

    Reaction 1 with HgO
    H2O + SO2 + HgO ? H2SO4 + Hg

    Reaction 2 with chlorine
    2H2O + SO2 + Cl2 ? H2SO4 + 2HCl

    (ii) Show, by writing a half-equation, that this oxidising agent in reaction 1 is an
    electron acceptor.
    ................................ ................................ ................................ ................................ ..
    (iii) Write a half-equation for the oxidation process occurring in reaction 2.
    ................................ ................................ ................................ ................................ ..
    (iv) Write a half-equation for the reduction process occurring in reaction 2.
    ................................ ................................ ................................ ................................ ..

    the answer for ii is Hg 2+ + 2e- -------> Hg

    the answer for iii is 2H2O + SO2 ? H2SO4 + 2e-

    why dosnt the oxygen on the Hg come into the half equation?

    Thanks
    pleaseeeeeeeeeee help meeeeeeeeeeeeeee
    Offline

    0
    ReputationRep:
    Has anyone bothered going over unit two stuff like halides and extraction of metals ect?
 
 
 
  • 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.