Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 13, Problem 2E
Interpretation Introduction
(a)
Interpretation:
Whether the solubility of carbon dioxide gas in soda increases or decreases by increasing temperature from
Concept introduction:
The solubility of a gas is inversely proportional to the temperature. If the temperature increases the solubility decreases. If the temperature decreases the solubility of the gas increases.
Interpretation Introduction
(b)
Interpretation:
Whether the solubility of carbon dioxide gas in soda increases or decreases when the partial pressure of
Concept introduction:
The solubility of the gas in a solvent is directly proportional to the partial pressure of the gas above the solvent. This effect is found by the scientist William Henry during his studies on the solubility of gases in liquids and is known as Henry’s law.
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Using reaction free energy to predict equilibrium composition
Consider the following equilibrium:
N2 (g) + 3H2 (g) = 2NH3 (g) AG⁰ = -34. KJ
Now suppose a reaction vessel is filled with 8.06 atm of nitrogen (N2) and 2.58 atm of ammonia (NH3) at 106. °C. Answer the following questions about this
system:
?
rise
Under these conditions, will the pressure of N2 tend to rise or fall?
☐ x10
fall
Is it possible to reverse this tendency by adding H₂?
In other words, if you said the pressure of N2 will tend to rise, can that be
changed to a tendency to fall by adding H₂? Similarly, if you said the
pressure of N2 will tend to fall, can that be changed to a tendency to rise
by adding H₂?
If you said the tendency can be reversed in the second question, calculate
the minimum pressure of H₂ needed to reverse it.
Round your answer to 2 significant digits.
yes
no
☐
atm
☑
5
00.
18
Ar
Chapter 13 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 13 - Prob. 1CECh. 13 - Prob. 2CECh. 13 - Prob. 3CECh. 13 - Prob. 4CECh. 13 - Prob. 5CECh. 13 - Prob. 6CECh. 13 - Prob. 7CECh. 13 - Prob. 8CECh. 13 - Prob. 9CECh. 13 - Prob. 10CE
Ch. 13 - Prob. 11CECh. 13 - Prob. 12CECh. 13 - Prob. 1KTCh. 13 - Prob. 2KTCh. 13 - Prob. 3KTCh. 13 - Prob. 4KTCh. 13 - Prob. 5KTCh. 13 - Prob. 6KTCh. 13 - Prob. 7KTCh. 13 - Prob. 8KTCh. 13 - Prob. 9KTCh. 13 - Prob. 10KTCh. 13 - Prob. 11KTCh. 13 - Prob. 12KTCh. 13 - Prob. 13KTCh. 13 - Prob. 14KTCh. 13 - Prob. 15KTCh. 13 - Prob. 16KTCh. 13 - Prob. 17KTCh. 13 - Prob. 18KTCh. 13 - Prob. 19KTCh. 13 - Prob. 20KTCh. 13 - Prob. 1ECh. 13 - Prob. 2ECh. 13 - Prob. 3ECh. 13 - Prob. 4ECh. 13 - Prob. 5ECh. 13 - Prob. 6ECh. 13 - Prob. 7ECh. 13 - Prob. 8ECh. 13 - Prob. 9ECh. 13 - Prob. 10ECh. 13 - Prob. 11ECh. 13 - Prob. 12ECh. 13 - Prob. 13ECh. 13 - Prob. 14ECh. 13 - Prob. 15ECh. 13 - Prob. 16ECh. 13 - Prob. 17ECh. 13 - Prob. 18ECh. 13 - Prob. 19ECh. 13 - Prob. 20ECh. 13 - Prob. 21ECh. 13 - Prob. 22ECh. 13 - Prob. 23ECh. 13 - Prob. 24ECh. 13 - Prob. 25ECh. 13 - Prob. 26ECh. 13 - Prob. 27ECh. 13 - Prob. 28ECh. 13 - Prob. 29ECh. 13 - Prob. 30ECh. 13 - Prob. 31ECh. 13 - Prob. 32ECh. 13 - Prob. 33ECh. 13 - Prob. 34ECh. 13 - Prob. 35ECh. 13 - Prob. 36ECh. 13 - Prob. 37ECh. 13 - Prob. 38ECh. 13 - Prob. 39ECh. 13 - Prob. 40ECh. 13 - Prob. 41ECh. 13 - Prob. 42ECh. 13 - Prob. 43ECh. 13 - Prob. 44ECh. 13 - Prob. 45ECh. 13 - Prob. 46ECh. 13 - Prob. 47ECh. 13 - Prob. 48ECh. 13 - Prob. 49ECh. 13 - Prob. 50ECh. 13 - Prob. 51ECh. 13 - Prob. 52ECh. 13 - Prob. 53ECh. 13 - Prob. 54ECh. 13 - Prob. 55ECh. 13 - Prob. 56ECh. 13 - Prob. 57ECh. 13 - Prob. 58ECh. 13 - Prob. 59ECh. 13 - Prob. 60ECh. 13 - Prob. 61ECh. 13 - Prob. 62ECh. 13 - Prob. 63ECh. 13 - Prob. 64ECh. 13 - Prob. 65ECh. 13 - Prob. 66ECh. 13 - Prob. 67ECh. 13 - Prob. 68ECh. 13 - Prob. 70ECh. 13 - Prob. 71ECh. 13 - Prob. 72ECh. 13 - Prob. 73ECh. 13 - Prob. 74ECh. 13 - Prob. 75ECh. 13 - Prob. 76ECh. 13 - Prob. 77ECh. 13 - Prob. 78ECh. 13 - Prob. 79ECh. 13 - Prob. 80ECh. 13 - Prob. 81ECh. 13 - Prob. 82ECh. 13 - Prob. 83ECh. 13 - Prob. 84ECh. 13 - Prob. 1STCh. 13 - Prob. 2STCh. 13 - Prob. 3STCh. 13 - Prob. 4STCh. 13 - Prob. 5STCh. 13 - Prob. 6STCh. 13 - Prob. 7STCh. 13 - Prob. 8STCh. 13 - Prob. 9STCh. 13 - Prob. 10STCh. 13 - Prob. 11STCh. 13 - Prob. 12STCh. 13 - Prob. 13STCh. 13 - Prob. 14STCh. 13 - Prob. 15STCh. 13 - Prob. 16ST
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- i need help with the followingarrow_forwardUsing reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NO(g) +Cl₂ (g) = 2NOC1 (g) AGº = -41. kJ Now suppose a reaction vessel is filled with 8.90 atm of chlorine (C12) and 5.71 atm of nitrosyl chloride (NOC1) at 1075. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of NOCI tend to rise or fall? x10 fall Is it possible to reverse this tendency by adding NO? In other words, if you said the pressure of NOCI will tend to rise, can that be changed to a tendency to fall by adding NO? Similarly, if you said the pressure of NOCI will tend to fall, can that be changed to a tendency to rise by adding NO? yes no If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO needed to reverse it. Round your answer to 2 significant digits. atm ☑ 18 Ararrow_forwardIdentifying the major species in weak acid or weak base equilibria The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HCN is a weak acid. acids: 0.29 mol of NaOH is added to 1.0 L of a 1.2M HCN solution. bases: ☑ other: 0.09 mol of HCl is added to acids: 1.0 L of a solution that is bases: 0.3M in both HCN and KCN. other: 0,0,... ? 00. 18 Ar 日arrow_forward
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY