Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Chapter U3.16, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The number of molecules of each gas, H2, N2 and CO2 needs to be calculated at STP.

Concept Introduction:

STP is a short form of standard temperature and pressure. The value for standard pressure and temperature is 1 atm and 273.15 K respectively.

At STP 1 mol of any gas occupies 22.4 L.

(a)

Expert Solution
Check Mark

Answer to Problem 4E

Number of molecules in 22.4 L of each gas is 6.023×1023 .

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol. Since, 1 mol gas must contain Avogadro’s number of molecules and the given gases, H2, N2 and CO2 are of 1 mole.

Thus, all the gas must contain 6.023×1023 number of molecules.

b)

Interpretation Introduction

Interpretation :

The number of density of atoms of each gas, H2, N2 and CO2 needs to be calculated at STP.

Concept Introduction:

Number density of a gas sample is the number of gas molecules present in certain volume of gas sample. It can be also expressed in mol/L.

b)

Expert Solution
Check Mark

Answer to Problem 4E

Number density for H2, N2 and CO2 will be same which is 0.045 mol/L.

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol.

The number of density of atoms is calculated using formula:

  number of density of atoms = nV

Where n is number of moles and V is volume.

Since, the number of moles of each gas is 1 and volume is 22.4 L so,

density (n/V) =122.4 mol/L =0.045 mol/L.

c)

Interpretation Introduction

Interpretation:

The sample that contains largest mass needs to be determined amongst the following:

H2, N2 and CO2

Concept Introduction:

The formula used to determine the mass of a sample is:

  Mass of a sample= number of mol of sample × Molecular weight of sample 

c)

Expert Solution
Check Mark

Answer to Problem 4E

22.4 L CO2 will have the largest mass.

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol.

Molecular weight of H2, N2 and CO2 are 2 g/mol, 28 g/mol and 144 g/mol respectively.

Thus, mass of 1 mol H2=1 mol×2 g/mol=2 g .

Mass of 1 mol N2=1 mol×28  g/mol=28 g .

Mass of 1 mol CO2=1 mol×44  g/mol=44 g .

d)

Interpretation Introduction

Interpretation :

The sample that contains largest number of atoms needs to be determined amongst the following:

H2, N2 and CO2

Concept Introduction :

Number of molecules per mol of substance is equal to Avogadro’s number which is equal to 6.023×1023 .

Atomicity is the number of atoms per molecules of an element.

Number of atoms =atomicity×number of molecules

d)

Expert Solution
Check Mark

Answer to Problem 4E

CO2 has the largest number of atoms.

Explanation of Solution

At STP 22.4 L H2, N2 and CO2 gas is equal to 1 mol.

Number of molecules in each gas sample is 6.023×1023 .

H2, N2 and CO2have 2,2 and 3 atoms per molecule respectively.

Number of atoms of H2=2×6.023×1023.

Number of atoms of N2=2×6.023×1023.

Number of atoms of CO2=3×6.023×1023.

e)

Interpretation Introduction

Interpretation:

The sample that contains largest density needs to be determined amongst the following:

H2, N2 and CO2

Concept Introduction:

The ratio of mass and volume of a substance is said to be its density. Mathematically,

  Mass density = mass(m)volume(V)

e)

Expert Solution
Check Mark

Answer to Problem 4E

CO2 has maximum mass density.

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol.

Molecular weight of H2, N2 and CO2 are 2 g/mol, 28 g/mol and 144 g/mol respectively.

Thus, mass of 1 mol H2=1 mol×2 g/mol=2 g .

Mass density =222.4 g/L=0.089 g/L

Mass of 1 mol N2=1 mol×28  g/mol=28 g .

Mass density =2822.4 g/L=1.25 g/L

Mass of 1 mol CO2=1 mol×44  g/mol=44 g

Mass density =4422.4 g/L=1.96 g/L .

Chapter U3 Solutions

Living By Chemistry: First Edition Textbook

Ch. U3.2 - Prob. 4ECh. U3.2 - Prob. 5ECh. U3.2 - Prob. 6ECh. U3.2 - Prob. 7ECh. U3.3 - Prob. 1TAICh. U3.3 - Prob. 1ECh. U3.3 - Prob. 2ECh. U3.3 - Prob. 3ECh. U3.3 - Prob. 4ECh. U3.3 - Prob. 5ECh. U3.3 - Prob. 6ECh. U3.3 - Prob. 7ECh. U3.3 - Prob. 8ECh. U3.3 - Prob. 9ECh. U3.4 - Prob. 1TAICh. U3.4 - Prob. 1ECh. U3.4 - Prob. 2ECh. U3.4 - Prob. 3ECh. U3.4 - Prob. 4ECh. U3.4 - Prob. 5ECh. U3.4 - Prob. 6ECh. U3.4 - Prob. 7ECh. U3.4 - Prob. 8ECh. U3.4 - Prob. 9ECh. U3.5 - Prob. 1TAICh. U3.5 - Prob. 1ECh. U3.5 - Prob. 2ECh. U3.5 - Prob. 3ECh. U3.5 - Prob. 4ECh. U3.5 - Prob. 5ECh. U3.5 - Prob. 6ECh. U3.5 - Prob. 7ECh. U3.5 - Prob. 8ECh. U3.5 - Prob. 9ECh. U3.5 - Prob. 10ECh. U3.5 - Prob. 11ECh. U3.5 - Prob. 12ECh. U3.6 - Prob. 1TAICh. U3.6 - Prob. 1ECh. U3.6 - Prob. 2ECh. U3.6 - Prob. 3ECh. U3.6 - Prob. 4ECh. U3.6 - Prob. 5ECh. U3.6 - Prob. 6ECh. U3.6 - Prob. 7ECh. U3.6 - Prob. 8ECh. U3.7 - Prob. 1TAICh. U3.7 - Prob. 1ECh. U3.7 - Prob. 2ECh. U3.7 - Prob. 3ECh. U3.7 - Prob. 4ECh. U3.7 - Prob. 5ECh. U3.7 - Prob. 6ECh. U3.8 - Prob. 1TAICh. U3.8 - Prob. 1ECh. U3.8 - Prob. 2ECh. U3.8 - Prob. 3ECh. U3.8 - Prob. 4ECh. U3.8 - Prob. 5ECh. U3.8 - Prob. 6ECh. U3.8 - Prob. 7ECh. U3.8 - Prob. 8ECh. U3.8 - Prob. 9ECh. U3.8 - Prob. 10ECh. U3.9 - Prob. 1TAICh. U3.9 - Prob. 1ECh. U3.9 - Prob. 2ECh. U3.9 - Prob. 3ECh. U3.9 - Prob. 4ECh. U3.9 - Prob. 5ECh. U3.9 - Prob. 6ECh. U3.9 - Prob. 7ECh. U3.9 - Prob. 8ECh. U3.9 - Prob. 10ECh. U3.10 - Prob. 1TAICh. U3.10 - Prob. 1ECh. U3.10 - Prob. 2ECh. U3.10 - Prob. 4ECh. U3.10 - Prob. 5ECh. U3.10 - Prob. 6ECh. U3.10 - Prob. 7ECh. U3.11 - Prob. 1TAICh. U3.11 - Prob. 1ECh. U3.11 - Prob. 2ECh. U3.11 - Prob. 3ECh. U3.11 - Prob. 4ECh. U3.11 - Prob. 5ECh. U3.12 - Prob. 1TAICh. U3.12 - Prob. 1ECh. U3.12 - Prob. 2ECh. U3.12 - Prob. 3ECh. U3.12 - Prob. 4ECh. U3.12 - Prob. 5ECh. U3.12 - Prob. 6ECh. U3.12 - Prob. 7ECh. U3.12 - Prob. 8ECh. U3.13 - Prob. 1TAICh. U3.13 - Prob. 1ECh. U3.13 - Prob. 2ECh. U3.13 - Prob. 3ECh. U3.13 - Prob. 4ECh. U3.13 - Prob. 5ECh. U3.13 - Prob. 6ECh. U3.13 - Prob. 7ECh. U3.14 - Prob. 1TAICh. U3.14 - Prob. 1ECh. U3.14 - Prob. 2ECh. U3.15 - Prob. 1TAICh. U3.15 - Prob. 1ECh. U3.15 - Prob. 2ECh. U3.15 - Prob. 3ECh. U3.15 - Prob. 4ECh. U3.15 - Prob. 5ECh. U3.15 - Prob. 6ECh. U3.15 - Prob. 7ECh. U3.15 - Prob. 8ECh. U3.16 - Prob. 1TAICh. U3.16 - Prob. 1ECh. U3.16 - Prob. 2ECh. U3.16 - Prob. 3ECh. U3.16 - Prob. 4ECh. U3.16 - Prob. 5ECh. U3.16 - Prob. 6ECh. U3.16 - Prob. 7ECh. U3.17 - Prob. 1TAICh. U3.17 - Prob. 1ECh. U3.17 - Prob. 2ECh. U3.17 - Prob. 3ECh. U3.17 - Prob. 4ECh. U3.17 - Prob. 5ECh. U3.17 - Prob. 6ECh. U3.17 - Prob. 7ECh. U3.17 - Prob. 8ECh. U3.18 - Prob. 1TAICh. U3.18 - Prob. 1ECh. U3.18 - Prob. 2ECh. U3.18 - Prob. 3ECh. U3.18 - Prob. 4ECh. U3.18 - Prob. 5ECh. U3.18 - Prob. 6ECh. U3.18 - Prob. 7ECh. U3.18 - Prob. 8ECh. U3.18 - Prob. 9ECh. U3.18 - Prob. 10ECh. U3.18 - Prob. 11ECh. U3.18 - Prob. 12ECh. U3.19 - Prob. 1TAICh. U3.19 - Prob. 1ECh. U3.19 - Prob. 2ECh. U3.19 - Prob. 4ECh. U3 - Prob. SI1RECh. U3 - Prob. SI2RECh. U3 - Prob. SI3RECh. U3 - Prob. SI4RECh. U3 - Prob. SI5RECh. U3 - Prob. SII1RECh. U3 - Prob. SII2RECh. U3 - Prob. SII3RECh. U3 - Prob. SII4RECh. U3 - Prob. SII5RECh. U3 - Prob. SIII1RECh. U3 - Prob. SIII2RECh. U3 - Prob. SIII3RECh. U3 - Prob. SIII4RECh. U3 - Prob. SIII5RECh. U3 - Prob. 1RECh. U3 - Prob. 2RECh. U3 - Prob. 3RECh. U3 - Prob. 4RECh. U3 - Prob. 5RECh. U3 - Prob. 6RECh. U3 - Prob. 7RECh. U3 - Prob. 8RECh. U3 - Prob. 9RECh. U3 - Prob. 10RECh. U3 - Prob. 11RECh. U3 - Prob. 12RECh. U3 - Prob. 13RE
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