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Chemistry unit 6 June 2016

Can we please discuss upcoming chem unit 6 here? 'cuz I couldn't find a thread for that!

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Hii. Yess 😊

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Yeah sure!

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Reply 3
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baebae
7
does anyone have any notes omg :O
guys please help me lol
Reply 4
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Aimen.
OP
7
I found this on the internet! Seems kinda helpful but we actually need to revise experiments from AS level aswell
And yea my chem teacher said they'll probably give us boiling point apparatus this time!
a2organicreactions-130110030254-phpapp01.doc59.4KB
Reply 5
Avatar for Ayman!
Ayman!
15
Checking in :hello:
Original post by Aimen.
I found this on the internet! Seems kinda helpful but we actually need to revise experiments from AS level aswell
And yea my chem teacher said they'll probably give us boiling point apparatus this time!


What is a GRIGNARD and should we know it? :s-smilie:
Do any of you guys have the George facer Edexcel As / A2 chemistry unit 3 & 6: Chemistry Laboratory Skills in pdf?
Original post by Ayman!
Checking in :hello:


Welcome on board :h:

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Reply 9
Avatar for Ayman!
Ayman!
15
Could someone explain 3(b)(iv) to me on this paper? I just don't get how to do it. @samb1234, mate mind taking a look? I know you don't do U6 but this is U5 redox really
Original post by Ayman!
Could someone explain 3(b)(iv) to me on this paper? I just don't get how to do it. @samb1234, mate mind taking a look? I know you don't do U6 but this is U5 redox really


Of course. I have worked through it now and checked the answer is right, where have you got up to?
Reply 11
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Ayman!
15
Original post by samb1234
Of course. I have worked through it now and checked the answer is right, where have you got up to?


Okay, so here's my train of thought - the oxidation number of V in VO3- is +5. There are 2.5 x 10-3 mol of VO3- ions in the initial solution, so same moles of V5+. Now we know that moles of MnO4- 1.5 x 10-3 and MnO4- : e = 1:5, so moles of e is 5 x 1.5 x 10-3 = 7.5 x 10-3 mol. Now is it just as simple as comparing the molar ratio of our initial ion and the electrons lost? I know by observation that the answer is most likely V2+.

On another note, I'm surprised with the boundaries - 35/50 for an A*! I thought I did quite poorly because I got 43.
Original post by Ayman!
Okay, so here's my train of thought - the oxidation number of V in VO3- is +5. There are 2.5 x 10-3 mol of VO3- ions in the initial solution, so same moles of V5+. Now we know that moles of MnO4- 1.5 x 10-3 and MnO4- : e = 1:5, so moles of e is 5 x 1.5 x 10-3 = 7.5 x 10-3 mol. Now is it just as simple as comparing the molar ratio of our initial ion and the electrons lost? I know by observation that the answer is most likely V2+.

On another note, I'm surprised with the boundaries - 35/50 for an A*! I thought I did quite poorly because I got 43.


That's what I did. The managanate needs 7.5x1-^-3 moles of electrons, and there are 2.5x10^-3 moles of vanadium so each individual vanadium must donate (7.5x10^-3 /2.5x10^-3) 3 electrons. Therefore if the final charge is +5, and each vanadium has lost 3 e- as part of the process, than it must have been +2 before
Original post by Ayman!
Okay, so here's my train of thought - the oxidation number of V in VO3- is +5. There are 2.5 x 10-3 mol of VO3- ions in the initial solution, so same moles of V5+. Now we know that moles of MnO4- 1.5 x 10-3 and MnO4- : e = 1:5, so moles of e is 5 x 1.5 x 10-3 = 7.5 x 10-3 mol. Now is it just as simple as comparing the molar ratio of our initial ion and the electrons lost? I know by observation that the answer is most likely V2+.

On another note, I'm surprised with the boundaries - 35/50 for an A*! I thought I did quite poorly because I got 43.


and those boundaries are almost physics unit 6 level haha
A question, do we have to know about Vanadium for Unit 6? (Just for confirmation I'm asking)
Reply 15
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Ayman!
15
Original post by samb1234
That's what I did. The managanate needs 7.5x1-^-3 moles of electrons, and there are 2.5x10^-3 moles of vanadium so each individual vanadium must donate (7.5x10^-3 /2.5x10^-3) 3 electrons. Therefore if the final charge is +5, and each vanadium has lost 3 e- as part of the process, than it must have been +2 before


Thanks a lot man! I understand it now. I was trying to find an alternative using something other than ratios. Guess it isn't meant to be. :lol:
Original post by Ayman!
Thanks a lot man! I understand it now. I was trying to find an alternative using something other than ratios. Guess it isn't meant to be. :lol:


you could do a similar thing just by using the ratio of mno4 to v
Original post by Ayman!
Okay, so here's my train of thought - the oxidation number of V in VO3- is +5. There are 2.5 x 10-3 mol of VO3- ions in the initial solution, so same moles of V5+. Now we know that moles of MnO4- 1.5 x 10-3 and MnO4- : e = 1:5, so moles of e is 5 x 1.5 x 10-3 = 7.5 x 10-3 mol. Now is it just as simple as comparing the molar ratio of our initial ion and the electrons lost? I know by observation that the answer is most likely V2+.

On another note, I'm surprised with the boundaries - 35/50 for an A*! I thought I did quite poorly because I got 43.


I'm a bit confused, why did you multiply 1.5 x 10-3 by 5 ?
Reply 18
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Ayman!
15
Original post by Adorable98
I'm a bit confused, why did you multiply 1.5 x 10-3 by 5 ?


Mole ratios?
Original post by Ayman!
Mole ratios?


I see thnx, do you know how to answer 3 (d)

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