AQA A-level Chemistry paper 2 7405/2 Unofficial Markscheme

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Reply 100

woahexams

lol i got everything wrong so y'all have done better than some ppl :P

Reply 101

RinneTensigan

Original post by Lolexamswhat

Same! But I think it's bc they gave us lnA and not A or something
It's shouldn't be tooo many marks

No they didnt they gave us what A is equal to not ln a

Reply 102

qwertyuytrfdesw

Original post by emilyvrl

Step 1 :
H2SO4 + HNO3 --> H2NO3 + HSO4

Step 2 :
Methylbenzene + HNO3 --> methylnitrobenzene + H2O ( conc H2SO4 catalyst )

Step 3 :
MethylNitrobenzene + 6[H] --> methylphenylamime + 2H2O ( Sn/HCl catalysts , room température )

Okay so that's all I did for the very last question . It makes sense to me but I have no clue if it's right . Any opinions ?

i dont think so as it wanted phenylethane amine so i used free radical substitution of chlorine then added a nitrile group then H2

Reply 103

TSRP3

Original post by brumtown0121

Was I the only one who talked about distinguishing K L M N separately???? I gave each letter its own test, cos I swear question said they were in separate test tubes and to my knowledge it wasn't a mixture, it was just separate solutions of each letter, one of u clever clogs lemme know what's up

Nah I did that

Reply 104

alevelsarecrap

Original post by brumtown0121

I did that too. I said K was an optical isomer as it has a chiral carbon, would I still get the mark?

Reply 105

darsh shah

https://www.thestudentroom.co.uk/showthread.php?t=4791640

What did you all think of the first 2 exams? Answer the poll and see how others did to.

Reply 106

username2021515

Original post by Sapereaude45

Most of my classmates,me included, got around 130mg give or take depending on rounding.
Here was my method:
Work out volume of gas lost as CO2 by subtracting intial volume from final volume

Use ideal gas equation to determine moles of CO2 present in gas.

Use the fact that the combustion took place in a 6:1 ratio(Can't remember exactly)

Multiple by Mr of the orginal compound.I also carried all sig figs unitl the end
Also would I be marked down for naming it as N-ethylpropanamide as opposed to ethylpropanamide?I know it is techically the exact same thing but as we know AQA pedantry knows no bounds.

Any help peeps?

Reply 107

peacheee

Original post by Khanman123

I got EGF for the benzene ring bit

Me too

Reply 108

_.lukeg._

They removed coursewrok so doubt it'd be 80% for A*, more like 75%

Original post by Khanman123

Grade boundaries?? I think they'll be similar to the old spec, 75% for an A 80% for A*

Reply 109

peacheee

Original post by p____

I put LiAlH4 on the last question is that right or was it meant to be HCl?

LiAlH4 is right for reduction of the nitrile compound

Reply 110

peacheee

Did anyone else notice mistakes in the paper?

Reply 111

HHH7

equilibrium question what did everyone get

Reply 112

Adz777

A last year for AS was 60%. Paper 1 was pretty difficult too, although paper 2 was decent. I think considering this and the fact that we're doing a new specification, A will be around the 65-70% for paper 1 and around 70-75% for paper 2. I'd expect A* to be about 10% above this.

Reply 113

username2021515

Original post by RinneTensigan

No they didnt they gave us what A is equal to not ln a

It was lnA as I remember.I re-read this part multiple times as I was lose marks to those silly mistakes...Hoping I didn't make too many on this paper-need an A/A*.

Reply 114

Balthznar

i believe this was the question for the mg of p.
C6H10O2 +15/2O2 -> 6CO2 +5H2O
N=PV/RT
P=105000 pa
V= 335 x10^-6 m^3 (total volume of gas products)
= 155 x10^-6 m^3 (total volume after CO2 was reacted off)
R =8.31
T= 25+273 =298k
How much C6H10O2 in mg was used?

(edited 6 years ago)

Reply 115

username2021515

What did everyone get for that cyclic compund question using the data sheet?
One of my friends and I both got something like a cyclic ester-I can't remember exactly.

Reply 116

peacheee

Did anyone notice the mistakes in the paper?

Reply 117

username2021515

Original post by Balthznar

i believe this was the question for the mg of p.
C6H10O2 +17/2O2 -> 6CO2 +5H2O
N=PV/RT
P=105000 pa
V= 335 x10^-6 m^3 (total volume of gas products)
= 155 x10^-6 m^3 (total volume after CO2 was reacted off)
R =8.31
T= 25+273 =298k
How much C6H10O2 in mg was used?

Yeah, looks good to me.I still don't understand how people got 145mg.I think they may have forgot to subtract the two volumes of gas to find CO2, unless it was a silly error on my part!

Reply 118

tsr120

Original post by Sapereaude45

Yeah, looks good to me.I still don't understand how people got 145mg.I think they may have forgot to subtract the two volumes of gas to find CO2, unless it was a silly error on my part!