precipitation question help please? thanks

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ah4p
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exam today

why C
? thanks Image

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charco
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(Original post by ah4p)
exam today

why C
? thanks Image

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calculate moles of barium sulphate 2.334/233 = 0.01

hence 0.01 mol of metal sulphate was added (they are all 1:1 reactions)

0.01 mol of magnesium sulphate = 1.20 g
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BioStudentx
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(Original post by ah4p)
exam today

why C
? thanks Image

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Work out number of moles of barium sulphate by doing 2.334/(16+16+16+16+32)=0.01
You know there is 1:1 ratio and there is 1.204 grams of the Metal Sulphate so you do 1.204/0.01=120.4
120.4 is the mR of the whole metal sulphate so you minus the sulphate
120.4-16-16-16-16-32.065=24.4

Hope this helped
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ah4p
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(Original post by charco)
calculate moles of barium sulphate 2.334/233 = 0.01

hence 0.01 mol of metal sulphate was added (they are all 1:1 reactions)

0.01 mol of magnesium sulphate = 1.20 g
ahh thanks so much

would D be a 1:1 reaction?
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charco
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(Original post by ah4p)
ahh thanks so much

would D be a 1:1 reaction?
yes

Na2SO4 + BaCl2 --> BaSO4 + 2NaCl
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ah4p
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(Original post by charco)
yes

Na2SO4 + BaCl2 --> BaSO4 + 2NaCl
oops thanks

do you know this one
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can anyone explain the answer is -0.81V

i thought xidation occurs at anode (positive electrodfe)

and voltage produced = reduction - oxidation
1.03 = r - 0.49
r = 1.52V

but this is wrong Image

thnx for any help
exam today Image
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charco
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(Original post by ah4p)
oops thanks

do you know this one
ImageName:  2002 8.PNG
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can anyone explain the answer is -0.81V

i thought xidation occurs at anode (positive electrodfe)

and voltage produced = reduction - oxidation
1.03 = r - 0.49
r = 1.52V

but this is wrong Image

thnx for any help
exam today Image
You must be careful with the two types of cell:

In all cells oxidation occurs at the anode and reduction at the cathode (redcat)

BUT

The cathode is positive in an electrochemical cell and negative in an electrolytic cell.

In this case you have an electrochemical cell (generates electricity). Hence the negative side produces electrons (which flow around the external circuit to the positive electrode) and oxidation occurs, i.e. it is the anode.
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ah4p
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(Original post by charco)
You must be careful with the two types of cell:

In all cells oxidation occurs at the anode and reduction at the cathode (redcat)

BUT

The cathode is positive in an electrochemical cell and negative in an electrolytic cell.

In this case you have an electrochemical cell (generates electricity). Hence the negative side produces electrons (which flow around the external circuit to the positive electrode) and oxidation occurs, i.e. it is the anode.
thank you

something like this came up in my exam today and also the question about entropy of a perfect crystal being 0 at 0K came up too

thanks for all your help
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charco
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(Original post by ah4p)
thank you

something like this came up in my exam today and also the question about entropy of a perfect crystal being 0 at 0K came up too

thanks for all your help
Good luck with your results ...
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