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Find out the volume of Cl_(2) at STP produced by the action of 100cm^(3) of 0.2 N HCl on excess of MnO_(2). |
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Answer» `"HCl PRESENT in "100cm^(3)" of 0.2 N HCl"=(0.2)/(1000)xx"100g eq.= 0.02g eq."xx"36.5 g = 0.73 g"` |
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