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# Chemistry F334 Calculations watch

1. Please can someone explain to me how question 2.) C.) ii.) works.

Much appreciated!
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2. F334 JAN 2010.pdf (196.2 KB, 99 views)
3. (Original post by thelion0)
Please can someone explain to me how question 2.) C.) ii.) works.

Much appreciated!
It's just the difference between the electrode potentials of the two half cells
4. (Original post by charco)
It's just the difference between the electrode potentials of the two half cells
5. (Original post by thelion0)
ahhh yes, so I did

2Cii) Work out the moles of thiosulphate: 0.0205 * 0.2

Using the equations:

mol thiosulphate/2 = moles of iodine
moles of iodine * 2 = moles of copper

Hence moles of thiosulphate = moles of copper ions.

To find the total moles of copper ions multiply by 10 to scale up to the 250ml solution.

Then moles of copper ==> mass of copper

Then percentage = 100 * mass of copper/mass of sample
6. (Original post by charco)
ahhh yes, so I did

2Cii) Work out the moles of thiosulphate: 0.0205 * 0.2

Using the equations:

mol thiosulphate/2 = moles of iodine
moles of iodine * 2 = moles of copper

Hence moles of thiosulphate = moles of copper ions.

To find the total moles of copper ions multiply by 10 to scale up to the 250ml solution.

Then moles of copper ==> mass of copper

Then percentage = 100 * mass of copper/mass of sample
haha! thank you very much I understand it now. its the calculations that get me each time..

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