The partial pressure of the C O (reactant) more C O 2 (products) partial pressure of the C O (reactant) more C O 2 (products)

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Answer 1

Question

Chapter 14, Problem 14.100SP

a)

Interpretation Introduction

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

b)

Interpretation Introduction

To determine:

To find the partial pressure will be equal to its rate of decomposition.

c)

Interpretation Introduction

To determine:

To find the forward reaction and reverse reaction at equilibrium.

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Can you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.

Part 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answer

Part 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00

Answer 2

Question

Chapter 14, Problem 14.100SP

a)

Interpretation Introduction

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

b)

Interpretation Introduction

To determine:

To find the partial pressure will be equal to its rate of decomposition.

c)

Interpretation Introduction

To determine:

To find the forward reaction and reverse reaction at equilibrium.

Expert Solution & Answer

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Answer 3

Question

Chapter 14, Problem 14.100SP

a)

Interpretation Introduction

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

b)

Interpretation Introduction

To determine:

To find the partial pressure will be equal to its rate of decomposition.

c)

Interpretation Introduction

To determine:

To find the forward reaction and reverse reaction at equilibrium.

Expert Solution & Answer

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Answer 4

Question

Chapter 14, Problem 14.100SP

a)

Interpretation Introduction

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

b)

Interpretation Introduction

To determine:

To find the partial pressure will be equal to its rate of decomposition.

c)

Interpretation Introduction

To determine:

To find the forward reaction and reverse reaction at equilibrium.

Expert Solution & Answer

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Answer 5

Chapter 14, Problem 14.100SP

a)

Interpretation Introduction

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

b)

Interpretation Introduction

To determine:

To find the partial pressure will be equal to its rate of decomposition.

c)

Interpretation Introduction

To determine:

To find the forward reaction and reverse reaction at equilibrium.

Answer 6

Chapter 14, Problem 14.100SP

a)

Interpretation Introduction

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

Answer 7

Chapter 14, Problem 14.100SP

a)

Interpretation Introduction

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

Answer 8

b)

Interpretation Introduction

To determine:

To find the partial pressure will be equal to its rate of decomposition.

Answer 9

b)

Interpretation Introduction

To determine:

To find the partial pressure will be equal to its rate of decomposition.

Answer 10

c)

Interpretation Introduction

To determine:

To find the forward reaction and reverse reaction at equilibrium.

Answer 11

c)

Interpretation Introduction

To determine:

To find the forward reaction and reverse reaction at equilibrium.

Answer 12

Expert Solution & Answer

Answer 13

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Answer 14

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Answer 15

Ch. 14 - Prob. 14.1P Ch. 14 - APPLY 14.2 Nitrogen dioxide, a pollutant that...Ch. 14 - Prob. 14.3P Ch. 14 - APPLY 14.4 Lactic acid, which builds up in muscle...Ch. 14 - Conceptual PRACTICE 14.5 The following pictures...Ch. 14 - Conceptual APPLY 14.6 The equilibrium constant...Ch. 14 - PRACTICE 14.7 In the industrial synthesis of...Ch. 14 - APPLY 14.8 At 25 °C, Kp = 25 for the reaction...Ch. 14 - Prob. 14.9P Ch. 14 - APPLY 14.10For the reaction...

The partial pressure of the C O (reactant) more C O 2 (products) partial pressure of the C O (reactant) more C O 2 (products)

Videos

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

To determine:

To find the partial pressure will be equal to its rate of decomposition.

To determine:

To find the forward reaction and reverse reaction at equilibrium.

Want to see the full answer?

Chapter 14 Solutions

The partial pressure of the C O (reactant) more C O 2 (products) partial pressure of the C O (reactant) more C O 2 (products)

Videos

To determine: The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

To determine: To find the partial pressure will be equal to its rate of decomposition.

To determine: To find the forward reaction and reverse reaction at equilibrium.

Want to see the full answer?

Chapter 14 Solutions

To determine:

The partial pressure of the CO (reactant) more CO2 (products) partial pressure of the CO (reactant) more CO2 (products)

To determine:

To find the partial pressure will be equal to its rate of decomposition.

To determine:

To find the forward reaction and reverse reaction at equilibrium.