Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card
Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card
5th Edition
ISBN: 9781305367487
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 18, Problem 74QRT

(a)

Interpretation Introduction

Interpretation:

Balanced nuclear equation has to be given for the below equation,

    U92238+N714+5n01

Concept Introduction:

Nuclear reaction is the one where the nucleus of the atom is involved in the reaction.  This can be represented in form of a nuclear equation.  Missing particle in a nuclear equation can be identified by using the mass number and atomic number balance.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given nuclear equation is,

    U92238+N714+5n01

The unknown particle can be assigned as XZA.  Therefore, the given nuclear equation can be rewritten as,

  U92238+N714XZA+5n01

Mass number of the particle formed:

    238+14 = A+5 A = 252-5 A = 247

Atomic number of the particle formed:

    92+7 = Z+0Z = 99

Therefore, the atomic number of the particle is 99 and mass number is 247.  The particle with atomic number 99 is Einsteinium.  Therefore, the complete balanced nuclear equation can be given as,

    U92238+N714E99247s+5n01

(b)

Interpretation Introduction

Interpretation:

Balanced nuclear equation has to be given for the below equation,

    U92238+F100249m+5n01

Concept Introduction:

Refer part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given nuclear equation is,

    U92238+F100249m+5n01

The unknown particle can be assigned as XZA.  Therefore, the given nuclear equation can be rewritten as,

  U92238+XZAF100249m+5n01

Mass number of the particle formed:

    238+A = 249+5 A = 254-238 A = 16

Atomic number of the particle formed:

    92+Z = 100+0Z = 8

Therefore, the atomic number of the particle is 8 and mass number is 16.  The particle with atomic number 8 is Oxygen.  Therefore, the complete balanced nuclear equation can be given as,

    U92238+O816F100249m+5n01

(c)

Interpretation Introduction

Interpretation:

Balanced nuclear equation has to be given for the below equation,

    E99253s+M101256d+n01

Concept Introduction:

Refer part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given nuclear equation is,

    E99253s+M101256d+n01

The unknown particle can be assigned as XZA.  Therefore, the given nuclear equation can be rewritten as,

  E99253s+XZAM101256d+n01

Mass number of the particle formed:

    253+A = 256+1 A = 257-253 A = 4

Atomic number of the particle formed:

    99+Z = 101+0Z = 2

Therefore, the atomic number of the particle is 2 and mass number is 4.  The particle with atomic number 2 is Helium.  Therefore, the complete balanced nuclear equation can be given as,

    E99253s+H24eM101256d+n01

(d)

Interpretation Introduction

Interpretation:

Balanced nuclear equation has to be given for the below equation,

    C96246m+N102254o+4n01

Concept Introduction:

Refer part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Given nuclear equation is,

    C96246m+N102254o+4n01

The unknown particle can be assigned as XZA.  Therefore, the given nuclear equation can be rewritten as,

  C96246m+XZAN102254o+4n01

Mass number of the particle formed:

    246+A = 254+4 A = 258-246 A = 12

Atomic number of the particle formed:

    96+Z = 102+0Z = 6

Therefore, the atomic number of the particle is 6 and mass number is 12.  The particle with atomic number 6 is Carbon.  Therefore, the complete balanced nuclear equation can be given as,

    C96246m+C612N102254o+4n01

(e)

Interpretation Introduction

Interpretation:

Balanced nuclear equation has to be given for the below equation,

    C98252f+L103257r+5n01

Concept Introduction:

Refer part (a).

(e)

Expert Solution
Check Mark

Explanation of Solution

Given nuclear equation is,

    C98252f+L103257r+5n01

The unknown particle can be assigned as XZA.  Therefore, the given nuclear equation can be rewritten as,

  C98252f+XZAL103257r+5n01

Mass number of the particle formed:

    252+A = 257+5 A = 262-252 A = 10

Atomic number of the particle formed:

    98+Z = 103+0Z = 5

Therefore, the atomic number of the particle is 5 and mass number is 10.  The particle with atomic number 5 is Boron.  Therefore, the complete balanced nuclear equation can be given as,

    C98252f+B510L103257r+5n01

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Chapter 18 Solutions

Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card

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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