`44g` of a sample on complete combustion given `88g CO_2 and 36g` of `H_2 O.` The molecular formula of the compound may be :

Complete combustion of 0.858 g of compound X given 2.64g of CO_2 and 1.26 g of H_(2)O. The lowest molecular mass X can have:

0.246 g of an organic compound gave 0.198 g of CO_2 and 0.101 g of H_2O on complete combustion calculate the percentage composition of Carbon and Hydrogen.

Write the empirical formula of the compounds having the molecular formulas : H2O2

10g of CaCO3 on heating gave 4.4g of CO2 and X g of CaO. Applying law of conservation of mass , calculate the mass of CaO.

11.6g of an organic compound having formula C_n H_(2n+2) is burnt in excess of O_2(g) initially taken in a 22.41 litre steel vessel. Before reactio0n the gaseous mixture was at 273K with pressure reading 2 atm. Aftercomplete combustion and loss of considerable amount of heat, the mixture of product and excess of O_2 had a temperature of 546K and 4.6 atm pressure. The formula of organic compound is :

A gascous mixture contains 56 g N_2, 44 g CO_(2) and 16 g CH_(4) . The total pressure of mixture is 720 mm. What is partial pressure of CH_(4) ?

For the reaction O_(3(g)) + O_((g)) rarr 2O_(2(g)) , if the rate law expression is, rate = K[O_3] [O] , the molecularity and order of the reaction are respectively