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Edexcel IGCSE Chemistry 18th May/June 2017 Thread

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Original post by weirdperson7878
what did people get for the compounds that went through the filter paper? I got water and magnesium sulfate but other people are saying that sulfuric acid went through the filter paper.
I got the same
Yeah but it's an observation and theoretically it's right:/.
Reply 182
Could anyone who remembers the marks please put them on the unofficial mark scheme
what about the question on Carbon reacting with 0xygen? i wrote that it reacts to form Carbon dioxide which then reacts with C to to guve the reducing agent carbon monoxide
also what about the last question? i wrote that calcium carbide particles have more kinectic energy so frequency of successful collisions increases and the rate of reaction increases. I also mentioned about activation energy
Any predictions for paper 2???
How did it go? :smile:
Thought it was fine

Electrolysis calculation was 1.27g
The 3 gases were hydrogen, chlorine and oxygen
The kj/mol was 403.2
The last question was -535
Reply 188
Oh my days that was a maths exam 😳
What did you get for the faradays question? I got 0.02g
Original post by Gneale12345
Thought it was fine

Electrolysis calculation was 1.27g
The 3 gases were hydrogen, chlorine and oxygen
The kj/mol was 403.2
The last question was -535


Mate i got 405 what did you do?
Original post by Starbox359
What did you get for the faradays question? I got 0.02g



I got 127 g because you multiply the 0.02 by 63.5 I think
Original post by solark
Mate i got 405 what did you do?



I also got 403.2

You did 18.9/(3/64)
Original post by solark
Mate i got 405 what did you do?

I got 18900 for the calculation above it then i found the mass of ethanol by taking away the mass after combustion from before which was 1.5g
The moles was 1.5g/32 which gave me 3/64 moles.

3/64 moles = 18900J
18900/ 3/64 gave you one mole which was 403200J
/1000 gave u 403.2
Reply 194
Original post by Gneale12345
I got 18900 for the calculation above it then i found the mass of ethanol by taking away the mass after combustion from before which was 1.5g
The moles was 1.5g/32 which gave me 3/64 moles.

3/64 moles = 18900J
18900/ 3/64 gave you one mole which was 403200J
/1000 gave u 403.2

Are u able to start an unofficial mark scheme
Original post by Gneale12345
I got 18900 for the calculation above it then i found the mass of ethanol by taking away the mass after combustion from before which was 1.5g
The moles was 1.5g/32 which gave me 3/64 moles.

3/64 moles = 18900J
18900/ 3/64 gave you one mole which was 403200J
/1000 gave u 403.2


Yeah i rounded up 18900 to 19kj thats why :wink: thanks
Original post by solark
How did it go? :smile:


Was quite hard in some parts. Too much organic chemistry.
Original post by Starbox359
I also got 403.2

You did 18.9/(3/64)


I think 403.2 was wrong. The reaction was exothermic, and therefore you had to use a minus for the temperature change.
(edited 6 years ago)
Original post by Gneale12345
Thought it was fine

Electrolysis calculation was 1.27g
The 3 gases were hydrogen, chlorine and oxygen
The kj/mol was 403.2
The last question was -535


I did not get the electrolysis calculation.
I got 3 gases as hydrogen (for negative electrode), water and chlorine gas for positive electrode.

3 half equations:
2Cl- + 2e- -> Cl2
2H+ + 2e- -> H2
H+ + OH- -> H2O

I got something negative for the kj/mol, because you had to use negative temperature change (exothermic reaction)

Can't remember what I got for the last question.
Reply 199
Original post by JMR2017
I did not get the electrolysis calculation.
I got 3 gases as hydrogen (for negative electrode), water and chlorine gas for positive electrode.

3 half equations:
2Cl- + 2e- -> Cl2
2H+ + 2e- -> H2
H+ + OH- -> H2O

I got something negative for the kj/mol, because you had to use negative temperature change (exothermic reaction)

Can't remember what I got for the last question.


The three gases were hydrogen, chlorine and oxygen in pretty sure

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