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The volume of 10 vol `H_(2)O_(2)` required to liberate 500 `cm^(3)` of `O_(2)` at STPA. 50 mlB. `5*0 ml`C. 15 mlD. 100 ml. |
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Answer» Correct Answer - A 10 ml of `O_(2)` is obtained at STP from `H_(2)O_(2)=1` ml 500 ml of `O_(2)` is obtained at STP from `H_(2)O_(2)=1//10xx500=50` mL. |
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