s.v
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Hey guys,
I don't understand question 6(e) and question 8 (c)
here's the paper,
http://www.ocr.org.uk/Images/144762-...d-elements.pdf

Could someone please explain it in the easiest way possible???
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Borek
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XSO4 - what is the valence of the X? Apply Faraday's law of electrolysis to find the charge, then to find the equivalent mass of X.

The other question is about finding moles and number of electrons exchanged. You need to write half reactions first (the one for vanadium will have one unknown - number of electrons).
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s.v
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Im sorry I dont understand that method.. could you explain how you would answer each question plz?

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Billpaid
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(Original post by s.v)
Hey guys,
I don't understand question 6(e) and question 8 (c)
here's the paper,
http://www.ocr.org.uk/Images/144762-...d-elements.pdf

Could someone please explain it in the easiest way possible???
Cr ions become Cr metal by gaining 3 electrons per chromium.
X must be a metal with 2+ ions because it is bezzy mates with a sulphate ion.These become metal ions by losing two electrons.


So when two chromium ions turn into chromium metal, they needed to react with three X atoms to provide the necessary electrons.


So the molar ratio is 2 of Cr to 3 of X ( or 1 to 1.5)


How to answer question?
1. Work out moles of chromium using mass divided by Ar of Chromium (52)
2. Multiply by 1.5 to get moles of X
3. Use Mr = m/n to get Mr of X. (You were given mass of X)
4. Go to Periodic Table and see which group 2 metal it is!


8c is simply because both species v2+ and the manganate are the same colour , ( violet).
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s.v
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Thanks for that
could you help with the part 2 of question 8c please.. I didnt make it clear.. it was that part which I found confusing..

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(Original post by s.v)
Thanks for that
could you help with the part 2 of question 8c please.. I didnt make it clear.. it was that part which I found confusing..

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1. Work out the moles of manganate using vol and conc.

2. work out moles of vanadium metal using mass and Ar of vanadium. Then
divide this by 5 as they only used 10cm3, ( a fifth of the sample)

3. You then need the ratio of Vanadium to manganate so do answer to part 2 divided by answer to part 1. This comes out as 1.67 ( meaning 1.67:1 or 5:3)

Explainy bit, read slowly. This means every 5 Vanadiums react with 3 manganates. As every manganate uses 5 electrons ( see half equation) that means 15 electrons are donated from the 5 vanadiums ( 3 from each). So if each vanadium has lost 3 electrons to become V5+ they must have been V2+ originally! n=2

.... and breathe.

5V2+(aq) + 3MnO4–(aq) + 3H2O(l) makes 5VO3–(aq) + 3Mn2+(aq) + 6H+(aq)

Not a pleasant question when under pressure in an exam but always use the technique of trying to use the numbers given to do whatever you can with them to at least get some marks. Steps 1, 2 and 3 would get you 5 of the 7 marks.
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