Chemistry and Chemical Reactivity - AP Edition
Chemistry and Chemical Reactivity - AP Edition
10th Edition
ISBN: 9781337399203
Author: Kotz
Publisher: CENGAGE L
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Chapter 19, Problem 64GQ

(a)

Interpretation Introduction

Interpretation:

The element which is the weakest oxidizing agent has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is a decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(a)

Expert Solution
Check Mark

Answer to Problem 64GQ

Se- Weakest oxidizing agent.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

From the given standard electrode potentials Se has low value.

Therefore, it is weakest oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The ion or HX which is the weakest oxidising agent has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(b)

Expert Solution
Check Mark

Answer to Problem 64GQ

F-- Weakest reducing agent.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

From the given standard electrode potentials F- has high value.

Therefore, it is a weakest reducing agent.

(c)

Interpretation Introduction

Interpretation:

The element which is capable of oxidizing H2O to O2 has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(c)

Expert Solution
Check Mark

Answer to Problem 64GQ

F2 and Cl2 are capable of H2O to O2

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

From the given standard electrode potentials F2 and Cl2 has high E0 value.

Therefore, these are strongest oxidizing agents have capable of able to oxidized to water.

(d)

Interpretation Introduction

Interpretation:

The element which is capable of oxidizing H2O to S has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(d)

Expert Solution
Check Mark

Answer to Problem 64GQ

F2 ,Cl2,Br2, I2 and O2 are able to oxidize H2S to S.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

F2 ,Cl2,Br2, I2 and O2 are able to oxidize H2S to S.

Because the reduction potential values of F2 ,Cl2,Br2, I2 and O2 are higher than the H2.

(e)

Interpretation Introduction

Interpretation:

It has to be identified whether O2 is capable of oxidizing I- to I2 in acid solution.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(e)

Expert Solution
Check Mark

Answer to Problem 64GQ

O2 has capable to oxidize I- to I2.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

Only the stronger oxidizing agent has capable to oxidize I- to I2.

From the given non-metals O2 is a strong oxidizing agent. Therefore, O2 has capable to oxidize I- to I2.

(f)

Interpretation Introduction

Interpretation:

It has to be identified whether S is capable of oxidizing I- to I2.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(f)

Expert Solution
Check Mark

Answer to Problem 64GQ

S cannot oxidize  I- to I2

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

Only the stronger oxidizing agent has capable to oxidize I- to I2.

S has low reduction potential. Therefore, S cannot oxidize  I- to I2

(g)

Interpretation Introduction

Interpretation:

It has to be identified whether the reaction H2S(aq) + Se(s) H2Se(aq) + S(s) is product –favoured at equilibrium.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(g)

Expert Solution
Check Mark

Answer to Problem 64GQ

H2S(aq) + Se(s) H2Se(aq) + S(s) not a product –favoured at equilibrium.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

H2S(aq) + Se(s) H2Se(aq) + S(s) from the reaction H2S being oxidized to S and Se being reduced to H2Se.

To oxidize H2S, the compound must be a stronger oxidizing agent. But Se has a lower reduction potential. Therefore, H2S(aq) + Se(s) H2Se(aq) + S(s) not a product –favoured at equilibrium.

(h)

Interpretation Introduction

Interpretation:

It has to be identified whether the reaction H2S(aq) + I2(s) 2H+(aq) + 2I-(aq) is product –favoured at equilibrium.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive E0 are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative E0 values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

(h)

Expert Solution
Check Mark

Answer to Problem 64GQ

H2S(aq) + I2(s) 2H+(aq) + 2I-(aq) product favoured at equilibrium.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

F2+2e-  2F-                ; E0= + 2.87 VCl2+2e- 2Cl-             ; E0= + 1.36 VBr2+2e- 2Br-             ; E0= + 1.08 VI2+2e- 2I-                  ; E0= + 0.535 V

Further,

O2+4H++ 4e- 2H2O ; E0= + 1.229 VS +2H++ 2e- H2S     ; E0= + 0.14 VSe + 2H++2e- H2Se  ; E0 = - 0.40 V

H2S(aq) + I2(s) 2H+(aq) + 2I-(aq) from the reaction H2S is being oxidized to S and I2 is being reduced to I-.

To oxidize H2S, the compound must be a stronger oxidizing agent. This means that it must be easier to reduce .I2 has a higher reduction potential. Therefore, H2S(aq) + I2(s) 2H+(aq) + 2I-(aq) from the reaction H2S is being oxidized to S and I2 is  being reduced to I-.

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Chapter 19 Solutions

Chemistry and Chemical Reactivity - AP Edition

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In each, one...Ch. 19 - The following half-cells are available: (i)...Ch. 19 - Prob. 67GQCh. 19 - Prob. 68GQCh. 19 - A potential of 0.142 V is recorded (under standard...Ch. 19 - Prob. 70GQCh. 19 - The standard potential, E, for the reaction of...Ch. 19 - An electrolysis cell for aluminum production...Ch. 19 - Electrolysis of molten NaCl is done in cells...Ch. 19 - A current of 0.0100 A is passed through a solution...Ch. 19 - A current of 0.44 A is passed through a solution...Ch. 19 - Prob. 76GQCh. 19 - Prob. 77GQCh. 19 - Prob. 78GQCh. 19 - The products formed in the electrolysis of aqueous...Ch. 19 - Predict the products formed in the electrolysis of...Ch. 19 - Prob. 81GQCh. 19 - The metallurgy of aluminum involves electrolysis...Ch. 19 - Prob. 83GQCh. 19 - Prob. 84GQCh. 19 - Prob. 85GQCh. 19 - Prob. 86GQCh. 19 - Two Ag+(aq) | Ag(s) half-cells are constructed....Ch. 19 - Calculate equilibrium constants for the following...Ch. 19 - Prob. 89GQCh. 19 - Use the table of standard reduction potentials...Ch. 19 - Prob. 91GQCh. 19 - Prob. 92GQCh. 19 - Prob. 93GQCh. 19 - A voltaic cell is constructed in which one...Ch. 19 - An expensive but lighter alternative to the lead...Ch. 19 - The specifications for a lead storage battery...Ch. 19 - Manganese may play an important role in chemical...Ch. 19 - Prob. 98GQCh. 19 - Iron(II) ion undergoes a disproportionation...Ch. 19 - Copper(I) ion disproportionates to copper metal...Ch. 19 - Prob. 101GQCh. 19 - Prob. 102GQCh. 19 - Can either sodium or potassium metal be used as a...Ch. 19 - Galvanized steel pipes are used in the plumbing of...Ch. 19 - Consider an electrochemical cell based on the...Ch. 19 - Prob. 106ILCh. 19 - A silver coulometer (Study Question 106) was used...Ch. 19 - Four metals, A, B, C, and D, exhibit the following...Ch. 19 - Prob. 109ILCh. 19 - The amount of oxygen, O2, dissolved in a water...Ch. 19 - Prob. 111SCQCh. 19 - The free energy change for a reaction, rG, is the...Ch. 19 - Prob. 113SCQCh. 19 - (a) Is it easier to reduce water in acid or base?...Ch. 19 - Prob. 115SCQ
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ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
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Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
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Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
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Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY