Bundle: Chemistry: An Atoms First Approach, 2nd, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card
2nd Edition
ISBN: 9781305717633
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 18, Problem 44E
Interpretation Introduction
Interpretation: The binding energy per nucleon of
Concept introduction: The sum of masses of the component nucleons and the actual mass of a nucleus is known as the mass defect and it can be used to calculate the nuclear binding energy.
To determine: The binding energy per nucleon of
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Please correct answer and don't used hand raiting
(11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the
molecule depicted below.
Bond B
Bond A
Bond C
a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in
appropriate boxes.
Weakest
Bond
Strongest
Bond
b. (4pts) Consider the relative stability of all cleavage products that form when bonds A,
B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B,
and C are all carbon radicals.
i. Which ONE cleavage product is the most stable? A condensed or bond line
representation is fine.
ii. Which ONE cleavage product is the least stable? A condensed or bond line
representation is fine.
c. (5pts) Use principles discussed in lecture, supported by relevant structures, to
succinctly explain the why your part b (i) radical is more stable than your part b(ii)
radical. Written explanation can be no more than one-two succinct sentence(s)!
.
3°C
with
TH
12. (10pts total) Provide the major product for each reaction depicted below. If no reaction
occurs write NR. Assume heat dissipation is carefully controlled in the fluorine reaction.
3H
24 total (30)
24
21
2h
• 6H total
● 8H total
34
래
Br2
hv
major product
will be most
Substituted
12
hv
Br
NR
I too weak of a
participate in P-1
F₂
hv
Statistically
most favored
product will be
major
=
most subst =
thermo favored
hydrogen atom abstractor to
LL
F
Chapter 18 Solutions
Bundle: Chemistry: An Atoms First Approach, 2nd, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card
Ch. 18 - Prob. 1RQCh. 18 - Prob. 2RQCh. 18 - Prob. 3RQCh. 18 - Prob. 4RQCh. 18 - Prob. 5RQCh. 18 - Prob. 6RQCh. 18 - Prob. 7RQCh. 18 - Prob. 8RQCh. 18 - Prob. 9RQCh. 18 - Prob. 10RQ
Ch. 18 - Prob. 1QCh. 18 - Prob. 2QCh. 18 - Prob. 3QCh. 18 - Prob. 4QCh. 18 - Prob. 5QCh. 18 - Prob. 6QCh. 18 - Prob. 7QCh. 18 - Prob. 8QCh. 18 - Prob. 9QCh. 18 - Prob. 10QCh. 18 - Prob. 11ECh. 18 - Prob. 12ECh. 18 - Prob. 13ECh. 18 - Prob. 14ECh. 18 - Prob. 15ECh. 18 - Prob. 16ECh. 18 - Prob. 17ECh. 18 - Prob. 18ECh. 18 - Prob. 19ECh. 18 - Prob. 20ECh. 18 - Prob. 21ECh. 18 - Prob. 22ECh. 18 - Prob. 23ECh. 18 - Prob. 24ECh. 18 - Prob. 25ECh. 18 - Prob. 26ECh. 18 - Prob. 27ECh. 18 - Prob. 28ECh. 18 - Prob. 29ECh. 18 - Prob. 30ECh. 18 - Prob. 31ECh. 18 - Prob. 32ECh. 18 - Prob. 33ECh. 18 - Prob. 34ECh. 18 - Prob. 35ECh. 18 - Prob. 36ECh. 18 - Prob. 37ECh. 18 - Prob. 38ECh. 18 - Prob. 39ECh. 18 - Prob. 40ECh. 18 - Prob. 41ECh. 18 - Prob. 42ECh. 18 - Prob. 43ECh. 18 - Prob. 44ECh. 18 - Prob. 45ECh. 18 - Prob. 46ECh. 18 - Prob. 47ECh. 18 - Prob. 48ECh. 18 - Prob. 49ECh. 18 - Prob. 50ECh. 18 - Prob. 51ECh. 18 - Prob. 52ECh. 18 - Prob. 53ECh. 18 - A chemist studied the reaction mechanism for the...Ch. 18 - Prob. 55ECh. 18 - Prob. 56ECh. 18 - Prob. 57ECh. 18 - Prob. 58ECh. 18 - Prob. 59AECh. 18 - Prob. 60AECh. 18 - Prob. 61AECh. 18 - Prob. 62AECh. 18 - Prob. 63AECh. 18 - Prob. 64AECh. 18 - Prob. 65AECh. 18 - Prob. 66AECh. 18 - Prob. 67AECh. 18 - Prob. 68AECh. 18 - Prob. 69AECh. 18 - Prob. 70AECh. 18 - Prob. 71AECh. 18 - Prob. 72AECh. 18 - Prob. 73CWPCh. 18 - Prob. 74CWPCh. 18 - Prob. 75CWPCh. 18 - Prob. 76CWPCh. 18 - Prob. 77CWPCh. 18 - Prob. 78CWPCh. 18 - Prob. 79CPCh. 18 - Prob. 80CPCh. 18 - Prob. 81CPCh. 18 - Prob. 82CPCh. 18 - Prob. 83CPCh. 18 - Prob. 84CPCh. 18 - Prob. 85CPCh. 18 - Prob. 86CPCh. 18 - Prob. 87IPCh. 18 - Prob. 88IP
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- Five chemistry project topic that does not involve practicalarrow_forwardPlease correct answer and don't used hand raitingarrow_forwardQ2. Consider the hydrogenation of ethylene C2H4 + H2 = C2H6 The heats of combustion and molar entropies for the three gases at 298 K are given by: C2H4 C2H6 H2 AH comb/kJ mol¹ -1395 -1550 -243 Sº / J K¹ mol-1 220.7 230.4 131.1 The average heat capacity change, ACP, for the reaction over the temperature range 298-1000 K is 10.9 J K¹ mol¹. Using these data, determine: (a) the standard enthalpy change at 800 K (b) the standard entropy change at 800 K (c) the equilibrium constant at 800 K.arrow_forward
- 13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B Bond A Bond C a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest Bond Strongest Bond b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! Googlearrow_forwardPrint Last Name, First Name Initial Statifically more chances to abstract one of these 6H 11. (10pts total) Consider the radical chlorination of 1,3-diethylcyclohexane depicted below. 4 4th total • 6H total 래 • 4H total 21 total ZH 2H Statistical H < 3° C-H weakest - product abstraction here bund leads to thermo favored a) (6pts) How many unique mono-chlorinated products can be formed and what are the structures for the thermodynamically and statistically favored products? Product 6 Number of Unique Mono-Chlorinated Products Thermodynamically Favored Product Statistically Favored Product b) (4pts) Draw the arrow pushing mechanism for the FIRST propagation step (p-1) for the formation of the thermodynamically favored product. Only draw the p-1 step. You do not need to include lone pairs of electrons. No enthalpy calculation necessary H H-Cl Waterfoxarrow_forward10. (5pts) Provide the complete arrow pushing mechanism for the chemical transformation → depicted below Use proper curved arrow notation that explicitly illustrates all bonds being broken, and all bonds formed in the transformation. Also, be sure to include all lone pairs and formal charges on all atoms involved in the flow of electrons. CH3O II HA H CH3O-H H ①arrow_forward
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