EBK INTRODUCTION TO CHEMISTRY
EBK INTRODUCTION TO CHEMISTRY
5th Edition
ISBN: 9781260162165
Author: BAUER
Publisher: MCGRAW HILL BOOK COMPANY
Question
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Chapter 15, Problem 117QP

(a)

Interpretation Introduction

Interpretation:

The balanced nuclear reaction for the given nuclear reaction is to be determined.

(a)

Expert Solution
Check Mark

Explanation of Solution

In the given nuclear reaction shown below, the mass number is not balanced because the total mass number on the left side is not equal to the total mass number on the right side.

C612+H11N714+n01

To balance the given reaction, nitrogen isotope N714 is changed to nitrogen isotope N712 . The balanced nuclear reaction is shown below.

C612+H11N712+n01

(b)

Interpretation Introduction

Interpretation:

The balanced nuclear reaction for the given nuclear reaction is to be determined.

(b)

Expert Solution
Check Mark

Explanation of Solution

In the given nuclear reaction shown below, the atomic number is not balanced because the total atomic number on the left side is not equal to the total atomic number on the right side.

H24e+B49eC612+H11

To balance the given reaction, C612 is replaced with B512 . The balanced nuclear reaction is shown below.

H24e+B49eB512+H11

(c)

Interpretation Introduction

Interpretation:

From the given nuclear reaction, the correct balanced nuclear reaction is to be identified.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is as shown below:

H24e+N714N714+2H12

To determine whether the mass number is conserved, the sum of mass numbers of the reactants must be equal to the sum of mass numbers of the products in accordance with the law of conservation of mass.

AHe+AN=AN+2AH ……(1)

The values for mass number of helium AHe , nitrogen AN , and hydrogen AH are substituted as 4, 14, and 2 , respectively in the equation (1).

4+14=14 + 2218 = 18

Similarly, to determine whether the atomic number is conserved, the sum of atomic numbers of the reactants must be equal to the sum of atomic numbers of the products.

ZHe+ZN=ZN+2ZH ……(2)

The values for atomic number of helium ZHe , nitrogen ZN , and hydrogen ZH are substituted as 2, 7, and 1 , respectively in the equation (2).

2+7=7 + 219 = 9

Therefore, the given nuclear reaction shown below is balanced because the total atomic number and the mass number on the left side is equal to the total atomic number and mass number on the right side.

H24e+N714N714+2H12

(d)

Interpretation Introduction

Interpretation:

The balanced nuclear reaction for the given nuclear reaction is to be determined.

(d)

Expert Solution
Check Mark

Explanation of Solution

In the given nuclear reaction shown below, atomic number is not balanced because the total atomic number on the left side is not equal to the total atomic number on the right side.

O816C916+β+10

To balance the given reaction, β+10 is replaced with e10 . The balanced nuclear reaction is shown below.

O816C916+e10

(e)

Interpretation Introduction

Interpretation:

The balanced nuclear reaction for the given nuclear reaction is to be determined.

(e)

Expert Solution
Check Mark

Explanation of Solution

In the given nuclear reaction shown below, the atomic number is not balanced because the total atomic number on the left side is not equal to the total atomic number on the right side.

N714+e10O814

To balance the given reaction, O814 is replaced with C614 . The balanced nuclear reaction is shown below.

N714+e10C614

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Chapter 15 Solutions

EBK INTRODUCTION TO CHEMISTRY

Ch. 15 - Prob. 5PPCh. 15 - Prob. 6PPCh. 15 - Prob. 7PPCh. 15 - Prob. 8PPCh. 15 - Prob. 9PPCh. 15 - Prob. 10PPCh. 15 - Prob. 11PPCh. 15 - Prob. 1QPCh. 15 - Prob. 2QPCh. 15 - Prob. 3QPCh. 15 - Prob. 4QPCh. 15 - Prob. 5QPCh. 15 - Prob. 6QPCh. 15 - Prob. 7QPCh. 15 - Prob. 8QPCh. 15 - Prob. 9QPCh. 15 - Prob. 10QPCh. 15 - Prob. 11QPCh. 15 - Prob. 12QPCh. 15 - Prob. 13QPCh. 15 - Prob. 14QPCh. 15 - Prob. 15QPCh. 15 - Prob. 16QPCh. 15 - Prob. 17QPCh. 15 - Prob. 18QPCh. 15 - Prob. 19QPCh. 15 - Prob. 20QPCh. 15 - Prob. 21QPCh. 15 - Prob. 22QPCh. 15 - Prob. 23QPCh. 15 - Prob. 24QPCh. 15 - Prob. 25QPCh. 15 - Prob. 26QPCh. 15 - Prob. 27QPCh. 15 - Prob. 28QPCh. 15 - Prob. 29QPCh. 15 - Prob. 30QPCh. 15 - Prob. 31QPCh. 15 - Prob. 32QPCh. 15 - Prob. 33QPCh. 15 - Prob. 34QPCh. 15 - Prob. 35QPCh. 15 - Prob. 36QPCh. 15 - Prob. 37QPCh. 15 - Prob. 38QPCh. 15 - Prob. 39QPCh. 15 - Prob. 40QPCh. 15 - Prob. 41QPCh. 15 - Prob. 42QPCh. 15 - Prob. 43QPCh. 15 - Prob. 44QPCh. 15 - Prob. 45QPCh. 15 - Prob. 46QPCh. 15 - Prob. 47QPCh. 15 - Prob. 48QPCh. 15 - Prob. 49QPCh. 15 - Prob. 50QPCh. 15 - Prob. 51QPCh. 15 - Prob. 52QPCh. 15 - Prob. 53QPCh. 15 - Prob. 54QPCh. 15 - Prob. 55QPCh. 15 - Prob. 56QPCh. 15 - Prob. 57QPCh. 15 - Prob. 58QPCh. 15 - Prob. 59QPCh. 15 - Prob. 60QPCh. 15 - Prob. 61QPCh. 15 - Prob. 62QPCh. 15 - Prob. 63QPCh. 15 - Prob. 64QPCh. 15 - Prob. 65QPCh. 15 - Prob. 66QPCh. 15 - Prob. 67QPCh. 15 - Prob. 68QPCh. 15 - Prob. 69QPCh. 15 - Prob. 70QPCh. 15 - Prob. 73QPCh. 15 - Prob. 74QPCh. 15 - Prob. 75QPCh. 15 - Prob. 76QPCh. 15 - Prob. 77QPCh. 15 - Prob. 78QPCh. 15 - Prob. 79QPCh. 15 - Prob. 80QPCh. 15 - Prob. 81QPCh. 15 - Prob. 82QPCh. 15 - Prob. 83QPCh. 15 - Prob. 84QPCh. 15 - Prob. 85QPCh. 15 - Prob. 86QPCh. 15 - Prob. 87QPCh. 15 - Prob. 88QPCh. 15 - Prob. 89QPCh. 15 - Prob. 90QPCh. 15 - Prob. 91QPCh. 15 - Prob. 92QPCh. 15 - Prob. 93QPCh. 15 - Prob. 94QPCh. 15 - Prob. 95QPCh. 15 - Prob. 96QPCh. 15 - Prob. 97QPCh. 15 - Prob. 98QPCh. 15 - Prob. 99QPCh. 15 - Prob. 100QPCh. 15 - Prob. 101QPCh. 15 - Prob. 102QPCh. 15 - Prob. 103QPCh. 15 - Prob. 104QPCh. 15 - Prob. 105QPCh. 15 - Prob. 106QPCh. 15 - Prob. 107QPCh. 15 - Prob. 108QPCh. 15 - Prob. 109QPCh. 15 - Prob. 110QPCh. 15 - Prob. 111QPCh. 15 - Prob. 112QPCh. 15 - Prob. 113QPCh. 15 - Prob. 114QPCh. 15 - Prob. 115QPCh. 15 - Prob. 116QPCh. 15 - Prob. 117QPCh. 15 - Prob. 118QPCh. 15 - Prob. 119QPCh. 15 - Prob. 120QPCh. 15 - Prob. 121QPCh. 15 - Prob. 122QPCh. 15 - Prob. 123QP
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