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# AQA CHEM5 A2 Chemistry - 19th June 2013 watch

1. (Original post by AspiringDoctor)
Cool, it's just I don't see it much in the mark scheme but should be okay Thanks

Another thing, just double checking that if state symbols aren't explicitly asked for, do we still need to put them? Don't want to make a mistake and get penalised if I didn't need to put them in the first place...
If they dont ask for them I wouldn't put them in as the mark scheme usually says 'ignore state symbols'
However if you put them in and you get them wrong you could loose the mark
2. can someone guide me through percentage mass calculation as im finding them difficult. Thanks in advance,
3. okay, just got it, nevermind
4. (Original post by choco.chip)
can anyone help me with june 2012 question 6dii please?

why do you times the moles by 5/3 not 3/5?
MnO4-:FeC2O4
3:5

So once you've found the moles of MnO4- you divide by 3 then times by 5, which is the same as multiplying by 5/3
5. (Original post by choco.chip)
okay, just got it, nevermind
explain pleasseee
6. (Original post by Kaii_Smith)
If they dont ask for them I wouldn't put them in as the mark scheme usually says 'ignore state symbols'
However if you put them in and you get them wrong you could loose the mark
Great, thank you!
7. (Original post by Bootala)
explain pleasseee

(Original post by crc290)
MnO4-:FeC2O4
3:5

So once you've found the moles of MnO4- you divide by 3 then times by 5, which is the same as multiplying by 5/3
8. (Original post by darryalar)
can someone guide me through percentage mass calculation as im finding them difficult. Thanks in advance,
This might be a bit unclear doing it blind, but I'll have a shot. You want to draw yourself a little grid, with the constituent elements of the compound going horizontally. Then, underneath, you do the following lines
1) Percentage of each. Either given to you, or left to work out (ie, 70% this, 20% that, and something else)
2) If you assume, temporarily, that the %s are a mass relative to 100, then divide each percentage by the Atomic Mass to get the relative moles of each
3) Divide each amount of moles by the smallest molar value, so you should have at least one value of 1 with everything else in a nice ratio of nearly whole numbers

This gives you the empirical formula, ie the simplest one. Extrapolate up to match a given Mr or mass or what-have-you if one is given.
Also what are the equations in alkaline conditions?
10. (Original post by crc290)
soooooo, why is it 3:5 in the first place? i'm gonnna fail so hard :/
11. If the thermodynamics section tomorrow is like the one in Jan 13! i will blow up :L
12. (Original post by crc290)
MnO4-:FeC2O4
3:5

So once you've found the moles of MnO4- you divide by 3 then times by 5, which is the same as multiplying by 5/3
Omg I'm just such an idiot, just realised I swapped the ratio around, no wonder I got the wrong answer. Thank you for clarifying it
13. (Original post by FreshPrince987)
This exam along with CHEM 4 completely screwed me over. I am defo getting a U in chem this year, but for my retakes next year could anyone here who's confident in Chem suggest a way to revise, because i can literally answer one question and nothing else, no matter how much i seem to revise and answer questions.
Hello mate. Ive got some very good advice for you as a mature student. Email me personally if you like. [email protected]. cheers.
This might be a bit unclear doing it blind, but I'll have a shot. You want to draw yourself a little grid, with the constituent elements of the compound going horizontally. Then, underneath, you do the following lines
1) Percentage of each. Either given to you, or left to work out (ie, 70% this, 20% that, and something else)
2) If you assume, temporarily, that the %s are a mass relative to 100, then divide each percentage by the Atomic Mass to get the relative moles of each
3) Divide each amount of moles by the smallest molar value, so you should have at least one value of 1 with everything else in a nice ratio of nearly whole numbers

This gives you the empirical formula, ie the simplest one. Extrapolate up to match a given Mr or mass or what-have-you if one is given.
I think i wasn't very clear in what i was asking for my friend, Im talking about percentage mass in titrations, the question you normally get in chem5 papers.
15. Can anyone tell me if this equation is right or if I've gone wrong somewhere?

2Cr(OH)63- + 3H2O2 ---> 2CrO42- + 8H2O + 2OH-
16. (Original post by Bootala)
soooooo, why is it 3:5 in the first place? i'm gonnna fail so hard :/
You need to balance out the electrons.
So 15 electrons...
Mn04- 5e-x3= 15 electrons
Bottom two equation combined 3e-x5=15 electrons

ratio= 3:5
17. I do not understand June 2011 7ci) It is only bonded to 4 ligands but it says the coordination number is 6?
18. (Original post by Bootala)
soooooo, why is it 3:5 in the first place? i'm gonnna fail so hard :/

(Original post by crc290)
You have to combine the bottom two half equations first

Fe2+ Fe3+ + e-
C2O42- 2CO2 + 2e-

FeC2O4 Fe3+ + 2CO2 + 3e-

Then combine this equation with the top equation

3MnO4- + 5FeC2O4 + 24H+ 3Mn2+ + 5Fe3+ + 10CO2 + 12H2O

So for every 3 moles of MnO4- which react, 5 moles of FeC2O4 react i.e. a 3:5 ratio
19. (Original post by Atz23)
I do not understand June 2011 7ci) It is only bonded to 4 ligands but it says the coordination number is 6?
C2O42- is a bidentate ligand, so it forms two coordinate bonds, giving a total of 6 in the complex
20. Jan 2013 mark scheme anyone?

Also mental block, If something is partially dissociated what does this mean?

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