Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined. Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

Question

6. The equilibrium constant for the reaction 2 HBr (g) → H2(g) + Br2(g) Can be expressed by the empirical formula 11790 K In K-6.375 + 0.6415 In(T K-¹) - T Use this formula to determine A,H as a function of temperature. Calculate A,-H at 25 °C and at 100 °C.

Answer 1

Question

Chapter 20, Problem 19E

(a)

Interpretation Introduction

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

(b)

Interpretation Introduction

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

(c)

Interpretation Introduction

Interpretation: The time required to decrease the radioactivity below 25% of the initial measured value needs to be determined.

Concept Introduction : The rate law equation for first order reaction is as follows:

ln(CC0) = -kt

Where;

C0 = initial radioactivity at t= 0

C = radioactivity at time t

k = rate constant

t = time

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

6. The equilibrium constant for the reaction 2 HBr (g) → H2(g) + Br2(g) Can be expressed by the empirical formula 11790 K In K-6.375 + 0.6415 In(T K-¹) - T Use this formula to determine A,H as a function of temperature. Calculate A,-H at 25 °C and at 100 °C.

3. Nitrosyl chloride, NOCI, decomposes according to 2 NOCI (g) → 2 NO(g) + Cl2(g) Assuming that we start with no moles of NOCl (g) and no NO(g) or Cl2(g), derive an expression for Kp in terms of the equilibrium value of the extent of reaction, Seq, and the pressure, P. Given that K₂ = 2.00 × 10-4, calculate Seq/ of 29/no when P = 0.080 bar. What is the new value по ƒª/ at equilibrium when P = 0.160 bar? Is this result in accord with Le Châtelier's Principle?

Consider the following chemical equilibrium: 2SO2(g) + O2(g) = 2SO3(g) • Write the equilibrium constant expression for this reaction. Now compare it to the equilibrium constant expression for the related reaction: • . 1 SO2(g) + O2(g) = SO3(g) 2 How do these two equilibrium expressions differ? What important principle about the dependence of equilibrium constants on the stoichiometry of a reaction can you learn from this comparison?

Answer 2

Question

Chapter 20, Problem 19E

(a)

Interpretation Introduction

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

(b)

Interpretation Introduction

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

(c)

Interpretation Introduction

Interpretation: The time required to decrease the radioactivity below 25% of the initial measured value needs to be determined.

Concept Introduction : The rate law equation for first order reaction is as follows:

ln(CC0) = -kt

Where;

C0 = initial radioactivity at t= 0

C = radioactivity at time t

k = rate constant

t = time

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 3

Question

Chapter 20, Problem 19E

(a)

Interpretation Introduction

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

(b)

Interpretation Introduction

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

(c)

Interpretation Introduction

Interpretation: The time required to decrease the radioactivity below 25% of the initial measured value needs to be determined.

Concept Introduction : The rate law equation for first order reaction is as follows:

ln(CC0) = -kt

Where;

C0 = initial radioactivity at t= 0

C = radioactivity at time t

k = rate constant

t = time

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 4

Question

Chapter 20, Problem 19E

(a)

Interpretation Introduction

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

(b)

Interpretation Introduction

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

(c)

Interpretation Introduction

Interpretation: The time required to decrease the radioactivity below 25% of the initial measured value needs to be determined.

Concept Introduction : The rate law equation for first order reaction is as follows:

ln(CC0) = -kt

Where;

C0 = initial radioactivity at t= 0

C = radioactivity at time t

k = rate constant

t = time

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 5

Chapter 20, Problem 19E

(a)

Interpretation Introduction

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

(b)

Interpretation Introduction

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

(c)

Interpretation Introduction

Interpretation: The time required to decrease the radioactivity below 25% of the initial measured value needs to be determined.

Concept Introduction : The rate law equation for first order reaction is as follows:

ln(CC0) = -kt

Where;

C0 = initial radioactivity at t= 0

C = radioactivity at time t

k = rate constant

t = time

Answer 6

Chapter 20, Problem 19E

(a)

Interpretation Introduction

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

Answer 7

Chapter 20, Problem 19E

(a)

Interpretation Introduction

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

Answer 8

(b)

Interpretation Introduction

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

Answer 9

(b)

Interpretation Introduction

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

Answer 10

(c)

Interpretation Introduction

Interpretation: The time required to decrease the radioactivity below 25% of the initial measured value needs to be determined.

Concept Introduction : The rate law equation for first order reaction is as follows:

ln(CC0) = -kt

Where;

C0 = initial radioactivity at t= 0

C = radioactivity at time t

k = rate constant

t = time

Answer 11

(c)

Interpretation Introduction

Interpretation: The time required to decrease the radioactivity below 25% of the initial measured value needs to be determined.

Concept Introduction : The rate law equation for first order reaction is as follows:

ln(CC0) = -kt

Where;

C0 = initial radioactivity at t= 0

C = radioactivity at time t

k = rate constant

t = time

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 15

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 16

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 17

Trending nowThis is a popular solution!

Answer 18

Trending nowThis is a popular solution!

Answer 19

Trending nowThis is a popular solution!

Answer 20

Trending now

Answer 21

This is a popular solution!

Answer 22

See solution

Check out sample textbook solution

Answer 23

Ch. 20 - Prob. 1E Ch. 20 - Prob. 2E Ch. 20 - Prob. 3E Ch. 20 - Prob. 4E Ch. 20 - Prob. 5E Ch. 20 - Prob. 6E Ch. 20 - Prob. 7E Ch. 20 - Prob. 8E Ch. 20 - Prob. 9E Ch. 20 - Prob. 10E

Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined. Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

Solution Summary: The author explains how the value of k in s-1 is determined based on the half-life of the radioactive copper-64.

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

Interpretation: Asample having 28.0mg ⁶⁴Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is ⁶⁴Cu 64.0 u

Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

Chapter 20 Solutions

Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined. Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

Solution Summary: The author explains how the value of k in s-1 is determined based on the half-life of the radioactive copper-64.

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined. Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.

Interpretation: Asample having 28.0mg 64Cu then decay events produced in first second is to be calculated. Considering the atomic mass of is 64Cu 64.0 u Concept Introduction:Spontaneous breakdown of an atomic nucleus and results in release of energy from nucleus is called radioactive decay.

Chapter 20 Solutions

Interpretation : Radioactive copper-64 decays with a half-life of 12.8 days. Then the value of k in s-1 is to be determined.

Concept Introduction : The time required to disintegrate exactly half of the radioactive substance is called half-life period.