OWLV2 FOR MASTERTON/HURLEY'S CHEMISTRY:
OWLV2 FOR MASTERTON/HURLEY'S CHEMISTRY:
8th Edition
ISBN: 9781305079304
Author: Hurley
Publisher: IACCENGAGE
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Chapter 18, Problem 51QAP
Interpretation Introduction

(a)

Interpretation:

The balanced chemical equation for the decay of U-235 due to neutrons bombardment needs to be determined.

Concept introduction:

A nuclear chemical equation is said to be balanced if there are same number of protons and neutrons are present on both side of the reaction arrow. The nuclear reactions take place by emission of alpha, beta and gamma particles from the different isotopes of the atoms.

Interpretation Introduction

(b)

Interpretation:

The value of change in energy for fission of 1 g of U-235 needs to be determined.

Concept introduction:

The mass defect for any nuclear reaction is the change in masses of the species involved in the process.

For a general nuclear reaction as follows:

AB+C

The mass defect can be calculated as follows:

mB+mCmA

Here, m is atomic mass of A, B and C atoms.

From the mass defect, the change in energy can be calculated as follows:

ΔE=Δmc2

Here, Δm is change in mass and c is speed of light. If the value is negative, the energy is released, if it is positive the energy is required for the reaction.

Interpretation Introduction

(c)

Interpretation:

The mass of TNT needed to produce the energy same as produced from 1 mg of U-235 needs to be determined.

Concept introduction:

The mass defect for any nuclear reaction is the change in masses of the species involved in the process.

For a general nuclear reaction as follows:

AB+C

The mass defect can be calculated as follows:

mB+mCmA

Here, m is atomic mass of A, B and C atoms.

From the mass defect, the change in energy can be calculated as follows:

ΔE=Δmc2

Here, Δm is change in mass and c is speed of light. If the value is negative, the energy is released, if it is positive the energy is required for the reaction.

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Chapter 18 Solutions

OWLV2 FOR MASTERTON/HURLEY'S CHEMISTRY:

Ch. 18 - Prob. 11QAPCh. 18 - Prob. 12QAPCh. 18 - Prob. 13QAPCh. 18 - Prob. 14QAPCh. 18 - Prob. 15QAPCh. 18 - Prob. 16QAPCh. 18 - Prob. 17QAPCh. 18 - Prob. 18QAPCh. 18 - Balance the following equations by filling in the...Ch. 18 - Prob. 20QAPCh. 18 - Prob. 21QAPCh. 18 - Prob. 22QAPCh. 18 - Prob. 23QAPCh. 18 - Prob. 24QAPCh. 18 - Prob. 25QAPCh. 18 - Prob. 26QAPCh. 18 - Prob. 27QAPCh. 18 - Prob. 28QAPCh. 18 - Prob. 29QAPCh. 18 - Prob. 30QAPCh. 18 - Prob. 31QAPCh. 18 - Prob. 32QAPCh. 18 - Prob. 33QAPCh. 18 - Prob. 34QAPCh. 18 - Prob. 35QAPCh. 18 - Prob. 36QAPCh. 18 - Prob. 37QAPCh. 18 - Prob. 38QAPCh. 18 - Prob. 39QAPCh. 18 - Prob. 40QAPCh. 18 - Prob. 41QAPCh. 18 - Prob. 42QAPCh. 18 - Prob. 43QAPCh. 18 - Prob. 44QAPCh. 18 - Prob. 45QAPCh. 18 - Prob. 46QAPCh. 18 - Prob. 47QAPCh. 18 - Prob. 48QAPCh. 18 - Prob. 49QAPCh. 18 - Prob. 50QAPCh. 18 - Prob. 51QAPCh. 18 - Prob. 52QAPCh. 18 - Prob. 53QAPCh. 18 - Prob. 54QAPCh. 18 - Prob. 55QAPCh. 18 - Prob. 56QAPCh. 18 - Prob. 57QAPCh. 18 - Prob. 58QAPCh. 18 - Prob. 59QAPCh. 18 - Prob. 60QAPCh. 18 - Prob. 61QAPCh. 18 - Prob. 62QAPCh. 18 - Prob. 63QAPCh. 18 - Prob. 64QAPCh. 18 - Prob. 65QAPCh. 18 - Prob. 66QAPCh. 18 - Prob. 67QAPCh. 18 - Prob. 68QAPCh. 18 - Prob. 69QAPCh. 18 - Prob. 70QAPCh. 18 - Prob. 71QAPCh. 18 - Prob. 72QAPCh. 18 - Fill in the following table:Ch. 18 - Prob. 74QAPCh. 18 - Prob. 75QAPCh. 18 - Prob. 76QAPCh. 18 - Prob. 77QAPCh. 18 - Prob. 78QAPCh. 18 - Prob. 79QAPCh. 18 - Carbon-14 (C-14) with a half-life of 5730 years...Ch. 18 - Prob. 81QAPCh. 18 - Prob. 82QAP
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