How
Much Electric CURRENT WILL BE PASSED IN KI
Solution. TO LIBERATE 10 gm I² in one hour. .
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=2.11ampere
Explanation:
127 g iodine (1 g equivalent ) is liberated by =96.500
coulomb
∴ 10 g of iodine is liberated by 12796,500×10Coulomb
Let the current strength be I
Time in Second =1×60×60
Quantity of electricity Q=I×timeinseconds
Current strength I=tQ=127×60×6096,500×10
=2.11ampere
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