World of Chemistry, 3rd edition

3rd Edition

ISBN: 9781133109655

Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste

Publisher: Brooks / Cole / Cengage Learning

See similar textbooks

Concept explainers

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can b…

Videos

Question

Chapter 9, Problem 15A

(a)

Interpretation Introduction

Interpretation: The mass in grams in $2.21 ×10-4$ moles of calcium carbonate needs to be calculated.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

$Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023$

(a)

Expert Solution

Answer to Problem 15A

$2.21 ×10-2 g$

Explanation of Solution

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate= $2.21 ×10-4$

Thus, there are 100 g in 1 mol.

Calculate mass in grams:

$2.21 ×10-4moles ×100 g1 mole = 2.21 ×10-2 g$

(b)

Interpretation Introduction

Interpretation: The mass in grams in 2.75 moles of Helium needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

$Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023$

(b)

Expert Solution

Answer to Problem 15A

$11.0 g$

Explanation of Solution

Molar mass of Helium = 4.00 g/mol

Moles of Helium = 2.75 moles

Thus, 1 mol of He has 4.00 g.

Calculate mass in grams:

$2.75 moles ×4.00 g1 mole = 11.0 g$

(c)

Interpretation Introduction

Interpretation: The mass in grams in 0.00975 moles of oxygen gas needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

$Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023$

(c)

Expert Solution

Answer to Problem 15A

$0.312 g$

Explanation of Solution

Molar mass of oxygen gas = 32.0 g/mol

Moles of oxygen gas = 0.00975 moles

Thus, 1 mol of oxygen gas has 32 g.

Calculate mass in grams:

$0.00975 moles ×32.0 g1 mole = 0.312 g$

(d)

Interpretation Introduction

Interpretation: The mass in grams in $7.21 ×10-3$ moles of carbon dioxide needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

$Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023$

(d)

Expert Solution

Answer to Problem 15A

$317×10-3g$

Explanation of Solution

Molar mass of carbon dioxide = 44.0 g/mol

Moles of carbon dioxide = $7.21 ×10-3$ moles

1 mol of carbon dioxide has 44 g.

Calculate mass in grams:

$7.21 ×10-3moles ×44.0 g1 mole = 317×10-3g$

(e)

Interpretation Introduction

Interpretation: The mass in grams in 0.835 moles of iron (II) sulphide needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

$Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023$

(e)

Expert Solution

Answer to Problem 15A

Mass = $73.4 g$

Explanation of Solution

Molar mass of iron (II) sulfide= 87.9 g/mol

Moles of iron (II) sulfide= 0.835 moles

Thus, 1 mol of iron sulphide has 87.9 g.

Calculate mass in grams:

$0.835 moles ×87.9 g1 mole = 73.4 g$

(f)

Interpretation Introduction

Interpretation: The mass in grams in 4.01 moles of potassium hydroxide needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

$Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023$

(f)

Expert Solution

Answer to Problem 15A

Mass = 225 g

Explanation of Solution

Molar mass of potassium hydroxide= 56.1 g/mol

Moles of potassium hydroxide= 4.01 moles

1 mol of potassium hydroxide has 56.1 g.

Calculate mass in grams:

$4.01 moles ×56.1 g1 mole = 225 g$

(g)

Interpretation Introduction

Interpretation: The mass in grams in 0.0219 moles of hydrogen gas needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

$Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023$

(g)

Expert Solution

Answer to Problem 15A

Mass = $0.044 g$

Explanation of Solution

Molar mass of hydrogen gas = 2.01 g/mol

Moles of hydrogen gas= 0.0219 moles

1 mole of hydrogen gas has 2.01 g.

Calculate mass in grams:

$0.0219 moles ×2.01 g1 mole = 0.044 g$

Chapter 9, Problem 14A

Chapter 9, Problem 16A

Chapter 9 Solutions

World of Chemistry, 3rd edition

Ch. 9.1 - Prob. 1RQ Ch. 9.1 - Prob. 2RQ Ch. 9.1 - Prob. 3RQ Ch. 9.1 - Prob. 4RQ Ch. 9.1 - Prob. 5RQ Ch. 9.2 - Prob. 1RQ Ch. 9.2 - Prob. 2RQ Ch. 9.2 - Prob. 3RQ Ch. 9.2 - Prob. 4RQ Ch. 9.2 - Prob. 5RQ

Ch. 9.2 - Prob. 6RQ Ch. 9.3 - Prob. 1RQ Ch. 9.3 - Prob. 2RQ Ch. 9.3 - Prob. 3RQ Ch. 9.3 - Prob. 4RQ Ch. 9.3 - Prob. 5RQ Ch. 9 - Prob. 1A Ch. 9 - Prob. 2A Ch. 9 - Prob. 3A Ch. 9 - Prob. 4A Ch. 9 - Prob. 5A Ch. 9 - Prob. 6A Ch. 9 - Prob. 7A Ch. 9 - Prob. 8A Ch. 9 - Prob. 9A Ch. 9 - Prob. 10A Ch. 9 - Prob. 11A Ch. 9 - Prob. 12A Ch. 9 - Prob. 13A Ch. 9 - Prob. 14A Ch. 9 - Prob. 15A Ch. 9 - Prob. 16A Ch. 9 - Prob. 17A Ch. 9 - Prob. 18A Ch. 9 - Prob. 19A Ch. 9 - Prob. 20A Ch. 9 - Prob. 21A Ch. 9 - Prob. 22A Ch. 9 - Prob. 23A Ch. 9 - Prob. 24A Ch. 9 - Prob. 25A Ch. 9 - Prob. 26A Ch. 9 - Prob. 27A Ch. 9 - Prob. 28A Ch. 9 - Prob. 29A Ch. 9 - Prob. 30A Ch. 9 - Prob. 31A Ch. 9 - Prob. 32A Ch. 9 - Prob. 33A Ch. 9 - Prob. 34A Ch. 9 - Prob. 35A Ch. 9 - Prob. 36A Ch. 9 - Prob. 37A Ch. 9 - Prob. 38A Ch. 9 - Prob. 39A Ch. 9 - Prob. 40A Ch. 9 - Prob. 41A Ch. 9 - Prob. 42A Ch. 9 - Prob. 43A Ch. 9 - Prob. 44A Ch. 9 - Prob. 45A Ch. 9 - Prob. 46A Ch. 9 - Prob. 47A Ch. 9 - Prob. 48A Ch. 9 - Prob. 49A Ch. 9 - Prob. 50A Ch. 9 - Prob. 51A Ch. 9 - Prob. 52A Ch. 9 - Prob. 53A Ch. 9 - Prob. 54A Ch. 9 - Prob. 55A Ch. 9 - Prob. 56A Ch. 9 - Prob. 57A Ch. 9 - Prob. 58A Ch. 9 - Prob. 59A Ch. 9 - Prob. 60A Ch. 9 - Prob. 61A Ch. 9 - Prob. 62A Ch. 9 - Prob. 63A Ch. 9 - Prob. 64A Ch. 9 - Prob. 65A Ch. 9 - Prob. 66A Ch. 9 - Prob. 67A Ch. 9 - Prob. 68A Ch. 9 - Prob. 69A Ch. 9 - Prob. 70A Ch. 9 - Prob. 71A Ch. 9 - Prob. 1STP Ch. 9 - Prob. 2STP Ch. 9 - Prob. 3STP Ch. 9 - Prob. 4STP Ch. 9 - Prob. 5STP Ch. 9 - Prob. 6STP Ch. 9 - Prob. 7STP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.