Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
Question
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Chapter 18, Problem 76CWP
Interpretation Introduction

Interpretation: The half life of Rubidium- 87 is given. A rock containing 109.7μg of 87Rb and 3.1μg of 87Sr is given. Assuming no 87Sr was originally present; the age of the rock is to be calculated.

Concept introduction: A process through which an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.

To determine: The age of the rock.

Expert Solution & Answer
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Answer to Problem 76CWP

Solution

The age of rock is 1.9×109years_ .

Explanation of Solution

Explanation

Given

The value of half life of 87Rb is 4.5×109years .

The decay constant is calculated by the formula,

λ=0.693t1/2

Where

  • t1/2 is the half life of 87Rb .
  • λ is decay constant.

Substitute the value of half life in the above formula.

λ=0.6934.7×1010years1=1.4×10-11years-1_ .

The decay constant is 1.4×10-11years-1_ .

Explanation

The number of moles of 87Rb is 1.262×10-6 mol_ .

The number of moles of 87Sr is 0.035×10-6 mol_ .

Given

The mass of 87Rb in the rock sample is 109.7μg

The mass of 87Sr in the rock sample is 3.1μg

The conversion of μg into g is done as,

1μg=106g

Therefore, the conversion of 109.7μg into g is,

109.7μg=109.7×106g

Similarly, the conversion of 3.1μg is done as,

3.1μg=3.1×106g

Molar mass of 87Rb is 86.90919 g/mol .

Number of moles of 87Rb is calculated by the formula,

Number of moles=Given massMolar mass

Substitute the value of given mass and molar mass of 87Rb in the above equation.

Number of moles=109.7×106g86.90919 g/mol=1.262×10-6 mol_

Molar mass of 87Sr is 86.90888 g/mol .

Number of moles of 87Sr is calculated by the formula,

Number of moles=Given massMolar mass

Substitute the value of given mass and molar mass of 87Sr in the above equation.

Number of moles=3.1×106g86.90888 g/mol=0.035×10-6 mol_

Explanation

The age of rock is 1.9×109years_ .

One mole of 87Rb decays to one mole of 87Sr in the beta decay, therefore, the total amount of 87Rb in the rock sample initially present (N0) is calculated by the formula,

Amount of87Rb and87Sr present initially=Amount of  87Rb+Amount of 87Sr 

Substitute value of amount of 87Rb and 87Sr in the above equation.

Amount of87Rb and87Sr present initially=1.262×106 mol+0.035×106 mol=1.297×106 mol

The amount of 87Rb present now (N) is 1.262×106 mol .

The age of the rock is calculated by the formula,

ln(NN0)=λt

Where,

  • N is the amount of 87Rb and 87Sr present now.
  • N0 is the amount of 87Rb present originally.
  • t is the decay time.
  • λ is the decay constant.

Substitute the values of N , N0 , λ in the above equation.

ln(NN0)=λtln1.262×1061.297×106=1.4×1011×t0.027=1.4×1011×tt=1.9×109years_

Therefore, the age of rock is 1.9×109years_ .

Conclusion

Conclusion

The age of rock is 1.9×109years_ .

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Chapter 18 Solutions

Chemistry: An Atoms First Approach

Ch. 18 - Prob. 1QCh. 18 - Prob. 2QCh. 18 - Prob. 3QCh. 18 - Prob. 4QCh. 18 - Prob. 5QCh. 18 - Prob. 6QCh. 18 - Prob. 7QCh. 18 - Prob. 8QCh. 18 - Prob. 9QCh. 18 - Prob. 10QCh. 18 - Prob. 11ECh. 18 - Prob. 12ECh. 18 - Prob. 13ECh. 18 - Prob. 14ECh. 18 - Prob. 15ECh. 18 - Prob. 16ECh. 18 - Prob. 17ECh. 18 - Prob. 18ECh. 18 - Prob. 19ECh. 18 - Prob. 20ECh. 18 - Prob. 21ECh. 18 - Prob. 22ECh. 18 - Prob. 23ECh. 18 - Prob. 24ECh. 18 - Prob. 25ECh. 18 - Prob. 26ECh. 18 - Prob. 27ECh. 18 - Prob. 28ECh. 18 - Prob. 29ECh. 18 - Prob. 30ECh. 18 - Prob. 31ECh. 18 - Prob. 32ECh. 18 - Prob. 33ECh. 18 - Prob. 34ECh. 18 - Prob. 35ECh. 18 - Prob. 36ECh. 18 - Prob. 37ECh. 18 - Prob. 38ECh. 18 - Prob. 39ECh. 18 - Prob. 40ECh. 18 - Prob. 41ECh. 18 - Prob. 42ECh. 18 - Prob. 43ECh. 18 - Prob. 44ECh. 18 - Prob. 45ECh. 18 - Prob. 46ECh. 18 - Prob. 47ECh. 18 - Prob. 48ECh. 18 - Prob. 49ECh. 18 - Prob. 50ECh. 18 - Prob. 51ECh. 18 - Prob. 52ECh. 18 - Prob. 53ECh. 18 - A chemist studied the reaction mechanism for the...Ch. 18 - Prob. 55ECh. 18 - Prob. 56ECh. 18 - Prob. 57ECh. 18 - Prob. 58ECh. 18 - Prob. 59AECh. 18 - Prob. 60AECh. 18 - Prob. 61AECh. 18 - Prob. 62AECh. 18 - Prob. 63AECh. 18 - Prob. 64AECh. 18 - Prob. 65AECh. 18 - Prob. 66AECh. 18 - Prob. 67AECh. 18 - Prob. 68AECh. 18 - Prob. 69AECh. 18 - Prob. 70AECh. 18 - Prob. 71AECh. 18 - Prob. 72AECh. 18 - Prob. 73CWPCh. 18 - Prob. 74CWPCh. 18 - Prob. 75CWPCh. 18 - Prob. 76CWPCh. 18 - Prob. 77CWPCh. 18 - Prob. 78CWPCh. 18 - Prob. 79CPCh. 18 - Prob. 80CPCh. 18 - Prob. 81CPCh. 18 - Prob. 82CPCh. 18 - Prob. 83CPCh. 18 - Prob. 84CPCh. 18 - Prob. 85CPCh. 18 - Prob. 86CPCh. 18 - Prob. 87IPCh. 18 - Prob. 88IP
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