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An electric discharge is passed through a mixture containing 50cc of O_(2) and 50cc of H_(2). The volume of the gases formed at 110^(@)C will be |
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Answer» 50cc `(50)/(2)lt(50)/(1)` `implies H_(2)` is the L.R `2ml H_(2)RARR 2ml H_(2)O` `50ml H_(2)rarr50mlH_(2)O` VOLUME of `O_(2)` used = 25ml implies leftover 25ml `THEREFORE` TOTAL volume of gases = `50+25=75ml` |
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