General, Organic, and Biological Chemistry - 4th edition
General, Organic, and Biological Chemistry - 4th edition
4th Edition
ISBN: 9781259883989
Author: by Janice Smith
Publisher: McGraw-Hill Education
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Chapter 10.6, Problem 10.8PP
Interpretation Introduction

(a)

Interpretation:

The value of X in the following reaction should be predicted.

  11H+X12H++10e

Concept Introduction:

The reactions involving the formation of new nucleus with emission of some radiation along with it from an original nucleus are known as nuclear reactions.

Nuclear reaction is of two types:

  1. Nuclear fusion:

    The combination of two lighter nuclei to form a larger nucleus is known as nuclear fusion.

  2. Nuclear fission:

    The splitting of a heavy nucleus into lighter nucleus and neutrons is known as nuclear fission.

Interpretation Introduction

(b)

Interpretation:

The value of Y in the following reaction should be predicted.

  11H+12HY

Concept Introduction:

The reactions involving the formation of new nucleus with emission of some radiation along with it from an original nucleus are known as nuclear reactions.

Nuclear reaction is of two types :

  1. Nuclear fusion:

    The combination of two lighter nuclei to form a larger nucleus is known as nuclear fusion.

  2. Nuclear fission:

    The splitting of a heavy nucleus into lighter nucleus and neutrons is known as nuclear fission.

Interpretation Introduction

(c)

Interpretation:

The value of Z in the following reaction should be predicted.

  11H+23He24He+Z

Concept Introduction:

The reactions involving the formation of new nucleus with emission of some radiation along with it from an original nucleus are known as nuclear reactions.

Nuclear reaction is of two types :

  1. Nuclear fusion:

    The combination of two lighter nuclei to form a larger nucleus is known as nuclear fusion.

  2. Nuclear fission:

    The splitting of a heavy nucleus into lighter nucleus and neutrons is known as nuclear fission.

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How did you get this? k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs   0.693 a 102.0 nanogram(ng) sample of sodium-24 was stored in a lead-lined cabinet for 2.5 days How much sodium-24 remained? The half-life os sodium -24 is 15 hours    _________ ng sodium -24 remain  check_circle Expert Answer thumb_up   thumb_down Step 1 Radioactive decay is a first order reaction. Conversion of days into hours: 1 day2.5 days===24 hrs2.5×2460 hrs1 day=24 hrs2.5 days=2.5×24=60 hrs Calculation of k: k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs-1 Step 2 Radioactive decay is a first order reaction, therefore: [A]=[Ao] e−kt[A]=[Ao] e-kt where [A] is the final amount remained at time t, [Ao] is the initial amount. Putting values: [A]====102.0 ng×e−0.0462 hrs−1×60 hrs102.0 ng×e−2.772102.0 ng×0.062546.38 ng[A]=102.0 ng×e-0.0462 hrs-1×60 hrs=102.0 ng×e-2.772=102.0 ng×0.06254=6.38 ng Thus 6.38 ng of sodium-24 remained.
A 7.5 gram sample of Sr-90 was formed in 1960. The half life is 28 years. In what year will only 0.9357 grams remain? Show work
Lead-194 has a half-life of 11 minutes. A sample of lead-194 with a mass of 0.149g is allowed to decay for 66 minutes. What mass of lead-194 in grams remains in the sample after 66 minutes? a. 2.33 x 10 -3 b. 2.48 x 10 -4 c. 2.25 x 10 -4 d. 1.35 x 10 -2 e. no correct response

Chapter 10 Solutions

General, Organic, and Biological Chemistry - 4th edition

Ch. 10.2 - Prob. 10.3PPCh. 10.2 - Prob. 10.9PCh. 10.2 - Prob. 10.10PCh. 10.2 - Prob. 10.11PCh. 10.3 - Prob. 10.4PPCh. 10.3 - Prob. 10.12PCh. 10.3 - Prob. 10.13PCh. 10.3 - Prob. 10.14PCh. 10.4 - To treat a thyroid tumor, a patient must be given...Ch. 10.4 - A sample of iodine-131 (t1/2=8.0 days) has an...Ch. 10.4 - Prob. 10.15PCh. 10.5 - Prob. 10.16PCh. 10.5 - Prob. 10.17PCh. 10.5 - Prob. 10.18PCh. 10.6 - Prob. 10.7PPCh. 10.6 - Prob. 10.8PPCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43PCh. 10 - Prob. 44PCh. 10 - Prob. 45PCh. 10 - Prob. 46PCh. 10 - Prob. 47PCh. 10 - Prob. 48PCh. 10 - Prob. 49PCh. 10 - Prob. 50PCh. 10 - Prob. 51PCh. 10 - Prob. 52PCh. 10 - Prob. 53PCh. 10 - Prob. 54PCh. 10 - Prob. 55PCh. 10 - Prob. 56PCh. 10 - Prob. 57PCh. 10 - Prob. 58PCh. 10 - Prob. 59PCh. 10 - Prob. 60PCh. 10 - Prob. 61PCh. 10 - Prob. 62PCh. 10 - Prob. 63PCh. 10 - Prob. 64PCh. 10 - Prob. 65PCh. 10 - Prob. 66PCh. 10 - Prob. 67PCh. 10 - Prob. 68PCh. 10 - Prob. 69PCh. 10 - Prob. 70PCh. 10 - Prob. 71PCh. 10 - Prob. 72PCh. 10 - Prob. 73PCh. 10 - Prob. 74PCh. 10 - Prob. 75PCh. 10 - Prob. 76PCh. 10 - Prob. 77PCh. 10 - Prob. 78PCh. 10 - Prob. 79PCh. 10 - Prob. 80PCh. 10 - Prob. 81PCh. 10 - Prob. 82PCh. 10 - Prob. 83PCh. 10 - Prob. 84PCh. 10 - Prob. 85PCh. 10 - Prob. 86PCh. 10 - Prob. 87PCh. 10 - Prob. 88PCh. 10 - Prob. 89CPCh. 10 - Prob. 90CP
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