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

1. (Original post by sabahshahed294)
Thanks a lot!
Haha no problem but just a quick note, I accidentally got the reverse and normal phase HPLCs the wrong way round.

Normal phase HPLC = polar STATIONARY phase and non-polar MOBILE phase.

Reverse phase HPLC = non-polar STATIONARY phase and polar MOBILE phase.

EDIT: Okay I just checked it and I was right the first time round anyway XD I have confused myself twice in less than 5 minutes great hahahaha
2. (Original post by Don Pedro K.)
Haha no problem but just a quick note, I accidentally got the reverse and normal phase HPLCs the wrong way round.

Normal phase HPLC = polar STATIONARY phase and non-polar MOBILE phase.

Reverse phase HPLC = non-polar STATIONARY phase and polar MOBILE phase.
Ah okay. Np! Got what you meant.
3. Does anyone know the grade boundaries for the IAL Jan 2016 paper?

4. can someone help me with (b)? how do we solve it?
5. (Original post by mizzyportia)
Does anyone know the grade boundaries for the IAL Jan 2016 paper?
http://qualifications.pearson.com/co...Boundaries.pdf

For Unit 4, A* was at 80.
6. (Original post by ihaspotato)

can someone help me with (b)? how do we solve it?
It's just a normal Kp calculation - it's given you the mole fraction of N2O4, so from that you can calculate the partial pressure of N2O4 as mol fraction x total pressure. Since the mole fraction of N2O4 is 0.25, what can you deduce about the mole fraction (and hence the partial pressure) of NO2?
Spoiler:
Show
Hint: mole fractions must add up to what....
Then just plug into your Kp equation.
7. (Original post by ihaspotato)

can someone help me with (b)? how do we solve it?

so you have 0.25 n2o4, and 0.75 no2
work out the mole fractions (total moles =1)
then work out partial pressures (given total pressure =3)
put into Kp expression
the answer should be B i think?
8. (Original post by ihaspotato)

can someone help me with (b)? how do we solve it?
Mols at Equib: N2O4= 0.25, NO2= 0.75
Mole Fraction: N2O4= 0.25, NO2= 0.75
partial pressure: N2O4= 0.25*3, NO2= 0.75*3
N2O4= 0.75 and NO2= 2.25

Kp= p(NO2)^2/p(N2O4)
= (2.25)^2/(0.75)
= 6.75 atm

9. anyone understand why the answer is B?
thanks!
10. http://www.physicsandmathstutor.com/...dexcel-unit-4/

Should I read the notes or solve the papers and make mistakes and learn? is anyone using these notes?

And thank you Neha and blueynuey for your time!
11. (Original post by Neha121101)

anyone understand why the answer is B?
thanks!
We know that thermodynamically stable means that the total entropy is less then 0
The entropy of system for this reaction is positive as it is going from a solid --> gas
Therefore stability must come from the entropy of surrounding being more negative then the system, which must mean that enthalpy is positive for surrounding to be negative (look at the formula).

activation energy affects kinetic stability not thermodynamic stability therefore answer can only be B.
12. Hey, I'm sooo stressed about these NMR spectrum questions. I get soo confused and have no idea how you get the final product. I can't believe the exams tomorrow and I'm weak on so many areas. If anyone could explain NMR I'd really appreciate it. Thanks

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13. (Original post by blueynuey)
We know that thermodynamically stable means that the total entropy is less then 0
The entropy of system for this reaction is positive as it is going from a solid --> gas
Therefore stability must come from the entropy of surrounding being more negative then the system, which must mean that enthalpy is positive for surrounding to be negative (look at the formula).

activation energy affects kinetic stability not thermodynamic stability therefore answer can only be B.
would it be right to say thermodynamically stable and thermodynamically feasible are are opposites?

15. Can someone show how to solve this...
16. (Original post by ayvaak)

The question tells you that the concentrations of both the silver ions and iron (II) ions are the same. Therefore, to work out the concentration of one of them you simply need to work out the moles of one of the 2 (lets say Ag+) that were initially present to start off with, and then work out what the new concentration in 50cm3, (because you are taking your reaction mixture samples from the 50cm3 of solution formed)

Therefore:

1. Moles Ag+ initially present = 0.1 x 0.025 = 0.0025mol

2. Concentration of Ag+ present in 10.0cm3 = 0.0025mol/0.05dm-3 = 0.05 moldm-3
17. Could someone explain this? I put D because I thought the less the molecule interacts with the eluent, the faster it travels through. However, the answer is B. Could someone please explain?

In one type of high-performance liquid chromatography, the stationary phase is non-polar and a polar solvent is used as the eluent. Which of the following would travel through the chromatography column most quickly?
a) tetrachloromethane
b) chloromethane
c) iodomethane
d) hexane

Also, please can someone explain how the attack by 2,4-dinitrophenylhydrazine could be by a nucleophile? I thought the 2,4-dnp is attracted to the C=O group of the carbonyl (which would mean it is an electrophile).

18. (Original post by ihaspotato)

Can someone show how to solve this...
It's simply hydration enthalpy of Sr2+ + 2 x hydration enthalpy of F-
19. (Original post by Funky_Giraffe)
Please can someone explain how the attack by 2,4-dinitrophenylhydrazine could be by a nucleophile? I thought the 2,4-dnp is attracted to the C=O group of the carbonyl (which would mean it is an electrophile).

Thanks!
The 2-4 dinitrophenylhydrazine is attracted to the partial positive charge on the carbon atom of the carbonyl group so it therefore reacts as a nucleophile.
20. (Original post by setarcos)
The 2-4 dinitrophenylhydrazine is attracted to the partial positive charge on the carbon atom of the carbonyl group so it therefore reacts as a nucleophile.
Ahh I see. thanks

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