We are asked to write the formula for polyatomic ions and name each of the following compounds. Concept Introduction: The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion. To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example NaCl e. Here Na + 1 is cation and Cl − 1 is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the NaCl compound is ‘sodium chloride’. To determine: Thename each of the following compounds: a. M n C O 3 b. A u 2 S O 4 c. M g 3 ( P O 4 ) 2 d. F e ( H C O 3 ) 3

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Answer 1

Question

Chapter 6.4, Problem 6.44QAP

Interpretation Introduction

Interpretation:

We are asked to write the formula for polyatomic ions and name each of the following compounds.

Concept Introduction:

The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion.

To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example $NaCl$ e. Here ${Na}^{+ 1}$ is cation and ${Cl}^{- 1}$ is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the $NaCl$ compound is ‘sodium chloride’.

To determine:

Thename each of the following compounds:

a. $M n C O_{3}$

b. $A u_{2} S O_{4}$

c. $M g_{3} {(P O_{4})}_{2}$

d. $F e {(H C O_{3})}_{3}$

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Answer 2

Question

Chapter 6.4, Problem 6.44QAP

Interpretation Introduction

Interpretation:

We are asked to write the formula for polyatomic ions and name each of the following compounds.

Concept Introduction:

The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion.

To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example $NaCl$ e. Here ${Na}^{+ 1}$ is cation and ${Cl}^{- 1}$ is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the $NaCl$ compound is ‘sodium chloride’.

To determine:

Thename each of the following compounds:

a. $M n C O_{3}$

b. $A u_{2} S O_{4}$

c. $M g_{3} {(P O_{4})}_{2}$

d. $F e {(H C O_{3})}_{3}$

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Answer 3

Question

Chapter 6.4, Problem 6.44QAP

Interpretation Introduction

Interpretation:

We are asked to write the formula for polyatomic ions and name each of the following compounds.

Concept Introduction:

The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion.

To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example $NaCl$ e. Here ${Na}^{+ 1}$ is cation and ${Cl}^{- 1}$ is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the $NaCl$ compound is ‘sodium chloride’.

To determine:

Thename each of the following compounds:

a. $M n C O_{3}$

b. $A u_{2} S O_{4}$

c. $M g_{3} {(P O_{4})}_{2}$

d. $F e {(H C O_{3})}_{3}$

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Answer 4

Question

Chapter 6.4, Problem 6.44QAP

Interpretation Introduction

Interpretation:

We are asked to write the formula for polyatomic ions and name each of the following compounds.

Concept Introduction:

The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion.

To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example $NaCl$ e. Here ${Na}^{+ 1}$ is cation and ${Cl}^{- 1}$ is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the $NaCl$ compound is ‘sodium chloride’.

To determine:

Thename each of the following compounds:

a. $M n C O_{3}$

b. $A u_{2} S O_{4}$

c. $M g_{3} {(P O_{4})}_{2}$

d. $F e {(H C O_{3})}_{3}$

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Answer 5

Chapter 6.4, Problem 6.44QAP

Interpretation Introduction

Interpretation:

We are asked to write the formula for polyatomic ions and name each of the following compounds.

Concept Introduction:

The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion.

To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example $NaCl$ e. Here ${Na}^{+ 1}$ is cation and ${Cl}^{- 1}$ is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the $NaCl$ compound is ‘sodium chloride’.

To determine:

Thename each of the following compounds:

a. $M n C O_{3}$

b. $A u_{2} S O_{4}$

c. $M g_{3} {(P O_{4})}_{2}$

d. $F e {(H C O_{3})}_{3}$

Answer 6

Chapter 6.4, Problem 6.44QAP

Interpretation Introduction

Interpretation:

We are asked to write the formula for polyatomic ions and name each of the following compounds.

Concept Introduction:

The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion.

To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example $NaCl$ e. Here ${Na}^{+ 1}$ is cation and ${Cl}^{- 1}$ is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the $NaCl$ compound is ‘sodium chloride’.

To determine:

Thename each of the following compounds:

a. $M n C O_{3}$

b. $A u_{2} S O_{4}$

c. $M g_{3} {(P O_{4})}_{2}$

d. $F e {(H C O_{3})}_{3}$

Answer 7

Chapter 6.4, Problem 6.44QAP

Interpretation Introduction

Interpretation:

We are asked to write the formula for polyatomic ions and name each of the following compounds.

Concept Introduction:

The ionic compound is formed when the cation and anion come together. Ionic compound is formed by crisscross method. In this method the numerical value of ionic charges on cation and anion interchanged to become the subscript of the other ion.

To name the ionic compound first we name the cation that is name of the element followed by name of the anion. Consider an example $NaCl$ e. Here ${Na}^{+ 1}$ is cation and ${Cl}^{- 1}$ is anion, so the compound is named with Sodium followed by name of the anion that is chloride ion. Therefore the name of the $NaCl$ compound is ‘sodium chloride’.

To determine:

Thename each of the following compounds:

a. $M n C O_{3}$

b. $A u_{2} S O_{4}$

c. $M g_{3} {(P O_{4})}_{2}$

d. $F e {(H C O_{3})}_{3}$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 6.1 - State the number of electrons that must be lost by...Ch. 6.1 - Prob. 6.2QAP Ch. 6.1 - State the number of electrons lost or gained when...Ch. 6.1 - State the number of electrons lost or gained when...Ch. 6.1 - Prob. 6.5QAP Ch. 6.1 - Prob. 6.6QAP Ch. 6.1 - Prob. 6.7QAP Ch. 6.1 - Prob. 6.8QAP Ch. 6.1 - Prob. 6.9QAP Ch. 6.1 - Prob. 6.10QAP

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