Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 18, Problem 15QRT

(a)

Interpretation Introduction

Interpretation:

The missing particle in the nuclear equation has to be filled.

    ? N1022e + e+10

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  This is a natural transmutation reaction where the nuclide of one element is converted into nuclide of another element.  Radioactive decay happens naturally.  This can also be done artificially in the laboratory by means of bombardment reaction.  Bombardment reaction is the one where the target nuclei is hit by a small fast moving high-energy particle to give a daughter nuclide and a small particle such as proton or neutron.  This can be represented in form of a nuclear equation.  A balanced nuclear equation is the one in which the sum of subscripts on both sides are equal and sum of superscripts on both sides are equal.

(a)

Expert Solution
Check Mark

Answer to Problem 15QRT

The missing particle in the given equation is N1122a.

Explanation of Solution

Given nuclear equation is,

? N1022e + e+10

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 22 and the sum of subscript in the product side is 11.  This has to be balanced in the reactant side.  There is a short of 22 in the superscript and 11 in the subscript when comparing with the reactant side.  The particle with mass number 22 and atomic number 11 is Sodium.  Therefore, the missing particle is N1122a.  The complete nuclear equation can be given as,

N1122a N1022e + e+10

Conclusion

Missing particle in the given nuclear equation is identified as N1122a.

(b)

Interpretation Introduction

Interpretation:

The missing particle in the nuclear equation has to be filled.

    I53122 X54122e + ?

Concept Introduction:

Refer part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 15QRT

The missing particle in the given equation is e10.

Explanation of Solution

Given nuclear equation is,

I53122 X54122e + ?

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 122 and the sum of subscript in the product side is 54.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 122.  Sum of subscript is 53.  There is an excess of 1 in the subscript when comparing with the reactant side.  The particle with mass number 0 and atomic number -1 is electron.  Therefore, the missing particle is e10.  The complete nuclear equation can be given as,

I53122 X54122e + e10

Conclusion

Missing particle in the given nuclear equation is identified as e10.

(c)

Interpretation Introduction

Interpretation:

The missing particle in the nuclear equation has to be filled.

    P84210o ? + H24e

Concept Introduction:

Refer part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 15QRT

The missing particle in the given equation is P82206b.

Explanation of Solution

Given nuclear equation is,

P84210o ? + H24e

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 4 and the sum of subscript in the product side is 2.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 210.  Sum of subscript is 84.  There is a short of 206 in the superscript and 82 in the subscript when comparing with the product side.  The particle with mass number 206 and atomic number 82 is Lead.  Therefore, the missing particle is P82206b.  The complete nuclear equation can be given as,

P84210o P82206b + H24e

Conclusion

Missing particle in the given nuclear equation is identified as P82206b.

(d)

Interpretation Introduction

Interpretation:

The missing particle in the nuclear equation has to be filled.

    A79195u + ? P78195t

Concept Introduction:

Refer part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 15QRT

The missing particle in the given equation is e10.

Explanation of Solution

Given nuclear equation is,

A79195u + ? P78195t

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 195 and the sum of subscript in the product side is 78.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 195.  Sum of subscript is 79.  There is an excess of 1 in subscript on the reactant side when comparing with the product side.  The particle with mass number 0 and atomic number -1 is an electron.  Therefore, the missing particle is e10.  The complete nuclear equation can be given as,

A79195u + e10 P78195t

Conclusion

Missing particle in the given nuclear equation is identified as e10.

(e)

Interpretation Introduction

Interpretation:

The missing particle in the nuclear equation has to be filled.

    P94241u + O816 5n01 + ?

Concept Introduction:

Refer part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 15QRT

The missing particle in the given equation is N102252o.

Explanation of Solution

Given nuclear equation is,

P94241u + O816 5n01 + ?

In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal.  By looking into the above equation, the sum of superscript in the product side is 5 and the sum of subscript in the product side is 0.  This has to be balanced in the reactant side.  Sum of superscript in the reactant side is 257.  Sum of subscript is 102.  There is a short of 252 in the superscript and 102 in the subscript when comparing with the reactant side.  The particle with mass number 252 and atomic number 102 is Nobelium.  Therefore, the missing particle is N102252o.  The complete nuclear equation can be given as,

P94241u + O816 5n01 + N102252o

Conclusion

Missing particle in the given nuclear equation is identified as N102252o.

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Chapter 18 Solutions

Chemistry: The Molecular Science

Ch. 18.4 - Prob. 18.6PSPCh. 18.4 - Prob. 18.7PSPCh. 18.4 - Prob. 18.6ECh. 18.4 - Prob. 18.7CECh. 18.5 - Prob. 18.8ECh. 18.5 - Prob. 18.9CECh. 18.6 - Prob. 18.10ECh. 18.6 - Prob. 18.11ECh. 18.7 - Prob. 18.12ECh. 18.8 - Prob. 18.13ECh. 18.8 - Prob. 18.14ECh. 18.9 - Prob. 18.15ECh. 18 - Prob. 1SPCh. 18 - Prob. 2SPCh. 18 - Prob. 3SPCh. 18 - Prob. 4SPCh. 18 - Prob. 5SPCh. 18 - Prob. 1QRTCh. 18 - Prob. 2QRTCh. 18 - Prob. 3QRTCh. 18 - Prob. 4QRTCh. 18 - Prob. 5QRTCh. 18 - Prob. 6QRTCh. 18 - Prob. 7QRTCh. 18 - Prob. 8QRTCh. 18 - Prob. 9QRTCh. 18 - Complete the table.Ch. 18 - Prob. 11QRTCh. 18 - Prob. 12QRTCh. 18 - Prob. 13QRTCh. 18 - Prob. 14QRTCh. 18 - Prob. 15QRTCh. 18 - Prob. 16QRTCh. 18 - Prob. 17QRTCh. 18 - Prob. 18QRTCh. 18 - Prob. 19QRTCh. 18 - Prob. 20QRTCh. 18 - Prob. 21QRTCh. 18 - Prob. 22QRTCh. 18 - Prob. 23QRTCh. 18 - Prob. 24QRTCh. 18 - Prob. 25QRTCh. 18 - Prob. 26QRTCh. 18 - Prob. 27QRTCh. 18 - Prob. 28QRTCh. 18 - Prob. 29QRTCh. 18 - Prob. 30QRTCh. 18 - Prob. 31QRTCh. 18 - Prob. 32QRTCh. 18 - Prob. 33QRTCh. 18 - Prob. 34QRTCh. 18 - Prob. 35QRTCh. 18 - Prob. 36QRTCh. 18 - Prob. 37QRTCh. 18 - Prob. 38QRTCh. 18 - Prob. 39QRTCh. 18 - Prob. 40QRTCh. 18 - Prob. 41QRTCh. 18 - Prob. 42QRTCh. 18 - Prob. 43QRTCh. 18 - Prob. 44QRTCh. 18 - Prob. 45QRTCh. 18 - Prob. 46QRTCh. 18 - Prob. 47QRTCh. 18 - Prob. 48QRTCh. 18 - Prob. 49QRTCh. 18 - Prob. 50QRTCh. 18 - Prob. 51QRTCh. 18 - Prob. 52QRTCh. 18 - Prob. 53QRTCh. 18 - Prob. 54QRTCh. 18 - Prob. 55QRTCh. 18 - Prob. 56QRTCh. 18 - Prob. 57QRTCh. 18 - Prob. 58QRTCh. 18 - Prob. 59QRTCh. 18 - Prob. 60QRTCh. 18 - Prob. 61QRTCh. 18 - Prob. 62QRTCh. 18 - Prob. 63QRTCh. 18 - Prob. 64QRTCh. 18 - Prob. 65QRTCh. 18 - Prob. 66QRTCh. 18 - Prob. 67QRTCh. 18 - Prob. 68QRTCh. 18 - Prob. 69QRTCh. 18 - Prob. 70QRTCh. 18 - Prob. 71QRTCh. 18 - Prob. 72QRTCh. 18 - Prob. 73QRTCh. 18 - Prob. 74QRTCh. 18 - Prob. 75QRTCh. 18 - Prob. 76QRTCh. 18 - Prob. 77QRTCh. 18 - Prob. 78QRTCh. 18 - Prob. 79QRTCh. 18 - Prob. 80QRTCh. 18 - Prob. 81QRTCh. 18 - Prob. 82QRTCh. 18 - Prob. 83QRTCh. 18 - Prob. 84QRTCh. 18 - Prob. 85QRTCh. 18 - Prob. 86QRTCh. 18 - Prob. 87QRTCh. 18 - Prob. 88QRTCh. 18 - Prob. 89QRTCh. 18 - Prob. 91QRTCh. 18 - Prob. 92QRTCh. 18 - Prob. 93QRTCh. 18 - Prob. 94QRTCh. 18 - Prob. 95QRTCh. 18 - Prob. 96QRTCh. 18 - Prob. 18.ACPCh. 18 - Prob. 18.BCPCh. 18 - Prob. 18.CCPCh. 18 - Prob. 18.DCPCh. 18 - Prob. 18.ECP
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