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

Does anyone know the grade boundaries for the IAL Jan 2016 paper?

Reply 683

ihaspotato

can someone help me with (b)? how do we solve it? :s-smilie:

(edited 7 years ago)

Reply 684

sabahshahed294

12

Original post by mizzyportia

Does anyone know the grade boundaries for the IAL Jan 2016 paper?

http://qualifications.pearson.com/content/dam/pdf/Support/Grade-boundaries/International-A-level/1601-IAL-Grade-Boundaries.pdf

For Unit 4, A* was at 80.

Reply 685

Original post by ihaspotato

can someone help me with (b)? how do we solve it? :s-smilie:

It's just a normal K_p 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

Then just plug into your K_pequation.

Reply 686

Neha121101

1

Original post by ihaspotato

can someone help me with (b)? how do we solve it? :s-smilie:

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? :smile:

Reply 687

blueynuey

5

Original post by ihaspotato

can someone help me with (b)? how do we solve it? :s-smilie:

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
Therefore answer is B

(edited 7 years ago)

Reply 688

Neha121101

1

anyone understand why the answer is B?
thanks! :smile:

Reply 689

ihaspotato

http://www.physicsandmathstutor.com/chemistry-revision/edexcel-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! :tongue:

(edited 7 years ago)

Reply 690

blueynuey

5

Original post by Neha121101

anyone understand why the answer is B?
thanks! :smile:

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.

Reply 691

TeaAndTextbooks

9

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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Reply 692

imnoteinstein

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?

Reply 693

ayvaak

I dont understand why the conc is 0.05. Could someone please help?
Thanks in advance

Reply 694

ihaspotato

Can someone show how to solve this...

Reply 695

Original post by ayvaak

I dont understand why the conc is 0.05. Could someone please help?
Thanks in advance

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 50cm³of 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

Reply 696

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? :smile:

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).

(edited 7 years ago)

Reply 697

Original post by ihaspotato

Can someone show how to solve this...

It's simply hydration enthalpy of Sr²⁺ + 2 x hydration enthalpy of F^-

Reply 698

setarcos

8

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.

Reply 699