Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry (6th Edition)
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Chapter 17, Problem 76E
Interpretation Introduction

Interpretation:

The amount of the radioactive nuclide that would remain after 4.0 days, which has a half-life value of 12 hours 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:

Nt=N0×e(0.693×tk)

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 required to reach the value of Nt.

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The SN 1 mechanism starts with the rate-determining step which is the dissociation of the alkyl halide into a carbocation and a halide ion. The next step is the rapid reaction of the carbocation intermediate with the nucleophile; this step completes the nucleophilic substitution stage. The step that follows the nucleophilic substitution is a fast acid-base reaction. The nucleophile now acts as a base to remove the proton from the oxonium ion from the previous step, to give the observed product. Draw a curved arrow mechanism for the reaction, adding steps as necessary. Be sure to include all nonzero formal charges. Cl: Add/Remove step G Click and drag to start drawing a structure.

Chapter 17 Solutions

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry (6th Edition)

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