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    (Original post by medicdreamer)
    Does anyone understand Q8 OF JUNE 2011 F325?

    For step 2 , why is it help equations :

    Cu2+ + co3(2-) goes to cuc03.

    2H+ + CO3(2-) goes to H20 + C02.

    what about no3- and n02-?
    The NO3- and NO2 simply dont get involved right? Because the carbonate ion is CO3(2-) and you cant join together two negetively charged ion because of electrostatic repulsion. So because they aren't involved the NO3-, is just a spectator and NO2 is also still there

    Thats what i think anyway
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    Hi, can someone explain the buffer question on the June 2012 paper, question 3c (ii) please? I get a pH of 4.12 when the mark scheme answer states 4.22 as the answer
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    (Original post by master y)
    I am also confused a bout this , can any1 help?
    Hi there, the way I worked this out was as follows:

    Step 1: Simple, combining the 2 equations given as the copper is reacting with the HNO3.. so, u just balance and combine them,
    Cu(s) + 4HNO3(l) --> Cu2+(aq) + 2NO3-(aq) + 2NO2(g) + 2H20(l)

    Step 2: It says that the sodium carbonate, is neutralised by any acid..
    You should know that.. when acid + carbonate --> salt + CO2 + H20
    So, I did.. NaCO3 + 2HN03 (as any acid, this was given so I used this- it makes no difference.. btw, u see I am writing out the equation first just to make it a bit easier to understand.. from the equation, u can then deduce the ionic equation), so NaCO3 + HN03--> Na(NO3)2 + C02 + H20

    hence, ionic equation.. Na+ CO32- +2H+ + 2NO3- --> Na+ + 2NO3- + CO2 + H20

    cancelling, gives.. CO32- + 2H+ --> CO2 + H20

    u follow same format for equation 4, this is just how I worked it out.. try equation 4 and see if u get
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    Does anyone know the difference between the

    End-point

    vs the Equivalence Point?
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    Guys, do we just need to memorise the Alkali and Acid fuel cell reactions?

    They ask for them a few times in papers and is it just a simple recall question or is there a way to understand it so it's not just recall?
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    does anyone know how to work out question 7 on the February 2012 paper? its on Kc equillibrium and its so harrrrd!

    http://www.ocr.org.uk/Images/79734-q...d-elements.pdf - QP

    http://www.ocr.org.uk/Images/61819-m...ts-january.pdf - MS

    also on this paper can someone help me with question 4) iii). where do you get pH = -log(0.0129) from???

    this is the hardest paper iv done
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    (Original post by s24a)
    does anyone know how to work out question 7 on the February 2012 paper? its on Kc equillibrium and its so harrrrd!

    http://www.ocr.org.uk/Images/79734-q...d-elements.pdf - QP

    http://www.ocr.org.uk/Images/61819-m...ts-january.pdf - MS

    also on this paper can someone help me with question 4) iii). where do you get pH = -log(0.0129) from???

    this is the hardest paper iv done
    For your second question it is Ka times the conentration of the acid. You then square root this value becuase [H+] approx = [A-]

    Is this the only paper you've done?
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    (Original post by master y)
    For your second question it is Ka times the conentration of the acid. You then square root this value becuase [H+] approx = [A-]

    Is this the only paper you've done?
    i did do that and thats why i was confused! I have now realised though that i misread the Ka for 4.43x10^-14 rather than 4.43x10^-4

    No this isn't the first paper that i have done. I have done 4 others, all under timed conditions, and this for me was the hardest one
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    (Original post by Better)
    Does anyone know the difference between the

    End-point

    vs the Equivalence Point?
    End point = point at which indicator changes colour (pH)

    Equivalence point = point at which the number of moles of H+(?) = The number of moles of OH- (I think)
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    (Original post by Funtry)
    End point = point at which indicator changes colour (pH)

    Equivalence point = point at which the number of moles of H+(?) = The number of moles of OH- (I think)
    Okay thanks. I just did January 2013, and my ass was kicked. I need to mark it but Jesus christ, that was tough.

    The level of questions has SIGNIFICANTLY gotten harder.

    They through in so much AS-Stuff I had no clue about.

    Has anyone done it? It's going to be very hard to get an A* in Chemistry..... Jesus.

    On Physics and Maths I'm finally after all my hard work scoring above 90's but Chemistry 5, whats going on, Chem is supposed to be my strongest subject.
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    January 2013 might have been the hardest thing I've ever done in my life. I had no clue what to do for Question 8. If they make another dumb paper like that one, I will literally fight someone.
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    (Original post by Better)
    Okay thanks. I just did January 2013, and my ass was kicked. I need to mark it but Jesus christ, that was tough.

    The level of questions has SIGNIFICANTLY gotten harder.

    They through in so much AS-Stuff I had no clue about.

    Has anyone done it? It's going to be very hard to get an A* in Chemistry..... Jesus.

    On Physics and Maths I'm finally after all my hard work scoring above 90's but Chemistry 5, whats going on, Chem is supposed to be my strongest subject.
    Do you have a link? I'd like to do it
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    (Original post by zef1995)
    Spoiler:
    Show


    Thank you very much

    I've had another go at the question, and I've got the answer you posted (Yay!).

    Now we know that n=2, we can construct the equation for the reaction:

    MnO4- +8H+ +5e- -> Mn2+ +4H2O

    As n=2, vanadium is oxidised from V2+ to VO3-, and we need to construct a half equation for this reaction.

    First we can put down what we know: V2+ -> VO3- + 3e- (I've put 3e- because vanadium increases its oxidation state by 3)
    As you can see, VO3- contains oxygen - but where did these come from? As there are no oxygens in V2+, we can add oxygen by adding water on the left side of the equation:

    V2+ + H2O -> VO3- +3e-

    However, although we have oxygen on both sides, we have no hydrogens! We can sort this by adding H+ to the right hand side:

    V2+ + H2O -> VO3- +3e- + H+

    We now have all the components of the half equation, we just need to balance them:

    We know there are 3 oxygens on the right, so we have 3 H2O. As we have 3 H2O, there are 6 hydrogens (so 6H+).

    Final half equation: V2+ +3H2O -> VO3- + 6H+ +3e-

    Now we have both half equations, we can work out an equation for the whole reaction:

    V2+ +3H2O -> VO3- + 6H+ + 3e-
    MnO4- + 5e- + 8H+ -> Mn2+ + 4H2O

    3e- for Vanadium equation, and 5e- for the Manganese one. To balance out the electrons, multiply Vanadium equation by 5 and Manganese's by 3 to give 15e- in both equations.

    5V2+ + 15H2O -> 5VO3- + 30H+ + 15e-
    3MnO4- + 15e- + 24H+ -> 3Mn2+ + 12H2O

    Combined to give: 5V2+ + 15H2O + 3MnO4- +15e- +24H+ -> 3Mn2+ + 12H2O + 15e- + 30H+ + 5VO3-

    As you can see, we have 15e- on each side (these cancel), and water and protons on both sides. The equation can therefor be simplified to:

    5V2+ + 3MnO4- +3H2O -> 3Mn2+ + 6H+ + 5VO3-

    I know that was very long winded, I just wanted to make sure I covered everything....I hope that's useful!

    If you have any questions, please feel free to ask


    This Paper was so hard, thank you for taking the time to do this....think I should do some Maths and then mark this tomorrow can't stress over this for too long....
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    (Original post by Funtry)
    Do you have a link? I'd like to do it
    Found it bro!!

    Here you go

    (Original post by freetown)
    2013 January paper
    (F325)

    QP:
    https://docs.google.com/file/d/0Bwx_...it?usp=sharing
    MS:
    https://docs.google.com/file/d/0Bwx_...it?usp=sharing


    Credit to Wheeeto
    so if you see him around make sure you give him some rep
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    (Original post by Better)
    Found it bro!!

    Here you go
    Thanks! I can't be arsed to do any physics revision tonight, so I'll go try that now
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    (Original post by awilson008)
    I did the Jan 2013 paper as well; it was HARD, especially the last question.

    Anybody know what score out of 100 gets you full UMS?
    Yo bro I found it super hard too.

    Has anyone got any recommendations on what I should do to understand Unstructured Redox Titrations Questions.

    I am finding them fairly challenging?
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    (Original post by Better)
    January 2013 might have been the hardest thing I've ever done in my life. I had no clue what to do for Question 8. If they make another dumb paper like that one, I will literally fight someone.
    I took it in January... I got an E
    I was so gutted because I need a B for uni and I was so confident haha, it really kicked my ass!
    Hopefully this time it will be a lot easier!
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    (Original post by Funtry)
    Thanks! I can't be arsed to do any physics revision tonight, so I'll go try that now
    do you have the stretch and challenge questions from the CD to post on here please?
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    (Original post by Better)
    This Paper was so hard, thank you for taking the time to do this....think I should do some Maths and then mark this tomorrow can't stress over this for too long....
    I agree, it was a challenging paper! You're welcome, I'm glad that I can help!
    I'm guessing you've just done Jan 13 as a mock? You may have done better than you think! At least we have a difficult paper as a mock, it means we can see the type of stretch and challenge questions they can ask and learn how to tackle them
    I hope your revision is going well!
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    Hey Guys I just cracked redox titrations I was feeling the same helplessness and stress as you guys but now I love them. Personally I recommend everyone struggling with them to sit down with your teacher, personal tutor or even anyone you know who can help. What really helped me was going through it with someone step by step. I can't stress this enough. DRAW A DIAGRAMMATIC picture of whats going on EXACTLY. If the stem says ' 25cm^3 of solution was taken from the original sample of 250cm^3' DRAW IT IN YOUR diagram like an equation of whats going to what on a seperate sheet of paper show yourself that its been divided by ten so that when you get your moles you know to times it by ten to get you moles in 250 etc. I can't stress it enough about how important it is that you make it clear what the student in the question is doing. Once you do this it becomes so much easier. http://www.chemteam.info/Redox/WS-re...-problems.html This site also has some really good questions which I feel were challenging enough to warm you up for exam questions. Hope this helps. AND GOING THROUGH IT WITH SOMEONE WHO KNOWS EXACTLY WHATS GOING ON IS EQUALLY IMPORTANT. I feel it's something that needs to be taught to you in great detail.

    Now I have a problem myself. If anyone can show me some good note of what goes on in lattice enthalpy of hydration & solution questions I'd really appreciate it. I Just don't understand the concepts of why a metal higher then another in the group of the periodic table would have a higher lattice enthalpy of hydration. I thought it'd be the other way round. (This paragraph is probably is wrong in terms of the chemistry of it which shows how confused I am by the concepts).
 
 
 
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