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How much PCl_(5) must be added to a one litre vessel at 250^(@)C in order to obtain a 35 concentration of 0.1 mol of Cl_(2)? K_(c) for PCl_(5) hArr PCl_(3)+Cl_(2) is 0.0414 mol L^(-1) |
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Answer» `:. K_(c)=(0.1/1xx0.1/1)/((a-0.1)/1) ( :' "volume is" 1 L)` `0.0414=0.01/((a-0.1))` or `a=0.3415 "mol"` |
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