TexCalculate the weight of CO havin...
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TexCalculate the weight of CO having the same number of oxygen atoms as are present in 22g of carbondioxide.

JEE/Engineering Exams
Chemistry
Solution
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( 22 mathrm{m} ) of ( mathrm{co}_{2} ) ( frac{22}{44} times 2 times N_{A}=n_{0} cdot 900 ) moles of 0 : ( Rightarrow frac{22}{44} x^{2}=1 ) mol 80 molus ( y(0=1 ) ( I=frac{omega t}{28} ) ( (x=28 g m) ) gis it Right??
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