Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078129865
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
Question
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Chapter 3, Problem 3.57P

(a)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_Cu(NO3)2(aq)+_KOH(aq)_Cu(OH)2(s)+_KNO3(aq)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element. Generally, oxygen atoms are balanced in last.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(a)

Expert Solution
Check Mark

Answer to Problem 3.57P

The balanced chemical reaction is as follows:

Cu(NO3)2(aq)+2KOH(aq)Cu(OH)2(s)+2KNO3(aq)

Explanation of Solution

Start with Cu(NO3)2. The equation implies that 2 N atoms are present on the reactant side while only 1 N atom is present on the product side. Balance nitrogen (N) atoms first. To balance N, place the coefficient 2 in front of KNO3. The equation becomes,

_Cu(NO3)2(aq)+_KOH(aq)_Cu(OH)2(s)+2_KNO3(aq)

Nitrogen atoms are balanced. Balance potassium (K) atoms next. 1 K atom is present on the reactant side while 2 K atoms are present on the product side. To balance K atoms, place the coefficient 2 in front of KOH. The equation becomes,

_Cu(NO3)2(aq)+2_KOH(aq)_Cu(OH)2(s)+2_KNO3(aq)

Balance copper (Cu) atoms. 1 Cu atom is present on both sides. Place the coefficient 1 in front of Cu(NO3)2 and Cu(OH)2Cu. The equation becomes,

1_Cu(NO3)2(aq)+2_KOH(aq)1_Cu(OH)2(s)+2_KNO3(aq)

Check whether the equation is balanced or not as follows:

Reactants(1 Cu, 2 K, 2 N, 2 H, 8 O)Products(1 Cu, 2 K, 2 N, 2 H, 8 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

Cu(NO3)2(aq)+2KOH(aq)Cu(OH)2(s)+2KNO3(aq)

(b)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_BCl3(g)+_H2O(l)_H3BO4(s)+_HCl(g)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(b)

Expert Solution
Check Mark

Answer to Problem 3.57P

The balanced chemical reaction is as follows:

BCl3(g)+3H2O(l)H3BO4(s)+3HCl(g)

Explanation of Solution

The equation implies that 3 Cl atoms are present on the reactant side while only 1 Cl atom is present on the product side. Balance chlorine (Cl) atoms first. To balance Cl, place the coefficient 3 in front of HCl. The equation becomes,

_BCl3(g)+_H2O(l)_H3BO4(s)+3_HCl(g)

Chlorine atoms are balanced. Next, balance hydrogen (H) atoms. 2 H atoms are present on the reactant side while 6 H atoms are present on the product side. To balance H atoms, place the coefficient 3 in front of H2O. The equation becomes,

_BCl3(g)+3_H2O(l)_H3BO4(s)+3_HCl(g)

Hydrogen atoms are balanced. Place coefficient 1 in front of BCl3 and H3BO4.

1_BCl3(g)+3_H2O(l)1_H3BO4(s)+3_HCl(g)

Check whether the equation is balanced or not as follows:

Reactants(1 B, 3 Cl, 6 H, 3 O)Products(1 B, 3 Cl, 6 H, 3 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

BCl3(g)+3H2O(l)H3BO4(s)+3HCl(g)

(c)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_CaSiO3(s)+_HF(g)_SiF4(g)+_CaF2(s)+_H2O(l)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(c)

Expert Solution
Check Mark

Answer to Problem 3.57P

The balanced chemical reaction is as follows:

CaSiO3(s)+6HF(g)SiF4(g)+CaF2(s)+3H2O(l)

Explanation of Solution

The equation implies that 1 F atom is present on the reactant side while only 6 F atom is present on the product side. Balance fluorine (F) atoms first and oxygen (O) atoms in last. To balance F, place the coefficient 6 in front of HF. The equation becomes,

_CaSiO3(s)+6_HF(g)_SiF4(g)+_CaF2(s)+_H2O(l)

Fluorine atoms are balanced, next, balance hydrogen (H) atoms. 6 H atoms are present on the reactant side while 2 H atoms are present on the product side. To balance H atoms, place the coefficient 3 in front of H2O. The equation becomes,

_CaSiO3(s)+6_HF(g)_SiF4(g)+_CaF2(s)+3_H2O(l)

Hydrogen atoms are balanced. One calcium (Ca) atom, one silicon (Si) atom, and three oxygen (O) atoms. Place coefficient 1 in front of CaSiO3, SiF4, and CaF2 as,

1_CaSiO3(s)+6_HF(g)1_SiF4(g)+1_CaF2(s)+3_H2O(l)

Check whether the equation is balanced or not as follows:

Reactants(1 Ca, 1 Si, 6 H, 6 F, 3 O)Products(1 Ca, 1 Si, 6 H, 6 F, 3 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

CaSiO3(s)+6HF(g)SiF4(g)+CaF2(s)+3H2O(l)

(d)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_(CN)2(g)+_H2O(l)_H2C2O4(aq)+_H2O(g)+_NH3(g)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(d)

Expert Solution
Check Mark

Answer to Problem 3.57P

The balanced chemical reaction is as follows:

(CN)2(g)+4H2O(l)H2C2O4(aq)+2NH3(g)

Explanation of Solution

The equation implies that 2 N atoms is present on the reactant side while 1 N atom is present on the product side. Balance nitrogen (N) atoms first and oxygen (O) atoms in last. To balance N, place the coefficient 2 in front of NH3. The equation becomes,

_(CN)2(g)+_H2O(l)_H2C2O4(aq)+2_NH3(g)

Nitrogen atoms are balanced, hence, place the coefficient 1 in front of (CN)2. Next, balance oxygen (O) atoms. 1 O atom is present on the reactant side while 4 O atoms are present on the product side. Hence, place the coefficient 4 in front of H2O. The equation becomes,

_(CN)2(g)+4_H2O(l)_H2C2O4(aq)+2_NH3(g)

Oxygen atoms are balanced. Two carbon (C) atoms and 8 hydrogens (H) are present on both sides. Place coefficient 1 in front of (CN)2 and H2C2O4 as:

1_(CN)2(g)+4_H2O(l)1_H2C2O4(aq)+2_NH3(g)

Check whether the equation is balanced or not as follows:

Reactants(2 C, 2 N, 8 H, 4 O)Products(2 C, 2 N, 8 H, 4 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

(CN)2(g)+4H2O(l)H2C2O4(aq)+2NH3(g)

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Chapter 3 Solutions

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change

Ch. 3.1 - Prob. 3.6AFPCh. 3.1 - For many years, compounds known as...Ch. 3.1 - Use the information in Follow-up Problem 3.6A to...Ch. 3.1 - Prob. 3.7BFPCh. 3.2 - Prob. 3.8AFPCh. 3.2 - A sample of an unknown compound contains 6.80 mol...Ch. 3.2 - A sample of an unknown compound is found to...Ch. 3.2 - Prob. 3.9BFPCh. 3.2 - Prob. 3.10AFPCh. 3.2 - Prob. 3.10BFPCh. 3.2 - A dry-cleaning solvent (ℳ = 146.99 g/mol) that...Ch. 3.2 - Prob. 3.11BFPCh. 3.3 - Prob. 3.12AFPCh. 3.3 - Prob. 3.12BFPCh. 3.3 - Prob. 3.13AFPCh. 3.3 - Prob. 3.13BFPCh. 3.4 - Prob. 3.14AFPCh. 3.4 - The tarnish that forms on objects made of silver...Ch. 3.4 - Prob. 3.15AFPCh. 3.4 - In the reaction that removes silver tarnish (see...Ch. 3.4 - Prob. 3.16AFPCh. 3.4 - Prob. 3.16BFPCh. 3.4 - Prob. 3.17AFPCh. 3.4 - Prob. 3.17BFPCh. 3.4 - Prob. 3.18AFPCh. 3.4 - Prob. 3.18BFPCh. 3.4 - In the reaction in Follow-up Problem 3.18A, how...Ch. 3.4 - Prob. 3.19BFPCh. 3.4 - Prob. 3.20AFPCh. 3.4 - Prob. 3.20BFPCh. 3.4 - Marble (calcium carbonate) reacts with...Ch. 3.4 - Prob. 3.21BFPCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Why might the expression “1 mol of chlorine” be...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Calculate the molar mass of each of the...Ch. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: Mass...Ch. 3 - Calculate each of the following quantities: Amount...Ch. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Calculate each of the following: Mass % of I in...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Oxygen is required for the metabolic combustion of...Ch. 3 - Cisplatin (right), or Platinol, is used in the...Ch. 3 - Allyl sulfide (below) gives garlic its...Ch. 3 - Iron reacts slowly with oxygen and water to form a...Ch. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - The mineral galena is composed of lead(II) sulfide...Ch. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - List three ways compositional data may be given in...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - What is the molecular formula of each...Ch. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Find the empirical formula of each of the...Ch. 3 - An oxide of nitrogen contains 30.45 mass % N. (a)...Ch. 3 - Prob. 3.45PCh. 3 - A sample of 0.600 mol of a metal M reacts...Ch. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.75PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.77PCh. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - Prob. 3.95PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.97PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Prob. 3.101PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.104PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.107PCh. 3 - Prob. 3.108PCh. 3 - Prob. 3.109PCh. 3 - Prob. 3.110PCh. 3 - Prob. 3.111PCh. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - Prob. 3.121PCh. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.128PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 3 - Manganese is a key component of extremely hard...Ch. 3 - The human body excretes nitrogen in the form of...Ch. 3 - Aspirin (acetylsalicylic acid, C9H8O4) is made by...Ch. 3 - Prob. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Prob. 3.137PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.140P
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