Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 25, Problem 21PS

(a)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Concept Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(a)

Expert Solution
Check Mark

Answer to Problem 21PS

F2654e+H24e2n01+2856Ni.

Explanation of Solution

The radioactive isotope of iron-54 when irradiated with alpha particle forms 2856Ni by the emission of 2 neutrons. In 2856Ni has atomic number 28 and mass number 56.

(b)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(b)

Expert Solution
Check Mark

Answer to Problem 21PS

A1327l+H24eP1530+n0-1

Explanation of Solution

The radioactive isotope of Aluminium-27 when irradiated with alpha particle forms P1530 by the emission of one neutrons. In P1530 has atomic number 15 and mass number 30.

(c)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(c)

Expert Solution
Check Mark

Answer to Problem 21PS

S1632+n01H11+1532P

Explanation of Solution

The radioactive isotope of Sulphur-32 is irradiated with neutron forms P1532 by the emission of a proton. In P1532 has atomic number 15 and mass number 32.

(d)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(d)

Expert Solution
Check Mark

Answer to Problem 21PS

M4296o+H12n01+T4397c

Explanation of Solution

The radioactive isotope of M4296o is irradiated with deuterium forms T4397c by the emission of a neutron. In T4397c has atomic number 97 and mass number 43.

(e)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(e)

Expert Solution
Check Mark

Answer to Problem 21PS

M4298o+n01T4399c+β-10

Explanation of Solution

The radioactive isotope of M4298o is irradiated with neutron forms T4399c by the emission of a electron. In T4399c has atomic number 99 and mass number 43.

(f)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(f)

Expert Solution
Check Mark

Answer to Problem 21PS

F918O818+e+10

Explanation of Solution

The radioactive isotope of F918 is irradiated forms O818 by the emission of a positron. In O818 has atomic number 8 and mass number 18.

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Chapter 25 Solutions

Chemistry & Chemical Reactivity

Ch. 25.8 - Prob. 1.2ACPCh. 25.8 - Prob. 1.3ACPCh. 25.8 - Calculate the molar mass (atomic weight) of...Ch. 25.8 - Prob. 2.1ACPCh. 25.8 - Prob. 2.2ACPCh. 25.8 - Prob. 2.3ACPCh. 25.8 - Prob. 2.4ACPCh. 25.8 - Prob. 2.5ACPCh. 25.8 - Prob. 2.6ACPCh. 25.8 - Prob. 3.1ACPCh. 25.8 - Prob. 3.2ACPCh. 25.8 - Prob. 3.3ACPCh. 25.8 - Prob. 3.4ACPCh. 25 - Prob. 1PSCh. 25 - Prob. 2PSCh. 25 - Prob. 4PSCh. 25 - Prob. 5PSCh. 25 - Prob. 6PSCh. 25 - Prob. 7PSCh. 25 - Prob. 8PSCh. 25 - Prob. 9PSCh. 25 - Prob. 10PSCh. 25 - Prob. 11PSCh. 25 - Prob. 12PSCh. 25 - Prob. 13PSCh. 25 - Prob. 14PSCh. 25 - Prob. 15PSCh. 25 - Prob. 16PSCh. 25 - Prob. 17PSCh. 25 - Prob. 18PSCh. 25 - Prob. 19PSCh. 25 - Prob. 20PSCh. 25 - Prob. 21PSCh. 25 - Prob. 22PSCh. 25 - Prob. 23PSCh. 25 - Prob. 24PSCh. 25 - Prob. 25PSCh. 25 - Prob. 26PSCh. 25 - Prob. 27PSCh. 25 - Prob. 28PSCh. 25 - Prob. 29PSCh. 25 - Prob. 30PSCh. 25 - Prob. 31PSCh. 25 - Prob. 32PSCh. 25 - Prob. 33PSCh. 25 - Prob. 34PSCh. 25 - Prob. 35PSCh. 25 - Prob. 36PSCh. 25 - Prob. 37PSCh. 25 - Prob. 38PSCh. 25 - Prob. 39PSCh. 25 - Prob. 40PSCh. 25 - Prob. 41PSCh. 25 - Prob. 42PSCh. 25 - Prob. 43PSCh. 25 - Prob. 44PSCh. 25 - Prob. 45PSCh. 25 - Prob. 46PSCh. 25 - Prob. 47PSCh. 25 - Prob. 48PSCh. 25 - Prob. 49PSCh. 25 - Prob. 50PSCh. 25 - Prob. 51PSCh. 25 - Prob. 52PSCh. 25 - Prob. 53PSCh. 25 - Prob. 54PSCh. 25 - Prob. 55PSCh. 25 - Some of the reactions explored by Ernest...Ch. 25 - Prob. 57GQCh. 25 - Prob. 58GQCh. 25 - Prob. 59GQCh. 25 - Prob. 60GQCh. 25 - Prob. 61GQCh. 25 - Prob. 62GQCh. 25 - Prob. 63GQCh. 25 - Prob. 64GQCh. 25 - Prob. 65ILCh. 25 - Prob. 66ILCh. 25 - Prob. 67ILCh. 25 - Prob. 68ILCh. 25 - Prob. 69ILCh. 25 - Prob. 70ILCh. 25 - Prob. 71SCQCh. 25 - Prob. 72SCQCh. 25 - Prob. 73SCQCh. 25 - Prob. 74SCQCh. 25 - Prob. 76SCQCh. 25 - Prob. 77SCQCh. 25 - Prob. 78SCQCh. 25 - Prob. 79SCQ
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