third place

+no

Only 1/3 gas is left in the test tube.

Let the test tube volume be v.

The concentration is (v/22.4) * (2/3)/(2v/3) =1/22.4.

Molar/liter

4 NO2+O2+2h2o = = = 4

nitric acid

(O2 should be added to describe its slow introduction)

The concentration is (V/22.4)/(V)= 1/22.4.

Molar/liter