The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Aqa chem 4/ chem 5 june 2016 thread

Scroll to see replies

Original post by Anam
I put R as [cr(oh)6]3- and the reagent as NaOH, NH3 would give [cr(nh3)6]3+ wouldnt it?


Only in xs tho
http://www.thestudentroom.co.uk/showthread.php?t=4180942&p=66014002#post66014002

Unofficial ms
Original post by Suits101
Spot on.



Wasn't R Cr(H2O)3(OH)3? If so then yes.



That's the proper way to do it :smile:




I think i got the letter wrong i mean the solution formed after Cr (oh)3 (h2o)3
Original post by SuruthiG
I think i got the letter wrong i mean the solution formed after Cr (oh)3 (h2o)3


[Cr(OH)6]3-

Reagent is NaOH (excess)
Original post by potato cell
You shouldn't have divided by 2. The answer was -135 or something similar.


Thank you Mr Potato cell could star in Toy story 4
Original post by sim153
bumping this up because im wondering the same thing


I put no effect because it was just increasing pressure of one electrode :/ the number of electrons released from H2 would still be the same? And thus the flow of charge/current/emf would be the same?

I dont get why im wrong though pls someone explain
For the Cr3+ to Cr2+ soultion the reagents were Zn and HCl(or other acid) right? And the was blue.
Original post by Cadherin
Yay! :biggrin:


I agree
Was it Na2O + H2O = 2NaOH
or 2Na + 2H2o -> 2NaOH + H2
(edited 7 years ago)
For the equations for the catalyst in the contact process would you get the mark with the correct balanced equations but with equlibrium signs not single arrows as the original equation is an equlibrium equation? I mean it goes in both directions and the catalyst affects both routes of the equation equally. Thoughts as mark scheme just give single arrows but it doesn't say anywhere penalise equlibrium signs?
Reply 2910
Avatar for Fibsy
Fibsy
9
Original post by Jonnyss
Was it Na2O + H2O = 2NaOH
or 2Na + 2H2o -> 2NaOH + H2


I don't think the Q asked for the oxide

Posted from TSR Mobile
Original post by Kay Fearn
Lol has it ever been that high? The paper was easy but that's just how we found it.
A lot of people will much lower which always brings the grade boundaries down. I reckon 86 for an A* max


last year was 84 for A* and 76 for an A. The boundaries went up loads last year; there's no way they'll go up again that much. Id say 84/85 for A* and 79 for a A
What did everyone put for the oxidation product ?
For some reason on the first double page spread I identified the OXIDE they were discussing rather than the element! However I wrote the oxides of the correct elements e.g. I wrote P4O10 where you were meant to write P. Do you think this will be regarded a 'chemical error' and they won't mark my correct equations below, or do you think it'll only be a CE = O if you wrote the wrong element or the oxide of the wrong element??
Original post by Aethrell
Can someone please explain why increasing the pressure of oxygen would change that half cell equilibrium at all? I thought it was only if they were all gaseous?

So I put that there was no effect :frown:


Please can someone help me with this?
Original post by RME11
You dissolved the salt THEN added 250cm3 of water, you took 25cm3 of water from this - I almost made this mistake.


Ah fair enough then damn, thought about doing it your way but tried to be to clever by assuming the water was integral to the calculation
Original post by koolgurl14
What did everyone put for the oxidation product ?


oxygen 👀
Original post by FireBLue97
increase the surface area
then impurities poison the catalyst for 2 marks 2 differetn questions


You think i wouId get a mark for saying to increase the amount of active sites?
Any1 got an unofficial ms???
I am sure you can put either Phosphorus or Sulfur for the first question. Is there anyone who can confirm it?

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you