Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
Question
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Chapter 6, Problem 6K.3E

(a)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  Cl2(g)+S2O32-(aq)Cl(aq)+SO42-(aq)

Concept Introduction:

Net ionic equation:

Net ionic equation is defined as the specific species that only involves to a particular reaction. This type of equations is generally used in acid-base neutralization reactions and redox reactions.

Oxidizing agent:

The material which gains electron in a chemical reaction is called oxidizing agent. In this reaction, the oxidation number will be decreased.

Reducing agent:

The material, which loses electrons in a chemical reaction, is called reducing agent. In this reaction, the oxidation number will be increased.

(a)

Expert Solution
Check Mark

Answer to Problem 6K.3E

The balanced reaction of thiosulfate ion with chlorine gas is given below,

  S2O32-(aq)+4Cl2(g)+5H2O(l)8Cl(aq)+2SO42-(aq)+10H+

Here, the oxidizing agent is Cl2 and reducing agent is S2O32-.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  Cl2(g)+S2O32-(aq)Cl(aq)+SO42-(aq)

Oxidation half-reaction:

The oxidation number of S is increased from +2 to +6, therefore this is oxidation reaction.

  S2O32-(aq)SO42-(aq)

Balance the equation except H and O.

  S2O32-(aq)2SO42-(aq)

Balance the O atom by adding H2O left side

  S2O32-(aq)+5H2O(l)2SO42-(aq)

Balance the H atom by adding H+ ion on the right side

  S2O32-(aq)+5H2O(l)2SO42-(aq)+10H+

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 2 and the right side the net charge is +6, therefore it required 8 electrons on the right side to reduce the charge from +6 to 2.

  S2O32-(aq)+5H2O(l)2SO42-(aq)+10H++8e-

Therefore, the balanced oxidation half-reaction is

  S2O32-(aq)+5H2O(l)2SO42-(aq)+10H++8e-

Here, S2O32- act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Cl is decreased from 0 to 1, therefore this is reduction reaction.

  Cl2(g)Cl(aq)

Balance the equation except H and O.

  Cl2(g)2Cl(aq)

Balancing the net charges by adding e.

Here, in left side, the net charge is 0 and the right side the net charge is 2, therefore it required 22e on the left side to reduce the charge from 0 to 2.

  Cl2(g)+2e-2Cl(aq)

Therefore, the balanced reduction half-reaction is

  Cl2(g)+2e-2Cl(aq)

Here, the Cl2 act as gaining of electrons, so it is a oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 8 electrons are lost and in reduction half reaction 2 electron is gained. Therefore, multiply the oxidation half reaction by 2 and the reduction half reaction by 8.

  2S2O32-(aq)+10H2O(l)4SO42-(aq)+20H++16e-8Cl2(g)+16e-16Cl(aq)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  2S2O32-(aq)+8Cl2(g)+10H2O(l)16Cl(aq)+4SO42-(aq)+20H+

Divide by 2 each side of the arrow.

  S2O32-(aq)+4Cl2(g)+5H2O(l)8Cl(aq)+2SO42-(aq)+10H+

Therefore, the balanced net ionic equation the above reaction is

  S2O32-(aq)+4Cl2(g)+5H2O(l)8Cl(aq)+2SO42-(aq)+10H+

(b)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  MnO4-(aq)+H2SO3(aq)Mn2+(aq)+HSO4-(aq)

Concept introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 6K.3E

The balanced reaction of permanganate ion with sulfurous acid is given below,

5H2SO3(aq)+2MnO4-(aq)+H+5HSO4-(aq)+2Mn2+(aq)+3H2O(l)

Here, the oxidizing agent is MnO4- and reducing agent is H2SO3.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  MnO4-(aq)+H2SO3(aq)Mn2+(aq)+HSO4-(aq)

Oxidation half-reaction:

The oxidation number of S is increased from +4 to +6, therefore this is oxidation reaction.

  H2SO3(aq)HSO4-(aq)

Balance the equation except H and O.

  H2SO3(aq)HSO4-(aq)

Balance the O atom by adding H2O left side

  H2SO3(aq)+H2O(l)HSO4-(aq)

Balance the H atom by adding H+ ion on the right side

  H2SO3(aq)+H2O(l)HSO4-(aq)+3H+

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 0 and the right side the net charge is +2, therefore it required 2 electron on the right side to reduce the charge from +2 to 0.

  H2SO3(aq)+H2O(l)HSO4-(aq)+3H++2e-

Therefore, the balanced oxidation half-reaction is

  H2SO3(aq)+H2O(l)H2SO4-(aq)+3H++2e-

Here, H2SO3 lost 2 electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Mn is decreased from +7 to +2, therefore this is reduction reaction.

  MnO4-(aq)Mn2+(aq)

Balance the equation except H and O.

  MnO4-(aq)Mn2+(aq)

Balance the O atom by adding H2O right side

  MnO4-(aq)Mn2+(aq)+4H2O(l)

Balance the H atom by adding H+ ion on the right side

  MnO4-(aq)+8H+Mn2+(aq)+4H2O(l)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is +7 and the right side the net charge is +2, therefore it required 5 electrons on the left side to reduce the charge from +7 to +2.

  MnO4-(aq)+8H++5e-Mn2+(aq)+4H2O(l)

Here, the MnO4- gains 5 electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 2 electrons are lost and in reduction half reaction 5 electrons is gained. Therefore, multiply the oxidation half reaction by 5 and the reduction half reaction by 2.

  5H2SO3(aq)+5H2O(l)5HSO4-(aq)+15H++10e-2MnO4-(aq)+16H++10e-2Mn2+(aq)+8H2O(l)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  5H2SO3(aq)+2MnO4-(aq)+H+5HSO4-(aq)+2Mn2+(aq)+3H2O(l)

Therefore, the balanced net ionic equation the above reaction is

  5H2SO3(aq)+2MnO4-(aq)+H+5HSO4-(aq)+2Mn2+(aq)+3H2O(l)

(c)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  H2S(aq)+Cl2(g)S(s)+Cl-(aq)

Concept introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 6K.3E

The balanced reaction of hydrosulfuric acid with chlorine given below,

  H2S(aq)+Cl2(g)S(s)++2Cl-(aq)+2H+

Here, the oxidizing agent is Cl2 and reducing agent is H2S.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  H2S(aq)+Cl2(g)S(s)+Cl-(aq)

Oxidation half-reaction:

The oxidation number of S is increased from 2 to 0, therefore this is oxidation reaction.

  H2S(aq)S(s)

Balance the equation except H and O.

  H2S(aq)S(s)

Balance the H atom by adding H+ ion on the right side

  H2S(aq)S(s)+2H+

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 0 and the right side the net charge is +2, therefore it required 2 electrons on the right side to reduce the charge from +2 to 0.

  H2S(aq)S(s)+2H++2e-

Therefore, the balanced oxidation half-reaction is

  H2S(aq)S(s)+2H++2e-

Here, H2S act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Cl is decreased from 0 to 1, therefore this is reduction reaction.

  Cl2(g)Cl-(aq)

Balance the equation except H and O.

  Cl2(g)2Cl-(aq)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 0 and the right side the net charge is 2, therefore it required 2 electrons on the left side to equal the charge from 2 to 2.

  Cl2(g)+2e2Cl-(aq)

Here, the Cl2 act as gaining of electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 2 electrons are lost and in reduction half reaction 2 electrons is gained. Therefore, multiply the oxidation half reaction by 2 and the reduction half reaction by 2.

  2H2S(aq)2S(s)+4H++4e-2Cl2(g)+4e4Cl-(aq)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  2H2S(aq)+2Cl2(g)2S(s)++4Cl-(aq)+4H+

Divide by 2 each side of the arrow.

  H2S(aq)+Cl2(g)S(s)++2Cl-(aq)+2H+

Therefore, the balanced net ionic equation the above reaction is

  H2S(aq)+Cl2(g)S(s)++2Cl-(aq)+2H+

(d)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions, also the oxidizing agent and reducing agent has to be identified.

  H2S(aq)+Cl2(g)S(s)+Cl-(aq)

Concept introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 6K.3E

The balanced reaction of hydrosulfuric acid with chlorine given below,

  H2S(aq)+Cl2(g)S(s)++2Cl-(aq)+2H+

Here, the oxidizing agent is Cl2 and reducing agent is H2S.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  H2S(aq)+Cl2(g)S(s)+Cl-(aq)

Oxidation half-reaction:

The oxidation number of S is increased from 2 to 0, therefore this is oxidation reaction.

  H2S(aq)S(s)

Balance the equation except H and O.

  H2S(aq)S(s)

Balance the H atom by adding H+ ion on the right side

  H2S(aq)S(s)+2H+

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 0 and the right side the net charge is +2, therefore it required 2 electrons on the right side to reduce the charge from +2 to 0.

  H2S(aq)S(s)+2H++2e-

Therefore, the balanced oxidation half-reaction is

  H2S(aq)S(s)+2H++2e-

Here, H2S act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Cl is decreased from 0 to 1, therefore this is reduction reaction.

  Cl2(g)Cl-(aq)

Balance the equation except H and O.

  Cl2(g)2Cl-(aq)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 0 and the right side the net charge is 2, therefore it required 2 electrons on the left side to equal the charge from 2 to 2.

  Cl2(g)+2e2Cl-(aq)

Here, the Cl2 act as gaining of electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 2 electrons are lost and in reduction half reaction 2 electrons is gained. Therefore, multiply the oxidation half reaction by 2 and the reduction half reaction by 2.

  2H2S(aq)2S(s)+4H++4e-2Cl2(g)+4e4Cl-(aq)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  2H2S(aq)+2Cl2(g)2S(s)++4Cl-(aq)+4H+

Divide by 2 each side of the arrow.

  H2S(aq)+Cl2(g)S(s)++2Cl-(aq)+2H+

Therefore, the balanced net ionic equation the above reaction is

  H2S(aq)+Cl2(g)S(s)++2Cl-(aq)+2H+

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Chapter 6 Solutions

Chemical Principles: The Quest for Insight

Ch. 6 - Prob. 6A.5ECh. 6 - Prob. 6A.6ECh. 6 - Prob. 6A.7ECh. 6 - Prob. 6A.8ECh. 6 - Prob. 6A.9ECh. 6 - Prob. 6A.10ECh. 6 - Prob. 6A.11ECh. 6 - Prob. 6A.12ECh. 6 - Prob. 6A.13ECh. 6 - Prob. 6A.14ECh. 6 - Prob. 6A.15ECh. 6 - Prob. 6A.16ECh. 6 - Prob. 6A.17ECh. 6 - Prob. 6A.18ECh. 6 - Prob. 6A.19ECh. 6 - Prob. 6A.20ECh. 6 - Prob. 6A.21ECh. 6 - Prob. 6A.22ECh. 6 - Prob. 6A.23ECh. 6 - Prob. 6A.24ECh. 6 - Prob. 6B.1ASTCh. 6 - Prob. 6B.1BSTCh. 6 - Prob. 6B.2ASTCh. 6 - Prob. 6B.2BSTCh. 6 - Prob. 6B.3ASTCh. 6 - Prob. 6B.3BSTCh. 6 - Prob. 6B.1ECh. 6 - Prob. 6B.2ECh. 6 - Prob. 6B.3ECh. 6 - Prob. 6B.4ECh. 6 - Prob. 6B.5ECh. 6 - Prob. 6B.6ECh. 6 - Prob. 6B.7ECh. 6 - Prob. 6B.8ECh. 6 - Prob. 6B.9ECh. 6 - Prob. 6B.10ECh. 6 - Prob. 6B.11ECh. 6 - Prob. 6B.12ECh. 6 - Prob. 6C.1ASTCh. 6 - Prob. 6C.1BSTCh. 6 - Prob. 6C.2ASTCh. 6 - Prob. 6C.2BSTCh. 6 - Prob. 6C.3ASTCh. 6 - Prob. 6C.3BSTCh. 6 - Prob. 6C.1ECh. 6 - Prob. 6C.2ECh. 6 - Prob. 6C.3ECh. 6 - Prob. 6C.4ECh. 6 - Prob. 6C.5ECh. 6 - Prob. 6C.6ECh. 6 - Prob. 6C.7ECh. 6 - Prob. 6C.8ECh. 6 - Prob. 6C.9ECh. 6 - Prob. 6C.10ECh. 6 - Prob. 6C.11ECh. 6 - Prob. 6C.12ECh. 6 - Prob. 6C.13ECh. 6 - Prob. 6C.14ECh. 6 - Prob. 6C.15ECh. 6 - Prob. 6C.16ECh. 6 - Prob. 6C.17ECh. 6 - Prob. 6C.18ECh. 6 - Prob. 6C.19ECh. 6 - Prob. 6C.20ECh. 6 - Prob. 6C.21ECh. 6 - Prob. 6C.22ECh. 6 - Prob. 6D.1ASTCh. 6 - Prob. 6D.1BSTCh. 6 - Prob. 6D.2ASTCh. 6 - Prob. 6D.2BSTCh. 6 - Prob. 6D.3ASTCh. 6 - Prob. 6D.3BSTCh. 6 - Prob. 6D.4ASTCh. 6 - Prob. 6D.4BSTCh. 6 - Prob. 6D.5ASTCh. 6 - Prob. 6D.5BSTCh. 6 - Prob. 6D.6ASTCh. 6 - Prob. 6D.6BSTCh. 6 - Prob. 6D.1ECh. 6 - Prob. 6D.2ECh. 6 - Prob. 6D.3ECh. 6 - Prob. 6D.4ECh. 6 - Prob. 6D.5ECh. 6 - Prob. 6D.6ECh. 6 - Prob. 6D.7ECh. 6 - Prob. 6D.8ECh. 6 - Prob. 6D.9ECh. 6 - Prob. 6D.11ECh. 6 - Prob. 6D.12ECh. 6 - Prob. 6D.13ECh. 6 - Prob. 6D.14ECh. 6 - Prob. 6D.15ECh. 6 - Prob. 6D.16ECh. 6 - Prob. 6D.17ECh. 6 - Prob. 6D.18ECh. 6 - Prob. 6D.19ECh. 6 - Prob. 6D.20ECh. 6 - Prob. 6D.21ECh. 6 - Prob. 6D.22ECh. 6 - Prob. 6E.1ASTCh. 6 - Prob. 6E.1BSTCh. 6 - Prob. 6E.2ASTCh. 6 - Prob. 6E.2BSTCh. 6 - Prob. 6E.3ASTCh. 6 - Prob. 6E.1ECh. 6 - Prob. 6E.2ECh. 6 - Prob. 6E.3ECh. 6 - Prob. 6E.4ECh. 6 - Prob. 6E.5ECh. 6 - Prob. 6E.6ECh. 6 - Prob. 6E.7ECh. 6 - Prob. 6E.8ECh. 6 - Prob. 6E.9ECh. 6 - Prob. 6E.10ECh. 6 - Prob. 6E.11ECh. 6 - Prob. 6E.12ECh. 6 - Prob. 6E.13ECh. 6 - Prob. 6E.14ECh. 6 - Prob. 6E.15ECh. 6 - Prob. 6E.16ECh. 6 - Prob. 6E.17ECh. 6 - Prob. 6E.18ECh. 6 - Prob. 6F.1ASTCh. 6 - Prob. 6F.1BSTCh. 6 - Prob. 6F.2ASTCh. 6 - Prob. 6F.2BSTCh. 6 - Prob. 6F.1ECh. 6 - Prob. 6F.2ECh. 6 - Prob. 6F.3ECh. 6 - Prob. 6F.4ECh. 6 - Prob. 6F.5ECh. 6 - Prob. 6F.6ECh. 6 - Prob. 6F.7ECh. 6 - Prob. 6F.9ECh. 6 - Prob. 6F.10ECh. 6 - Prob. 6G.1ASTCh. 6 - Prob. 6G.1BSTCh. 6 - Prob. 6G.2ASTCh. 6 - Prob. 6G.2BSTCh. 6 - Prob. 6G.3ASTCh. 6 - Prob. 6G.3BSTCh. 6 - Prob. 6G.4ASTCh. 6 - Prob. 6G.4BSTCh. 6 - Prob. 6G.1ECh. 6 - Prob. 6G.3ECh. 6 - Prob. 6G.4ECh. 6 - Prob. 6G.5ECh. 6 - Prob. 6G.6ECh. 6 - Prob. 6G.7ECh. 6 - Prob. 6G.8ECh. 6 - Prob. 6G.9ECh. 6 - Prob. 6G.11ECh. 6 - Prob. 6G.12ECh. 6 - Prob. 6G.13ECh. 6 - Prob. 6G.14ECh. 6 - Prob. 6G.15ECh. 6 - Prob. 6G.16ECh. 6 - Prob. 6G.19ECh. 6 - Prob. 6G.20ECh. 6 - Prob. 6H.1ASTCh. 6 - Prob. 6H.1BSTCh. 6 - Prob. 6H.2ASTCh. 6 - Prob. 6H.2BSTCh. 6 - Prob. 6H.3ASTCh. 6 - Prob. 6H.3BSTCh. 6 - Prob. 6H.4ASTCh. 6 - Prob. 6H.4BSTCh. 6 - Prob. 6H.5ASTCh. 6 - Prob. 6H.5BSTCh. 6 - Prob. 6H.1ECh. 6 - Prob. 6H.3ECh. 6 - Prob. 6H.9ECh. 6 - Prob. 6H.10ECh. 6 - Prob. 6H.15ECh. 6 - Prob. 6H.16ECh. 6 - Prob. 6H.21ECh. 6 - Prob. 6H.22ECh. 6 - Prob. 6H.23ECh. 6 - Prob. 6H.24ECh. 6 - Prob. 6H.25ECh. 6 - Prob. 6H.26ECh. 6 - Prob. 6H.27ECh. 6 - Prob. 6H.28ECh. 6 - Prob. 6H.29ECh. 6 - Prob. 6H.30ECh. 6 - Prob. 6H.31ECh. 6 - Prob. 6H.32ECh. 6 - Prob. 6I.1ASTCh. 6 - Prob. 6I.1BSTCh. 6 - Prob. 6I.2ASTCh. 6 - Prob. 6I.2BSTCh. 6 - Prob. 6I.3ASTCh. 6 - Prob. 6I.3BSTCh. 6 - Prob. 6I.4ASTCh. 6 - Prob. 6I.4BSTCh. 6 - Prob. 6I.1ECh. 6 - Prob. 6I.2ECh. 6 - Prob. 6I.3ECh. 6 - Prob. 6I.4ECh. 6 - Prob. 6I.5ECh. 6 - Prob. 6I.6ECh. 6 - Prob. 6I.7ECh. 6 - Prob. 6I.8ECh. 6 - Prob. 6I.9ECh. 6 - Prob. 6I.10ECh. 6 - Prob. 6I.11ECh. 6 - Prob. 6I.12ECh. 6 - Prob. 6J.1ASTCh. 6 - Prob. 6J.1BSTCh. 6 - Prob. 6J.2ASTCh. 6 - Prob. 6J.2BSTCh. 6 - Prob. 6J.1ECh. 6 - Prob. 6J.2ECh. 6 - Prob. 6J.3ECh. 6 - Prob. 6J.4ECh. 6 - Prob. 6J.9ECh. 6 - Prob. 6J.10ECh. 6 - Prob. 6J.11ECh. 6 - Prob. 6J.15ECh. 6 - Prob. 6J.17ECh. 6 - Prob. 6K.1ASTCh. 6 - Prob. 6K.1BSTCh. 6 - Prob. 6K.2ASTCh. 6 - Prob. 6K.2BSTCh. 6 - Prob. 6K.1ECh. 6 - Prob. 6K.2ECh. 6 - Prob. 6K.3ECh. 6 - Prob. 6K.4ECh. 6 - Prob. 6K.5ECh. 6 - Prob. 6K.6ECh. 6 - Prob. 6K.7ECh. 6 - Prob. 6K.8ECh. 6 - Prob. 6L.1ASTCh. 6 - Prob. 6L.1BSTCh. 6 - Prob. 6L.2ASTCh. 6 - Prob. 6L.2BSTCh. 6 - Prob. 6L.3ASTCh. 6 - Prob. 6L.3BSTCh. 6 - Prob. 6L.1ECh. 6 - Prob. 6L.2ECh. 6 - Prob. 6L.3ECh. 6 - Prob. 6L.4ECh. 6 - Prob. 6L.5ECh. 6 - Prob. 6L.7ECh. 6 - Prob. 6L.9ECh. 6 - Prob. 6M.1ASTCh. 6 - Prob. 6M.1BSTCh. 6 - Prob. 6M.2ASTCh. 6 - Prob. 6M.2BSTCh. 6 - Prob. 6M.3ASTCh. 6 - Prob. 6M.3BSTCh. 6 - Prob. 6M.4ASTCh. 6 - Prob. 6M.4BSTCh. 6 - Prob. 6M.1ECh. 6 - Prob. 6M.2ECh. 6 - Prob. 6M.9ECh. 6 - Prob. 6M.10ECh. 6 - Prob. 6N.1ASTCh. 6 - Prob. 6N.1BSTCh. 6 - Prob. 6N.2ASTCh. 6 - Prob. 6N.2BSTCh. 6 - Prob. 6N.3BSTCh. 6 - Prob. 6N.4ASTCh. 6 - Prob. 6N.4BSTCh. 6 - Prob. 6N.1ECh. 6 - Prob. 6N.2ECh. 6 - Prob. 6N.5ECh. 6 - Prob. 6N.6ECh. 6 - Prob. 6N.7ECh. 6 - Prob. 6N.9ECh. 6 - Prob. 6N.10ECh. 6 - Prob. 6N.11ECh. 6 - Prob. 6N.12ECh. 6 - Prob. 6N.21ECh. 6 - Prob. 6N.23ECh. 6 - Prob. 6O.1ASTCh. 6 - Prob. 6O.1BSTCh. 6 - Prob. 6O.2ASTCh. 6 - Prob. 6O.2BSTCh. 6 - Prob. 6O.3ASTCh. 6 - Prob. 6O.3BSTCh. 6 - Prob. 6O.4ASTCh. 6 - Prob. 6O.4BSTCh. 6 - Prob. 6O.1ECh. 6 - Prob. 6O.2ECh. 6 - Prob. 6O.3ECh. 6 - Prob. 6O.4ECh. 6 - Prob. 6O.5ECh. 6 - Prob. 6O.6ECh. 6 - Prob. 6O.7ECh. 6 - Prob. 6O.8ECh. 6 - Prob. 6O.9ECh. 6 - Prob. 6O.10ECh. 6 - Prob. 6O.11ECh. 6 - Prob. 6O.12ECh. 6 - Prob. 6O.13ECh. 6 - Prob. 6O.14ECh. 6 - Prob. 6O.15ECh. 6 - Prob. 6O.16ECh. 6 - Prob. 6.1ECh. 6 - Prob. 6.3ECh. 6 - Prob. 6.4ECh. 6 - Prob. 6.5ECh. 6 - Prob. 6.6ECh. 6 - Prob. 6.8ECh. 6 - Prob. 6.9ECh. 6 - Prob. 6.10ECh. 6 - Prob. 6.11ECh. 6 - Prob. 6.12ECh. 6 - Prob. 6.13ECh. 6 - Prob. 6.14ECh. 6 - Prob. 6.25ECh. 6 - Prob. 6.40ECh. 6 - Prob. 6.41ECh. 6 - Prob. 6.43ECh. 6 - Prob. 6.45ECh. 6 - Prob. 6.46ECh. 6 - Prob. 6.47ECh. 6 - Prob. 6.51ECh. 6 - Prob. 6.53ECh. 6 - Prob. 6.65ECh. 6 - Prob. 6.75ECh. 6 - Prob. 6.77E
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