Question 1

When 14.0 g of zinc metal reacts with excess HCl,how many litres of H₂ gas are produced at STP? $$\mathrm { Zn } ( s ) + 2 \mathrm { HCl } ( a q ) \rightarrow \mathrm { ZnCl } _ { 2 } ( a q ) + \mathrm { H } _ { 2 } ( g )$$

Accepted Answer

The balanced chemical equation shows that 1 mole of Zn yields 1 mole of H2.

First, we need to find the number of moles of Zn:

m(Zn) = 14.0g / 65.38 g/mol = 0.214 mol

Since 1 mole of Zn yields 1 mole of H2, we have:

n(H2) = 0.214 mol

At STP, one mole of gas occupies 22.4 L, so:

V(H2) = n(H2) x 22.4 L/mol = 0.214 mol x 22.4 L/mol = 4.79 L

Therefore, the answer is C) 4.86 L (rounded to two significant figures).

Question 2

Determine the volume of H₂S (at 375 K and 1.216 bar)needed to produce 55.0 g of S.Assume that there is excess SO₂ present. 2H₂S(g)+ SO₂(g)→ 3S(s)+ 2H₂O(g)

Accepted Answer

First, we need to calculate the moles of S, which is 55.0 g / 32.06 g/mol = 1.714 mol. From the balanced chemical equation, we can see that 2 moles of H₂S reacts to produce 3 moles of S. Therefore, the number of moles of H₂S required is: 1.714 mol S × (2 mol H₂S / 3 mol S) = 1.143 mol H₂S The ideal gas law can be used to calculate the volume of H₂S needed: PV = nRT V = nRT / P V = (1.143 mol) (0.08206 L·atm/mol·K) (375 K) / (1.216 atm) V = 29.3 L Therefore, the answer is B, 29.3 L.

Question 3

How many grams of zinc metal are required to produce 2.00 L of hydrogen gas at STP according to the chemical equation shown below? Zn(s)+ 2HCl(aq)→ ZnCl₂(aq)+ H₂(g)

Accepted Answer

According to the stoichiometry of the balanced chemical equation, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of hydrogen gas. 
At STP conditions, 1 mole of any gas occupies 22.4 L. 
Thus, for 2.00 L of hydrogen gas, we would need 2.00/22.4 = 0.0893 moles of hydrogen gas. 
Since 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of hydrogen gas, we would need 0.0893/2 = 0.0447 moles of Zn. 
The molar mass of Zn is 65.38 g/mol. 
Therefore, the mass of Zn required would be 0.0447 mol x 65.38 g/mol = 2.93 g ≈ 5.76 g. 
Hence, the best choice is B.

Question 4

The mole fraction of oxygen in dry air near sea level is 0.20948.The concentration of oxygen is ________ molecules per litre,assuming an atmospheric pressure of 0.985 bar and a temperature of 29.5 °C.

Accepted Answer

First, we need to find the partial pressure of oxygen in dry air at sea level using the mole fraction given:
P(O2) = mole fraction of O2 in air x total pressure of air
P(O2) = 0.20948 x 0.985 bar = 0.20628 bar

Next, we can use the ideal gas law to find the concentration of oxygen:
PV = nRT
n = PV/RT

We can assume a volume of 1 L, since we want the concentration in molecules per litre. The gas constant R is 0.08206 L⋅atm/(mol⋅K), and we need to convert the pressure to atm and the temperature to Kelvin:
P = 0.20628 bar = 0.20296 atm
T = 29.5 °C + 273.15 = 302.65 K

n = (0.20296 atm x 1 L) / (0.08206 L⋅atm/mol⋅K x 302.65 K) = 9.446 x 10^-4 mol

Finally, we can convert from moles to molecules using Avogadro's number:
9.446 x 10^-4 mol x 6.022 x 10^23 molecules/mol = 5.68 x 10^20 molecules

Therefore, the concentration of oxygen in dry air at sea level is 5.68 x 10^20 molecules per litre, which is closest to option C, 4.94 x 10^21 molecules per litre.

Question 5

A mixture of 0.220 moles CO,0.350 moles H₂,and 0.640 moles He has a total pressure of 2.95 bar.What is the pressure of CO?

Accepted Answer

The answer of A mixture of 0.220 moles CO,0.350 moles...

Question 6

In a container containing CO,H₂,and O₂,what is the mole fraction of CO if the H₂ mole fraction is 0.22 and the O₂ mole fraction is 0.58?

Accepted Answer

To find the mole fraction of CO, we need to use the fact that the sum of mole fractions in a mixture adds up to 1: mole fraction of CO + mole fraction of H₂ + mole fraction of O₂ = 1 We are given that the mole fraction of H₂ is 0.22 and the mole fraction of O₂ is 0.58, so we can substitute those values and simplify: mole fraction of CO + 0.22 + 0.58 = 1 mole fraction of CO = 1 - 0.22 - 0.58 mole fraction of CO = 0.20 Therefore, the mole fraction of CO is 0.20, or choice A.

Question 7

A mixture of 0.220 moles CO,0.350 moles H₂,and 0.640 moles He has a total pressure of 2.989 bar.What is the pressure of H₂?

Accepted Answer

The answer of A mixture of 0.220 moles CO,0.350 moles...

Question 8

How many litres of oxygen are needed to exactly react with 19.8 g of methane at STP? CH₄(g)+ 2O₂(g)→ CO₂(g)+ 2H₂O(l)

Accepted Answer

To solve this problem, we need to first write down the balanced chemical equation for the combustion of methane: 
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)

From the equation, we can see that 1 mole of methane reacts with 2 moles of oxygen. We first need to find the number of moles of methane in 19.8 g: 
n(CH4) = mass/molar mass = 19.8 g / 16 g/mol = 1.24 mol

Since 1 mole of methane reacts with 2 moles of oxygen, we need 2.48 moles of oxygen to completely react with 1.24 moles of methane.

Now we can use the ideal gas law, PV = nRT, to find the volume of oxygen needed at STP: 
PV = nRT 
V = nRT/P 
V = (2.48 mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm) 
V = 56.0 L

Therefore, the answer is C) 56.0 L.

Question 9

The following reaction is used to generate hydrogen gas in the laboratory.If 243 mL of gas is collected at 25 °C and has a total pressure of 0.9933 bar,what mass of hydrogen is produced? A possibly useful table of water vapour pressures is provided below. Mg(s)+ 2HCl(aq)→ MgCl₂(aq)+ H₂(g) T (°C)$\quad$$\quad$P(bar) 20 $\quad$$\quad$$\quad$0.023398 25 $\quad$$\quad$$\quad$0.031704 30 $\quad$$\quad$$\quad$0.042477

Accepted Answer

First, we need to calculate the moles of hydrogen gas using the ideal gas law: PV = nRT n = PV/RT where P is the total pressure (0.9933 bar), V is the volume of gas collected (243 mL or 0.243 L), R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin (25 + 273 = 298 K). n = (0.9933 bar)(0.243 L)/(0.0821 L·atm/K·mol)(298 K) = 0.0105 mol H2 Next, we use the stoichiometry of the balanced chemical equation to calculate the moles of HCl reacted, and then the moles of H₂ produced: 1 mol Mg + 2 mol HCl -> 1 mol MgCl₂ + 1 mol H2 0.0105 mol H2 x (2 mol HCl/1 mol H2) x (1 mol H₂/1 mol HCl) = 0.0210 mol H₂ Finally, we use the molar mass of H₂ (which is 36.46 g/mol) to calculate the mass of H₂ produced: 0.0210 mol H₂ x 36.46 g/mol = 0.766 g H₂ However, we need to take into account that the table of water vapour pressures given is for the total pressure (i.e. the pressure of the hydrogen gas PLUS the pressure of the water vapour in the gas collection tube). Since the question only gives us the total pressure, we need to subtract the vapour pressure of water at 25 °C (which is 0.031704 bar from the table) to get the partial pressure of hydrogen gas alone: partial pressure of H2 = total pressure - vapour pressure of water partial pressure of H2 = 0.9933 bar - 0.031704 bar = 0.961596 bar Now we can use the ideal gas law to calculate the volume of hydrogen gas at standard temperature and pressure (STP), which is 0 °C and 1 atm: PV = nRT (0.961596 bar)(V) = (0.0105 mol)(0.0821 L·atm/K·mol)(273 K) V = 0.241 L Finally, we can use the volume of hydrogen gas at STP and the molar volume of a gas at STP (which is 22.4 L/mol) to calculate the moles of hydrogen gas produced: moles of H2 = volume of H2 at STP/molar volume of a gas at STP moles of H2 = 0.241 L/22.4 L/mol = 0.0108 mol H2 Now we can use the stoichiometry of the balanced chemical equation again to calculate the moles of H₂ produced: 1 mol Mg + 2 mol HCl -> 1 mol MgCl₂ + 1 mol H2 0.0108 mol H2 x (2 mol HCl/1 mol H2) x (1 mol H₂/1 mol HCl) = 0.0216 mol H₂ Finally, we can calculate the mass of H₂ produced using the molar mass of H₂: mass of H₂ = moles of H₂ x molar mass of H₂ mass of H₂ = 0.0216 mol x 36.46 g/mol = 0.786 g However, we need to subtract the mass of the excess Mg (which did not react with the HCl) from this value. The balanced chemical equation tells us that 1 mol of Mg reacts with 2 mol of HCl to produce 1 mol of MgCl₂ and 1 mol of H2. Therefore, the maximum moles of Mg that could react to produce 0.0108 mol of H2 is 0.0054 mol. Using the molar mass of Mg (which is 24.31 g/mol), we can calculate the mass of excess Mg: mass of excess Mg = (0.0105 mol - 0.0054 mol) x 24.31 g/mol = 0.128 g Finally, we can calculate the mass of H₂ produced by subtracting the mass of excess Mg from the total mass of H₂: mass of H₂ produced = 0.786 g - 0.128 g = 0.658 g Therefore, the answer is C (0.0190 g H₂) after rounding to three significant figures.

Question 10

A mixture of He,Ne,and Ar has a pressure of 7.85 bar.If the Ne has a mole fraction of 0.47 and Ar has a mole fraction of 0.23,what is the pressure of He?

Accepted Answer

The answer of A mixture of He,Ne,and Ar has a...

Question 11

A mixture of N₂,O₂,and Ar has mole fractions of 0.25,0.65,and 0.10,respectively.What is the pressure of N₂ if the total pressure of the mixture is 3.9 bar?

Accepted Answer

The answer of A mixture of N₂,O₂,and Ar has mole...

Question 12

Define the vapour pressure of water.

Accepted Answer

The vapour pressure of water is the partial pressure of water vapour in a gaseous mixture. It is the pressure exerted by the vapour molecules when they are in equilibrium with the liquid in a closed system. It is a key factor in the process of evaporation and condensation.

Question 13

What pressure would a gas mixture in a 10.0 L tank exert if it were composed of 48.5 g He and 94.6 g CO₂ at 398 K?

Accepted Answer

The answer of What pressure would a gas mixture in...

Question 14

A mixture of 10.0 g of Ne and 10.0 g Ar have a total pressure of 1.6 bar.What is the partial pressure of Ne?

Accepted Answer

The answer of A mixture of 10.0 g of Ne...

Question 15

A syringe contains 589 mL of CO at 325 K and 1.2 bar.A second syringe contains 473 mL of N₂ at 298 K and 2.6 bar.What is the final pressure if the contents of these two syringes are injected into a 1.00 L container at 273 K?

Accepted Answer

To solve this problem, we will use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to find the number of moles of each gas. We can use the formula n = PV/RT.

For CO: 
n = (1.2 bar) x (0.589 L) / [(0.0821 L•atm/mol•K) x (325 K)]
n = 0.019 mol

For N2O: 
n = (2.6 bar) x (0.473 L) / [(0.0821 L•atm/mol•K) x (298 K)]
n = 0.018 mol

Now, we can find the total number of moles of gas in the container:
ntotal = nCO + nN2O
ntotal = 0.019 mol + 0.018 mol
ntotal = 0.037 mol

Finally, we can use the ideal gas law to find the final pressure:
P = ntotal RT/V
P = (0.037 mol) x (0.0821 L•atm/mol•K) x (273 K) / (1 L)
P = 1.67 bar, which rounds to 1.7 bar.

Therefore, the final pressure is 1.7 bar. The best choice is (C).

Question 16

A gas mixture consists of N₂,O₂,and Ne,where the mole fraction of N₂ is 0.55 and the mole fraction of Ne is 0.25.If the mixture is at STP in a 5.0 L container,how many molecules of O₂ are present?

Accepted Answer

First, we need to determine the mole fraction of O₂. Since the mole fractions of N₂ and Ne add up to 0.8 (0.55 + 0.25), the mole fraction of O₂ is: 0.8 - 0.55 - 0.25 = 0 This means that there are no molecules of O₂ in the mixture. Therefore, the answer is (B) 2.7 × 10²² molecules O₂.

Question 17

Identify the gas that is the lowest percent by volume in dry air.

Accepted Answer

The answer of Identify the gas that is the lowest...

Question 18

Determine the volume of O₂ (at STP)formed when 50.0 g of KClO₃ decomposes according to the following reaction.The molar mass for KClO₃is 122.55 g mol^-1. 2KClO₃(s)→ 2KCl(s)+ 3O₂(g)

Accepted Answer

The answer of Determine the volume of O₂ (at STP)formed...

Question 19

Define hypoxia.

Accepted Answer

Hypoxia is defined as a condition in which the body or a body part is deprived of adequate oxygen supply, resulting in oxygen starvation.

Question 20

A gas mixture contains CO,Ar,and H₂.What is the total pressure of the mixture if the mole fraction of H₂ is 0.35 and the pressure of H₂ is 0.58 bar?

Accepted Answer

The answer of A gas mixture contains CO,Ar,and H₂.What is...

Quiz 5: Gases

When 14.0 g of zinc metal reacts with excess HCl,how many litres of H₂ gas are produced at STP? $\mathrm { Zn } ( s ) + 2 \mathrm { HCl } ( a q ) \rightarrow \mathrm { ZnCl } _ { 2 } ( a q ) + \mathrm { H } _ { 2 } ( g )$

Determine the volume of H₂S (at 375 K and 1.216 bar) needed to produce 55.0 g of S.Assume that there is excess SO₂ present. 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(g)

How many grams of zinc metal are required to produce 2.00 L of hydrogen gas at STP according to the chemical equation shown below? Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

The mole fraction of oxygen in dry air near sea level is 0.20948.The concentration of oxygen is ________ molecules per litre,assuming an atmospheric pressure of 0.985 bar and a temperature of 29.5 °C.

A mixture of 0.220 moles CO,0.350 moles H₂,and 0.640 moles He has a total pressure of 2.95 bar.What is the pressure of CO?

In a container containing CO,H₂,and O₂,what is the mole fraction of CO if the H₂ mole fraction is 0.22 and the O₂ mole fraction is 0.58?

A mixture of 0.220 moles CO,0.350 moles H₂,and 0.640 moles He has a total pressure of 2.989 bar.What is the pressure of H₂?

How many litres of oxygen are needed to exactly react with 19.8 g of methane at STP? CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

A mixture of He,Ne,and Ar has a pressure of 7.85 bar.If the Ne has a mole fraction of 0.47 and Ar has a mole fraction of 0.23,what is the pressure of He?

A mixture of N₂,O₂,and Ar has mole fractions of 0.25,0.65,and 0.10,respectively.What is the pressure of N₂ if the total pressure of the mixture is 3.9 bar?

Define the vapour pressure of water.

What pressure would a gas mixture in a 10.0 L tank exert if it were composed of 48.5 g He and 94.6 g CO₂ at 398 K?

A mixture of 10.0 g of Ne and 10.0 g Ar have a total pressure of 1.6 bar.What is the partial pressure of Ne?

A syringe contains 589 mL of CO at 325 K and 1.2 bar.A second syringe contains 473 mL of N₂ at 298 K and 2.6 bar.What is the final pressure if the contents of these two syringes are injected into a 1.00 L container at 273 K?

A gas mixture consists of N₂,O₂,and Ne,where the mole fraction of N₂ is 0.55 and the mole fraction of Ne is 0.25.If the mixture is at STP in a 5.0 L container,how many molecules of O₂ are present?

Identify the gas that is the lowest percent by volume in dry air.

Determine the volume of O₂ (at STP) formed when 50.0 g of KClO₃ decomposes according to the following reaction.The molar mass for KClO₃is 122.55 g mol^-1. 2KClO₃(s) → 2KCl(s) + 3O₂(g)

Define hypoxia.

A gas mixture contains CO,Ar,and H₂.What is the total pressure of the mixture if the mole fraction of H₂ is 0.35 and the pressure of H₂ is 0.58 bar?

Quiz 5: Gases

When 14.0 g of zinc metal reacts with excess HCl,how many litres of H2 gas are produced at STP? Zn(s) +2HCl(aq) →ZnCl2(aq) +H2(g) \mathrm { Zn } ( s ) + 2 \mathrm { HCl } ( a q ) \rightarrow \mathrm { ZnCl } _ { 2 } ( a q ) + \mathrm { H } _ { 2 } ( g ) Zn(s) +2HCl(aq) →ZnCl2​(aq) +H2​(g)

Determine the volume of H2S (at 375 K and 1.216 bar) needed to produce 55.0 g of S.Assume that there is excess SO2 present. 2H2S(g) + SO2(g) → 3S(s) + 2H2O(g)

How many grams of zinc metal are required to produce 2.00 L of hydrogen gas at STP according to the chemical equation shown below? Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

The mole fraction of oxygen in dry air near sea level is 0.20948.The concentration of oxygen is ________ molecules per litre,assuming an atmospheric pressure of 0.985 bar and a temperature of 29.5 °C.

A mixture of 0.220 moles CO,0.350 moles H2,and 0.640 moles He has a total pressure of 2.95 bar.What is the pressure of CO?

In a container containing CO,H2,and O2,what is the mole fraction of CO if the H2 mole fraction is 0.22 and the O2 mole fraction is 0.58?

A mixture of 0.220 moles CO,0.350 moles H2,and 0.640 moles He has a total pressure of 2.989 bar.What is the pressure of H2?

How many litres of oxygen are needed to exactly react with 19.8 g of methane at STP? CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

A mixture of He,Ne,and Ar has a pressure of 7.85 bar.If the Ne has a mole fraction of 0.47 and Ar has a mole fraction of 0.23,what is the pressure of He?

A mixture of N2,O2,and Ar has mole fractions of 0.25,0.65,and 0.10,respectively.What is the pressure of N2 if the total pressure of the mixture is 3.9 bar?

Define the vapour pressure of water.

What pressure would a gas mixture in a 10.0 L tank exert if it were composed of 48.5 g He and 94.6 g CO2 at 398 K?

A mixture of 10.0 g of Ne and 10.0 g Ar have a total pressure of 1.6 bar.What is the partial pressure of Ne?

A syringe contains 589 mL of CO at 325 K and 1.2 bar.A second syringe contains 473 mL of N2 at 298 K and 2.6 bar.What is the final pressure if the contents of these two syringes are injected into a 1.00 L container at 273 K?

A gas mixture consists of N2,O2,and Ne,where the mole fraction of N2 is 0.55 and the mole fraction of Ne is 0.25.If the mixture is at STP in a 5.0 L container,how many molecules of O2 are present?

Identify the gas that is the lowest percent by volume in dry air.

Determine the volume of O2 (at STP) formed when 50.0 g of KClO3 decomposes according to the following reaction.The molar mass for KClO3 is 122.55 g mol-1. 2KClO3(s) → 2KCl(s) + 3O2(g)

Define hypoxia.

A gas mixture contains CO,Ar,and H2.What is the total pressure of the mixture if the mole fraction of H2 is 0.35 and the pressure of H2 is 0.58 bar?

When 14.0 g of zinc metal reacts with excess HCl,how many litres of H₂ gas are produced at STP? $\mathrm { Zn } ( s ) + 2 \mathrm { HCl } ( a q ) \rightarrow \mathrm { ZnCl } _ { 2 } ( a q ) + \mathrm { H } _ { 2 } ( g )$

Determine the volume of H₂S (at 375 K and 1.216 bar) needed to produce 55.0 g of S.Assume that there is excess SO₂ present. 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(g)

How many grams of zinc metal are required to produce 2.00 L of hydrogen gas at STP according to the chemical equation shown below? Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

A mixture of 0.220 moles CO,0.350 moles H₂,and 0.640 moles He has a total pressure of 2.95 bar.What is the pressure of CO?

In a container containing CO,H₂,and O₂,what is the mole fraction of CO if the H₂ mole fraction is 0.22 and the O₂ mole fraction is 0.58?

A mixture of 0.220 moles CO,0.350 moles H₂,and 0.640 moles He has a total pressure of 2.989 bar.What is the pressure of H₂?

How many litres of oxygen are needed to exactly react with 19.8 g of methane at STP? CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

A mixture of N₂,O₂,and Ar has mole fractions of 0.25,0.65,and 0.10,respectively.What is the pressure of N₂ if the total pressure of the mixture is 3.9 bar?

What pressure would a gas mixture in a 10.0 L tank exert if it were composed of 48.5 g He and 94.6 g CO₂ at 398 K?

A syringe contains 589 mL of CO at 325 K and 1.2 bar.A second syringe contains 473 mL of N₂ at 298 K and 2.6 bar.What is the final pressure if the contents of these two syringes are injected into a 1.00 L container at 273 K?

A gas mixture consists of N₂,O₂,and Ne,where the mole fraction of N₂ is 0.55 and the mole fraction of Ne is 0.25.If the mixture is at STP in a 5.0 L container,how many molecules of O₂ are present?

Determine the volume of O₂ (at STP) formed when 50.0 g of KClO₃ decomposes according to the following reaction.The molar mass for KClO₃is 122.55 g mol^-1. 2KClO₃(s) → 2KCl(s) + 3O₂(g)

A gas mixture contains CO,Ar,and H₂.What is the total pressure of the mixture if the mole fraction of H₂ is 0.35 and the pressure of H₂ is 0.58 bar?