(a) Interpretation: The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined. Concept Introduction: The final mass of substance with initial mass m 1 and for n life time is calculated as- f i n a l m a s s = m 1 × ( 1 2 × 1 2 × ........... n )

Question

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Answer 1

Question

Chapter 10.3, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

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Answer 2

Question

Chapter 10.3, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Expert Solution & Answer

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Answer 3

Question

Chapter 10.3, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Expert Solution & Answer

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Answer 4

Question

Chapter 10.3, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Expert Solution & Answer

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Answer 5

Chapter 10.3, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Interpretation Introduction

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Answer 6

Chapter 10.3, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Answer 7

Chapter 10.3, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Answer 8

Interpretation Introduction

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Answer 9

Interpretation Introduction

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Answer 10

Interpretation Introduction

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Answer 11

Interpretation Introduction

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

Ch. 10.1 - Complete the following table for two isotopes of...Ch. 10.1 - Prob. 10.2P Ch. 10.1 - Prob. 10.3P Ch. 10.1 - Identify Q in each of the following symbols. a....Ch. 10.2 - Prob. 10.5P Ch. 10.2 - Prob. 10.6P Ch. 10.2 - Prob. 10.7P Ch. 10.2 - Prob. 10.8P Ch. 10.2 - Prob. 10.9P Ch. 10.2 - Prob. 10.10P

(a) Interpretation: The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined. Concept Introduction: The final mass of substance with initial mass m 1 and for n life time is calculated as- f i n a l m a s s = m 1 × ( 1 2 × 1 2 × ........... n )

Solution Summary: The author explains that the final mass of phosphorous-32 after two half-life is 0.25 g. The mass is determined by the multiplication of initial mass with half.

Concept explainers

(a)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

(b)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 4 half-life is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

(c)

Interpretation:

The amount of phosphorous-32 remained from a 1.00 g of sample after 8 half-lifes is to be determined.

Concept Introduction:

The final mass of substance with initial mass m₁ and for n life time is calculated as-

final mass=m1×(12×12×...........n)

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Chapter 10 Solutions

(a) Interpretation: The amount of phosphorous-32 remained from a 1.00 g of sample after 2 half-life is to be determined. Concept Introduction: The final mass of substance with initial mass m 1 and for n life time is calculated as- f i n a l m a s s = m 1 × ( 1 2 × 1 2 × ........... n )

Solution Summary: The author explains that the final mass of phosphorous-32 after two half-life is 0.25 g. The mass is determined by the multiplication of initial mass with half.

Concept explainers