If m is the mass of silver deposited at the cathode by 2A current flowing for 25min through a silver voltameter, then the mass deposited by 1.5A current flowing for 600s is _________.
m
3m
0.3m
None
A
0.3m
B
m
C
3m
D
None
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Solution
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m′m=I′t′It=1.5×6002×25×60=9003000=310=0.3
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