Question

A mixture of 2 g h2, 1g n2 .82 g ar find volume of container

Answer 1

(0.200 g H2) / (2.01588 g H2/mol) = 0.099212 mol H2

(1.00 g N2) / (28.01344 g N2/mol) = 0.035697 mol N2

(0.820 g Ar) / ( 39.948 g Ar/mol) = 0.020527 mol Ar

(0.099212 mol + 0.035697 mol + 0.020527 mol) x (22.414 L/mol) = 3.48 L

hope it will help you..

Answer 2

Explanation:

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