The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined. Concept Introduction: The integrated rate law for first order reaction is represented as follows: k = 2.303 t log [ A o ] [ A t ] Here, k is rate constant, [ A o ] is initial amount and [ A t ] is final amount of the sample.

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Answer 1

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Chapter 9, Problem 92P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

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

k=2.303tlog[Ao][At]

Here, k is rate constant, [Ao] is initial amount and [At] is final amount of the sample.

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Answer 2

Question

Chapter 9, Problem 92P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

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

k=2.303tlog[Ao][At]

Here, k is rate constant, [Ao] is initial amount and [At] is final amount of the sample.

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Answer 3

Question

Chapter 9, Problem 92P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

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

k=2.303tlog[Ao][At]

Here, k is rate constant, [Ao] is initial amount and [At] is final amount of the sample.

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Answer 4

Question

Chapter 9, Problem 92P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

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

k=2.303tlog[Ao][At]

Here, k is rate constant, [Ao] is initial amount and [At] is final amount of the sample.

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Answer 5

Chapter 9, Problem 92P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

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

k=2.303tlog[Ao][At]

Here, k is rate constant, [Ao] is initial amount and [At] is final amount of the sample.

Answer 6

Chapter 9, Problem 92P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

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

k=2.303tlog[Ao][At]

Here, k is rate constant, [Ao] is initial amount and [At] is final amount of the sample.

Answer 7

Chapter 9, Problem 92P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

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

k=2.303tlog[Ao][At]

Here, k is rate constant, [Ao] is initial amount and [At] is final amount of the sample.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 9.3 - Problem 9-1 Iodine-139 is a beta emitter. Write an...Ch. 9.3 - Problem 9-2 Thorium-223 is an alpha emitter. Write...Ch. 9.3 - Prob. 9.3QC Ch. 9.3 - Problem 9-4 Thallium-201, a radioisotope used to...Ch. 9.4 - Problem 9-5 Barium-122 has a half-life of 2...Ch. 9.5 - Problem 9-6 A radioactive isotope in a 9.0-mL vial...Ch. 9.5 - Prob. 9.7QC Ch. 9 - What is the relationship between frequency and...Ch. 9 - How does energy relate to frequency and...Ch. 9 - Prob. 3P

Solution Summary: The author explains that the half-life of phosphorus-32 is 14.3 days, the initial amount of sample is 1.2 mg, and the amount remaining after 28.6 days needs to be determined.

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Chapter 9 Solutions