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`10 mL` of a given solution of `H_(2)O_(2)` contains `0.91 g` of `H_(2)O_(2)`. Express its strength in volume. |
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Answer» `underset(68 g)(2H_(2)O_(2))to2H_(2)O+underset("22400 mL at NTP")(O_(2))` 68 g of `H_(2)O_(2)` produces 22400 mL of `O_(2)` at NTP. `:.` 0.91 g of `H_(2)O_(2)` produces `(22400xx0.91)/68`=300 mL of `O_(2)` at NTP Volume strength =`300/10=30` |
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