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A total charge of 97.4 kC passes through an electrolytic cell. Determine the quantity of substance produced in each of the foll...Asked by Sara
A total charge of 96.5 kC is passed through an electrolytic cell. Determine the quantity of substance produced in each of the following cases.
(a) the mass (in grams) of silver metal from a silver nitrate solution
(b) the volume (in liters at 273 K and 1.00 atm) of chlorine gas from a brine solution (concentrated aqueous sodium chloride solution)
(c) the mass of copper (in grams) from a copper(II) chloride solution
(a) the mass (in grams) of silver metal from a silver nitrate solution
(b) the volume (in liters at 273 K and 1.00 atm) of chlorine gas from a brine solution (concentrated aqueous sodium chloride solution)
(c) the mass of copper (in grams) from a copper(II) chloride solution
Answers
Answered by
DrBob222
a.
96,500 coulombs will deposit 1 equivalent of a metal. 1 equivalent of Ag is 107.9/1 electron = 107.9 grams. You have 94,500 C so.....
b.
2Cl^- ==> Cl2 + 2e
96,500 C will release molar mass Cl2/2 = ? grams and that will occupy 22.4 L at STP. You have 96,500 C so ......
c.
96,500 C will deposit 1 equivalent of Cu. 1 equivalent of Cu is 64.54/2 = ?g.
96,500 coulombs will deposit 1 equivalent of a metal. 1 equivalent of Ag is 107.9/1 electron = 107.9 grams. You have 94,500 C so.....
b.
2Cl^- ==> Cl2 + 2e
96,500 C will release molar mass Cl2/2 = ? grams and that will occupy 22.4 L at STP. You have 96,500 C so ......
c.
96,500 C will deposit 1 equivalent of Cu. 1 equivalent of Cu is 64.54/2 = ?g.
Answered by
amelia
I'm not getting the correct answers, sorry. Can you expand on the explanations?
Answered by
DrBob222
What are you supposed to use for C. Is it 96,485 or 96,500
Assuming 96,500, then you should deposit 107.9 g Ag which I would round to 108 for three significant figures.
For Cu I made a typo; it should be 63.55/2 and round that to 3 s.f.
For the Cl2, it should be 70.9/2 grams or 22.4L/2 if you are to use 22.4 for the molar volume.
Assuming 96,500, then you should deposit 107.9 g Ag which I would round to 108 for three significant figures.
For Cu I made a typo; it should be 63.55/2 and round that to 3 s.f.
For the Cl2, it should be 70.9/2 grams or 22.4L/2 if you are to use 22.4 for the molar volume.
Answered by
Akash
Define. The semipermiable membrane the freezing point of pure benzene is 278.4k calculate freezing point of the solution when 2 gms of a solute having molecular weight 100 is added to 100gms of benzene.