Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written. Concept introduction: Balanced Chemical equation: A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass. Balancing the equation: There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants. First write the skeletal reaction from the given information. Then count the number of atoms of each element in reactants as well as products. Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

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Answer 1

Question

Chapter 3, Problem 153QRT

(a)

Interpretation Introduction

Interpretation:

Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.

First write the skeletal reaction from the given information.

Then count the number of atoms of each element in reactants as well as products.

Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, net ionic equation has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(c)

Interpretation Introduction

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

(d)

Interpretation Introduction

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

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Answer 2

Question

Chapter 3, Problem 153QRT

(a)

Interpretation Introduction

Interpretation:

Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.

First write the skeletal reaction from the given information.

Then count the number of atoms of each element in reactants as well as products.

Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, net ionic equation has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(c)

Interpretation Introduction

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

(d)

Interpretation Introduction

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

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Answer 3

Question

Chapter 3, Problem 153QRT

(a)

Interpretation Introduction

Interpretation:

Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.

First write the skeletal reaction from the given information.

Then count the number of atoms of each element in reactants as well as products.

Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, net ionic equation has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(c)

Interpretation Introduction

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

(d)

Interpretation Introduction

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

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Answer 4

Question

Chapter 3, Problem 153QRT

(a)

Interpretation Introduction

Interpretation:

Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.

First write the skeletal reaction from the given information.

Then count the number of atoms of each element in reactants as well as products.

Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, net ionic equation has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(c)

Interpretation Introduction

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

(d)

Interpretation Introduction

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

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Answer 5

Chapter 3, Problem 153QRT

(a)

Interpretation Introduction

Interpretation:

Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.

First write the skeletal reaction from the given information.

Then count the number of atoms of each element in reactants as well as products.

Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, net ionic equation has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(c)

Interpretation Introduction

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

(d)

Interpretation Introduction

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

Answer 6

Chapter 3, Problem 153QRT

(a)

Interpretation Introduction

Interpretation:

Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.

First write the skeletal reaction from the given information.

Then count the number of atoms of each element in reactants as well as products.

Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

Answer 7

Chapter 3, Problem 153QRT

(a)

Interpretation Introduction

Interpretation:

Iron(III) chloride reaction with arsenate gives iron(III) arsenate for that, the balanced reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.

First write the skeletal reaction from the given information.

Then count the number of atoms of each element in reactants as well as products.

Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

Answer 8

(b)

Interpretation Introduction

Interpretation:

For the given reaction, net ionic equation has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

Answer 9

(b)

Interpretation Introduction

Interpretation:

For the given reaction, net ionic equation has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

Answer 10

(c)

Interpretation Introduction

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

Answer 11

(c)

Interpretation Introduction

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

Answer 12

(d)

Interpretation Introduction

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

Answer 13

(d)

Interpretation Introduction

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

Answer 14

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Answer 15

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Answer 16

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Answer 17

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Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

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Chapter 3 Solutions

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Interpretation: The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

Interpretation: 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.

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Chapter 3 Solutions

Interpretation:

The volume of 0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate has to be calculated.

Interpretation:

0.150-M iron(III) chloride required to precipitate 25.0 mL of a solution that contains 0.025-M arsenate. The mass of the iron(III) arsenate has to be calculated.