Solutions for Chemistry, The Molecular Nature of Matter and Change, 3rd Canadian Edition by Silberberg (All Chapters included)
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Course
Chemistry, The Molecular Nature of Matter 3e
Institution
Trent University (TrentU
)
Complete Solutions Manual for Chemistry, The Molecular Nature of Matter and Change, 3rd Canadian Edition by Martin Silberberg ; ISBN13: 9781259654855....(Full Chapters included and organized in reverse order from Chapter 25 to 1)...Chapter 1 Keys to the Study of Chemistry
Chapter 2 The Components ...
Chemistry, The Molecular Nature of
Matter and Change, 3rd Canadian
Edition by Martin Silberberg
Complete Chapter Solutions Manual
are included (Ch 1 to 25)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
Chapter 1 Keys to the Study of Chemistry
Chapter 2 The Components of Matter
Chapter 3 Stoichiometry and Chemical Equations
Chapter 4 Gases and the Kinetic-Molecular Theory
Chapter 5 Thermochemistry: Energy Flow and Chemical Change
Chapter 6 Quantum Theory and Atomic Structure
Chapter 7 Electron Configuration and Chemical Periodicity
Chapter 8 Models of Chemical Bonding
Chapter 9 The Shapes of Molecules
Chapter 10 Theories of Covalent Bonding
Chapter 11 Intermolecular Forces: Liquids, Solids, and Phase Changes
Chapter 12 The Properties of Mixtures: Solutions and Colloids
Chapter 13 Periodic Patterns in the Main-Group Elements
Chapter 14 Kinetics: Rates and Mechanisms of Chemical Reactions
Chapter 15 Equilibrium: The Extent of Chemical Reactions
Chapter 16 Acid-Base Equilibria
Chapter 17 Ionic Equilibria in Aqueous Systems
Chapter 18 Thermodynamics: Entropy, Gibbs Energy, and the Direction of
Chemical Reactions
Chapter 19 Electrochemistry: Chemical Change and Electrical Work
Chapter 20 Organic Compounds and the Atomic Properties of Carbon
Chapter 21 Organic Reaction Mechanisms
Chapter 22 Special Topics in Organic Chemistry
Chapter 23 The Elements in Nature and Industry
Chapter 24 Transition Elements and Their Coordination Compounds
Chapter 25 Nuclear Reactions and Their Applications
,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch25-1)
CHAPTER 25 NUCLEAR REACTIONS AND
THEIR APPLICATIONS
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B25.1 In the s-process, a nucleus captures a neutron sometime over a long period of time. Then the nucleus emits a beta
particle to form another element. The stable isotopes of most heavy elements up to 209Bi form by the s-process.
The r-process very quickly forms less stable isotopes and those with A greater than 230 by multiple neutron
captures, followed by multiple beta decays.
B25.2 Plan: Find the change in mass of the reaction by subtracting the mass of the products from the mass of the
reactants and convert the change in mass to energy with the conversion factor between u and MeV. Convert the
energy per atom to energy per mole by multiplying by Avogadro’s number.
Solution:
m = mass of reactants – mass of products
= [(4)(1.007825)]u – [4.00260 + (2)(5.48580x10–4)]u
= (4.031300 – 4.003697)u = 0.02760 u /4He atom = 0.02760284 g/mol 4He
0.02760284 u 4 He 931.5 MeV
Energy (MeV/atom) = = 25.7120 MeV/atom = 25.71 MeV/atom
1 atom 1u
Convert atoms to moles using Avogadro’s number.
25.7120 MeV 6.022x10 atoms
23
Energy = = 1.54838x1025 MeV/atom = 1.548x1025 MeV/mol
atom 1 mol
B25.3 The simultaneous fusion of three nuclei is a termolecular process. Termolecular processes have a very low
probability of occurring. The bimolecular fusion of 8Be with 4He is more likely.
B25.4 Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the
right side must be equal.
Solution:
83 Bi → 84 Po + −1
210 210 0 210
84 Po is nuclide A
210
84 Po → 82 Pb + 2
206 4 206
82 Pb is nuclide B
206
82 Pb + 3 01 n → 209
82 Pb
209
82 Pb is nuclide C
209
82 Pb → 83 Bi + −01
210 210
83 Bi is nuclide D
END–OF–CHAPTER PROBLEMS
25.1 a) Chemical reactions are accompanied by relatively small changes in energy while nuclear reactions are
accompanied by relatively large changes in energy.
b) Increasing temperature increases the rate of a chemical reaction but has no effect on a nuclear reaction.
c) Both chemical and nuclear reaction rates increase with higher reactant concentrations.
d) If the reactant is limiting in a chemical reaction, then more reactant produces more product and the yield
increases in a chemical reaction. The presence of more radioactive reagent results in more decay product, so a
higher reactant concentration increases the yield in a nuclear reaction.
, 25.2 a) The percentage of sulfur atoms that are sulfur-32 is 95.02%, the same as the relative abundance of 32S.
b) The atomic mass is larger than the isotopic mass of 32S. Sulfur-32 is the lightest isotope, as stated in the
problem, so the other 5% of sulfur atoms are heavier than 31.972070 u. The average mass of all the sulfur atoms
will therefore be greater than the mass of a sulfur-32 atom.
25.3 a) She found that the intensity of emitted radiation is directly proportional to the concentration of the element in
the various samples, not to the nature of the compound in which the element occurs.
b) She found that certain uranium minerals were more radioactive than pure uranium, which implied that they
contained traces of one or more as yet unknown, highly radioactive elements. Pitchblende is the principal ore of
uranium.
25.4 Plan: Radioactive decay that produces a different element requires a change in atomic number (Z, number of
protons).
Solution:
A
ZX A = mass number (protons + neutrons)
Z = number of protons (positive charge)
X = symbol for the particle
N = A – Z (number of neutrons)
a) Alpha decay produces an atom of a different element, i.e., a daughter with two less protons and two less
neutrons.
Z X → Z −2Y + 2 He
A A−4 4
2 fewer protons, 2 fewer neutrons
b) Beta decay produces an atom of a different element, i.e., a daughter with one more proton and one less neutron.
A neutron is converted to a proton and particle in this type of decay.
Z X → Z +1Y + −1
A A 0
1 more proton, 1 less neutron
c) Gamma decay does not produce an atom of a different element and Z and N remain unchanged.
A
Z X* → ZA X + 00 ( ZA X * = energy rich state), no change in number of protons or neutrons.
d) Positron emission produces an atom of a different element, i.e., a daughter with one less proton and one more
neutron. A proton is converted into a neutron and positron in this type of decay.
Z X → Z −1Y + +1
A A 0
1 less proton, 1 more neutron
e) Electron capture produces an atom of a different element, i.e., a daughter with one less proton and one more
neutron. The net result of electron capture is the same as positron emission, but the two processes are different.
Z X + −1e → Z −1Y
A 0 A
1 less proton, 1 more neutron
A different element is produced in all cases except (c).
25.5 The key factor that determines the stability of a nuclide is the ratio of the number of neutrons to the number of
protons, the N/Z ratio. If the N/Z ratio is either too high or not high enough, the nuclide is unstable and decays.
3
2 He N/Z = 1/2
2
2 He N/Z = 0/2, thus it is more unstable.
25.6 A neutron-rich nuclide decays to convert neutrons to protons while a neutron-poor nuclide decays to convert
protons to neutrons. The conversion of neutrons to protons occurs by beta decay:
0 n → 1 p + −1
1 1 0
The conversion of protons to neutrons occurs by either positron decay:
1 p → 0 n + 1
1 1 0
or electron capture:
1 p + −1e → 0 n
1 0 1
Neutron-rich nuclides, with a high N/Z, undergo decay. Neutron-poor nuclides, with a low N/Z, undergo
positron decay or electron capture.
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