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If a nitrogen gas sample occupies a volume of 2.55 l at 22.0 °c and 0.860 atm, what is the mass of the n2 gas? (r = 0.0821 atm•l/mol•k

Answer :

Kalahira
number of moles of gas = PV / RT = 0.860 x 2.55 / 0.821 x (22+273)
 
= 0.009054687 moles
 
Molar mass of N2 = 28g so we have 0.009054687 moles x 28g / mol = 0.253 g of N2

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