From a certain apparatus, the diffusion rate of hydrogen has an average value of `28.7 cm^(3) s^(-1)`. The diffusion of another gas under the same conditions is measured to have an average rate of `7.2 cm^(3) s^(-1)`. Identify the gas. [Hint : Use Graham's law of diffusion: `R_(1)//R_(2) = (M_(2) //M_(1) )^(1//2)`, where `R_(1), R_(2)`, are diffusion rates of gases 1 and 2, and `M_(1)` and `M_(2)` their respective molecular masses. The law is a simple consequence of kinetic theory.]
From a certain appartus the diffusion rate of hydrogen has an average value of 28.7 cm^2.s^-1 The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm^3.s^-1 Identify the gas.
For the diffusion of a gas at pressure P, the rate of diffusion is expressed by
Under the same conditions of temperature and pressure, the rate of diffusion of hydrogen gas is four times that of oxygen gas-explain.
Why are the rates of diffusion of N_(2)O and CO_(2) gases the same under identical set of conditions?
Under the same conditions, a gas diffuses sqrt(2) times as fast as SO_(2) gas. Find the molecular mass of the gas.
Find the ratio of molecular weight of two gaes when rate of diffusion is 16 times more for the 2nd one. State Grahams law diffusion.
The rate of diffusion of a gas is 2.92 times that of NH_(3) gas. Determine the molecular weight of that gas.
If a gas diffuses at the rate of one-half as fast as O_2 , find the molecular mass of the gas.
Prove: r_1/r_2 = sqrt(rho_2/rho_1) for Graham's law of diffusion.
If a gas diffuses at the rate of one quarter as fast as N_2 . Find the molecular mass.
