Chemistry: A Molecular Approach
Chemistry: A Molecular Approach
3rd Edition
ISBN: 9780321809247
Author: Nivaldo J. Tro
Publisher: Prentice Hall
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Chapter 19, Problem 47E
Interpretation Introduction

Interpretation:

The amount of the radioactive sample that would remain after 5.5 days, which has a half-life value of 3.8 days is to be determined.

Concept introduction:

The half-life of a substance is the numerical value in which the given radioactive substance is assumed to be reduced to half of its initial amount. The half-life for a given substance is represented by t1/2.

In case, the decay of a radioactive substance is exponential, it will remain constant for the life time of the substance.

After each half-life period, the amount of the substance is reduced to half of the initial number.

The time required for the decay of the substance to a given amount of substance can be calculated using the formula mentioned below:

ln(NtN0)=kt

In the above equation, ‘Nt’ represents the mass of the radioactive substance after a certain time interval t, ‘N0’ indicates the initial mass of the radioactive material, ‘k’ represents the decay constant and ‘t’ represents the time interval for the half-life (t1/2).

The half-life of the element is determined by the following formula:

t1/2=0.693k

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Chapter 19 Solutions

Chemistry: A Molecular Approach

Ch. 19 - Prob. 1ECh. 19 - Prob. 2ECh. 19 - Prob. 3ECh. 19 - Prob. 4ECh. 19 - Prob. 5ECh. 19 - Prob. 6ECh. 19 - Prob. 7ECh. 19 - Prob. 8ECh. 19 - Prob. 9ECh. 19 - Prob. 10ECh. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Prob. 59ECh. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65ECh. 19 - Prob. 66ECh. 19 - Prob. 67ECh. 19 - Prob. 68ECh. 19 - Prob. 69ECh. 19 - Prob. 70ECh. 19 - Prob. 71ECh. 19 - Prob. 72ECh. 19 - Prob. 73ECh. 19 - Prob. 74ECh. 19 - Prob. 75ECh. 19 - Prob. 76ECh. 19 - Prob. 77ECh. 19 - Prob. 78ECh. 19 - Prob. 79ECh. 19 - Prob. 80ECh. 19 - Prob. 81ECh. 19 - Prob. 82ECh. 19 - Prob. 83ECh. 19 - Prob. 84ECh. 19 - Prob. 85ECh. 19 - Prob. 86ECh. 19 - Prob. 87ECh. 19 - Prob. 88ECh. 19 - Prob. 89ECh. 19 - Prob. 90ECh. 19 - Prob. 91ECh. 19 - Prob. 92ECh. 19 - Prob. 93ECh. 19 - Prob. 94ECh. 19 - Prob. 95ECh. 19 - Prob. 96ECh. 19 - Prob. 97ECh. 19 - Prob. 98ECh. 19 - Prob. 99ECh. 19 - Prob. 100ECh. 19 - Prob. 101ECh. 19 - Prob. 102ECh. 19 - Prob. 103ECh. 19 - Prob. 104ECh. 19 - Prob. 105ECh. 19 - Prob. 106ECh. 19 - Prob. 107ECh. 19 - Prob. 108ECh. 19 - Prob. 109E
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