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precipitation question help please? thanks

exam today :frown:

why C
? thanks

2002 29.PNG2002 29.PNG
Original post by ah4p
exam today :frown:

why C
? thanks

2002 29.PNG2002 29.PNG


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
Original post by ah4p
exam today :frown:

why C
? thanks

2002 29.PNG2002 29.PNG

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 :smile:
Reply 3
Avatar for ah4p
ah4p
OP
16
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 :smile:

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

would D be a 1:1 reaction?


yes

Na2SO4 + BaCl2 --> BaSO4 + 2NaCl
Reply 5
Avatar for ah4p
ah4p
OP
16
Original post by charco
yes

Na2SO4 + BaCl2 --> BaSO4 + 2NaCl


oops thanks :smile:

do you know this one
2002 8.PNG

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

thnx for any help
exam today
Original post by ah4p
oops thanks :smile:

do you know this one
2002 8.PNG

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

thnx for any help
exam today


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.
Reply 7
Avatar for ah4p
ah4p
OP
16
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 :smile:

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

thanks for all your help :biggrin:
Original post by ah4p
thank you :smile:

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

thanks for all your help :biggrin:


Good luck with your results ...

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