CHEMISTRY-MASTERINGCHEMISTRY W/ETEXT
CHEMISTRY-MASTERINGCHEMISTRY W/ETEXT
8th Edition
ISBN: 9780135204634
Author: Robinson
Publisher: PEARSON
Question
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Chapter 20, Problem 20.16A
Interpretation Introduction

(a)

Interpretation:

The balanced nuclear reaction of the alpha decay of U92238 needs to be determined.

Concept introduction:

The nuclear reactions are a type of chemical process which leads to the formation of some new nuclei with the emission of certain particles. Usually, alpha or beta particles or gamma rays are emitted as a side product with some new daughter nuclei. The nuclear reactions follow the conservation of atomic number and mass number of nuclei. It means the sum of the mass number on both sides should be the same and the sum of the atomic number of nuclei should be the same on both sides.

Interpretation Introduction

(b)

Interpretation:

Themass change in g/atom and energy change in kJ/mol for the given nuclear reaction of the alpha decay of U92238 needs to be determined.

U92238 H24e  +  T90234h

Concept introduction:

The nuclear reactions are a type of chemical process which lead to the formation of some new nuclei with the emission of certain particles. Usually, alpha or beta particles or gamma rays are emitted as a side product with some new daughter nuclei.

The force between protons and neutrons holds these particles together. It can be measured by measuring the amount of energy necessary to break these forces.

It can be calculated with the help of Einstein’s mass − energy equation:

ΔE= m × c2

The mass defect can be calculated as the difference in the atomic mass and the sum of masses of all atomic particles.

Interpretation Introduction

(c)

Interpretation:

The energy released or absorbed during the radioactive decay of U92238 needs to be determined.

U92238 H24e  +  T90234h

Concept introduction:

The nuclear reactions are a type of chemical process which leads to the formation of some new nuclei with the emission of certain particles. Usually, alpha or beta particles or gamma rays are emitted as a side product with some new daughter nuclei.

The force between protons and neutrons holds these particles together. It can be measured by measuring the amount of energy necessary to break these forces.

It can be calculated with the help of Einstein’s mass-energy equation:

ΔE= m × c2

The mass defect can be calculated as the difference in the atomic mass and the sum of masses of all atomic particles.

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Chapter 20 Solutions

CHEMISTRY-MASTERINGCHEMISTRY W/ETEXT

Ch. 20 - Prob. 20.11PCh. 20 - Prob. 20.12ACh. 20 - Prob. 20.13PCh. 20 - Prob. 20.14ACh. 20 - Prob. 20.15PCh. 20 - Prob. 20.16ACh. 20 - Prob. 20.17PCh. 20 - Prob. 20.18ACh. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Prob. 20.21PCh. 20 - Prob. 20.22PCh. 20 - Prob. 20.23PCh. 20 - Prob. 20.24PCh. 20 - Prob. 20.25CPCh. 20 - Prob. 20.26SPCh. 20 - Prob. 20.27SPCh. 20 - Prob. 20.28SPCh. 20 - Prob. 20.29SPCh. 20 - Prob. 20.30SPCh. 20 - Prob. 20.31SPCh. 20 - Prob. 20.32SPCh. 20 - Prob. 20.33SPCh. 20 - Prob. 20.34SPCh. 20 - Prob. 20.35SPCh. 20 - Prob. 20.36SPCh. 20 - Prob. 20.37SPCh. 20 - Prob. 20.38SPCh. 20 - Prob. 20.39SPCh. 20 - Prob. 20.40SPCh. 20 - Prob. 20.41SPCh. 20 - Prob. 20.42SPCh. 20 - Prob. 20.43SPCh. 20 - Prob. 20.44SPCh. 20 - Prob. 20.45SPCh. 20 - Prob. 20.46SPCh. 20 - Prob. 20.47SPCh. 20 - Prob. 20.48SPCh. 20 - Prob. 20.49SPCh. 20 - The half-life of indium 111, a radioisotope used...Ch. 20 - The decay constant of plutonium 239 , a waste...Ch. 20 - Prob. 20.52SPCh. 20 - Plutonium 239 has a decay constant of 2.88105 year...Ch. 20 - Prob. 20.54SPCh. 20 - Prob. 20.55SPCh. 20 - A 1.0 mgsampleof79Sedecays initially atarate of...Ch. 20 - Prob. 20.57SPCh. 20 - A sample of 37Ar undergoes 8540...Ch. 20 - Prob. 20.59SPCh. 20 - Prob. 20.60SPCh. 20 - Prob. 20.61SPCh. 20 - Prob. 20.62SPCh. 20 - Prob. 20.63SPCh. 20 - Prob. 20.64SPCh. 20 - Prob. 20.65SPCh. 20 - Prob. 20.66SPCh. 20 - Prob. 20.67SPCh. 20 - Prob. 20.68SPCh. 20 - Prob. 20.69SPCh. 20 - Prob. 20.70SPCh. 20 - Prob. 20.71SPCh. 20 - Prob. 20.72SPCh. 20 - Prob. 20.73SPCh. 20 - Prob. 20.74SPCh. 20 - Prob. 20.75SPCh. 20 - Prob. 20.76SPCh. 20 - Prob. 20.77SPCh. 20 - Prob. 20.78SPCh. 20 - Prob. 20.79SPCh. 20 - Prob. 20.80SPCh. 20 - Prob. 20.81SPCh. 20 - Prob. 20.82SPCh. 20 - Prob. 20.83SPCh. 20 - Prob. 20.84SPCh. 20 - Prob. 20.85SPCh. 20 - Prob. 20.86SPCh. 20 - Prob. 20.87SPCh. 20 - Prob. 20.88SPCh. 20 - Prob. 20.89SPCh. 20 - Prob. 20.90SPCh. 20 - Prob. 20.91SPCh. 20 - Prob. 20.92SPCh. 20 - Prob. 20.93SPCh. 20 - Prob. 20.94SPCh. 20 - Prob. 20.95SPCh. 20 - Prob. 20.96SPCh. 20 - Prob. 20.97SPCh. 20 - Prob. 20.98SPCh. 20 - Prob. 20.99SPCh. 20 - Prob. 20.100SPCh. 20 - Prob. 20.101SPCh. 20 - Prob. 20.102SPCh. 20 - Prob. 20.103SPCh. 20 - Prob. 20.104SPCh. 20 - Prob. 20.105SPCh. 20 - Prob. 20.106SPCh. 20 - Prob. 20.107SPCh. 20 - Prob. 20.108SPCh. 20 - Prob. 20.109SPCh. 20 - Prob. 20.110SPCh. 20 - Prob. 20.111SPCh. 20 - Prob. 20.112SPCh. 20 - Prob. 20.113SPCh. 20 - Prob. 20.114MPCh. 20 - Prob. 20.115MPCh. 20 - Prob. 20.116MPCh. 20 - Prob. 20.117MPCh. 20 - Prob. 20.118MPCh. 20 - Prob. 20.119MPCh. 20 - Prob. 20.120MP
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