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One mole of nitrogen gas at 0.8atm takes...

One mole of nitrogen gas at `0.8atm` takes `38s` to diffuse through a pinhole, while `1 mol` of an unknown fluoride of xenon at `1.6 atm` takes `57 s` to diffuse through the same hole. Calculate the molecular formation of the compound.

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To solve the problem, we will use Graham's law of effusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. The formula can be expressed as: \[ \frac{R_1}{R_2} = \frac{P_1}{P_2} \cdot \frac{\sqrt{M_2}}{\sqrt{M_1}} \] Where: - \( R_1 \) and \( R_2 \) are the rates of diffusion of gas 1 and gas 2, respectively. ...
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One mole of nitrogen gas at 0.8atm takes 38s to diffuse through a pinhole, while 1 mol of an unknown fluoride of xenon at 1.6 atm takes 57 s to diffuse through the same hole. Calculate the value of n in the compound if formula is XeF_n .

One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57s to diffuse through the same hole. Calculate the molecular formula of the compound.

One mole of nitrogen gas at 0.8 atm takes 38 second of diffuse through a pin hole whereas one mole of an unknown compound of Xenon with fluorine at 1.6 atm takes 56.26 second to diffuse through at same hole, then the molecular formula of the compound is (Atomic mass of Xenon : 131.3 u)

One mole of He gas at 0.6 atm takes 40 s to diffuse through a pin hole. Calculate the time required for diffusion of half mole of CH_(4) gas at 1.2 atm through the same hole.

300 mL of nitrogen gas diffuse through a a porous membrane in 100 seconds. How long will it take for 400 mL of CO_(2) to diffuse through the same membrane under similar conditions of temperature and pressure ?

Under identical conditions of pressure and temperature, 4 L of gaseous moxture ( H_(2) and CH_(4) ) effuses through a hole in 5 min whereas 4 L of a gas X of molecular mass 36 takes to 10 min to effuse through the same hole. The mole ratio of H_(2):CH_(4) in the mixture is:

1.40L of an unknown gas requires 57 second to diffuse and the same volume of N_(2) gas takes 84 second to diffuse at the same temperature and pressure. What is the molecular mass of the unknown gas?

20 dm^3 of an unknown gas diffuse through a porous partition in 60 s, whereas 14.1 dm^3 of O_2 under similar conditions diffuse in 30 s. What is the molecular mass of the gas ?

A vessel with small opening contained equal volumes of oxygen and an unknown gas. Oxygen effused through the opening 1.8 times faster than the unknown gas. If the atomic mass of oxygen is 16, calculate the molecular mass of the unknown gas.

The composition of the equilibrium mixture for the equilibrium Cl_(2)hArr2Cl at 1470^(@)K , may be determined by the rate of diffusion of mixture through a pin hole. It is found that at 1470^(@)K , the mixture diffuses 1.16 times as fast as krypton (83.8) diffuses under the same conditions. Calculate the % degree of dissociation of Cl2 at equilibrium.

CENGAGE CHEMISTRY ENGLISH-STATES OF MATTER-Exercises (Ture False)
  1. One mole of nitrogen gas at 0.8atm takes 38s to diffuse through a pinh...

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  2. In the van der Waals equation (P + (n^(2)a)/(V^(2)))(V - nb) = nRT ...

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  3. Kinetic energy of a molecule is zero at 0^(@)C

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  4. Gas in a closed container will exert much higher pressure due to gravi...

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  5. The graph between PV vs P at constant temperature is linear parallel t...

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  6. Real gases show deviation from ideal behavior at low temperature and h...

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  7. All the molecules in a given sample of gas move with same speed.

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  8. Small value of a means, gas can be easily liqueifed.

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  9. Small value of a means, gas can be easily liqueifed.

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  10. Rate of diffusion is directly proportional to the square root of molec...

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  11. For ideal gases, Z = 1 at all temperature and pressure.

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  12. According to charles's law,

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  13. The pressure of moist gas is higher than pressure of dry gas.

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  14. Gases do not occupy volume and do not have force of attraction.

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  15. The van der Waal equation of gas is (P + (n^(2)a)/(V^(2))) (V - nb)...

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  16. Surface tension and surface energy have different dimensions.

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  17. The plot of PV vs P at particular temperature is called isovbar.

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  18. Give reasons for the following in one or two sentences. (a) A bottle...

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  19. Can a gas with a = 0 be liquefied?

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  20. The van der waals constants have same values for all the gases.

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  21. All the molecules in a given sample of gas move with same speed.

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