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**Graham's Law** states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

**Effusion** involves a gas escaping a container through an opening.

Concept #1: Understanding Graham's Law of Effusion.

**Transcript**

Welcome back guys. In this new video, we'll get to look at the effusion rates of different gases.

Tied with effusion is basically the concept of velocity and speed. Remember we use root mean square equations when we're dealing with the speed of one gas. We use effusion when we're comparing the speeds or velocities of multiple gases. That's when we use this concept.

Here we're going to say Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. What this is really saying is that the speed or rate at which a gas moves, so effusion is just seen as speed, the speed of a gas is the opposite of its weight.

What that means is if my speed is high, it's because the mass of my gas is low. Basically, the less you weigh as a gas, the faster you move. That's what effusion is trying to explain. Effusion looks and compares multiple gases to each other. Gas A moves faster than Gas B. Why is that? Well, Gas A must weigh less. That's why it's moving faster.

We're going to say when we're comparing the rates of two gases, we use this form of Graham's Law of Effusion. Now pay close attention. Basically, we're going to say the rate of Gas One is equal to its rate divided by the rate of Gas Two. We're going to say that equals the square root – look if Gas One's rate is on the top, then it's mass is on the bottom because remember Graham's Law says that rate and mass are inversely proportional. They're on opposite levels. Here if the rate of Gas Two is on the bottom, then its mass is on the top. We're going to have to take this approach to answer some of these questions.

When comparing the **rate or speed of two gases** then we must use **Graham’s Law of Effusion**.

Example #1: Calculate the ratio of the effusion rates of helium and methane (CH_{4}).

The gas named first will represent gas 1 and the second gas will represent gas 2. Once that is established we simply use Graham’s Law of Effusion.

Example #2: Rank the following in order of increasing rate of effusion:

O_{2} AlF_{5} CO_{2} Xe

0 of 3 completed

At a given temperature and pressure, a sample of Gas A is observed to diffuse twice as fast as a sample of a different gas, B. Based on this:
A. The molar mass of A is one fourth that of B
B. The molar mass of A is one half that of B
C. The molar mass of A is 0.707 times that of B
D. The molar mass of A is 1.414 times that of B
E. The molar mass of A is four times that of B

How many times faster will He gas pass through a pin hole than SO2 gas?
a) 4
b) 1/4
c) 16
d) 1/16
e) none of the given answers

An unknown gas Q diffuses at a RATE 1.65 times faster than that of propane, C 3H8. Which of these are most likely to be Q?
a. He
b. Ne
c. H2O
d. CH4
e. O2

The mass particle is 10% larger than that of particle B. If particle B travels 100 m during effusion, how far would particle A travel during effusion?
a) 90 m
b) 95 m
c) 100 m
d) 105 m
e) 110 m

Exactly 10 mL, of gas A (MM = 15.0 g/mol) effuse through an opening in 2.0 sec. Exactly 10 mL of gas B effuse through the same opening in 8.0 sec. What is the molar mass of gas B?
a) 60 g/mol
b) 480 g/mol
c) 240 g/mol
d) 120 g/mol

A sample of N2 gas is contaminated with a gas (A) of unknown molar mass. The partial pressure of each gas is known to be 200 Torr at 25°C. The gases are allowed to effuse through a pinhole, and it is found that gas A escapes at three times the rate of N2. The molar mass of gas A is:
a) 3.11
b) 252
c) 84.0
d) 9.33
e) none of these

Second-hand smoke consist of many molecules including nicotine. How much faster or slower does the nicotine diffuse, compared to carbon dioxide? (MW nicotine = 160g/mol)
1. About half as fast
2. As gases, the two molecules diffuse at the same rate
3. About 4 times faster
4. About a fourth as fast
5. About twice as fast

The gas for which instance will effuse faster?

Oxygen gas effuses at a rate that is 1.49 times that of an unknown gas (at the same temperature). What is the identity of the unknown gas?
A) Ar
B) HBr
C) SO2
D) Cl2
E) He

The mass of particle A is 10% larger than that of particle B. If particle B travels 100m during effusion, how far would particle A travel during effusion?
a. 90 m
b. 95 m
c. 100 m
d. 105 m
e. 110 m

At a particular temperature and pressure, it takes 4.55 min for a 1.5 L sample of He to effuse through a porous membrane. How long would it take for a 1.5 L of F2 to effuse under the same conditions?
a. 18 min
b. 14 min
c. 10 min
d. 5 min
e. 1.5 min

A sample of helium effuses through a porous container 7.70 times faster than does unknown gas X. What is the molar mass of the unknown gas?
1. 129.96
2. 153.76
3. 169.0
4. 112.36
5. 190.44
6. 237.16
7. 275.56
8. 392.04
9. 201.64
10. 368.64

At any given temperature, how much more quickly will He effuse than Xe?
1. 5.7 times more quickly
2. .03 times more quickly
3. 8.1 times more quickly
4. They will diffuse at the same rate
5. 0.12 times more quickly
6. 32.8 times more quickly

Carbon dioxide effuses through a pinhole at a rate of 0.232 ml/min at 25.0 °C. Another gas effuses at a rate of 0.363 ml/min. What is the molar mass of the gas?

NO2 gas effuses ________times faster than Cl 2 gas.
a. 1.54
b. 0.649
c. 1.24
d. 0.770
e. 2.37

Consider the chemical reaction: 2 SO2 (g) → SO(g) + SO3 (g)
Determine if the sentence below is true or false
Between the two products, SO3 would have the highest rate of effusion.

The rate of effusion of oxygen to an unknown gas is 0.935. What is the other gas?
A) Ar
B) F2
C) Ne
D) N2

Calculate the ratio of effusion rates for Ar and Kr.

A sample of neon effuses from a container in 76 seconds. The same amount of an unknown noble gas requires 155 seconds. Identify the gas.

A sample of N2O effuses from a container in 42 seconds. How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?

Using Graham's Law of effusion, calculate the ratio of the effusion rates of H 2 to CO through a pinhole.

Rank from the highest to lowest effusion rate. To rank items as equivalent, overlap them.H2, Ar, Ne, C4H8, CO

Given the gases Ne, He, H2, Kr, put them in order of their INCREASING rate of effusion.
1. Kr < Ne < He < H 2
2. H2 < He < Ne < Kr
3. He < H2 < Kr < Ne
4. Ne < Kr < H 2 < He

Which gas is the predominant nitrogen-containing pollutant in the atmosphere?a) N2b) NH3c) NO2d) N2O5

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of 12C16O, 12C17O, and 12C18O. Name some advantages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

The rate of effusion of two different gases is known asA) Avogadro's Law B) Graham's Law C) Charles's Law D) Boyle's LawE) Dalton's Law

Calculate the ratio of effusion rates of Cl2 to F2.

Consider a 1.0-L sample of helium gas and a 1.0-L sample of argon gas, both at room temperature and atmospheric pressure. d. Which gas sample would have the faster rate of effusion?

A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon.d. If a small hole were opened in the flask, which gas would effuse more quickly?

A compound contains only C, H, and N. It is 58.51% C and 7.37% H by mass. Helium effuses through a porous frit 3.20 times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

It took 4.5 minutes for 1.0 L helium to effuse through a porous barrier. How long will it take for 1.0 L Cl2 gas to effuse under identical conditions?

When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH4Cl forms where gaseous NH3 and gaseous HCl first come into contact. NH3 (g) + HCl(g) ⟶ NH4Cl(s) At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both NH3 and HCl, and find out how much faster NH3 diffuses than HCl.)

What is the ratio of effusion rates for O 2 and Kr?

Containers A, B, and C are attached by closed stopcocks of negligible volume.If each particle shown in the picture represents 106 particles, (a) How many blue particles and black particles are in B after the stopcocks are opened and the system reaches equilibrium?

A sample of Xe takes 75 seconds to effuse out of a container. An unknown gas takes 37 seconds to effuse out of the identical container under identical conditions.What is the most likely identity of the unknown gas?

You may want to reference (Pages 416 - 419) Section 10.8 while completing this problem.Which of the following statements are true?(a) O2 will effuse faster than Cl2.(b) Effusion and diffusion are different names for the same process.(c) Perfume molecules travel to your nose by the process of effusion.(d) The higher the density of a gas, the shorter the mean free path.

Rank the three halogens (F2, Cl2, and Br2) with respect to their rate of effusion.

A balloon filled with helium gas is found to take 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?

A mixture of gaseous disulfur difluoride, dinitrogen tetrafluoride, and sulfur tetrafluoride is placed in an effusion apparatus.(a) Rank the gases in order of increasing effusion rate.

Freon-12 is used as a refrigerant in central home air conditioners. The rate of effusion of Freon-12 to Freon-11 (molar mass = 137.4 g/mol) is 1.07:1. The formula of Freon-12 is one of the following: CF4, CF3Cl, CF2Cl2, CFCl3, or CCl4. Which formula is correct for Freon-12?

Which of the following gases diffuse more slowly than oxygen? F2, Ne, N2O, C2H2, NO, Cl2, H2S

What is the ratio of effusion rates for the lightest gas, H2, and the heaviest known gas, UF6?

The rate of effusion of NH3 is 2.40 mole/min. What would be the rate of effusion of He under the same conditions? a) 2.40 b) 10.2 c) 1.54 d) 4.95 e) 0.56

Which of the following gases diffuses slowest?(A) F2(B) CO (C) CO2(D) Cl2(E) N2

Calculate the ratio of effusion rates for Ar and Kr.

In an effusion experiment, it was determined that nitrogen gas, N2, effused at a rate 1.812 times faster than an unknown gas. What is the molar mass of the unknown gas?Express your answer to four significant figures and include the appropriate units.The rate of effusion of a gas, r, is inversely proportional to the square root of its molar mass, M. The relative rate of two different gases is expressed asr1r2=M2M1where r1 and r2 are the effusion rates of two gases and M1 and M2 are their respective molar masses.

Rank the following gases in order of decreasing rate of effusion.Rank from the highest to lowest effusion rate. To rank items as equivalent, overlap them.In a given sample of gas, the particles move at varying speeds. The root mean square speed (rms speed) of particles in a gas sample, u, is given by the formulau=3RTMwhere T is the Kelvin temperature, M is the molar mass in kg/mol, and R = 8.314 J/(mol ⋅ K) is the gas constant. Effusion is the escape of gas molecules through a tiny hole into a vacuum.The rate of effusion of a gas is directly related to the rms speed of the gas molecules, so it's inversely proportional to the square root of its mass. The rms speed is related to kinetic energy, rather than average speed, and is the speed of a molecule possessing a kinetic energy identical to the average kinetic energy of the sample.Given its relationship to the mass of the molecule, you can conclude that the lighter the molecules of the gas, the more rapidly it effuses. Mathematically, this can be expressed aseffusion rate∝1mThe relative rate of effusion can be expressed in terms of molecular masses mA and mB asrate of gas A effusionrate of gas B effusion=mBmA

How much faster do ammonia (NH3) molecules effuse than carbon monoxide (CO) molecules?Enter the ratio of the rates of effusion. Express your answer numerically using three significant figures.In a given sample of gas, the particles move at varying speeds. The root mean square speed (rms speed) of particles in a gas sample, u, is given by the formulau=3RTMwhere T is the Kelvin temperature, M is the molar mass in kg/mol, and R = 8.314 J/(mol ⋅ K) is the gas constant. Effusion is the escape of gas molecules through a tiny hole into a vacuum.The rate of effusion of a gas is directly related to the rms speed of the gas molecules, so it's inversely proportional to the square root of its mass. The rms speed is related to kinetic energy, rather than average speed, and is the speed of a molecule possessing a kinetic energy identical to the average kinetic energy of the sample.Given its relationship to the mass of the molecule, you can conclude that the lighter the molecules of the gas, the more rapidly it effuses. Mathematically, this can be expressed aseffusion rate∝1mThe relative rate of effusion can be expressed in terms of molecular masses mA and mB asrate of gas A effusionrate of gas B effusion=mBmA

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane (CH4) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

Suppose you have two 1-L flasks, one containing N2 at STP and the other containing CH4 at STP.You may want to reference (Pages 415 - 419)Section 10.8 while completing this problem.How do these systems compare with respect to the rate of effusion through a pinhole leak?

A sample of N2 O effuses from a container in 47 seconds.How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?

We separate U-235 from U-238 by fluorinating a sample of uranium to form UF6 (which is a gas) and then taking advantage of the different rates of effusion and diffusion for compounds containing the two isotopes.Calculate the ratio of effusion rates for 238 UF6 and 235 UF6. The atomic mass of U-235 is 235.054 amu and that of U-238 is 238.051 amu.

Consider a 1.0-L sample of helium gas and a 1.0-L sample of argon gas, both at room temperature and atmospheric pressure.Which gas sample would have the fastest rate of effusion?a) sample of helium b) sample of argon c) samples would have the same rate of effusion

Calculate the root-mean-square velocity and kinetic energy of CO, CO2, and SO3 at 286 K .Which gas has the greatest effusion rate?

Gas Y effuses half as fast as O2. What is the molar mass of gas Y?

O2 (g) effuses at a rate that is __?__ times that of Br2 (g) under the same conditions. (rate of O2)/(rate of Br2)=?

Find the ratio of effusion rates of hydrogen gas and krypton gas.

If 250 mL of methane, CH4, effuses through a small hole in 28 s, the time required for the same volume of helium to pass through the hole under the same conditions will be

It takes 125 minutes for 10 mL of Ar gas to effuse through a porous barrier. How many minutes would it take the same amount of Ne gas to effuse through the same barrier?A. 108.4 min B. 70 min C. 72.5 min D. 85.3 min E. 88.8 min

The rate of effusion of oxygen to an unknown gas is 0.935. What is the other gas?

CO(g) effuses at a rate that is ______ times that of Cl2(g) under the same conditions.

A thin glass tube 1 m long is filled with Ar gas at 1 atm, and the ends are stoppered with cotton plugs: HCl gas is introduced at one end of the tube, and simultaneously NH3 gas is introduced at the other end. When the two gases diffuse through the cotton plugs down the tube and meet, a white ring appears due to the formation of NH4Cl(s).At which location - a, b, or c - do you expect the ring to form?

The rate of effusion of SO2 gas through a porous barrier is observed to be 7.01 x 10-4 mol/h. Under the same conditions, the rate of effusion of CH4 gas would be __________ mol/h.

The rate of effusion of Xe gas through a porous barrier is to be 7.31 x 10-4 mol/h. Under the same conditions, the rate of effusion of O2 gas would be __________ mol/h.

As discussed in the “Chemistry Put to Work” box in Section 10.8 in the textbook, enriched uranium can be produced by effusion of gaseous UF6 across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, U(g).Compare the ratio of effusion rates for 235U and 238U to the ratio for UF6 given in the essay in the textbook.

You may want to reference (Pages 224 - 230) Sections 5.8 while completing this problem.A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon. Sort the conditions based on the gas described. Drag the appropriate items to their respective bins.

Arsenic(III) sulfide sublimes readily, even below its melting point of 320 oC. The molecules of the vapor phase are found to effuse through a tiny hole at 0.28 times the rate of effusion of Ar atoms under the same conditions of temperature and pressure.What is the molecular formula of arsenic(III) sulfide in the gas phase?

Calculate the relative rate of diffusion of 1H2 (molar mass 2.0 g/mol) compared to that of 2H2 (molar mass 4.0 g/mol) and the relative rate of diffusion of O2 (molar mass 32 g/mol) compared to that of O3 (molar mass 48 g/mol).

White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions. How many atoms are in a molecule of gaseous white phosphorus?

Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.

A sample of neon effuses from a container in 77 seconds. The same amount of an unknown noble gas requires 157 seconds.Identify the second gas.

In a given diffusion apparatus, 15.0 mL of HBr gas diffuses in 1.0 min. In the same apparatus and under the same conditions, 20.3 mL of an unknown gas diffuses in 1.0 min. The unknown gas is a hydrocarbon.Find its molecular formula.

A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.

Aluminum chloride is easily vaporized above 180°C. The gas escapes through a pinhole 0.122 times as fast as helium at the same conditions of temperature and pressure in the same apparatus. What is the molecular formula of aluminum chloride gas?

A sample of an unknown gas effuses in 11.1 min. An equal volume of H 2 in the same apparatus under the same conditions effuses in 2.42 min. What is the molar mass of the unknown gas?

Explain the difference between diffusion and effusion.

Describe how the molecules in a perfume bottle travel from the bottle to your nose.

Two identical balloons are filled to the same volume, one with air and one with helium. The next day, the volume of the air-filled balloon has decreased by 5.1 % .By what percent has the volume of the helium-filled balloon decreased? (Assume that the air is four-fifths nitrogen and one-fifth oxygen, and that the temperature did not change.)

How is the effusion rate of a gas related to its molar mass?

The rate of effusion of CH4 gas through a porous barrier is observed to be 5.48 x 10-4 mol/h. Under the same condition, the rate of effusion of Xe gas would be ____ mol/h.

The average molecular speed in a sample of O3 gas at a certain temperature is 386 m/s. The average molecular speed in a sample of Ar gas is temperature.

A sample of CH4 gas is observed to effuse through a porous barrier in 5.53 minutes. Under the same conditions, the same number of moles of an unknown gas requires 7.07 minutes to effuse through the same barrier. The molar mass of the unknown gas is g/mol.

How much faster do ammonia (NH 3) molecules effuse than carbon monoxide (CO) molecules?Enter the ratio of the rates of effusion. Express your answer numerically using three significant figures.

An illustration of Grahams law of effusion.Because pressure and temperature are constant in this figure but volume changes, which other quantity in the ideal-gas equation must also change?

The following graph shows the distribution of molecular velocities for two different molecules (A and B) at the same temperature. Which molecule would have the higher rate of effusion? a) A b) B c) not enough information to determine

Explain the difference between effusion and diffusion.

An organic compound containing only C, H, and N yields the following data.i. Complete combustion of 35.0 mg of the compound produced 33.5 mg CO2 and 41.1 mg H2O.ii. A 65.2-mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129), giving 35.6 mL of dry N2 at 740. torr and 25°C.iii. The effusion rate of the compound as a gas was measured and found to be 24.6 mL/min. The effusion rate of argon gas, under identical conditions, is 26.4 mL/min. What is the molecular formula of the compound?

A glass tube contains an equal number of moles of helium and argon. After five minutes, half of the particles escape through a small hole in the glass.What are the relative amounts of helium and argon in the tube at five minutes?Match the words in the left column to the appropriate blanks in the sentences on the right.The rate of effusion is ______ to the square root of the molar mass. Since the molar mass of helium is _______ than the molar mass of argon, helium will effuse _______ . Therefore, after five minutes when the half of particles escape, the number of moles of helium will be ________ than the number of moles of argon.

During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. Show the calculation that supports this value. The molar mass of 235UF6 = 235.043930 + 6 × 18.998403 = 349.034348 g/mol, and the molar mass of 238UF6 = 238.050788 + 6 × 18.998403 = 352.041206 g/ mol.

You may want to reference (Pages 416 - 419) Section 10.8 while completing this problem.At constant pressure, the mean free path (λ) of a gas molecule is directly proportional to temperature. At constant temperature, λ is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, λ is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it Rmfp, like the ideal-gas constant) and define units for Rmfp.

In the process known as osmosis, _____ moves through a semipermeable membrane into an area of _____ concentration. A) solute; lower solute B) solute; higher soluteC) solvent; lower solvent D) solvent; higher solvent

Hydrogen has two naturally occurring isotopes, 1H and 2H. Chlorine also has two naturally occurring isotopes, 35Cl and 37Cl. Thus, hydrogen chloride gas consists of four distinct types of molecules: 1H35Cl, 1H37Cl, 2H35Cl, and 2H37Cl. Place these four molecules in order of decreasing rate of effusion.

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 115 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 L of O2 gas to effuse.Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

As discussed in the “Chemistry Put to Work” box in Section 10.8 in the textbook, enriched uranium can be produced by effusion of gaseous UF6 across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, U(g).Calculate the ratio of effusion rates for 235U and 238U.

Consider two 1-L samples of gas, one H2 and one O2, both at 1 atm and 25°C. Compare the samples in terms of the characteristics listed. time for a given fraction of O2 molecules to effuse____time for a given fraction of H2 molecules to effuse a. is greater thanb. is less than c. is equal to d. is unrelated to e. None of these

The rate of effusion of SO2 gas through a porous barrier is observed to be 5.29 x 10-4 mol/h. Under the same conditions, the rate of effusion of NO2 gas would be mol/h.

1.560 x 10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 98.8 s.Under identical conditions, 1.749 x 10-4 mol of argon gas takes 84.9 s to effuse. What is the molar mass of the unidentified substance (in g/mol)? The percentage C by mass in the above unidentified substance is 53.0%. It may also contain H and/or O, but no other element. What is the molecular formula of the substance? In typing this formula use the convention of C first, H second and O last. For example, C6H8O2 would be C6H802.) Under identical conditions, how many moles of acetylene (C2H2) gas would effuse in 86.5 s?

A sample of O3 gas is observed to effuse through a pourous barrier in 1.07 minutes. Under the same conditions, the same number of moles of an unknown gas requires 1.44 minutes to effuse through the same barrier. The molar mass of the unknown gas is ___________ g/mol.

A sample of Br2(g) takes 34.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane?

The effusion rate of A gas molecules will be twice higher than one of B gas molecules when B molar mass is ____ higher than A molar mass. (a) None of the above (b) 4 times (c) 3 times (d) 2 times

Calculate the ratio of effusion rates for Ar and Kr.

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