Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed) Watch

Poll: How pumped up are you for this exam?-(warning)-(bad jokes arene this poll!)
"Titanium-I'm not going to corrode (even at high temperatures)" (A*) (22)
16.67%
"Benzene's my middle name, give me the paper in a week and I'll ace it!" (A) (27)
20.45%
"Yeah, I'm fairly electrophillic (positively charged) about the exam" (B) (27)
20.45%
"I'm in the middle of the salt bridge, but I will pass-eventually" (C) (21)
15.91%
"I'm feeling rather electroNegative about this exam" (D) (18)
13.64%
"Benzene, what's that?" (E) (6)
4.55%
"Chemistry, what's that?" (F) (11)
8.33%
This discussion is closed.
Lgambo
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Me going to the exam hall tomorrow will be like trying to put a cat in a cat carrier on the way to the vets.
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anony.mouse
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(Original post by bubblegummer)


Guess your teacher went nuts looking for the answer :P We have too many stuff to remember from unit 1 to 5 and sometimes i have no idea which one to apply
To be fair on him, he's never taught edexcel chemistry before, so isn't entirely familiar with the data book and thus didn't realise that that page is in it.
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lewiss111
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(Original post by ghostcommand99)
Does anyone know how to explain why there's 196g of Chromium (III) sulfate in Jan 2013 paper Question 2C? Shouldn't it 392g at 1mol/dm3?
its dichromate if you look in the question, so a 1 mol dm3 conc of chromium comes from half of 392
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doodledee1
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on the jan 2013 paper there are a couple of multi choice questions I am confused about, firstly question 2c (i thought was C, correct answer B) and also question 11 (I thought was B, correct answer D). Could someone please explain them , thanks!
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Bord3r
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(Original post by ghostcommand99)
Does anyone know how to explain why there's 196g of Chromium (III) sulfate in Jan 2013 paper Question 2C? Shouldn't it 392g at 1mol/dm3?
There's two chromium ions per chromium sulfate molecule so in order to get 1 moldm^-3 you need only half of 1 mole of chromium sulfate.
Hope this helps.


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clydeshen411
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(Original post by kabOOmm)
but they said in the mark scheme
Copper (formed (by disproportionation)) is oxidized(by nitric acid) must be stated in words
stand alone mark


does this mean that copper(pink solid) did form but got oxidized? why couldn't this happen with the sulphuric acid too?
Yeah I was wrong, CuI should form as well.


  1. [4H+(aq) + SO􏰋-(aq)], [H2SO3(aq) + H2O(l)]|Pt , E=+0.17V

    Cu2+(aq) + 2e- 􏰓 Cu(s) E=+0.34V

    0.17- 0.34 <0 so only disproportionation occurs with sulfric acid. But nitric acid is able to oxidise cu to cu2+ since +0.94>+0.34





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Lgambo
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You all sound so clever I wish I were all of you
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ChooChooCherry
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(Original post by ghostcommand99)
Does anyone know how to explain why there's 196g of Chromium (III) sulfate in Jan 2013 paper Question 2C? Shouldn't it 392g at 1mol/dm3?
can u please send me the link for the jan 2013 paper and ms? thank yoouu
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clydeshen411
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(Original post by ChooChooCherry)
can u please send me the link for the jan 2013 paper and ms? thank yoouu
Links are on page 80
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paddyroddy
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(Original post by anony.mouse)
To be fair on him, he's never taught edexcel chemistry before, so isn't entirely familiar with the data book and thus didn't realise that that page is in it.
the reason scientifically (not just the data book ) is that there is a big gap between 7th & 8th which suggests this removing the 3p6 electron which is a much higher energy than the 3d.

the other one with a big gap was A between 1st & 2nd which would mean not transition metal as needs at least 3 from 4s2 rd1

and B has 2nd and 3rd which again is the same reason as A
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posthumus
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(Original post by clydeshen411)
I think the colour should be lighter. Greater splitting means more energy absorbed, so light with shorter wavelength (darker colour) is absorbed, then the complementary colour should be lighter maybe?
You make a really good point, now I am unsure!

Hope they give us a color wheel in the exam
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Bubblezzzz
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(Original post by LeaX)
thank you so much so does the anhydrous calcium chloride remove the ether too?
no it doesn't, CaCl2 only removes water, the ether can then be distilled off by gentle heating (it will very often have a boiling point much less than that of the organic substance you want) x
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orange94
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Planning on recapping on all the organic mechanism for the whole of chemistry unit 1 - unit 5! And going over transition complexes and colours! And some nasty titration questions and weird error questions! Good approach for today?


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Bubblezzzz
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(Original post by posthumus)
You make a really good point, now I am unsure!

Hope they give us a color wheel in the exam
Yhh that would be extremely helpful, but we all know edexcel, they will try and make this the worst exam experience of our lives
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Bubblezzzz
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(Original post by clydeshen411)
Yeah I was wrong, CuI should form as well.


  1. [4H+(aq) + SO􏰋-(aq)], [H2SO3(aq) + H2O(l)]|Pt , E=+0.17V

    Cu2+(aq) + 2e- 􏰓 Cu(s) E=+0.34V

    0.17- 0.34 <0 so only disproportionation occurs with sulfric acid. But nitric acid is able to oxidise cu to cu2+ since +0.94>+0.34





yhh this question was really weird because I was trying to understand what role sulfuric acid played in the initial reaction they gave us, i.e. the disproportionation. As far as I can see, disproportionation always occurs, provided the two half-equations when combined, give a positive E cell value. The acid then is involved in the second step, as an oxidising agent, which in the case of nitric is able to convert the Cu back to Cu2+ - Here Cu+ isn't reformed since the nitric is sufficiently positive to make the reaction 'go all the way' back to Cu2+ resulting in the blue solution.
Sulfuric acid however, has a more negative half cell value, so is unable to oxidise the Cu solid formed.
Hope that provides some clarification x
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orange94
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HELP HELP HELP NEE CONFORMATION!! B

E cell is directly proportional to LnK so if you were to be Given the e cell value can you work out K from it?

Is it like the same in the entropy calculations when Stotal is directly proportional to LnK and you can work out K by rearranging the equation!


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clydeshen411
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(Original post by Bubblezzzz)
yhh this question was really weird because I was trying to understand what role sulfuric acid played in the initial reaction they gave us, i.e. the disproportionation. As far as I can see, disproportionation always occurs, provided the two half-equations when combined, give a positive E cell value. The acid then is involved in the second step, as an oxidising agent, which in the case of nitric is able to convert the Cu back to Cu2+ - Here Cu+ isn't reformed since the nitric is sufficiently positive to make the reaction 'go all the way' back to Cu2+ resulting in the blue solution.
Sulfuric acid however, has a more negative half cell value, so is unable to oxidise the Cu solid formed.
Hope that provides some clarification x
yeah this is a much clearer explanation! My english sucks
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clydeshen411
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(Original post by orange94)
HELP HELP HELP NEE CONFORMATION!! B

E cell is directly proportional to LnK so if you were to be Given the e cell value can you work out K from it?

Is it like the same in the entropy calculations when Stotal is directly proportional to LnK and you can work out K by rearranging the equation!


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No we dont know the ratio between E cell and lnk, while Stotal = R lnk where R = 8.31 (a constant). I think they will only test u on how will k change if Ecell changes.
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LeaX
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(Original post by Bubblezzzz)
no it doesn't, CaCl2 only removes water, the ether can then be distilled off by gentle heating (it will very often have a boiling point much less than that of the organic substance you want) x
(Original post by bubblegummer)
It removes iodine from ether so it forms solid calcium iodide. ether will be seperated by it. You're welcome
thank you both.
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orange94
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(Original post by clydeshen411)
No we dont know the ratio between E cell and lnk, while Stotal = R lnk where R = 8.31 (a constant). I think they will only test u on how will k change if Ecell changes.
Lol silly me! So since ecell is directly proportional to entropy that means as ecell increases K increases too and vice-versa


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