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