AQA CHEM5 A2 Chemistry - 19th June 2013

there was no rounding involved !

there was no rounding involved !

you get the answer when delta g = 0 as t = 462.7, thus at 463 it will not be feasible, hence it is only feasible when t is less than or equal to 462.7 not 463, so they used 462 as a whole number.

Ok, I'll try the calculation later

Ok, I'll try the calculation later

you get the answer when delta g = 0 as t = 462.7, thus at 463 it will not be feasible, hence it is only feasible when t is less than or equal to 462.7 not 463, so they used 462 as a whole number.

Could someone tell me what the Isa is out of? The total marks? The total marks for the paper? The total marks for the practical ? The total Psa marks?


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anyone here done the transition metal isa, im doing it this week.

that is OK ! another person on here tried it and I understand now!

Hello,
You know when we are using colorimeter and we make samples of known concentrations, why do we add the ligand (in the mark schemes they say 'to intensify the colour'. But I don;t totally understand this! THANKS !

Total=50 raw marks = 60 UMS
ISA practical=8 raw marks
ISA written exam=30 raw marks
PSA=12 raw marks

Ok.

Some ligands have a more intense colour so are used to replace the normal ligand e.g. water, so the colour change is more quantitatively identifiable.

This is my last exam.
When I put my pen down at 10.45am on 19th of June, I KNOW it will be the best feeling in the world

Hello, pls can you help me with understanding where I went wrong in this question :

Ammonium iron (II) sulphate crystals have the following formula:
(NH4)2SO4.FeSO4.nH2O. In an experiment to determine n, 8.492g of the salt were dissolved and made up to 250 cm3 of solution with distilled water and dilute sulphuric acid. A 25 cm3 portion of the solution was further acidified and titrated against potassium manganate (VII) solution of concentration 0.0150 moldm-3. A volume of 22.5 cm3 was required. Determine n.



This was my working out:

Moles of MnO4- ions= 0.0003375 moles
Moles of Fe2+ = Ans*5= 0.0016875 moles
Moles in 250 cm3 = 0.016875 moles
Moles = Mass / Mr so (8.492/0.016875)= 503.2..
Then, work out the Mr of the thing from the question and I get 302. Substract 302 from the Mr --> 201.2
The divide this by 18.
I got 11 as the nearest whole number.


But mark scheme is saying 12. What did i do wrong? thanks for all the help in advance!

This was a pretty simple mistake, but a really common one. Basically you've tripped up on the easy bit of getting the Mr of the crystal wrong, its 284.1 and this taken away from 503.23 = 219
219/18 = 12

This was a pretty simple mistake, but a really common one. Basically you've tripped up on the easy bit of getting the Mr of the crystal wrong, its 284.1 and this taken away from 503.23 = 219
219/18 = 12

This was a pretty simple mistake, but a really common one. Basically you've tripped up on the easy bit of getting the Mr of the crystal wrong, its 284.1 and this taken away from 503.23 = 219
219/18 = 12