Active mass of 5.6 lit N_2 at STP

1.	Active mass of 5.6 lit N_2 at STP
Answer» 22.4M 0.25M `1/(22.4)M` 4M Solution :ACTIVE mass = `(N)/(V//"(LIT)")=((5.6)/(22.4))/(5.6)=(1)/(22.4)M`

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