Asked by precious
A current of 45A is passing through a solution of gold salt for 1:45 minutes
a)calculate the mass of AU deposited
b )number of mass deposited
c) if the same current is use find the time taken for 5.5 g of the AU deposited
a)calculate the mass of AU deposited
b )number of mass deposited
c) if the same current is use find the time taken for 5.5 g of the AU deposited
Answers
Answered by
DrBob222
coulombs = amperes x seconds or
coulombs = 45 x 1.75 hrs x (60 min/hr) x (60 sec/min) = 283,500
96,485 coulombs will deposit about 197/3 g Au But you can use closer numbers than that.
So g Au deposited = (197/3) x (283,500/96,485) = ? g Au.
I suppose part b is mols and not mass so g Au/molar mass = mols Au.
C. Follow the steps in part A to do C and find the time.
Post your work if you get stuck.
coulombs = 45 x 1.75 hrs x (60 min/hr) x (60 sec/min) = 283,500
96,485 coulombs will deposit about 197/3 g Au But you can use closer numbers than that.
So g Au deposited = (197/3) x (283,500/96,485) = ? g Au.
I suppose part b is mols and not mass so g Au/molar mass = mols Au.
C. Follow the steps in part A to do C and find the time.
Post your work if you get stuck.
Answered by
Patricia
Please where did you get the 96,485 from.
Answered by
Anonymous
How is it 96485 and where did the 197/3 come from
Answered by
Esther
what the correct answer?
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