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When water is electrolysed, hydrogen and oxygen gas are produced. If 1.008 g of `H_(2)` is liberated at the cathode. What mass of `O_(2)` is formed at the anode? |
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Answer» `because(w_(1))/(w_(2))=(E_(1))/(E_(2)),` where `E_(1)` and `E_(2)` are eq.wt. of `H_(2)` and `O_(2)` respectively `therefore(1.008)/(w_(2))=(1.008)/(8)rArrw_(2)=8g` |
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