The mass of Xe in the given room needs to be calculated. Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is: PV = nRT - (1) Where R is Universal gas constant.

Question

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Answer 1

Question

Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

a.

Expert Solution

Answer to Problem 121AE

$0.177 g$ is the mass of Xe in the given room.

Explanation of Solution

Given:

The volume of Xe in atmosphere = $9.0×10−6 %$ at $25o C$ and $1 atm$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

$Volume = 7.26 m×8.80 m×5.26 mVolume = = 336.05 m3$

Converting volume from m³to L as:

$1 m3=103 L$

$= 336.05 m3 = 336.05 × 103 L$

The amount of Xe present in the atmosphere is:

$9.0 × 10−6 % of Xe = ( 9.0 × 10 −6 100)= 9.0 × 10−8$

So, $9.0 × 10−8$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

$Volume of Xenon = volume of room × amount of Xenon in airVolume of Xenon = 336.05 × 103 L × 9.0 × 10−8Volume of Xenon = 3.02 × 10−2 L$

Now, the number of moles of Xe present in the room is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 3.3 × 10−2 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (3.3× 10 −2 L) (0 .08206 Latm K −1 mol −1) (298 K)n= 3.3× 10 −224.44 moln = 1.35×10−3 mol$

Now, the mass of Xe is determined using formula:

$number of moles, n = mass, mmolar mass, M$

Rearranging as:

$Mass = n × M$

Molar mass of Xe is $131.3 g/mol$ .

Substituting the values:

$Mass = 1.35 × 10−3 mol × 131.3 g /molMass = 0.177 g$

Thus, the mass of Xe in the given room is $0.177 g$ .

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

b.

Expert Solution

Answer to Problem 121AE

$5×1015 atoms$ is the number of Xe atoms inhaled in each breath.

Explanation of Solution

The amount of Xe in $2 L$ air is:

$Volume of Xe in 2 L air = Xe per L × given amountVolume of Xe in 2 L air = 9.0 × 10−8 × 2 LVolume of Xe in 2 L air = 1.8 × 10−7 L$

Now, the number of moles of Xe present in the breath is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 1.8 × 10−7 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (1.8 × 10 −7 L) (0 .08206 Latm K −1 mol −1) (298 K)n = 7.4 × 10−9 mol$

Now, the number of atoms ofXe is determined using formula:

$Number of atoms = n × NA$

Where $NA$ is Avogadro’s constant having value $6.022 × 1023 atoms/mol$ .

$Number of atoms = 7.4×10−9 mol × 6.022×1023 atoms/molNumber of atoms = 4.5×1015 atomsNumber of atoms ≃ 5×1015 atoms$

Thus, the number of Xe atoms inhaled in each breath is $5×1015 atoms$ .

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1. Identify the following alkenes as E or Z NH₂ Br 2. Draw the structures based on the IUPAC names (3R,4R)-3-bromo-4-fluoro- 1-hexene (Z)-4-bromo-2-iodo-3-ethyl- 3-heptene تر 3. For the following, predict all possible elimination product(s) and circle the major product. HO H₂SO4 Heat 80 F4 OH H2SO4 Heat 어요 F5 F6 1 A DII 4 F7 F8 F9 % & 5 6 7 * ∞ 8 BAB 3 E R T Y U 9 F D G H J K O A F11 F10

Draw the major product of this reaction. Ignore inorganic byproducts. ○ O 1. H₂O, pyridine 2. neutralizing work-up a N W X 人 P

Answer 2

Question

Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

a.

Expert Solution

Answer to Problem 121AE

$0.177 g$ is the mass of Xe in the given room.

Explanation of Solution

Given:

The volume of Xe in atmosphere = $9.0×10−6 %$ at $25o C$ and $1 atm$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

$Volume = 7.26 m×8.80 m×5.26 mVolume = = 336.05 m3$

Converting volume from m³to L as:

$1 m3=103 L$

$= 336.05 m3 = 336.05 × 103 L$

The amount of Xe present in the atmosphere is:

$9.0 × 10−6 % of Xe = ( 9.0 × 10 −6 100)= 9.0 × 10−8$

So, $9.0 × 10−8$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

$Volume of Xenon = volume of room × amount of Xenon in airVolume of Xenon = 336.05 × 103 L × 9.0 × 10−8Volume of Xenon = 3.02 × 10−2 L$

Now, the number of moles of Xe present in the room is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 3.3 × 10−2 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (3.3× 10 −2 L) (0 .08206 Latm K −1 mol −1) (298 K)n= 3.3× 10 −224.44 moln = 1.35×10−3 mol$

Now, the mass of Xe is determined using formula:

$number of moles, n = mass, mmolar mass, M$

Rearranging as:

$Mass = n × M$

Molar mass of Xe is $131.3 g/mol$ .

Substituting the values:

$Mass = 1.35 × 10−3 mol × 131.3 g /molMass = 0.177 g$

Thus, the mass of Xe in the given room is $0.177 g$ .

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

b.

Expert Solution

Answer to Problem 121AE

$5×1015 atoms$ is the number of Xe atoms inhaled in each breath.

Explanation of Solution

The amount of Xe in $2 L$ air is:

$Volume of Xe in 2 L air = Xe per L × given amountVolume of Xe in 2 L air = 9.0 × 10−8 × 2 LVolume of Xe in 2 L air = 1.8 × 10−7 L$

Now, the number of moles of Xe present in the breath is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 1.8 × 10−7 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (1.8 × 10 −7 L) (0 .08206 Latm K −1 mol −1) (298 K)n = 7.4 × 10−9 mol$

Now, the number of atoms ofXe is determined using formula:

$Number of atoms = n × NA$

Where $NA$ is Avogadro’s constant having value $6.022 × 1023 atoms/mol$ .

$Number of atoms = 7.4×10−9 mol × 6.022×1023 atoms/molNumber of atoms = 4.5×1015 atomsNumber of atoms ≃ 5×1015 atoms$

Thus, the number of Xe atoms inhaled in each breath is $5×1015 atoms$ .

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Answer 3

Question

Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

a.

Expert Solution

Answer to Problem 121AE

$0.177 g$ is the mass of Xe in the given room.

Explanation of Solution

Given:

The volume of Xe in atmosphere = $9.0×10−6 %$ at $25o C$ and $1 atm$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

$Volume = 7.26 m×8.80 m×5.26 mVolume = = 336.05 m3$

Converting volume from m³to L as:

$1 m3=103 L$

$= 336.05 m3 = 336.05 × 103 L$

The amount of Xe present in the atmosphere is:

$9.0 × 10−6 % of Xe = ( 9.0 × 10 −6 100)= 9.0 × 10−8$

So, $9.0 × 10−8$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

$Volume of Xenon = volume of room × amount of Xenon in airVolume of Xenon = 336.05 × 103 L × 9.0 × 10−8Volume of Xenon = 3.02 × 10−2 L$

Now, the number of moles of Xe present in the room is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 3.3 × 10−2 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (3.3× 10 −2 L) (0 .08206 Latm K −1 mol −1) (298 K)n= 3.3× 10 −224.44 moln = 1.35×10−3 mol$

Now, the mass of Xe is determined using formula:

$number of moles, n = mass, mmolar mass, M$

Rearranging as:

$Mass = n × M$

Molar mass of Xe is $131.3 g/mol$ .

Substituting the values:

$Mass = 1.35 × 10−3 mol × 131.3 g /molMass = 0.177 g$

Thus, the mass of Xe in the given room is $0.177 g$ .

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

b.

Expert Solution

Answer to Problem 121AE

$5×1015 atoms$ is the number of Xe atoms inhaled in each breath.

Explanation of Solution

The amount of Xe in $2 L$ air is:

$Volume of Xe in 2 L air = Xe per L × given amountVolume of Xe in 2 L air = 9.0 × 10−8 × 2 LVolume of Xe in 2 L air = 1.8 × 10−7 L$

Now, the number of moles of Xe present in the breath is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 1.8 × 10−7 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (1.8 × 10 −7 L) (0 .08206 Latm K −1 mol −1) (298 K)n = 7.4 × 10−9 mol$

Now, the number of atoms ofXe is determined using formula:

$Number of atoms = n × NA$

Where $NA$ is Avogadro’s constant having value $6.022 × 1023 atoms/mol$ .

$Number of atoms = 7.4×10−9 mol × 6.022×1023 atoms/molNumber of atoms = 4.5×1015 atomsNumber of atoms ≃ 5×1015 atoms$

Thus, the number of Xe atoms inhaled in each breath is $5×1015 atoms$ .

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Answer 4

Question

Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

a.

Expert Solution

Answer to Problem 121AE

$0.177 g$ is the mass of Xe in the given room.

Explanation of Solution

Given:

The volume of Xe in atmosphere = $9.0×10−6 %$ at $25o C$ and $1 atm$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

$Volume = 7.26 m×8.80 m×5.26 mVolume = = 336.05 m3$

Converting volume from m³to L as:

$1 m3=103 L$

$= 336.05 m3 = 336.05 × 103 L$

The amount of Xe present in the atmosphere is:

$9.0 × 10−6 % of Xe = ( 9.0 × 10 −6 100)= 9.0 × 10−8$

So, $9.0 × 10−8$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

$Volume of Xenon = volume of room × amount of Xenon in airVolume of Xenon = 336.05 × 103 L × 9.0 × 10−8Volume of Xenon = 3.02 × 10−2 L$

Now, the number of moles of Xe present in the room is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 3.3 × 10−2 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (3.3× 10 −2 L) (0 .08206 Latm K −1 mol −1) (298 K)n= 3.3× 10 −224.44 moln = 1.35×10−3 mol$

Now, the mass of Xe is determined using formula:

$number of moles, n = mass, mmolar mass, M$

Rearranging as:

$Mass = n × M$

Molar mass of Xe is $131.3 g/mol$ .

Substituting the values:

$Mass = 1.35 × 10−3 mol × 131.3 g /molMass = 0.177 g$

Thus, the mass of Xe in the given room is $0.177 g$ .

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

b.

Expert Solution

Answer to Problem 121AE

$5×1015 atoms$ is the number of Xe atoms inhaled in each breath.

Explanation of Solution

The amount of Xe in $2 L$ air is:

$Volume of Xe in 2 L air = Xe per L × given amountVolume of Xe in 2 L air = 9.0 × 10−8 × 2 LVolume of Xe in 2 L air = 1.8 × 10−7 L$

Now, the number of moles of Xe present in the breath is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 1.8 × 10−7 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (1.8 × 10 −7 L) (0 .08206 Latm K −1 mol −1) (298 K)n = 7.4 × 10−9 mol$

Now, the number of atoms ofXe is determined using formula:

$Number of atoms = n × NA$

Where $NA$ is Avogadro’s constant having value $6.022 × 1023 atoms/mol$ .

$Number of atoms = 7.4×10−9 mol × 6.022×1023 atoms/molNumber of atoms = 4.5×1015 atomsNumber of atoms ≃ 5×1015 atoms$

Thus, the number of Xe atoms inhaled in each breath is $5×1015 atoms$ .

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Answer 5

Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

a.

Expert Solution

Answer to Problem 121AE

$0.177 g$ is the mass of Xe in the given room.

Explanation of Solution

Given:

The volume of Xe in atmosphere = $9.0×10−6 %$ at $25o C$ and $1 atm$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

$Volume = 7.26 m×8.80 m×5.26 mVolume = = 336.05 m3$

Converting volume from m³to L as:

$1 m3=103 L$

$= 336.05 m3 = 336.05 × 103 L$

The amount of Xe present in the atmosphere is:

$9.0 × 10−6 % of Xe = ( 9.0 × 10 −6 100)= 9.0 × 10−8$

So, $9.0 × 10−8$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

$Volume of Xenon = volume of room × amount of Xenon in airVolume of Xenon = 336.05 × 103 L × 9.0 × 10−8Volume of Xenon = 3.02 × 10−2 L$

Now, the number of moles of Xe present in the room is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 3.3 × 10−2 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (3.3× 10 −2 L) (0 .08206 Latm K −1 mol −1) (298 K)n= 3.3× 10 −224.44 moln = 1.35×10−3 mol$

Now, the mass of Xe is determined using formula:

$number of moles, n = mass, mmolar mass, M$

Rearranging as:

$Mass = n × M$

Molar mass of Xe is $131.3 g/mol$ .

Substituting the values:

$Mass = 1.35 × 10−3 mol × 131.3 g /molMass = 0.177 g$

Thus, the mass of Xe in the given room is $0.177 g$ .

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

b.

Expert Solution

Answer to Problem 121AE

$5×1015 atoms$ is the number of Xe atoms inhaled in each breath.

Explanation of Solution

The amount of Xe in $2 L$ air is:

$Volume of Xe in 2 L air = Xe per L × given amountVolume of Xe in 2 L air = 9.0 × 10−8 × 2 LVolume of Xe in 2 L air = 1.8 × 10−7 L$

Now, the number of moles of Xe present in the breath is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 1.8 × 10−7 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (1.8 × 10 −7 L) (0 .08206 Latm K −1 mol −1) (298 K)n = 7.4 × 10−9 mol$

Now, the number of atoms ofXe is determined using formula:

$Number of atoms = n × NA$

Where $NA$ is Avogadro’s constant having value $6.022 × 1023 atoms/mol$ .

$Number of atoms = 7.4×10−9 mol × 6.022×1023 atoms/molNumber of atoms = 4.5×1015 atomsNumber of atoms ≃ 5×1015 atoms$

Thus, the number of Xe atoms inhaled in each breath is $5×1015 atoms$ .

Answer 6

Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

a.

Expert Solution

Answer to Problem 121AE

$0.177 g$ is the mass of Xe in the given room.

Explanation of Solution

Given:

The volume of Xe in atmosphere = $9.0×10−6 %$ at $25o C$ and $1 atm$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

$Volume = 7.26 m×8.80 m×5.26 mVolume = = 336.05 m3$

Converting volume from m³to L as:

$1 m3=103 L$

$= 336.05 m3 = 336.05 × 103 L$

The amount of Xe present in the atmosphere is:

$9.0 × 10−6 % of Xe = ( 9.0 × 10 −6 100)= 9.0 × 10−8$

So, $9.0 × 10−8$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

$Volume of Xenon = volume of room × amount of Xenon in airVolume of Xenon = 336.05 × 103 L × 9.0 × 10−8Volume of Xenon = 3.02 × 10−2 L$

Now, the number of moles of Xe present in the room is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 3.3 × 10−2 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (3.3× 10 −2 L) (0 .08206 Latm K −1 mol −1) (298 K)n= 3.3× 10 −224.44 moln = 1.35×10−3 mol$

Now, the mass of Xe is determined using formula:

$number of moles, n = mass, mmolar mass, M$

Rearranging as:

$Mass = n × M$

Molar mass of Xe is $131.3 g/mol$ .

Substituting the values:

$Mass = 1.35 × 10−3 mol × 131.3 g /molMass = 0.177 g$

Thus, the mass of Xe in the given room is $0.177 g$ .

Answer 7

Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

Answer 8

a.

Expert Solution

Answer to Problem 121AE

$0.177 g$ is the mass of Xe in the given room.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The volume of Xe in atmosphere = $9.0×10−6 %$ at $25o C$ and $1 atm$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

$Volume = 7.26 m×8.80 m×5.26 mVolume = = 336.05 m3$

Converting volume from m³to L as:

$1 m3=103 L$

$= 336.05 m3 = 336.05 × 103 L$

The amount of Xe present in the atmosphere is:

$9.0 × 10−6 % of Xe = ( 9.0 × 10 −6 100)= 9.0 × 10−8$

So, $9.0 × 10−8$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

$Volume of Xenon = volume of room × amount of Xenon in airVolume of Xenon = 336.05 × 103 L × 9.0 × 10−8Volume of Xenon = 3.02 × 10−2 L$

Now, the number of moles of Xe present in the room is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 3.3 × 10−2 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (3.3× 10 −2 L) (0 .08206 Latm K −1 mol −1) (298 K)n= 3.3× 10 −224.44 moln = 1.35×10−3 mol$

Now, the mass of Xe is determined using formula:

$number of moles, n = mass, mmolar mass, M$

Rearranging as:

$Mass = n × M$

Molar mass of Xe is $131.3 g/mol$ .

Substituting the values:

$Mass = 1.35 × 10−3 mol × 131.3 g /molMass = 0.177 g$

Thus, the mass of Xe in the given room is $0.177 g$ .

Answer 13

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

b.

Expert Solution

Answer to Problem 121AE

$5×1015 atoms$ is the number of Xe atoms inhaled in each breath.

Explanation of Solution

The amount of Xe in $2 L$ air is:

$Volume of Xe in 2 L air = Xe per L × given amountVolume of Xe in 2 L air = 9.0 × 10−8 × 2 LVolume of Xe in 2 L air = 1.8 × 10−7 L$

Now, the number of moles of Xe present in the breath is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 1.8 × 10−7 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (1.8 × 10 −7 L) (0 .08206 Latm K −1 mol −1) (298 K)n = 7.4 × 10−9 mol$

Now, the number of atoms ofXe is determined using formula:

$Number of atoms = n × NA$

Where $NA$ is Avogadro’s constant having value $6.022 × 1023 atoms/mol$ .

$Number of atoms = 7.4×10−9 mol × 6.022×1023 atoms/molNumber of atoms = 4.5×1015 atomsNumber of atoms ≃ 5×1015 atoms$

Thus, the number of Xe atoms inhaled in each breath is $5×1015 atoms$ .

Answer 14

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$PV = nRT$ - (1)

Where R is Universal gas constant.

Answer 15

b.

Expert Solution

Answer to Problem 121AE

$5×1015 atoms$ is the number of Xe atoms inhaled in each breath.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The amount of Xe in $2 L$ air is:

$Volume of Xe in 2 L air = Xe per L × given amountVolume of Xe in 2 L air = 9.0 × 10−8 × 2 LVolume of Xe in 2 L air = 1.8 × 10−7 L$

Now, the number of moles of Xe present in the breath is determined using equation (1):

$PV = nRT$

The values to be substituted are:

$P = 1.0 atm$ , $V = 1.8 × 10−7 L$ , $R = 0.08206 Latm K−1mol−1$ and $T = 25+273 = 298 K$

$n = PVRTn = (1.0 atm) (1.8 × 10 −7 L) (0 .08206 Latm K −1 mol −1) (298 K)n = 7.4 × 10−9 mol$

Now, the number of atoms ofXe is determined using formula:

$Number of atoms = n × NA$

Where $NA$ is Avogadro’s constant having value $6.022 × 1023 atoms/mol$ .

$Number of atoms = 7.4×10−9 mol × 6.022×1023 atoms/molNumber of atoms = 4.5×1015 atomsNumber of atoms ≃ 5×1015 atoms$

Thus, the number of Xe atoms inhaled in each breath is $5×1015 atoms$ .

Answer 20

Ch. 18 - Prob. 1E Ch. 18 - Prob. 2E Ch. 18 - Prob. 3E Ch. 18 - Prob. 4E Ch. 18 - Prob. 5E Ch. 18 - Prob. 6E Ch. 18 - Prob. 7E Ch. 18 - Prob. 8E Ch. 18 - Prob. 9E Ch. 18 - The electrolysis of aqueous sodium chloride...

The mass of Xe in the given room needs to be calculated. Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is: PV = nRT - (1) Where R is Universal gas constant.

Answer to Problem 121AE

Explanation of Solution

Answer to Problem 121AE

Explanation of Solution

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