Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 19, Problem 38E
Interpretation Introduction
Interpretation:
The redox equation for the given redox reactants is to be stated. The direction in which the reaction will be favored at the equilibrium is to be stated.
Concept introduction:
The equilibrium is a state of the
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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 19 Solutions
Introductory Chemistry: An Active Learning Approach
Ch. 19 - Prob. 1ECh. 19 - Prob. 2ECh. 19 - Classify each of the following half-reaction...Ch. 19 - Prob. 4ECh. 19 - Prob. 5ECh. 19 - Prob. 6ECh. 19 - Prob. 7ECh. 19 - Prob. 8ECh. 19 - Prob. 9ECh. 19 - Prob. 10E
Ch. 19 - Prob. 11ECh. 19 - Identify each of the following half-reaction as...Ch. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - In this section, each equation identifies an...Ch. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - As an example of an electrolytic cell, the text...Ch. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 19.1TCCh. 19 - Prob. 19.2TCCh. 19 - Prob. 19.3TCCh. 19 - Prob. 1CLECh. 19 - Prob. 2CLECh. 19 - Prob. 3CLECh. 19 - Prob. 4CLECh. 19 - Prob. 5CLECh. 19 - Prob. 1PECh. 19 - Prob. 2PECh. 19 - Prob. 3PECh. 19 - Prob. 4PECh. 19 - Prob. 5PECh. 19 - Prob. 6PECh. 19 - Consider the reaction of copper and nitric acid:...Ch. 19 - Prob. 8PECh. 19 - Prob. 9PECh. 19 - Prob. 10PECh. 19 - Prob. 11PECh. 19 - Aqueous chromate ion, CrO42(aq), and hydrogen...Ch. 19 - Prob. 13PE
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- Q1: Answer the questions for the reaction below: ..!! Br OH a) Predict the product(s) of the reaction. b) Is the substrate optically active? Are the product(s) optically active as a mix? c) Draw the curved arrow mechanism for the reaction. d) What happens to the SN1 reaction rate in each of these instances: 1. Change the substrate to Br "CI 2. Change the substrate to 3. Change the solvent from 100% CH3CH2OH to 10% CH3CH2OH + 90% DMF 4. Increase the substrate concentration by 3-fold.arrow_forwardExperiment 27 hates & Mechanisms of Reations Method I visual Clock Reaction A. Concentration effects on reaction Rates Iodine Run [I] mol/L [S₂082] | Time mo/L (SCC) 0.04 54.7 Log 1/ Time Temp Log [ ] 13,20] (time) / [I] 199 20.06 23.0 30.04 0.04 0.04 80.0 22.8 45 40.02 0.04 79.0 21.6 50.08 0.03 51.0 22.4 60-080-02 95.0 23.4 7 0.08 0-01 1970 23.4 8 0.08 0.04 16.1 22.6arrow_forward(15 pts) Consider the molecule B2H6. Generate a molecular orbital diagram but this time using a different approach that draws on your knowledge and ability to put concepts together. First use VSEPR or some other method to make sure you know the ground state structure of the molecule. Next, generate an MO diagram for BH2. Sketch the highest occupied and lowest unoccupied MOs of the BH2 fragment. These are called frontier orbitals. Now use these frontier orbitals as your basis set for producing LGO's for B2H6. Since the BH2 frontier orbitals become the LGOS, you will have to think about what is in the middle of the molecule and treat its basis as well. Do you arrive at the same qualitative MO diagram as is discussed in the book? Sketch the new highest occupied and lowest unoccupied MOs for the molecule (B2H6).arrow_forward
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