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

Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread Watch

    Offline

    6
    ReputationRep:
    Hey guys please can someone explain why D is the correct answer here? Trying to work it through but my brain won't allow me to visualise the graph! :s


    Offline

    12
    ReputationRep:
    Oh. Okay. Thanks..
    Offline

    6
    ReputationRep:
    (Original post by Supermanxxxxxx)
    Anyone know if we're meant to know all This


    Posted from TSR Mobile
    Well it's on a past edexcel paper...
    Offline

    3
    ReputationRep:
    (Original post by Funky_Giraffe)
    Well it's on a past edexcel paper...
    Well I meant to say I can't see it anywhere in the spec so wondering how it ties in to anything


    Posted from TSR Mobile
    Offline

    12
    ReputationRep:
    (Original post by Funky_Giraffe)
    Hey guys please can someone explain why D is the correct answer here? Trying to work it through but my brain won't allow me to visualise the graph! :s


    Rate against Concentration graph for a second order reaction is curved. It means that if we double concentration, rate increases by 4 times so in the question, it's mentioned "the square of the concentration" meaning if we are already taking the square of concentration then now it should be directly proportional to the rate hence it's a straight line graph and even the word "second" is written in bold so answer is D.
    Offline

    9
    ReputationRep:
    I've just been doing some multiple choice Qs and am slightly unsure about these 2 in particular

    With the first one, I'm just not sure how to do! I worked out the nitrogen going from +5 to -3 and aluminium being reduced by -3 but wasn't sure how to choose between C and D and in the end got the 50:50 choice wrong

    On the second question, I guess it's just a matter of principle that I need to learn but I thought X ray diffraction only gave you the bond lengths? Can someone tell me why I'm wrong here?!

    EDIT: not sure why there's a third attachment... please ignore!!

    Thanks in advance guys!
    Attached Images
       
    Offline

    12
    ReputationRep:
    (Original post by gabby07)
    I've just been doing some multiple choice Qs and am slightly unsure about these 2 in particular

    With the first one, I'm just not sure how to do! I worked out the nitrogen going from +5 to -3 and aluminium being reduced by -3 but wasn't sure how to choose between C and D and in the end got the 50:50 choice wrong

    On the second question, I guess it's just a matter of principle that I need to learn but I thought X ray diffraction only gave you the bond lengths? Can someone tell me why I'm wrong here?!

    EDIT: not sure why there's a third attachment... please ignore!!

    Thanks in advance guys!
    Hi!
    For the first one, Aluminium is being oxidised actually, not reduced. In such questions, construct the oxidation and reduction equations and balance the electrons here. As Aluminium is being oxidised from 0 to +3, this means that it has lost 3 electrons to gain that charge and for nitrogen, we know it has been reduced to -3 from +5..so this means that nitrogen has gained 8 electrons! To cancel out the electrons, you have to multiply the oxidation half by 8 and reduction half by 3 to get 24 electrons. Then your work becomes easy
    For the second one, I'm not exactly sure myself why it's B but I believe there is some link between bond lengths and the angles(and I really dk what that link is tbh )
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by gabby07)
    I've just been doing some multiple choice Qs and am slightly unsure about these 2 in particular

    With the first one, I'm just not sure how to do! I worked out the nitrogen going from +5 to -3 and aluminium being reduced by -3 but wasn't sure how to choose between C and D and in the end got the 50:50 choice wrong

    On the second question, I guess it's just a matter of principle that I need to learn but I thought X ray diffraction only gave you the bond lengths? Can someone tell me why I'm wrong here?!

    EDIT: not sure why there's a third attachment... please ignore!!

    Thanks in advance guys!
    X ray diffraction essentially gives you a 3d map of the electron density, which you can use to work out both the bond lengths as well as the angles
    Offline

    12
    ReputationRep:
    (Original post by samb1234)
    X ray diffraction essentially gives you a 3d map of the electron density, which you can use to work out both the bond lengths as well as the angles
    If you don't mind me asking but how does that actually help us work out the angles? I mean bond lengths I got but angles?
    Offline

    3
    ReputationRep:
    (Original post by sabahshahed294)
    If you don't mind me asking but how does that actually help us work out the angles? I mean bond lengths I got but angles?
    Surely it's just process of elimination as you know the other 3 don't show bond angles


    Posted from TSR Mobile
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by sabahshahed294)
    If you don't mind me asking but how does that actually help us work out the angles? I mean bond lengths I got but angles?
    Because if you have a 3d map of the molecule you can easily work out the angle between the bonds as you know where they are in relation to each other
    Offline

    12
    ReputationRep:
    (Original post by samb1234)
    Because if you have a 3d map of the molecule you can easily work out the angle between the bonds as you know where they are in relation to each other
    Ah okay! Thanks a lot! Sorry for the stupid question though lol...U5 is not my strong point tbh
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by sabahshahed294)
    Ah okay! Thanks a lot! Sorry for the stupid question though lol...U5 is not my strong point tbh
    It's fine don't worry about it
    Offline

    2
    ReputationRep:
    From my understanding of the below attached electrodes if they underwent redox; Mg would be undergo oxidation and the Cu reduction.. But I am uncertain as how to name the electrodes. Is the Mg electrode surrounded by cations a negatively charged electrode (thus the negative electrode) and therefore the cathode?

    I guess what I'm trying to say is that is it a positive electrode if itself is charged positively or it a positive electrode if its surrounded by positive ions?

    This is probably a simple question but Q has been driving me insane the more I think about?
    Thanks in advance
    Attached Images
     
    Offline

    12
    ReputationRep:
    (Original post by ayvaak)
    From my understanding of the below attached electrodes if they underwent redox; Mg would be undergo oxidation and the Cu reduction.. But I am uncertain as how to name the electrodes. Is the Mg electrode surrounded by cations a negatively charged electrode (thus the negative electrode) and therefore the cathode?

    I guess what I'm trying to say is that is it a positive electrode if itself is charged positively or it a positive electrode if its surrounded by positive ions?

    This is probably a simple question but Q has been driving me insane the more I think about?
    Thanks in advance
    The place where reduction takes place is the cathode. In the cathode, electrons are deficient so it is the positive electrode whereas the place where oxidation takes place is the anode and in the anode, electrons are in abundance so it is the negative electrode. And from my knowledge, yes it is the electrodes that are charged. Hope it helps!
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by ayvaak)
    From my understanding of the below attached electrodes if they underwent redox; Mg would be undergo oxidation and the Cu reduction.. But I am uncertain as how to name the electrodes. Is the Mg electrode surrounded by cations a negatively charged electrode (thus the negative electrode) and therefore the cathode?

    I guess what I'm trying to say is that is it a positive electrode if itself is charged positively or it a positive electrode if its surrounded by positive ions?

    This is probably a simple question but Q has been driving me insane the more I think about?
    Thanks in advance
    (Original post by sabahshahed294)
    The place where reduction takes place is the cathode. In the cathode, electrons are deficient so it is the positive electrode whereas the place where oxidation takes place is the anode and in the anode, electrons are in abundance so it is the negative electrode. And from my knowledge, yes it is the electrodes that are charged. Hope it helps!
    It is neither. Both electrodes, technically speaking are negative. Voltage is relative, so essentially the 'positive' electrode is actually negative just not as negative as the other electrode so relative to that electrode is positive
    Offline

    12
    ReputationRep:
    (Original post by samb1234)
    It is neither. Both electrodes, technically speaking are negative. Voltage is relative, so essentially the 'positive' electrode is actually negative just not as negative as the other electrode so relative to that electrode is positive
    So, it's by the voltage this judgement is made...and not by the gain or loss in electrons?
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by sabahshahed294)
    So, it's by the voltage this judgement is made...and not by the gain or loss in electrons?
    well no it is still the fact that the 'negative' eleectrode is going to have electrons flowing to the 'positive' electrode, but the charge of both electrodes is technically negative
    Offline

    12
    ReputationRep:
    (Original post by samb1234)
    well no it is still the fact that the 'negative' eleectrode is going to have electrons flowing to the 'positive' electrode, but the charge of both electrodes is technically negative
    Oh I see. My bad. Thank you very much once again.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by sabahshahed294)
    Oh I see. My bad. Thank you very much once again.
    no problem, I'm just procrastinating to avoid finishing my chem notes anyway haha so glad i could help
 
 
 
  • 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
    Would you like to hibernate through the winter months?
  • 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

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