Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781305581982
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 20, Problem 6E

A

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

A

Expert Solution
Check Mark

Answer to Problem 6E

Hence the missing particle will be beta particle.

Explanation of Solution

Given information : The given reaction is as follows,

  G73aG73e+?

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  G73aG73e+XZA , now equate the atomic mass and number of the both sides. The following will be the result:-

  73=73+A31=32+Z

Hence A=0 and Z= -1. Hence the missing particle will be beta particle and it is represented by e10 . Now the final equation will be,

  G73aG73e+e10

B

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

B

Expert Solution
Check Mark

Answer to Problem 6E

The missing particle will be alpha particle.

Explanation of Solution

Given information : The equation is given,

  P192tO188s+?

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  P192tO188s+XZA , now equate the atomic mass and number of the both sides. The following will be the result:-

  192=188+A78=76+Z

From the above equation, A= 4 and Z=2. Hence the missing particle is alpha particle which is represented by the H24e . Now the equation will be,

  P192tO188s+H24e

C

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

C

Expert Solution
Check Mark

Answer to Problem 6E

Hence the missing particle is positron.

Explanation of Solution

Given information : The given equation is as follows,

  B205iP205b+?

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  B205iP205b+XZA , now equate the atomic mass and number of the both sides. The following will be the result:-

  205=205+A83=82+Z

From the equation, A= 0, and Z= 1, hence this particle will be positron which is represented by the e10 .

D

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

D

Expert Solution
Check Mark

Answer to Problem 6E

Hence the missing particle is beta particle.

Explanation of Solution

Given information : The given equation is given below,

  C241m+?A241m

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  C241m+XZAA241m , now equate the atomic mass and number of the both sides. The following will be the result:-

  241+A=24196+Z=95

From the equation A=0 and Z= -1. Hence the missing particle is beta particle and represented by e10 . Now the equation become,

  C241m+e10A241m

E

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

E

Expert Solution
Check Mark

Answer to Problem 6E

Hence the missing particle is beta particle.

Explanation of Solution

Given information : The given equation is given below,

  C60oN60i+?

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  C60oN60i+XZA , now equate the atomic mass and number of the both sides. The following will be the result:-

  60=60+A27=28+Z

From the equation, A= 0 and Z= -1. Hence the missing particle is beta particle and represented by e10 . Now the equation will be,

  C60oN60i+e10

F

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

F

Expert Solution
Check Mark

Answer to Problem 6E

Hence the missing particle is beta particle.

Explanation of Solution

Given information : The given equation is given below,

  T97c+?M97o

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  T97c+XZAM97o , now equate the atomic mass and number of the both sides. The following will be the result:-

  97+A=9743+Z=42

From the equation, A= 0 and Z= -1. Hence the missing particle is beta particle and represented by e10 . Now the equation will be,

  T97c+e10M97o

G

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

G

Expert Solution
Check Mark

Answer to Problem 6E

Hence the missing particle is beta particle.

Explanation of Solution

Given information : The given equation is given below,

  T99cR99u+?

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  T99cR99u+XZA , now equate the atomic mass and number of the both sides. The following will be the result:-

  99=99+A43=44+Z ,

From the equation, A= 0 and Z= -1. Hence the missing particle is beta particle and represented by e10 . Now the equation will be,

  T99cR99u+e10

H

Interpretation Introduction

Interpretation : The given equation needs to be completed.

Concept Introduction : The radioactive decay process is the process in which element is decayed by releasing a, ß, and γ rays.

H

Expert Solution
Check Mark

Answer to Problem 6E

Hence the missing particle is alpha particle.

Explanation of Solution

Given information : The given equation is given below,

  P239uU235+?

Let the missing particle is XZA , where A represents the atomic mass and Z represents the atomic number. Now the equation looks like this,

  P239uU235+XZA , now equate the atomic mass and number of the both sides. The following will be the result:-

  239=235+A94=92+Z ,

From the above equation, A= 4 and Z=2. Hence the missing particle is alpha particle which is represented by the H24e . Now the equation will be:

  P239uU235+H24e .

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Chapter 20 Solutions

Chemical Principles

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