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What weight of `AgCl` would be precipitated if `10 mL HCl` gas `12^(@)C` and `750 mm` pressure were passed into excess of silver nitrate? |
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Answer» Meq. Of `AgCl =` Meq. of `HCl =` milli-mole of `HCl_((g))` `= (PV)/(RT) xx 10^(3)` `(w)/(143.5) xx 1000 = (750)/(760) xx (10)/(1000) xx (10^(3))/(0.0821 xx 285) = 0.422` `:. w_(AgCl) = (0.422 xx 143.5)/(1000) = 0.0605 g` |
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