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Calculate the current ( in `Ma)` required to deposit `0.195 g` of platinum metal in `5.0` hours from a solution of `PtCl_(6)^(2-) : (` Atomic weight `: pt = 195 )`A. 310B. 31C. 21.44D. 5.26 |
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Answer» Correct Answer - 3 `(W)/(E)=(Ixxt)/(96500),(0.195)/(195)xx4=(5xx60xx60xxI)/(96500)` `I=21.44` |
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