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 21, Problem 104A

(a)

Interpretation Introduction

Interpretation: The oxidation number of each atom in CaCr2O7 needs to be determined.

Concept Introduction: An oxidation number of an element is defined as the number assigned to it in any chemical composition representing number of electrons gained or lost by an atom of that element in forming the chemical compound.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given substance is CaCr2O7 . This can be represented as follow:

  Ca2+ and Cr2O72

Thus, oxidation number of Ca is +2. Now, general oxidation number of oxygen is -2 thus, oxidation number of Cr is calculated as follows:

  2x+72=22x14=22x=2+142x=12x=6

Therefore, oxidation number of Cr is +6.

The oxidation number of Ca, Cr and O atom is +2, +6 and -2 respectively.

(b)

Interpretation Introduction

Interpretation: The oxidation number of each atom in KMnO4 needs to be determined.

Concept Introduction: An oxidation number of an element is defined as the number assigned to it in any chemical composition representing number of electrons gained or lost by an atom of that element in forming the chemical compound.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given substance is KMnO4 . This can be represented as follows:

  K+ and MnO4

Thus, the oxidation number of K is +1. Now, general oxidation number of O is -2 thus, oxidation number of Mn is calculated as follows:

  x+42=1x8=1x=1+8x=+7

Therefore, oxidation number of Mn is +7.

The oxidation number of K, Mn and O is +1, +7 and -2 respectively.

(c)

Interpretation Introduction

Interpretation: The oxidation number of each atom in CaNO32 needs to be determined.

Concept Introduction: An oxidation number of an element is defined as the number assigned to it in any chemical composition representing number of electrons gained or lost by an atom of that element in forming the chemical compound.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given substance is CaNO32

This can be represented as Ca2+ and NO32

Thus, the oxidation number of Ca is +2. The general oxidation number of O is -2. Thus, the oxidation number of N can be calculated as follows:

  x+32=2x6=2x=2+6x=+4

Thus, oxidation number of N is +4.

The oxidation number of Ca, N and O is +2, +4 and -2 respectively.

(d)

Interpretation Introduction

Interpretation: The oxidation number of each atom in AlOH3 needs to be determined.

Concept Introduction: An oxidation number of an element is defined as the number assigned to it in any chemical composition representing number of electrons gained or lost by an atom of that element in forming the chemical compound.

(d)

Expert Solution
Check Mark

Explanation of Solution

The given substance is AlOH3

This can be represented as follows:

  Al3+ and 3OH

Thus, oxidation number of Al is +3. The general oxidation number of O and H is -2 and 1 respectively.

Therefore, the oxidation number of Al, O and H is +3, -2 and +1 respectively.

Chapter 21 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 21.2 - Prob. 11SPCh. 21.2 - Prob. 12SPCh. 21.2 - Prob. 13SPCh. 21.2 - Prob. 14LCCh. 21.2 - Prob. 15LCCh. 21.2 - Prob. 16LCCh. 21.2 - Prob. 17LCCh. 21.2 - Prob. 18LCCh. 21.2 - Prob. 19LCCh. 21.3 - Prob. 20LCCh. 21.3 - Prob. 21LCCh. 21.3 - Prob. 22LCCh. 21.3 - Prob. 23LCCh. 21.3 - Prob. 24LCCh. 21.3 - Prob. 25LCCh. 21 - Prob. 26ACh. 21 - Prob. 27ACh. 21 - Prob. 28ACh. 21 - Prob. 29ACh. 21 - Prob. 30ACh. 21 - Prob. 31ACh. 21 - Prob. 32ACh. 21 - Prob. 33ACh. 21 - Prob. 34ACh. 21 - Prob. 35ACh. 21 - Prob. 36ACh. 21 - Prob. 37ACh. 21 - Prob. 38ACh. 21 - Prob. 39ACh. 21 - Prob. 40ACh. 21 - Prob. 41ACh. 21 - Prob. 42ACh. 21 - Prob. 43ACh. 21 - Prob. 44ACh. 21 - Prob. 45ACh. 21 - Prob. 46ACh. 21 - Prob. 47ACh. 21 - Prob. 48ACh. 21 - Prob. 49ACh. 21 - Prob. 50ACh. 21 - Prob. 51ACh. 21 - Prob. 52ACh. 21 - Prob. 53ACh. 21 - Prob. 54ACh. 21 - Prob. 55ACh. 21 - Prob. 56ACh. 21 - Prob. 57ACh. 21 - Prob. 58ACh. 21 - Prob. 59ACh. 21 - Prob. 60ACh. 21 - Prob. 61ACh. 21 - Prob. 62ACh. 21 - Prob. 63ACh. 21 - Prob. 64ACh. 21 - Prob. 65ACh. 21 - Prob. 66ACh. 21 - Prob. 67ACh. 21 - Prob. 68ACh. 21 - Prob. 69ACh. 21 - Prob. 70ACh. 21 - Prob. 71ACh. 21 - Prob. 72ACh. 21 - Prob. 73ACh. 21 - Prob. 74ACh. 21 - Prob. 75ACh. 21 - Prob. 76ACh. 21 - Prob. 77ACh. 21 - Prob. 78ACh. 21 - Prob. 79ACh. 21 - Prob. 80ACh. 21 - Prob. 81ACh. 21 - Prob. 82ACh. 21 - Prob. 83ACh. 21 - Prob. 84ACh. 21 - Prob. 85ACh. 21 - Prob. 86ACh. 21 - Prob. 87ACh. 21 - Prob. 88ACh. 21 - Prob. 89ACh. 21 - Prob. 90ACh. 21 - Prob. 91ACh. 21 - Prob. 92ACh. 21 - Prob. 93ACh. 21 - Prob. 94ACh. 21 - Prob. 95ACh. 21 - Prob. 96ACh. 21 - Prob. 97ACh. 21 - Prob. 98ACh. 21 - Prob. 99ACh. 21 - Prob. 100ACh. 21 - Prob. 101ACh. 21 - Prob. 102ACh. 21 - Prob. 103ACh. 21 - Prob. 104ACh. 21 - Prob. 105ACh. 21 - Prob. 106ACh. 21 - Prob. 1STPCh. 21 - Prob. 2STPCh. 21 - Prob. 3STPCh. 21 - Prob. 4STPCh. 21 - Prob. 5STPCh. 21 - Prob. 6STPCh. 21 - Prob. 7STPCh. 21 - Prob. 8STPCh. 21 - Prob. 9STPCh. 21 - Prob. 10STPCh. 21 - Prob. 11STPCh. 21 - Prob. 12STPCh. 21 - Prob. 13STP
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