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

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Original post by samb1234
no problem, I'm just procrastinating to avoid finishing my chem notes anyway haha so glad i could help


You are procrastinating :tongue: ? lol, tbh you don't sound like one haha. I'm a bigger one than you :tongue:
Reply 321
Original post by samb1234
It is neither. Both electrodes, technically speaking are negative. Voltage is relative, so essentially the 'positive' electrode is actually negative just not as negative as the other electrode so relative to that electrode is positive



so just to confirm when you say 'not as negative' you're talking about the charge on the metal electrode itself?
Guys can someone tell me if this could be called 2-bromo-1-methylbenzene?Capture.JPG On wikipedia it said 1-bromo-2-methylbenzene but I don't understand why bromo is on the first carbon?
Can someone please explain question 8(a) and (b). I feel like it's a very simple question but i'm just not seeing something:

http://qualifications.pearson.com/content/dam/pdf/A-Level/Chemistry/2013/Exam-materials/6CH04_01_que_20150610.pdf
Reply 324
Original post by sabahshahed294
You are procrastinating :tongue: ? lol, tbh you don't sound like one haha. I'm a bigger one than you :tongue:


Haha i cant bring myseof to finish them theyre too long (at roughly 200 pages atm lol) so yeah
Original post by samb1234
Haha i cant bring myseof to finish them theyre too long (at roughly 200 pages atm lol) so yeah


Unit 5 is long tbh. 4 is better in this regard. Although I hate NMR Spectroscopy.. :')
Original post by HaydenMoussa
Can someone please explain question 8(a) and (b). I feel like it's a very simple question but i'm just not seeing something:

http://qualifications.pearson.com/content/dam/pdf/A-Level/Chemistry/2013/Exam-materials/6CH04_01_que_20150610.pdf


In 8(a), you just add up the enthalpy of both the ions whereas in (b), apply the formula...Enthalpy of Solution= -Lattice Enthalpy plus Enthlapy of Hydration.
@samb1234 why is the answer to this D?
lol.JPG
Reply 328
Original post by Don Pedro K.
@samb1234 why is the answer to this D?
lol.JPG


cant be first 2 due to position of the o, c clearly is wrong
Original post by samb1234
cant be first 2 due to position of the o, c clearly is wrong


Position of which O?
Reply 330
Original post by Don Pedro K.
Position of which O?


the O is always to the right of the carbonyl group - if it was a standard carboxylic or acyl chloride you would expect one to be to the right and to the left at the other end
Original post by samb1234
the O is always to the right of the carbonyl group - if it was a standard carboxylic or acyl chloride you would expect one to be to the right and to the left at the other end


Oh wow I just realised I read D as being a dicarboxylic acid when it is actually a hydroxycarboxylic acid (if that's what they're called) xD It makes sense now hahah.

Oh and also, you can't have two Os at either end of the repeat unit can you?
Reply 332
Original post by Don Pedro K.
Oh wow I just realised I read D as being a dicarboxylic acid when it is actually a hydroxycarboxylic acid (if that's what they're called) xD It makes sense now hahah.

Oh and also, you can't have two Os at either end of the repeat unit can you?


no you can't as far as i know
Original post by samb1234
no you can't as far as i know


Okay cool, thanks :smile:
Just checking in here, finished all the past papers now just trying to tide my revision over to the exams
Original post by Whizbox
Just checking in here, finished all the past papers now just trying to tide my revision over to the exams


how were your scores?
Reply 336
Original post by Whizbox
Just checking in here, finished all the past papers now just trying to tide my revision over to the exams



Wow already! Have you tried many IAL papers?
Picture9.jpg

is there a easy/ quick way of identifying which reactions are not bronsted lowry acid and bases ? please help x
Hi. Would anyone help out in 18(b)? I never really get this part right so a little elaborate explanation would be helpful. :smile: Thanks in Advance!

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