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

Atomic Structure

The basic structure of an atom is defined as the component-level of atomic structure of an atom. Precisely speaking an atom consists of three major subatomic particles which are protons, neutrons, and electrons. Many theories have been stated for explain…

Shape of the D Orbital

Shapes of orbitals are an approximate representation of boundaries in space for finding electrons occupied in that respective orbital. D orbitals are known to have a clover leaf shape or dumbbell inside where electrons can be found.

Videos

Question

Chapter 3, Problem 29A

(a)

Interpretation Introduction

Interpretation: The number of protons and neutrons present in the nucleus of 244Pu94 needs to be determined.

Concept Introduction :

The number of protons in an atom is called the atomic number. The addition of the number of protons and neutrons, is equal to the mass number. When the numbers of neutrons are different the mass number of element is changes but atomic number remains unchanged that consider as an isotope of the element.

(a)

Expert Solution

Answer to Problem 29A

The number of protons= 94

The number of neutrons= 150

The number of electrons = 94

Explanation of Solution

The number of protons= 94

For the number of neutrons= Mass number − number of

=244−94=150

The number of neutrons= 150

When the number of electron is equal to the number of protons, then atom is uncharged.

So the number of electrons = 94

(b)

Interpretation Introduction

Interpretation: The number of protons and neutrons are contained in the nucleus of A243m95 needs to be determined.

Concept Introduction :

The number of protons in an atom is called the atomic number. The sum of the number of protons and neutrons, is equal to the mass number. When the numbers of neutrons are different the mass number of element is changes but atomic number remains unchanged that consider as a isotope of the element.

(b)

Expert Solution

Answer to Problem 29A

The number of protons = 95

The number of neutrons= 148

The number of electrons= 95

Explanation of Solution

The number of protons = 95

For the number of neutrons= Mass number − number of protons

=243−95 = 148

The number of neutrons= 148

When the number of electron is equal to the number of protons, then atom is uncharged.

So the number of electrons= 95

(c)

Interpretation Introduction

Interpretation: The number of protons and neutrons are contained in the nucleus of 277Ac89 needs to be determined.

Concept Introduction :

The number of protons in an atom is called the atomic number. The sum of the number of protons and neutrons, is equal to the mass number. When the numbers of neutrons are different the mass number of element is changes but atomic number remains unchanged that consider as a isotope of the element.

(c)

Expert Solution

Answer to Problem 29A

The number of protons = 89

The number of neutrons =188

The number of electrons= 89

Explanation of Solution

The number of protons = 89

For the number of neutrons= Mass number − number of protons

=277−89=188

The number of neutrons =188

When the number of electron is equal to the number of protons, then atom is uncharged.

So the number of electrons= 89

(d)

Interpretation Introduction

Interpretation: The number of protons and neutrons are contained in the nucleus of 132Cs 55 needs to be determined.

Concept Introduction :

The number of protons in an atom is called the atomic number. The sum of the number of protons and neutrons, is equal to the mass number. When the numbers of neutrons are different the mass number of element is changes but atomic number remains unchanged that consider as a isotope of the element.

(d)

Expert Solution

Answer to Problem 29A

The number of protons = 55

The number of neutrons =77

The number of electrons= 55

Explanation of Solution

The number of protons = 55

For the number of neutrons= Mass number − number of protons

=132−55=77

The number of neutrons =77

When the number of electron is equal to the number of protons, then atom is uncharged.

So the number of electrons= 55

(e)

Interpretation Introduction

Interpretation: The number of protons and neutrons contained in the nucleus of 192Ir77 needs to be determined.

Concept Introduction :

The number of protons in an atom is called the atomic number. The sum of the number of protons and neutrons, is equal to the mass number. When the numbers of neutrons are different the mass number of element is changes but atomic number remains unchanged that consider as a isotope of the element.

(e)

Expert Solution

Answer to Problem 29A

The number of protons = 77

The number of neutrons =115

The number of electrons= 77

Explanation of Solution

The number of protons = 77

For the number of neutrons= Mass number − number of protons

=192−77=115

The number of neutrons =115

When the number of electron is equal to the number of protons, then atom is uncharged.

So the number of electrons= 77

(f)

Interpretation Introduction

Interpretation: The number of protons and neutrons contained in the nucleus of 54Mn25 needs to be determined.

Concept Introduction :

The number of protons in an atom is called the atomic number. The sum of the number of protons and neutrons, is equal to the mass number. When the numbers of neutrons are different the mass number of element is changes but atomic number remains unchanged that consider as a isotope of the element.

(f)

Expert Solution

Answer to Problem 29A

The number of protons = 25

The number of neutrons= 29

The number of electrons= 25

Explanation of Solution

The number of protons = 25

For the number of neutrons= Mass number − number of protons

=54−25=29

The number of neutrons= 29

When the number of electron is equal to the number of protons, then atom is uncharged.

So the number of electrons= 25

Chapter 3, Problem 28A

Chapter 3, Problem 30A

Chapter 3 Solutions

World of Chemistry, 3rd edition

Ch. 3.1 - Prob. 1RQ Ch. 3.1 - Prob. 2RQ Ch. 3.1 - Prob. 3RQ Ch. 3.1 - Prob. 4RQ Ch. 3.1 - Prob. 5RQ Ch. 3.1 - Prob. 6RQ Ch. 3.2 - Prob. 1RQ Ch. 3.2 - Prob. 2RQ Ch. 3.2 - Prob. 3RQ Ch. 3.2 - Prob. 4RQ

Ch. 3.2 - Prob. 5RQ Ch. 3.2 - Prob. 6RQ Ch. 3.3 - Prob. 1RQ Ch. 3.3 - Prob. 2RQ Ch. 3.3 - Prob. 3RQ Ch. 3.3 - Prob. 4RQ Ch. 3.3 - Prob. 5RQ Ch. 3.3 - Prob. 6RQ Ch. 3.3 - Prob. 7RQ Ch. 3.4 - Prob. 1RQ Ch. 3.4 - Prob. 2RQ Ch. 3.4 - Prob. 3RQ Ch. 3.4 - Prob. 4RQ Ch. 3.4 - Prob. 5RQ Ch. 3.4 - Prob. 6RQ Ch. 3.4 - Prob. 7RQ Ch. 3.4 - Prob. 8RQ Ch. 3.5 - Prob. 1RQ Ch. 3.5 - Prob. 2RQ Ch. 3.5 - Prob. 3RQ Ch. 3.5 - Prob. 4RQ Ch. 3.5 - Prob. 5RQ Ch. 3 - Prob. 1A Ch. 3 - Prob. 2A Ch. 3 - Prob. 3A Ch. 3 - Prob. 4A Ch. 3 - Prob. 5A Ch. 3 - Prob. 6A Ch. 3 - Prob. 7A Ch. 3 - Prob. 8A Ch. 3 - Prob. 9A Ch. 3 - Prob. 10A Ch. 3 - Prob. 11A Ch. 3 - Prob. 12A Ch. 3 - Prob. 13A Ch. 3 - Prob. 14A Ch. 3 - Prob. 15A Ch. 3 - Prob. 16A Ch. 3 - Prob. 17A Ch. 3 - Prob. 18A Ch. 3 - Prob. 19A Ch. 3 - Prob. 20A Ch. 3 - Prob. 21A Ch. 3 - Prob. 22A Ch. 3 - Prob. 23A Ch. 3 - Prob. 24A Ch. 3 - Prob. 25A Ch. 3 - Prob. 26A Ch. 3 - Prob. 27A Ch. 3 - Prob. 28A Ch. 3 - Prob. 29A Ch. 3 - Prob. 30A Ch. 3 - Prob. 31A Ch. 3 - Prob. 32A Ch. 3 - Prob. 33A Ch. 3 - Prob. 34A Ch. 3 - Prob. 35A Ch. 3 - Prob. 36A Ch. 3 - Prob. 37A Ch. 3 - Prob. 38A Ch. 3 - Prob. 39A Ch. 3 - Prob. 40A Ch. 3 - Prob. 41A Ch. 3 - Prob. 42A Ch. 3 - Prob. 43A Ch. 3 - Prob. 44A Ch. 3 - Prob. 45A Ch. 3 - Prob. 46A Ch. 3 - Prob. 47A Ch. 3 - Prob. 48A Ch. 3 - Prob. 49A Ch. 3 - Prob. 50A Ch. 3 - Prob. 51A Ch. 3 - Prob. 52A Ch. 3 - Prob. 53A Ch. 3 - Prob. 54A Ch. 3 - Prob. 55A Ch. 3 - Prob. 56A Ch. 3 - Prob. 57A Ch. 3 - Prob. 58A Ch. 3 - Prob. 59A Ch. 3 - Prob. 60A Ch. 3 - Prob. 61A Ch. 3 - Prob. 1STP Ch. 3 - Prob. 2STP Ch. 3 - Prob. 3STP Ch. 3 - Prob. 4STP Ch. 3 - Prob. 5STP Ch. 3 - Prob. 6STP Ch. 3 - Prob. 7STP Ch. 3 - Prob. 8STP Ch. 3 - Prob. 9STP Ch. 3 - Prob. 10STP

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.