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

1. (Original post by _H_V)
Doesn't the sodium thiosulphate react with iodine to produce iodide ions and not the other way round?

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Yeah sodium thiosulphate reacts with iodine to form iodide ions
2. (Original post by Ssttll)
The pH of the 10cm^3 HCl is 0 as -log(1)=0. adding 10 cm^3 to 990cm^3 gives a volume of 1000cm^3. this increase in volume essentially dilutes the acid conc by a factor of 2. Because the sol of HCl has been diluted by factor 2, you add 2 to the initial pH, This gives an overall pH of the 1000cm^3 HCl sol of: 2

Hope that helps!
But why does it decrease by a factor of 2?
3. Can anyone please show me how to solve q 24 a iii from Jan 2015 IAL?
4. Okay so have I got this right?
In a reaction where iodine is produced, sodium thiosulphate and starch is added to measure rate of reaction. The iodine initially produced is reduced to iodide ions by sodium thiosulphate.
Once all the sodium thiosulphate has reacted, the next iodine produced reacts with starch which produces a blue/black solution.
The rate of reaction can be measured by measuring the time it takes for colour to go from colourless to blue/black.
Could you correct any mistakes, much appreciated

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5. The pH of 3 solutions with concentration 1.0mol dm-3 was measured.
Solution 1 NH3
Solution 2 CH3COONa
Solution 3 NH4Cl
Which of the following shows the three solutions in order of increasing Ph?

A 1 2 3
B 3 2 1
C 3 1 2
D 2 3 1 Why is the correct answer B and not D?
6. Work out moles of hcl first which is (10/1000 x 1) which is 0.01
This is added to 990cm^3, so new volume is 1000cm^3, you have to work work out concentration of hcl in new volume (n/v=c)
This is (1000/1000) x 0.01 which is still 0.01
And - log(0.01) is = 2.

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7. Hey guys, does anyone have a link to the June 2015 Paper and Mark Scheme? Thank you in advance
8. (Original post by _H_V)
Work out moles of hcl first which is (10/1000 x 1) which is 0.01
This is added to 990cm^3, so new volume is 1000cm^3, you have to work work out concentration of hcl in new volume (n/v=c)
This is (1000/1000) x 0.01 which is still 0.01
And - log(0.01) is = 2.

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I have understood. Thanks!
9. Pulling an all nighter
10. 3.16 i m still alive

11. Attachment 549573549575

can someone please tell how you find mol at eqilibrium for H2?
Attached Images

12. Good luck everyone with cramming and then the actual exam ofc
13. (Original post by Rahatara Sadique)
The pH of 3 solutions with concentration 1.0mol dm-3 was measured.
Solution 1 NH3
Solution 2 CH3COONa
Solution 3 NH4Cl
Which of the following shows the three solutions in order of increasing Ph?

A 1 2 3
B 3 2 1
C 3 1 2
D 2 3 1 Why is the correct answer B and not D?
Solution 3 dissociates in water:
NH4CL + H20 -> HCL + NH3
and HCL is acidic
14. Goodluck for exam guys x)
15. Can someone tell me if the mark scheme will be posted by the end of the day please, thanks xxx
16. I really want to go into my next exams confident!! if i do well today i know i can do well in unit 5 !
17. (Original post by ihaspotato)

Attachment 549573549575

can someone please tell how you find mol at eqilibrium for H2?
We know that all of the H2 has had to react to form the CH3OH; since CH3OH = 38.5 moles at equilibrium, H2 must be 77.5 - 38.5 = 39 moles
18. (Original post by natcupine)
We know that all of the H2 has had to react to form the CH3OH; since CH3OH = 38.5 moles at equilibrium, H2 must be 77.5 - 38.5 = 39 moles
I'm afraid you forgot to do something natcupine!!

ihaspotato You would have to double the 38.5 sine the H2 is in a 2:1 molar ratio to the CH3OH So it's 77.5 - 77 = 0.5 moles
19. Does the sn2 or sn1 mechanism form a racemic mixture
20. (Original post by Supermanxxxxxx)
Does the sn2 or sn1 mechanism form a racemic mixture
Sn1 produces a carbocation intermediate which then allows for the formation of a racemic mixture.

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