[" The relative density of a mixture of "CO_(2)(g)&H_(2)O(g)" w.r.t.gaseous hydrogen atoms is "30." Then "],[" Calculate mol percentage of "CO_(2)]

The relative density of a mixture of CO_(2)(g) and H_(2)O(g) w.r.t. gaseous hydrogen atoms is 30. The mol% of the CO_(2) is

A gaseous mixture contains CO_(2) (g) and N_(2)O(g) in 2:5 ratio by mass. The ratio of the number of molecules of CO_(2)(g) and N_(2)O(g) is:

A gaseous mixture contains CO_(2) (g) and N_(2)O (g) in a 2 : 5 ratio by mass. The ratio of the number of molecules of CO_(2) (g) and N_(2)O is

2 g of a mixture of CO and CO_2 on reaction with excess I_(2)O_(5) produced 2.54g of I_2 . What will be the mass percentage of CO_(2) in the original mixture ?

A gaseous mixture of H_(2) and CO_(2) gas contains 88% by mass of CO_(2) . The vapour density of the mixture is :

A gaseous mixture of H_(2) and CO_(2) gas contains 66 mass % of CO_(2) . The vapour density of the mixture is :

0.765 g of an acid gives 0.535 g of CO_(2) and 0138 g of H_(2)O . Then the ratio of the percentage of carbon and hydrogen is