Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN: 9781337398909
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

(a)

Expert Solution
Check Mark

Answer to Problem 14.102PAE

Solution:

3.063×1014 decays per second

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×107 yr1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×107 yr1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×1014 s1

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

(b)

Expert Solution
Check Mark

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of  M 26g+Mass of positron)Mass of A26l|=|(25.982593+0.0005486)25.986892|=3.75×103 amu

Since, 1amu =931.7 MeV

Thus,

3.75×103 amu=3.75×103×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of 26Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

(c)

Expert Solution
Check Mark

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×1014 s1( 60 s 1 min)=1.84×1012 min1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol1×1.837×1012 min1=3.87×1012 MeV mol1 min1

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Chapter 14 Solutions

Chemistry for Engineering Students

Ch. 14 - Prob. 14.1PAECh. 14 - Prob. 14.2PAECh. 14 - Prob. 14.3PAECh. 14 - Prob. 14.4PAECh. 14 - (a) How does 14C enter a living plant? (b) Write...Ch. 14 - Prob. 14.6PAECh. 14 - Prob. 14.7PAECh. 14 - Prob. 14.8PAECh. 14 - Prob. 14.9PAECh. 14 - Prob. 14.10PAECh. 14 - Prob. 14.11PAECh. 14 - Prob. 14.12PAECh. 14 - Prob. 14.13PAECh. 14 - Prob. 14.14PAECh. 14 - Prob. 14.15PAECh. 14 - Prob. 14.16PAECh. 14 - Prob. 14.17PAECh. 14 - Prob. 14.18PAECh. 14 - Prob. 14.19PAECh. 14 - Prob. 14.20PAECh. 14 - Prob. 14.21PAECh. 14 - Prob. 14.22PAECh. 14 - Prob. 14.23PAECh. 14 - Prob. 14.24PAECh. 14 - Prob. 14.25PAECh. 14 - Prob. 14.26PAECh. 14 - Prob. 14.27PAECh. 14 - Prob. 14.28PAECh. 14 - Prob. 14.29PAECh. 14 - Prob. 14.30PAECh. 14 - Prob. 14.31PAECh. 14 - Prob. 14.32PAECh. 14 - Prob. 14.33PAECh. 14 - Prob. 14.34PAECh. 14 - Prob. 14.35PAECh. 14 - Prob. 14.36PAECh. 14 - Prob. 14.37PAECh. 14 - Prob. 14.38PAECh. 14 - Prob. 14.39PAECh. 14 - Prob. 14.40PAECh. 14 - Prob. 14.41PAECh. 14 - Prob. 14.42PAECh. 14 - Prob. 14.43PAECh. 14 - Prob. 14.44PAECh. 14 - Prob. 14.45PAECh. 14 - Prob. 14.46PAECh. 14 - Prob. 14.47PAECh. 14 - Prob. 14.48PAECh. 14 - Prob. 14.49PAECh. 14 - Prob. 14.50PAECh. 14 - Prob. 14.51PAECh. 14 - Prob. 14.52PAECh. 14 - Prob. 14.53PAECh. 14 - Prob. 14.54PAECh. 14 - Prob. 14.55PAECh. 14 - Prob. 14.56PAECh. 14 - Prob. 14.57PAECh. 14 - Prob. 14.58PAECh. 14 - Prob. 14.59PAECh. 14 - Prob. 14.60PAECh. 14 - Prob. 14.61PAECh. 14 - Prob. 14.62PAECh. 14 - Prob. 14.63PAECh. 14 - Prob. 14.64PAECh. 14 - Prob. 14.65PAECh. 14 - Prob. 14.66PAECh. 14 - Prob. 14.67PAECh. 14 - Prob. 14.68PAECh. 14 - Prob. 14.69PAECh. 14 - Prob. 14.70PAECh. 14 - Prob. 14.71PAECh. 14 - Prob. 14.72PAECh. 14 - Prob. 14.73PAECh. 14 - Prob. 14.74PAECh. 14 - Prob. 14.75PAECh. 14 - Prob. 14.76PAECh. 14 - Prob. 14.77PAECh. 14 - Prob. 14.78PAECh. 14 - Prob. 14.79PAECh. 14 - Prob. 14.80PAECh. 14 - Prob. 14.81PAECh. 14 - Prob. 14.82PAECh. 14 - Prob. 14.83PAECh. 14 - Prob. 14.84PAECh. 14 - Prob. 14.85PAECh. 14 - Prob. 14.86PAECh. 14 - Prob. 14.87PAECh. 14 - Prob. 14.88PAECh. 14 - Prob. 14.89PAECh. 14 - Prob. 14.90PAECh. 14 - Prob. 14.91PAECh. 14 - Prob. 14.92PAECh. 14 - Prob. 14.93PAECh. 14 - Prob. 14.94PAECh. 14 - Prob. 14.95PAECh. 14 - Prob. 14.96PAECh. 14 - Prob. 14.97PAECh. 14 - Prob. 14.98PAECh. 14 - Prob. 14.99PAECh. 14 - Prob. 14.100PAECh. 14 - Prob. 14.101PAECh. 14 - Prob. 14.102PAECh. 14 - Prob. 14.103PAECh. 14 - Prob. 14.104PAECh. 14 - Prob. 14.105PAECh. 14 - Prob. 14.106PAECh. 14 - Prob. 14.107PAE
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