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Question Number 62923 by peter frank last updated on 26/Jun/19
Answered by peter frank last updated on 27/Jun/19
Q=8Ct=20snoofionZn=?Zn2++2e→Zn1F=96500C2F=xx=2×96500Cfrom1mole=6.02×1023particle(ions)2×96500C→6.02×1023ions8C→yy=2.5×1019ionsnoofionofZn=2.5×1019ionI=Qt=820=2.4A
1)fromreactionZnZn2++2e→ZnQ=It=nen=Qe=81.6×10−19=2.5×1019
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