Hydrogen effusion rates

We can use Graham s law to determine the rate of effusion of an unknown gas knowing the rate of a known one or we can use it to determine the molecular mass of an unknown gas. For example, suppose you wanted to find the molar mass of an unknown gas. You measure its rate of effusion versus a known gas, H2. The rate of hydrogen effusion was 3.728 mL/s, while the rate of the unknown gas was 1.000 mL/s. The molar mass of H2 is 2.016 g/mol. Substituting into the Graham s law equation gives ... 

Hydrogen gas would effuse through a pinhole 3.728 times as fast as nitrogen gas. The answer is reasonable, since the lower the molecular mass, the faster the gas is moving. Sometimes we measure the effusion rates of a known gas and an unknown gas, and use Graham s law to calculate the molecular mass of the unknown gas. 

Calculate the ratio of effusion rates of oxygen (02) to hydrogen (H2). 

The effusion rate of hydrogen sulfide (H2S) is 1.50 mol/s. Another gas under similar conditions effuses at a rate of 1.25 mol/s. What is the molar mass of the second gas ... 

C is correct From Graham s law we know that the effusion rate for hydrogen is four times that of oxygen. 

Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF0), a gas used in the enrichment process to produce fuel for nuclear reactors (see Fig. 5.23). 

What is the ratio of the rate of effusion of the most abundant gas, nitrogen, to the lightest gas, hydrogen ... 

Balloon B could be filled with hydrogen since air would effuse through the balloon membrane holes at the same rate that hydrogen would effuse out as long as the two gases were at the same temperature. Balloon A could be filled with sulfur hexafluoride because it is heavier than air and would cause the gas to escape through membrane holes through the process of osmosis. 

The answer you get to this problem is 3.16. Putting this number over 1 can help you understand your answer. The ratio 3.16/1 means is that for every 3.16 mol of hydrogen gas that effuses, 1.00 mol of neon gas will effuse. This ratio is designed to compare rates. So hydrogen gas effuses 3.16 times faster than neon. 

Hydrogen, H2. The question states that the ratio of the rates is 4.0. Oxygen gas is a diatomic element, so it s written as O2 and has a molar mass of 32.00 g/mol. Substitute these known values into Graham s law to determine the molar mass of the unknown gas. The problem states that the unknown gas effuses at a rate 4.0 times faster than oxygen, so put the unknown gas over oxygen for the ratio. (In short, the unknown gas is A, and the oxygen is B). 

Hydrogen chloride effuses through a hole (under prescribed conditions of temperature and pressure) at the rate of 2 70 ml/mm At what velocity will helium effuse through the same hole under the same conditions7... 

D) Wrong Graham s law of effusion states that the rates of effusion are inversely related to the square root of the molar mass. This means that smaller particles will effuse more quickly than larger particles. All three balloons will get smaller at different rates, which means they will be different sizes on the next day (hydrogen will be smallest, followed by helium, and then sulfur hexafluoride). 

Figure 3. Molecular hydrogen flux vs temperature for sputter deposited ZnO thin films. The experiments were performed with a heating rate of 20 K/min. For clarity only three effusion spectra are plotted.

Accordingly the kinetic theory requires that the rate of effusion of a gas through a small hole be inversely proportional to the square root of its molecular weight. This law was discovered experimentally before the development of the kinetic theory— it was observed that hydrogen diffuses through a pcwous plate four times as rapidly as oxygen. 

Would deuterium (atomic weight 2.0147) effuse through a porous plate more rapidly or less rapidly than hydrogen Calculate the relative rates of e%ision of the two molecules. What would be the relative rate of effusion of a molecule made of one light hydrogen atom and one deuteriun) atom ... 

Compare the rate of effusion of carbon dioxide with that if hydrogen chloride at the same temperature and pressure. 

You are given the molar masses for ammonia and hydrogen chloride. To find the ratio of the diffusion rates for ammonia and hydrogen chloride, use the equation for Graham s law of effusion. 

A tiny hole forms in the side of the containei allowing a small amount of gas to effuse out. Analysis of the escaping gases shows that the rate of escape of the hydrogen is 33 times the rate of escape of the methanol. Calculate the equilibrium constant for the preceding reaction at 250°C. 

Hydrogen has two naturally occurring isotopes, H and H. Chlorine also has two naturally occurring isotopes, C1 and CL Thus, hydrogen chloride gas consists of four distinct types of molecules W Cl, W Cl, H Cl, and U CL Place these four molecules in order of increasing rate of effusion. 

The hydrogen fountain, shown in Figure 5.29, is dependent on the differences in rates of effusion of gases. (Can you explain why a heUum-filled balloon loses pressure after... 

If it takes 4.67 times as long for a particular gas to effuse as it takes hydrogen under the same conditions, what is the molecular weight of the gas (Note that the rate of effusion is inversely proportional to the time it takes for a gas to effuse.)... 

Hydrogen has two stable isotopes, H and H, with atomic masses of 1.0078 amu and 2.0141 amu, respectively. Ordinary hydrogen gas, H2, is a mixture consisting mostly of H2 and H H. Calculate the ratio of rates of effusion of H2 and H H under the same conditions. 

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