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Calculate the amount of H_(2)O_(2) present in 10 mL of 25 volume H_(2)O_(2) solution. |
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Answer» Solution :10 mL of 25 volume `H_(2)O_(2)` LIBERATE `O_(2)=10xx25=250` mL at NTP Now,`""underset(68g)(2H_(2)O_(2))to 2H_(2)O+underset(22400 mL at NTP)(O_(2))` `THEREFORE` Amount of `H_(2)O_(2)` that will liberate 250 mL of `O_(2)` at NTP `=(68xx250)/(22400)=0.759 g` |
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