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!
I got 149 mg as I used the value for 155 volume then divided by 5 as the only volume left then did that times 114 (mr of p) then x1000 from g to mg Pretty sure that is my error. Most people done: (105000x335x10^-6)/(8.31x298) N=1.42x10^-2 Then divide by 11 (total number of moles of gaseous products) Mass = NxMr Mass = 1.29x10^-2 x 114 Mass= 1.47x10^-1 grams x1000 Mass = 147mg (3sf) I feel like that's right but then again I haven't used the volume of 155 x10^-6 however the various methods that I've used to work it out has given the answer roughly around 147 mg but it could as well be something else
did the exact same thing but its wrong i didnt read the damn question such an easy 5 marks
If you had seen the structure properly it had no methyl group at the end so something must-have been tested with methyl group hence do cl2 or br2 in if light to get ch2cl which can be reached with KCN and then reduced to amine
I got 149 mg as I used the value for 155 volume then divided by 5 as the only volume left then did that times 114 (mr of p) then x1000 from g to mg Pretty sure that is my error. Most people done: (105000x335x10^-6)/(8.31x298) N=1.42x10^-2 Then divide by 11 (total number of moles of gaseous products) Mass = NxMr Mass = 1.29x10^-2 x 114 Mass= 1.47x10^-1 grams x1000 Mass = 147mg (3sf) I feel like that's right but then again I haven't used the volume of 155 x10^-6 however the various methods that I've used to work it out has given the answer roughly around 147 mg but it could as well be something else
It does make sense but why would you subtract the volumes?The volume lost was the CO2 removed after combustion.I probably being an idiot though.
It does make sense but why would you subtract the volumes?The volume lost was the CO2 removed after combustion.I probably being an idiot though.
Tbh I don't know why it gave us both volumes as I believe we only hat to use one of them, whether you used the 335 and divided by 11 or used 155 and divided by 5 should theoretically give you the same values but ofcouse they didn't. But like you say I have no idea why you would minus the volumes as that will give you the volume left at which 5H2O was moles of product left.
335-155=180cm^3 Use the same method with this new volume and divide by 6 moles You get 145mg of P
Tbh I don't know why it gave us both volumes as I believe we only hat to use one of them, whether you used the 335 and divided by 11 or used 155 and divided by 5 should theoretically give you the same values but ofcouse they didn't. But like you say I have no idea why you would minus the volumes as that will give you the volume left at which 5H2O was moles of product left.
335-155=180cm^3 Use the same method with this new volume and divide by 6 moles You get 145mg of P
I did use 180cm^3.I must have made a substitution error.Ah, well maybe 2 marks lost? At least it's all clearcup now
Unfortunately the correct answer was 148 mg.did you use that total 330cm3 gas volume. They reacted the gas with excess NaOH and i believe it was something like 180cm3 left after that. So u needed to do 330 Minus that value left after reaction to work out the volume of CO2.
For the 1,3 dinitrobenzene where it asked you for electrophilic substitution, does anyone remember the exact question? Did they ask you to form 1/3 dinitrobenzene from benzene or 1-nitrobenzene?
Thanks- so what do u think grade boundaries will be like for this paper?
I think this one was a tiny bit harder than paper one because of all that chromatography and NMR stuff. Maybe around: 75%A* 65%A 55%B 45%C What do others think?
For the 1,3 dinitrobenzene where it asked you for electrophilic substitution, does anyone remember the exact question? Did they ask you to form 1/3 dinitrobenzene from benzene or 1-nitrobenzene?