EBK INTRODUCTION TO CHEMISTRY
5th Edition
ISBN: 9781260162165
Author: BAUER
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 15, Problem 83QP
Interpretation Introduction
Interpretation:
The reason so as to why the heat of fission reaction is used directly to create steam to operate a turbine is to be stated.
Concept Introduction:
Some of the heavy radioactive nuclides are very unstable. They undergo decay in two ways, one is alpha decay and another is splitting of the nucleus and releasing a large amount of energy. Again, some of the heavy nuclides do not undergo fission spontaneously, but will undergo if irradiated or bombarded with suitable particle like neutron. Splitting of heavy radioactive nuclides into two or more nucleus with some number of neutrons is called fission reaction. The nuclear waste is the material that nuclear fuel becomes after its use in a fission reactor.
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Students have asked these similar questions
5
What would the complete ionic reaction be if aqueous solutions of potassium sulfate and barium acetate were mixed?
ed
of
Select one:
O a
2 K SO4 + Ba2 +2 C₂H3O21
K+SO4 + Ba2+ + 2 C2H3O21
K+SO42 + Ba2 +2 C2H3O2
BaSO4 +2 K+ + 2 C2H3O
estion
Ob.
O c.
Od.
2 K SO4 +Ba2 +2 C₂H₂O₂
BaSO4 + K+ + 2 C2H3O
BaSO4 + K + 2 C2H301
→Ba² +SO42 +2 KC2H3O
s page
(28 pts.) 7. Propose a synthesis for each of the following transformations. You must include the
reagents and product(s) for each step to receive full credit. The number of steps is provided.
(OC 4)
4 steps
4 steps
OH
b.
LTS
Solid:
AT=Te-Ti
Trial 1
Trial 2
Trial 3
Average
ΔΗ
Mass water, g
24.096
23.976
23.975
Moles of solid, mol
0.01763
001767
0101781
Temp. change, °C
2.9°C
11700
2.0°C
Heat of reaction, J
-292.37J -170.473
-193.26J
AH, kJ/mole
16.58K 9.647 kJ 10.85 kr
16.58K59.64701
KJ
mol
12.35k
Minimum AS,
J/mol K
41.582
mol-k
Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn &
Show your calculations for:
AH in J and then in kJ/mole for Trial 1:
qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J
qsin =
qrxn =
292.35 292.37J
AH in J = 292.375 0.2923kJ
0.01763m01
=1.65×107
AH in kJ/mol =
=
16.58K
0.01763mol
mol
qrx
Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con
Kelvin.)
AS=AHIT
(1.65×10(9.64×103) + (1.0
Jimai
Chapter 15 Solutions
EBK INTRODUCTION TO CHEMISTRY
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