calculate the volume occupied by 0.23g of nitrogen dioxide at S.T.P.[N=14,O=16].
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Answered by
14
Mass of one mole of nitrogen dioxide = 14 + 16 + 16 = 46g
No. of moles in 0.23g on NO2 = 0.005
volume occupied by 1 mole = 22.4 L
volume occupied by 0.005 mole = 0.112 L OR 112 mL:
No. of moles in 0.23g on NO2 = 0.005
volume occupied by 1 mole = 22.4 L
volume occupied by 0.005 mole = 0.112 L OR 112 mL:
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Answered by
9
NO2=14+32
NO2=46
.'.NO2 OCCUPIES 46g AT 22.4DM³
NO2 OCCUPIES 0.23G ST X DM3
X=22.4*23/4600
X=0.112DM³
NO2=46
.'.NO2 OCCUPIES 46g AT 22.4DM³
NO2 OCCUPIES 0.23G ST X DM3
X=22.4*23/4600
X=0.112DM³
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