Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
Question
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Chapter 7, Problem 115A
Interpretation Introduction

(a)

Interpretation:

The mistakes in the formula name of copper acetate need to be determined.

Concept introduction:

There are many rules to be followed while naming the ionic compounds. In naming of compound consisting transition metal, a Roman numeral must be given to indicate the oxidation state of the transition metal.

Expert Solution
Check Mark

Answer to Problem 115A

The Roman numeral to indicate the oxidation state of metal must be included in the formula name.

Explanation of Solution

The given formula name is copper acetate. Copper is a transition metal present in d-block of the periodic table. The transition metals have the capability of acquiring more than one oxidation states. Copper is capable of showing +2 and +1 oxidation states. In naming of an ionic compound of transition metal, a roman numeral in parenthesis must follow the metal name to indicate the oxidation state of the metal. So, in copper acetate, no roman numeral is given which is incorrect.

(b)

Interpretation Introduction

Interpretation:

The mistakes in the formula Mg2O2 needs to be determined.

Concept introduction:

In writing the name of compound, the valency of the both the cation and anion should be interchanged. If both anion and cation contains equal and opposite charge, they combine in 1:1 ratio.

(b)

Expert Solution
Check Mark

Answer to Problem 115A

The correct formula is MgO.

Explanation of Solution

Magnesium is a metal which forms +2 cation to attain stable electronic configuration and oxygen is a non-metal which can accept two electrons to form -2 anion to acquire stable electronic configuration. Magnesium and oxygen has opposite and equal charge and thus forms 1:1 ionic compound and the compound is written as MgO.

(c)

Interpretation Introduction

Interpretation:

The mistake in the formula Pb2O5 needs to be determined.

Concept introduction:

Transition elements are the elements present in the d-block of the periodic table. In writing the naming of compound, the valency of the both the cation and anion should be interchanged.

(c)

Expert Solution
Check Mark

Answer to Problem 115A

The correct formula of compound of lead and oxygen is PbO and PbO2.

Explanation of Solution

The given formula is Pb2O5. Lead (Pb) is a transition metal of d-block of the periodic table. It has four valence electrons and can acquire oxidation state +2 or +4. Oxygen can form -2 anion to acquire stable electronic configuration. If the Pb is +2 cation the formula would be PbO and if the Pb is +4 cation, the formula would be PbO2.

(d)

Interpretation Introduction

Interpretation:

The mistake in the chemical formula name disodium oxide needs to be determined.

Concept introduction:

In the chemical naming of compounds, cation should be named first and then the anion and there is no need to give the number of atoms in the compound.

(d)

Expert Solution
Check Mark

Answer to Problem 115A

Correct formula name is sodium oxide.

Explanation of Solution

From the formula name sodium dioxide, the chemical symbol is Na2O. The sodium belong to group 1 and it forms +1 cation to attain stable electronic configuration. Oxygen is a group 16 element and it can form -2 anion. In naming of the compound, Na2O, there is no need to mention the number of sodium atoms. So, the correct name is sodium oxide.

(e)

Interpretation Introduction

Interpretation:

The mistake in the Al2SO43 formula needs to be determined.

Concept introduction:

Polyatomic ions are the ions which is constituted by two or more atoms. Some examples of polyatomic ions are sulphate (SO4), carbonate (CO3) etc. In writing the chemical formula of the compound, a parenthesis should be used to separate the polyatomic ion from the cation.

(e)

Expert Solution
Check Mark

Answer to Problem 115A

The correct formula is Al2(SO4)3.

Explanation of Solution

In the compound Al2SO43, Aluminium (Al) is the cation of charge +3 and sulphate (SO4) is the anion of charge -2. The sulphate is the polyatomic ion which is to be surrounded by a parenthesis in the formula. So, the correct formula is Al2(SO4)3

Chapter 7 Solutions

Chemistry: Matter and Change

Ch. 7.2 - Prob. 11PPCh. 7.2 - Prob. 12SSCCh. 7.2 - Prob. 13SSCCh. 7.2 - Prob. 14SSCCh. 7.2 - Prob. 15SSCCh. 7.2 - Prob. 16SSCCh. 7.2 - Prob. 17SSCCh. 7.2 - Prob. 18SSCCh. 7.3 - Prob. 19PPCh. 7.3 - Prob. 20PPCh. 7.3 - Prob. 21PPCh. 7.3 - Prob. 22PPCh. 7.3 - Prob. 23PPCh. 7.3 - Prob. 24PPCh. 7.3 - Prob. 25PPCh. 7.3 - Prob. 26PPCh. 7.3 - Prob. 27PPCh. 7.3 - Prob. 28PPCh. 7.3 - Prob. 29PPCh. 7.3 - Prob. 30PPCh. 7.3 - Prob. 31PPCh. 7.3 - Prob. 32PPCh. 7.3 - Prob. 33PPCh. 7.3 - Prob. 34SSCCh. 7.3 - Prob. 35SSCCh. 7.3 - Prob. 36SSCCh. 7.3 - Prob. 37SSCCh. 7.3 - Prob. 38SSCCh. 7.3 - Prob. 39SSCCh. 7.4 - Prob. 40SSCCh. 7.4 - Prob. 41SSCCh. 7.4 - Prob. 42SSCCh. 7.4 - Prob. 43SSCCh. 7.4 - Prob. 44SSCCh. 7.4 - Prob. 45SSCCh. 7 - How do positive ions and negative ions form?Ch. 7 - Prob. 47ACh. 7 - Prob. 48ACh. 7 - Prob. 49ACh. 7 - Prob. 50ACh. 7 - Prob. 51ACh. 7 - Prob. 52ACh. 7 - Prob. 53ACh. 7 - Prob. 54ACh. 7 - Prob. 55ACh. 7 - Prob. 56ACh. 7 - Prob. 57ACh. 7 - Prob. 58ACh. 7 - Prob. 59ACh. 7 - Prob. 60ACh. 7 - Prob. 61ACh. 7 - Prob. 62ACh. 7 - Prob. 63ACh. 7 - Prob. 64ACh. 7 - Prob. 65ACh. 7 - Prob. 66ACh. 7 - Prob. 67ACh. 7 - Prob. 68ACh. 7 - Prob. 69ACh. 7 - Prob. 70ACh. 7 - Prob. 71ACh. 7 - Prob. 72ACh. 7 - Prob. 73ACh. 7 - Prob. 74ACh. 7 - Prob. 75ACh. 7 - Prob. 76ACh. 7 - Prob. 77ACh. 7 - Prob. 78ACh. 7 - Prob. 79ACh. 7 - Prob. 80ACh. 7 - Prob. 81ACh. 7 - Prob. 82ACh. 7 - Prob. 83ACh. 7 - Prob. 84ACh. 7 - Prob. 85ACh. 7 - Prob. 86ACh. 7 - Prob. 87ACh. 7 - Prob. 88ACh. 7 - Prob. 89ACh. 7 - Prob. 90ACh. 7 - Prob. 91ACh. 7 - Prob. 92ACh. 7 - Prob. 93ACh. 7 - Prob. 94ACh. 7 - Prob. 95ACh. 7 - Prob. 96ACh. 7 - Prob. 97ACh. 7 - Prob. 98ACh. 7 - Prob. 99ACh. 7 - Prob. 100ACh. 7 - Prob. 101ACh. 7 - Prob. 102ACh. 7 - Prob. 103ACh. 7 - Prob. 104ACh. 7 - Prob. 105ACh. 7 - Prob. 106ACh. 7 - Prob. 107ACh. 7 - Prob. 108ACh. 7 - Prob. 109ACh. 7 - Prob. 110ACh. 7 - Prob. 111ACh. 7 - Prob. 112ACh. 7 - Prob. 113ACh. 7 - Prob. 114ACh. 7 - Prob. 115ACh. 7 - Prob. 116ACh. 7 - Prob. 117ACh. 7 - Prob. 118ACh. 7 - Prob. 119ACh. 7 - Prob. 120ACh. 7 - Prob. 121ACh. 7 - Prob. 122ACh. 7 - Prob. 123ACh. 7 - Prob. 124ACh. 7 - Prob. 125ACh. 7 - Prob. 126ACh. 7 - Prob. 127ACh. 7 - Prob. 128ACh. 7 - Prob. 129ACh. 7 - Prob. 130ACh. 7 - Prob. 131ACh. 7 - Prob. 132ACh. 7 - Prob. 133ACh. 7 - Prob. 134ACh. 7 - Prob. 1STPCh. 7 - Prob. 2STPCh. 7 - Prob. 3STPCh. 7 - Prob. 4STPCh. 7 - Prob. 5STPCh. 7 - Prob. 6STPCh. 7 - Prob. 7STPCh. 7 - Prob. 8STPCh. 7 - Prob. 9STPCh. 7 - Prob. 10STPCh. 7 - Prob. 11STPCh. 7 - Prob. 12STPCh. 7 - Prob. 13STPCh. 7 - Prob. 14STPCh. 7 - Prob. 15STPCh. 7 - Prob. 16STPCh. 7 - Prob. 17STPCh. 7 - Prob. 18STPCh. 7 - Prob. 19STPCh. 7 - Prob. 20STP
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