Asked by Joy

A current of 45A is passed through a solution of gold salt for 1 hour 45 minutes calculate (a) the mass of gold deposited (b) the numbers of moles of gold deposited (c) if the same current is used, find the time taken for 5.5g of gold to be deposited (Ah=197, 1F=96500C
Answered by DrBob222
I assume this is AuCl3 or similar with Au in the +3 state. Also I assume you made a typo with that Ah when you meant to type Au for gold.
1 hr 45 minutes = 60 min + 45 min = 105 min
Coulombs = amperes x seconds = 45 x (105 min x 60 sec/min) = 283,500 C
96,500 C will deposit 197/3 = 65.7 g Au so
65.7 g Au x (283,500/96,500) = ? g Au deposited.
mols Au = grams Au/atomic mass Au = ?
(c) coulombs needed is
96,500 C will deposit 65.7 g Au. To deposit 4.5 g Au will require
96,500 C x 5.5 g/65.7 g = 8,078 C needed, then
C = amperes x seconds.
8078 C = 45 A x seconds. Solve for seconds.
Post your work if you get stuck.
Answered by Takela
0.334G
Answered by Dax
Not exactly what I got in my solvings
Answered by Benjamin Harris
2700sec
Answered by Benjamin Harris
127575
Answered by Favour
578.8 of Au
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