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From 200mg of CO_(2),10^(21) molecuels are removed. How many moles of CO_(2) are left? |
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Answer» Solution :Gram-molecular mass of `CO_(2)=44g` Mass of `10^(21)` molecuels of `CO_(2)=(44)/(6.02xx10^(23))xx10^(21)=0.073g` Mass of `CO_(2)` left `=(0.2-0.073)=0.127g` NUMBER of MOLES of `CO_(2)` left `=(0.127)/(44)=2.88xx10^(-3)` |
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