Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 25, Problem 67A

(a)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for beta emission by isotope H13 is to be written.

Interpretation:

Concept introduction:

A proton can change into a neutron or vice versa inside the radioactive sample's nucleus through a process known as beta decay. The radioactive sample's nucleus can approach the ideal neutron/proton ratio as closely as feasible through processes like beta decay and alpha decay.

(a)

Expert Solution
Check Mark

Answer to Problem 67A

Balanced reaction H13H23e+e10

Explanation of Solution

Beta decay involves the emission of a beta particle, to establish the balanced nuclear reaction for this decay, the atomic and mass number must be conserved from the initial isotope to the product of the decay. The element produced can then be identified from the atomic number. In this decay, the mass number stays the same while the atomic number increases by 1.

(b)

Interpretation Introduction

Interpretation:

The balanced nuclear equation for beta emission by M1228g is to be written.

Concept introduction:

A proton can change into a neutron or vice versa inside the radioactive sample's nucleus through a process known as beta decay. The radioactive sample's nucleus can approach the ideal neutron/proton ratio as closely as feasible through processes like beta decay and alpha decay.

(b)

Expert Solution
Check Mark

Answer to Problem 67A

Balanced reaction M1228gA1328l+e10

Explanation of Solution

Beta decay involves the emission of a beta particle, to establish the balanced nuclear reaction for this decay, the atomic and mass number must be conserved from the initial isotope to the product of the decay. The element produced can then be identified from the atomic number. In this decay, the mass number stays the same while the atomic number increases by 1.

(c)

Interpretation Introduction

Interpretation:

The balanced nuclear equation for beta emission by the I53131 is to be written.

Concept introduction:

A proton can change into a neutron or vice versa inside the radioactive sample's nucleus through a process known as beta decay. The radioactive sample's nucleus can approach the ideal neutron/proton ratio as closely as feasible through processes like beta decay and alpha decay.

(c)

Expert Solution
Check Mark

Answer to Problem 67A

  I53131X54131e+e10

Explanation of Solution

Beta decay involves the emission of a beta particle, to establish the balanced nuclear reaction for this decay, the atomic and mass number must be conserved from the initial isotope to the product of the decay. The element produced can then be identified from the atomic number. In this decay, the mass number stays the same while the atomic number increases by 1.

(d)

Interpretation Introduction

Interpretation:

The balanced nuclear equation for beta emission by S3475e is to be written.

Concept introduction:

A proton can change into a neutron or vice versa inside the radioactive sample's nucleus through a process known as beta decay. The radioactive sample's nucleus can approach the ideal neutron/proton ratio as closely as feasible through processes like beta decay and alpha decay.

(d)

Expert Solution
Check Mark

Answer to Problem 67A

  S3475eB3575r+e10

Explanation of Solution

Beta decay involves the emission of a beta particle, to establish the balanced nuclear reaction for this decay, the atomic and mass number must be conserved from the initial isotope to the product of the decay. The element produced can then be identified from the atomic number. In this decay, the mass number stays the same while the atomic number increases by 1

Chapter 25 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 25.2 - Prob. 11LCCh. 25.2 - Prob. 12LCCh. 25.2 - Prob. 13LCCh. 25.2 - Prob. 14LCCh. 25.2 - Prob. 15LCCh. 25.2 - Prob. 16LCCh. 25.3 - Prob. 18LCCh. 25.3 - Prob. 19LCCh. 25.3 - Prob. 20LCCh. 25.3 - Prob. 21LCCh. 25.3 - Prob. 22LCCh. 25.3 - Prob. 23LCCh. 25.3 - Prob. 25LCCh. 25.4 - Prob. 26LCCh. 25.4 - Prob. 27LCCh. 25.4 - Prob. 28LCCh. 25.4 - Prob. 29LCCh. 25.4 - Prob. 30LCCh. 25.4 - Prob. 31LCCh. 25.4 - Prob. 32LCCh. 25.4 - Prob. 33LCCh. 25 - Prob. 34ACh. 25 - Prob. 35ACh. 25 - Prob. 36ACh. 25 - Prob. 37ACh. 25 - Prob. 38ACh. 25 - Prob. 39ACh. 25 - Prob. 40ACh. 25 - Prob. 41ACh. 25 - Prob. 42ACh. 25 - Prob. 43ACh. 25 - Prob. 44ACh. 25 - Prob. 45ACh. 25 - Prob. 46ACh. 25 - Prob. 47ACh. 25 - Prob. 48ACh. 25 - Prob. 49ACh. 25 - Prob. 50ACh. 25 - Prob. 51ACh. 25 - Prob. 52ACh. 25 - Prob. 53ACh. 25 - Prob. 54ACh. 25 - Prob. 55ACh. 25 - Prob. 56ACh. 25 - Prob. 57ACh. 25 - Prob. 58ACh. 25 - Prob. 59ACh. 25 - Prob. 60ACh. 25 - Prob. 61ACh. 25 - Prob. 62ACh. 25 - Prob. 63ACh. 25 - Prob. 64ACh. 25 - Prob. 65ACh. 25 - Prob. 66ACh. 25 - Prob. 67ACh. 25 - Prob. 68ACh. 25 - Prob. 69ACh. 25 - Prob. 70ACh. 25 - Prob. 71ACh. 25 - Prob. 72ACh. 25 - Prob. 73ACh. 25 - Prob. 74ACh. 25 - Prob. 75ACh. 25 - Prob. 76ACh. 25 - Prob. 77ACh. 25 - Prob. 78ACh. 25 - Prob. 79ACh. 25 - Prob. 80ACh. 25 - Prob. 81ACh. 25 - Prob. 82ACh. 25 - Prob. 83ACh. 25 - Prob. 84ACh. 25 - Prob. 85ACh. 25 - Prob. 86ACh. 25 - Prob. 87ACh. 25 - Prob. 88ACh. 25 - Prob. 89ACh. 25 - Prob. 90ACh. 25 - Prob. 91ACh. 25 - Prob. 92ACh. 25 - Prob. 93ACh. 25 - Prob. 94ACh. 25 - Prob. 95ACh. 25 - Prob. 96ACh. 25 - Prob. 97ACh. 25 - Prob. 98ACh. 25 - Prob. 99ACh. 25 - Prob. 100ACh. 25 - Prob. 101ACh. 25 - Prob. 102ACh. 25 - Prob. 103ACh. 25 - Prob. 104ACh. 25 - Prob. 105ACh. 25 - Prob. 106ACh. 25 - Prob. 107ACh. 25 - Prob. 108ACh. 25 - Prob. 109ACh. 25 - Prob. 110ACh. 25 - Prob. 111ACh. 25 - Prob. 112ACh. 25 - Prob. 113ACh. 25 - Prob. 1STPCh. 25 - Prob. 2STPCh. 25 - Prob. 3STPCh. 25 - Prob. 4STPCh. 25 - Prob. 5STPCh. 25 - Prob. 6STPCh. 25 - Prob. 7STPCh. 25 - Prob. 8STPCh. 25 - Prob. 9STPCh. 25 - Prob. 10STPCh. 25 - Prob. 11STPCh. 25 - Prob. 12STPCh. 25 - Prob. 13STPCh. 25 - Prob. 14STPCh. 25 - Prob. 15STPCh. 25 - Prob. 16STPCh. 25 - Prob. 17STP
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