The formula of a gaseous hydrocarbon which requires 6 times of its own volume of `O_2` for complete oxidation and produces 4 times its own volume of `CO_2` is `C_xH_y`. The value of Y is

The vapour of an organic compound requires three times its own volume of oxygen for complete combustion and produces twice its own volume of carbon dioxide . Which of the following compounds would give these results ?

8c.c. of gaseous hydrocarbon requires 40c.c. of O_(2) for complete combustion. Identify hydrocarbon.

A gaseous paraffin requires five times its volume of oxygen for complete combustion. How many carbon atoms are present in a molecule of that paraffin?

A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of CO_(2) produced in the reaction is thrice the volume of hydrocarbon under the same conditions What is the ratio of molecular weight to emperical formula weight of the hydrocarbon?

A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of CO_(2) produced in the reaction is thrice the volume of hydrocarbon under the same conditions How many grams of water is produced by combustion of 0.1 mol of the givenn hydrocarbon?

10 mL of a certain hydrocarbon require 25 mL of oxygen for complete combustion and the volume CO_(2) of product is 20 mL. What is the formula of hydrocarbon ?

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O_(2) by volume for complete combustion. After combustion, the gases occupy 330 mL . Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

Twenty millilitres of a gaseous hydrocarbon required 400 ml of iar for complete cumbustion. The air contains 20% by explosion and cooling was found to be 380 ml. Q. Volume of O_2 used is:

At STP , for complete combustion of 3 g C_(2)H_(6) the required volume of O_(2) will be -

At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O_(2) for complete combustion, and 40 mL of CO_(2) is formed. The formula of the hydrocarbon is: