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 U4.11, Problem 8E

(a)

Interpretation Introduction

Interpretation:

The number of atoms in the container A and B should be compared.

Concept Introduction:

The amount of the substance that contains the same number of particles or atoms or molecules is termed as mole.

(a)

Expert Solution
Check Mark

Answer to Problem 8E

The number of atoms in the container A are more.

Explanation of Solution

The number of atoms can be expressed as the product as the number of moles and Avogadro number. The expression for number of atoms is represented as follows:

  Number of atoms=Number of moles×Avogadro number

As container A contains a total of 2 mol of elements thus, it will have larger number of atoms.

(b)

Interpretation Introduction

Interpretation:

The number of gaseous molecules in the container A and B should be compared.

Concept Introduction:

The amount of the substance that contains the same number of particles or atoms or molecules is termed as mole.

(b)

Expert Solution
Check Mark

Answer to Problem 8E

The number of gaseous molecules in the container A and B are same.

Explanation of Solution

In container A 1 mol of carbon and 1 mol of oxygen gas is present whereas in container B 1 mol of carbon dioxide gas is present.

That means each container contains one mole of each of oxygen gas and carbon dioxide gas thus, both containers A and B will have same gaseous molecules.

(c)

Interpretation Introduction

Interpretation:

The mass of the elements or molecules present in the container A and B should be compared.

Concept Introduction:

The mass of the atoms that are present in the chemical formula is termed as Molar mass. It corresponds to the sum of the atomic masses of all atoms that are present in the formula of the any compound.

(c)

Expert Solution
Check Mark

Answer to Problem 8E

The mass of the elements or molecules present in the container A and B is same.

Explanation of Solution

In container A 1 mol of carbon and 1 mol of oxygen gas is present whereas in container B 1 mol of carbon dioxide gas is present.

Thus, the mass of elements present in the container A and B will be calculated as follows:

  MassofContainerA=1×C+2×O=(1×12.011+2×15.999)g/mol=44.009g/molMassofContainerB=1×C+2×O=(1×12.011+2×15.999)g/mol=44.009g/mol

As the mass of elements present in the container A and B comes out be equal thus, both containers have same mass.

(d)

Interpretation Introduction

Interpretation:

The pressure of the elements or molecules present in the container A and B should be compared.

Concept Introduction:

Gaseous molecules are highly compressible due to the presence of large intermolecular distance in between the gaseous particles.

(d)

Expert Solution
Check Mark

Answer to Problem 8E

The pressure of the elements or molecules present in the container A and B is same.

Explanation of Solution

In container A 1 mol of carbon and 1 mol of oxygen gas is present whereas in container B 1 mol of carbon dioxide gas is present. As both container consists of 1 mol of one gas that means the pressure will be exerted by 1 gas at a time in a respective container thus, both container A and B will have same gaseous pressure.

Chapter U4 Solutions

Living By Chemistry: First Edition Textbook

Ch. U4.2 - Prob. 3ECh. U4.2 - Prob. 4ECh. U4.2 - Prob. 5ECh. U4.3 - Prob. 1TAICh. U4.3 - Prob. 1ECh. U4.3 - Prob. 2ECh. U4.3 - Prob. 3ECh. U4.3 - Prob. 4ECh. U4.3 - Prob. 6ECh. U4.4 - Prob. 1TAICh. U4.4 - Prob. 1ECh. U4.4 - Prob. 2ECh. U4.4 - Prob. 3ECh. U4.4 - Prob. 4ECh. U4.4 - Prob. 5ECh. U4.4 - Prob. 6ECh. U4.4 - Prob. 7ECh. U4.5 - Prob. 1TAICh. U4.5 - Prob. 1ECh. U4.5 - Prob. 2ECh. U4.5 - Prob. 3ECh. U4.5 - Prob. 4ECh. U4.6 - Prob. 1TAICh. U4.6 - Prob. 1ECh. U4.6 - Prob. 2ECh. U4.6 - Prob. 3ECh. U4.6 - Prob. 5ECh. U4.6 - Prob. 6ECh. U4.6 - Prob. 7ECh. U4.7 - Prob. 1TAICh. U4.7 - Prob. 1ECh. U4.7 - Prob. 2ECh. U4.7 - Prob. 4ECh. U4.7 - Prob. 5ECh. U4.8 - Prob. 1TAICh. U4.8 - Prob. 1ECh. U4.8 - Prob. 2ECh. U4.8 - Prob. 4ECh. U4.8 - Prob. 5ECh. U4.8 - Prob. 6ECh. U4.8 - Prob. 7ECh. U4.8 - Prob. 8ECh. U4.8 - Prob. 9ECh. U4.8 - Prob. 10ECh. U4.8 - Prob. 11ECh. U4.9 - Prob. 1TAICh. U4.9 - Prob. 1ECh. U4.9 - Prob. 2ECh. U4.9 - Prob. 3ECh. U4.9 - Prob. 4ECh. U4.9 - Prob. 5ECh. U4.9 - Prob. 6ECh. U4.9 - Prob. 7ECh. U4.9 - Prob. 8ECh. U4.10 - Prob. 1TAICh. U4.10 - Prob. 1ECh. U4.10 - Prob. 2ECh. U4.10 - Prob. 3ECh. U4.10 - Prob. 4ECh. U4.10 - Prob. 5ECh. U4.10 - Prob. 6ECh. U4.10 - Prob. 7ECh. U4.10 - Prob. 8ECh. U4.11 - Prob. 1TAICh. U4.11 - Prob. 1ECh. U4.11 - Prob. 2ECh. U4.11 - Prob. 3ECh. U4.11 - Prob. 4ECh. U4.11 - Prob. 5ECh. U4.11 - Prob. 6ECh. U4.11 - Prob. 7ECh. U4.11 - Prob. 8ECh. U4.12 - Prob. 1TAICh. U4.12 - Prob. 1ECh. U4.12 - Prob. 2ECh. U4.12 - Prob. 3ECh. U4.12 - Prob. 4ECh. U4.13 - Prob. 1TAICh. U4.13 - Prob. 1ECh. U4.13 - Prob. 2ECh. U4.13 - Prob. 3ECh. U4.13 - Prob. 4ECh. U4.13 - Prob. 5ECh. U4.13 - Prob. 6ECh. U4.13 - Prob. 7ECh. U4.13 - Prob. 8ECh. U4.13 - Prob. 9ECh. U4.13 - Prob. 10ECh. U4.14 - Prob. 1TAICh. U4.14 - Prob. 1ECh. U4.14 - Prob. 2ECh. U4.14 - Prob. 3ECh. U4.14 - Prob. 4ECh. U4.14 - Prob. 5ECh. U4.14 - Prob. 6ECh. U4.14 - Prob. 7ECh. U4.14 - Prob. 8ECh. U4.14 - Prob. 9ECh. U4.15 - Prob. 1TAICh. U4.15 - Prob. 1ECh. U4.15 - Prob. 2ECh. U4.15 - Prob. 3ECh. U4.15 - Prob. 4ECh. U4.15 - Prob. 5ECh. U4.15 - Prob. 6ECh. U4.15 - Prob. 7ECh. U4.15 - Prob. 8ECh. U4.16 - Prob. 1TAICh. U4.16 - Prob. 1ECh. U4.16 - Prob. 2ECh. U4.16 - Prob. 3ECh. U4.16 - Prob. 4ECh. U4.16 - Prob. 5ECh. U4.16 - Prob. 6ECh. U4.16 - Prob. 7ECh. U4.17 - Prob. 1TAICh. U4.17 - Prob. 1ECh. U4.17 - Prob. 2ECh. U4.17 - Prob. 3ECh. U4.17 - Prob. 5ECh. U4.17 - Prob. 6ECh. U4.17 - Prob. 7ECh. U4.18 - Prob. 1TAICh. U4.18 - Prob. 1ECh. U4.18 - Prob. 2ECh. U4.18 - Prob. 3ECh. U4.18 - Prob. 4ECh. U4.18 - Prob. 5ECh. U4.18 - Prob. 6ECh. U4.18 - Prob. 7ECh. U4.18 - Prob. 8ECh. U4.19 - Prob. 1TAICh. U4.19 - Prob. 1ECh. U4.19 - Prob. 2ECh. U4.19 - Prob. 3ECh. U4.19 - Prob. 4ECh. U4.19 - Prob. 6ECh. U4.19 - Prob. 7ECh. U4.19 - Prob. 8ECh. U4.20 - Prob. 1TAICh. U4.20 - Prob. 1ECh. U4.20 - Prob. 2ECh. U4.20 - Prob. 3ECh. U4.20 - Prob. 4ECh. U4.20 - Prob. 5ECh. U4.20 - Prob. 6ECh. U4.20 - Prob. 7ECh. U4.20 - Prob. 8ECh. U4.21 - Prob. 1TAICh. U4.21 - Prob. 1ECh. U4.21 - Prob. 2ECh. U4.21 - Prob. 4ECh. U4.21 - Prob. 5ECh. U4.21 - Prob. 6ECh. U4.21 - Prob. 7ECh. U4.21 - Prob. 8ECh. U4.22 - Prob. 1TAICh. U4.22 - Prob. 1ECh. U4.22 - Prob. 2ECh. U4.22 - Prob. 3ECh. U4.22 - Prob. 4ECh. U4.22 - Prob. 5ECh. U4.22 - Prob. 6ECh. U4.23 - Prob. 1ECh. U4.23 - Prob. 2ECh. U4.23 - Prob. 3ECh. U4.23 - Prob. 4ECh. U4.23 - Prob. 5ECh. U4.23 - Prob. 6ECh. U4.23 - Prob. 7ECh. U4.24 - Prob. 1ECh. U4.24 - Prob. 2ECh. U4.24 - Prob. 3ECh. U4.24 - Prob. 5ECh. U4.24 - Prob. 6ECh. U4.25 - Prob. 1TAICh. U4.25 - Prob. 1ECh. U4.25 - Prob. 2ECh. U4.25 - Prob. 3ECh. U4.25 - Prob. 4ECh. U4.26 - Prob. 1TAICh. U4.26 - Prob. 1ECh. U4.26 - Prob. 2ECh. U4.26 - Prob. 4ECh. U4.26 - Prob. 5ECh. U4.26 - Prob. 6ECh. U4 - Prob. SI3RECh. U4 - Prob. SI4RECh. U4 - Prob. SII1RECh. U4 - Prob. SII2RECh. U4 - Prob. SII3RECh. U4 - Prob. SII5RECh. U4 - Prob. SII6RECh. U4 - Prob. SIII1RECh. U4 - Prob. SIII2RECh. U4 - Prob. SIII3RECh. U4 - Prob. SIII4RECh. U4 - Prob. SIII5RECh. U4 - Prob. SIII6RECh. U4 - Prob. SIII7RECh. U4 - Prob. SIII8RECh. U4 - Prob. SIV1RECh. U4 - Prob. SIV2RECh. U4 - Prob. SIV3RECh. U4 - Prob. SIV4RECh. U4 - Prob. SV1RECh. U4 - Prob. SV2RECh. U4 - Prob. SV3RECh. U4 - Prob. 1RECh. U4 - Prob. 4RECh. U4 - Prob. 5RECh. U4 - Prob. 6RECh. U4 - Prob. 7RECh. U4 - Prob. 8RECh. U4 - Prob. 9RECh. U4 - Prob. 10RECh. U4 - Prob. 11RECh. U4 - Prob. 12RE
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