Question 1

A sample of H₂ gas (5.0 mol)effused through a pinhole in 10.5 s.It will take ________ s for the same amount of O₂ to effuse under the same conditions.

Accepted Answer

The answer of A sample of H₂ gas (5.0 mol)effused...

Question 2

CO (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.

Accepted Answer

The correct answer is A because the total pressure in the container is the sum of the partial pressures of CO and CO2. The partial pressure of CO is 5.00 g / 28.01 g/mol * 0.0821 Latm/(molK) * 323 K / 750.0 mL * 1000 mL/L = 0.292 atm. The partial pressure of CO2 is 5.00 g / 44.01 g/mol * 0.0821 Latm/(molK) * 323 K / 750.0 mL * 1000 mL/L = 0.146 atm. The total pressure is 0.292 atm + 0.146 atm = 0.438 atm.

Question 3

A flask contains a mixture of N₂ and H₂ at a total pressure of 5.20 atm.If there are 1.00 mol of H₂ and 8.00 mol of N₂ in the flask,what is the partial pressure (atm)of N₂?

Accepted Answer

To find the partial pressure of N₂, we need to know its mole fraction in the mixture. Mole fraction of N₂ = (number of moles of N₂) / (total number of moles) Mole fraction of N₂ = 8.00 / (8.00 + 1.00) = 0.8889 Since the total pressure is 5.20 atm, we can use the formula: Partial pressure of N₂ = (mole fraction of N₂) x (total pressure) Partial pressure of N₂ = 0.8889 x 5.20 = 4.62 atm Therefore, the answer is B.

Question 4

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at is ________ atm. (a = 0.211 L²-atm/mol²,b = 0.0171 L/mol)

Accepted Answer

The answer of Using the van der Waals equation,the pressure...

Question 5

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 °C is ________ atm. (a = 6.49 L²-atm/mol²,b = 0.0562 L/mol)

Accepted Answer

The answer of Using the van der Waals equation,the pressure...

Question 6

A sample of He gas (2.0 mmol)effused through a pinhole in 53 s.The same amount of an unknown gas,under the same conditions,effused through the pinhole in 248 s.The molecular mass of the unknown gas is ________ g/mol.

Accepted Answer

The molecular mass of the unknown gas can be found using Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula derived from Graham's law is:$$ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} $$where Rate_1 and Rate_2 are the effusion rates of gas 1 and gas 2, respectively, and M_1 and M_2 are their molar masses. Given that helium (He) has a molar mass of 4 g/mol and effused in 53 s, and the unknown gas effused in 248 s, we can set up the equation as follows, with the rate being inversely proportional to time:$$ \frac{53}{248} = \sqrt{\frac{M_{\text{unknown}}}{4}} $$Solving for M_unknown gives a molar mass of approximately 88 g/mol.

Question 7

CO (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO₂ in the container was ________ atm.

Accepted Answer

The partial pressure of a gas is equal to the mole fraction of that gas multiplied by the total pressure. The mole fraction of CO2 is equal to the number of moles of CO2 divided by the total number of moles of gas in the container. The number of moles of CO2 is equal to the mass of CO2 divided by the molar mass of CO2. The molar mass of CO2 is 44.01 g/mol. The number of moles of CO is equal to the mass of CO divided by the molar mass of CO. The molar mass of CO is 28.01 g/mol. The total number of moles of gas in the container is equal to the number of moles of CO plus the number of moles of CO2. The total pressure is equal to the sum of the partial pressures of the two gases. The partial pressure of CO2 is equal to the mole fraction of CO2 multiplied by the total pressure. The mole fraction of CO2 is equal to 5.00 g / (5.00 g + 5.00 g) = 0.500. The total pressure is equal to 1 atm. The partial pressure of CO2 is equal to 0.500 atm * 1 atm = 0.500 atm.

Question 8

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm.Atmospheric pressure was 783 mm Hg.The pressure of the argon in the container was ________ mm Hg.

Accepted Answer

Since the manometer is closed-end, the pressure of the argon is equal to the difference in mercury levels, which is 9.60 cm or 96.0 mm Hg.

Question 9

A sample of CO₂ gas (3.0 mol)effused through a pinhole in 18.0 s.It will take ________ s for the same amount of H₂ to effuse under the same conditions.

Accepted Answer

The answer of A sample of CO₂ gas (3.0 mol)effused...

Question 10

The root-mean-square speed of CO at 113 °C is ________ m/s.

Accepted Answer

The root-mean-square speed of a gas molecule is given by the equation:

URMS = √(3RT/M)

where R is the gas constant, T is the temperature in kelvin, and M is the molar mass of the gas.

To solve for the root-mean-square speed of CO at 113°C, we first need to convert the temperature to kelvin:

T = 113°C + 273.15 = 386.15 K

The molar mass of CO is 28.01 g/mol. The gas constant R is 8.31 J/mol·K.

URMS = √(3RT/M) = √[(3*8.31*386.15)/(0.02801)] = 585.9 m/s ≈ 586 m/s

Therefore, the answer is choice C.

Question 11

CO (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO in the container was ________ atm.

Accepted Answer

The molar mass of CO is 28.01 g/mol. Therefore, 5.00 g of CO is equivalent to $ \frac{5.00 \, \text{g}}{28.01 \, \text{g/mol}} = 0.1785 \, \text{mol} $. Using the ideal gas law $ PV = nRT $, where $ P $ is the pressure in atm, $ V $ is the volume in liters (0.750 L), $ n $ is the number of moles of gas (0.1785 mol), $ R $ is the ideal gas constant (0.0821 L·atm/(mol·K)), and $ T $ is the temperature in Kelvin (50.0 °C = 323.15 K), we can solve for $ P $. $$ P = \frac{nRT}{V} = \frac{(0.1785 \, \text{mol})(0.0821 \, \text{L·atm/(mol·K)})(323.15 \, \text{K})}{0.750 \, \text{L}} = 6.29 \, \text{atm} $$

Question 12

A fixed amount of gas at 25.0 °C occupies a volume of 10.0 L when the pressure is 751 torr.Use Boyle's law to calculate the pressure (torr)when the volume is reduced to 7.25 L at a constant temperature of

Accepted Answer

The answer of A fixed amount of gas at 25.0...

Question 13

Molecular compounds of low molecular weight tend to be gases at room temperature.Which of the following is most likely not a gas at room temperature?

Accepted Answer

KBr is an ionic compound and therefore has a higher melting and boiling point than molecular compounds. It is likely to be a solid or liquid at room temperature.

Question 14

A mixture of He and Ne at a total pressure of 0.95 atm is found to contain 0.32 mol of He and 0.56 mol of Ne.The partial pressure of Ne is ________ atm.

Accepted Answer

The partial pressure of a gas in a mixture can be calculated using the formula: $P_{\text{gas}} = X_{\text{gas}} \times P_{\text{total}}$, where $X_{\text{gas}}$ is the mole fraction of the gas. The mole fraction of Ne is calculated as $\frac{0.56}{0.32 + 0.56} = \frac{0.56}{0.88}$. The partial pressure of Ne is then $0.636 \times 0.95 \approx 0.60$ atm.

Question 15

Sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas: NaH (s)+ H₂O (l)→ NaOH (aq)+ H₂ (g) A sample of NaH weighing ________ g will produce 982 mL of gas at 28.0 °C and 765 torr,when the hydrogen is collected over water.The vapor pressure of water at this temperature is 28 torr.

Accepted Answer

First, calculate the pressure of dry hydrogen gas: 765 torr - 28 torr = 737 torr. Convert this to atm: 737/760 = 0.970 atm. Use the ideal gas law (PV=nRT) to find moles of H2: n = PV/RT = (0.970 atm)(0.982 L)/(0.0821 L·atm/mol·K)(301 K) = 0.0397 mol H2. Since the reaction is 1:1, moles of NaH = moles of H2 = 0.0397 mol. The molar mass of NaH is 24.0 g/mol, so mass = 0.0397 mol × 24.0 g/mol = 0.9528 g, which rounds to 0.925 g.

Question 16

A gas mixture of N₂ and H₂ has a total pressure of 9.40 atm and contains 11.3 mol of gas.If the partial pressure of N₂ is 4.89 atm,how many moles of H₂ are in the mixture?

Accepted Answer

The answer of A gas mixture of N₂ and H₂...

Question 17

SO₂ (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO₂ in the container was ________ atm.

Accepted Answer

The answer of SO₂ (5.00 g)and CO₂ (5.00 g)were placed...

Question 18

SO₂ (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.

Accepted Answer

The answer of SO₂ (5.00 g)and CO₂ (5.00 g)were placed...

Question 19

Sulfur dioxide (12.4 g)and carbon dioxide (12.4 g)are placed in a 500.0 mL container at 45.0 °C.What is the partial pressure (atm)of sulfur dioxide in the container?

Accepted Answer

To find the partial pressure of sulfur dioxide, use the ideal gas law: PV = nRT. First, calculate the moles of sulfur dioxide (SO2) using its molar mass (64.07 g/mol): n = 12.4 g / 64.07 g/mol = 0.1935 mol. Then, use the ideal gas law: P = nRT/V. R = 0.0821 L·atm/mol·K, T = 45.0 °C = 318.15 K, V = 0.500 L. P = (0.1935 mol)(0.0821 L·atm/mol·K)(318.15 K) / 0.500 L = 10.1 atm.

Question 20

A pressure of 0.500 atm is the same as a pressure of ________ of mm Hg.

Accepted Answer

To convert atm to mm Hg, we use the following conversion factor: 
1 atm = 760 mm Hg

Therefore, 
0.500 atm x 760 mm Hg/atm = 380 mm Hg

So the correct choice is (C).

Quiz 10: Gases

A sample of H₂ gas (5.0 mol) effused through a pinhole in 10.5 s.It will take ________ s for the same amount of O₂ to effuse under the same conditions.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.

A flask contains a mixture of N₂ and H₂ at a total pressure of 5.20 atm.If there are 1.00 mol of H₂ and 8.00 mol of N₂ in the flask,what is the partial pressure (atm) of N₂?

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at is ________ atm. (a = 0.211 L²-atm/mol²,b = 0.0171 L/mol)

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 °C is ________ atm. (a = 6.49 L²-atm/mol²,b = 0.0562 L/mol)

A sample of He gas (2.0 mmol) effused through a pinhole in 53 s.The same amount of an unknown gas,under the same conditions,effused through the pinhole in 248 s.The molecular mass of the unknown gas is ________ g/mol.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO₂ in the container was ________ atm.

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm.Atmospheric pressure was 783 mm Hg.The pressure of the argon in the container was ________ mm Hg.

A sample of CO₂ gas (3.0 mol) effused through a pinhole in 18.0 s.It will take ________ s for the same amount of H₂ to effuse under the same conditions.

The root-mean-square speed of CO at 113 °C is ________ m/s.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO in the container was ________ atm.

A fixed amount of gas at 25.0 °C occupies a volume of 10.0 L when the pressure is 751 torr.Use Boyle's law to calculate the pressure (torr) when the volume is reduced to 7.25 L at a constant temperature of

Molecular compounds of low molecular weight tend to be gases at room temperature.Which of the following is most likely not a gas at room temperature?

A mixture of He and Ne at a total pressure of 0.95 atm is found to contain 0.32 mol of He and 0.56 mol of Ne.The partial pressure of Ne is ________ atm.

A gas mixture of N₂ and H₂ has a total pressure of 9.40 atm and contains 11.3 mol of gas.If the partial pressure of N₂ is 4.89 atm,how many moles of H₂ are in the mixture?

SO₂ (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO₂ in the container was ________ atm.

SO₂ (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.

Sulfur dioxide (12.4 g) and carbon dioxide (12.4 g) are placed in a 500.0 mL container at 45.0 °C.What is the partial pressure (atm) of sulfur dioxide in the container?

A pressure of 0.500 atm is the same as a pressure of ________ of mm Hg.

Quiz 10: Gases

A sample of H2 gas (5.0 mol) effused through a pinhole in 10.5 s.It will take ________ s for the same amount of O2 to effuse under the same conditions.

CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.

A flask contains a mixture of N2 and H2 at a total pressure of 5.20 atm.If there are 1.00 mol of H2 and 8.00 mol of N2 in the flask,what is the partial pressure (atm) of N2?

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at is ________ atm. (a = 0.211 L2-atm/mol2,b = 0.0171 L/mol)

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 °C is ________ atm. (a = 6.49 L2-atm/mol2,b = 0.0562 L/mol)

A sample of He gas (2.0 mmol) effused through a pinhole in 53 s.The same amount of an unknown gas,under the same conditions,effused through the pinhole in 248 s.The molecular mass of the unknown gas is ________ g/mol.

CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO2 in the container was ________ atm.

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm.Atmospheric pressure was 783 mm Hg.The pressure of the argon in the container was ________ mm Hg.

A sample of CO2 gas (3.0 mol) effused through a pinhole in 18.0 s.It will take ________ s for the same amount of H2 to effuse under the same conditions.

The root-mean-square speed of CO at 113 °C is ________ m/s.

CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO in the container was ________ atm.

A fixed amount of gas at 25.0 °C occupies a volume of 10.0 L when the pressure is 751 torr.Use Boyle's law to calculate the pressure (torr) when the volume is reduced to 7.25 L at a constant temperature of

Molecular compounds of low molecular weight tend to be gases at room temperature.Which of the following is most likely not a gas at room temperature?

A mixture of He and Ne at a total pressure of 0.95 atm is found to contain 0.32 mol of He and 0.56 mol of Ne.The partial pressure of Ne is ________ atm.

A gas mixture of N2 and H2 has a total pressure of 9.40 atm and contains 11.3 mol of gas.If the partial pressure of N2 is 4.89 atm,how many moles of H2 are in the mixture?

SO2 (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO2 in the container was ________ atm.

SO2 (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.

Sulfur dioxide (12.4 g) and carbon dioxide (12.4 g) are placed in a 500.0 mL container at 45.0 °C.What is the partial pressure (atm) of sulfur dioxide in the container?

A pressure of 0.500 atm is the same as a pressure of ________ of mm Hg.

A sample of H₂ gas (5.0 mol) effused through a pinhole in 10.5 s.It will take ________ s for the same amount of O₂ to effuse under the same conditions.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.

A flask contains a mixture of N₂ and H₂ at a total pressure of 5.20 atm.If there are 1.00 mol of H₂ and 8.00 mol of N₂ in the flask,what is the partial pressure (atm) of N₂?

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at is ________ atm. (a = 0.211 L²-atm/mol²,b = 0.0171 L/mol)

Using the van der Waals equation,the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 °C is ________ atm. (a = 6.49 L²-atm/mol²,b = 0.0562 L/mol)

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO₂ in the container was ________ atm.

A sample of CO₂ gas (3.0 mol) effused through a pinhole in 18.0 s.It will take ________ s for the same amount of H₂ to effuse under the same conditions.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO in the container was ________ atm.

A gas mixture of N₂ and H₂ has a total pressure of 9.40 atm and contains 11.3 mol of gas.If the partial pressure of N₂ is 4.89 atm,how many moles of H₂ are in the mixture?

SO₂ (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The partial pressure of CO₂ in the container was ________ atm.

SO₂ (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0 °C.The total pressure in the container was ________ atm.