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Homework 11 Chapter 11
Homework 11 Chapter 11
Due: 11:59pm on Wednesday, November 23, 2016
You will receive no credit for items you complete after the assignment is due. Grading Policy
Problem 11.11
Part A
As a metal such as lead melts, what happens to the average kinetic energy of the atoms?
ANSWER:
The average kinetic energy decreases.
The average kinetic energy increases.
The average kinetic energy does not change.
Correct
Part B
As a metal such as lead melts, what happens to the average distance between the atoms?
ANSWER:
The average distance between the atoms increases.
The average distance between the atoms decreases.
The average distance between the atoms does not change.
Correct
Go Figure 11.10
Hydrogen bonding. Hydrogen bonding can occur when an H atom is bonded to an N, O, or F atom.
Part A
To form a hydrogen bond, what must the non­hydrogen atom (N, O, or F) involved in the bond possess?
ANSWER:
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Homework 11 Chapter 11
a bonding electron pair
a nonbonding electron pair
a low electronegativity value
an unpaired electron
Correct
Problem 11.16
Part A
Which is generally stronger, intermolecular interactions or intramolecular interactions?
ANSWER:
Intermolecular interactions are generally stronger.
Intramolecular interactions are generally stronger.
These interactions are equally strong.
Correct
Part B
Which of these kinds of interactions are broken when a liquid is converted to a gas?
ANSWER:
Intermolecular interactions are broken when a liquid is converted to a gas.
Intramolecular interactions are broken when a liquid is converted to a gas.
Both of these interactions are broken.
Correct
Problem 11.17
Describe the intermolecular forces that must be overcome to convert each of the following from a liquid or solid to a gas.
Part A
SO2
Check all that apply.
ANSWER:
Hydrogen­bonding.
London dispersion forces.
Dipole­dipole bonding.
Ion­dipole bonding.
Correct
Part B
CH CH OH
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Homework 11 Chapter 11
CH 3 CH 2 OH
Check all that apply.
ANSWER:
Hydrogen­bonding.
Dipole­dipole bonding.
Ion­dipole bonding.
London dispersion forces.
Correct
Part C
H 2 Se
Check all that apply.
ANSWER:
London dispersion forces.
Ion­dipole bonding.
Hydrogen­bonding.
Dipole­dipole bonding.
Correct
Problem 11.26
Rationalize the difference in boiling points between the members of the following pairs of substances.
Part A
HF
(20 ∘ C) and HCl (­85 ∘ C),
ANSWER:
HF has the higher boiling point because hydrogen bonding is weaker than dipole­dipole forces.
HF has the higher boiling point because hydrogen bonding is stronger than dipole­dipole forces.
HF has the higher boiling point because of ionic bonding.
HF has the higher boiling point because HF molecules are more polar.
Correct
Part B
CHCl3
(61 ∘ C) and CHBr3 (150 ∘ C),
ANSWER:
CHCl3 has the lower boiling point because hydrogen bonding is weaker than dipole­dipole forces.
CHBr3 has the higher boiling point because it has the higher molar mass, which leads to greater polarizability and stronger dispersion forces.
CHBr3 has the higher boiling point because CHBr3 molecules are less polar.
CHBr3 has the higher boiling point because hydrogen bonding is stronger than dipole­dipole forces.
Correct
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Homework 11 Chapter 11
Part C
Br2
(59 ∘ C) and ICl (97 ∘ C).
ANSWER:
ICl has the higher boiling point because of weak dispersion forces.
Br2 has the lower boiling point because of strong dispersion forces.
ICl has the higher boiling point because it is a polar molecule.
ICl has the higher boiling point because of ion­dipole­bounds.
Correct
Problem 11.38 with feedback
You may want to reference (
pages 455 ­ 456) Section 11.3 while completing this problem.
Part A
Compare the viscosity of n­pentane, CH3 CH2 CH2 CH2 CH3 , to the viscosity of n­hexane, CH3 CH2 CH2 CH2 CH2 CH3 .
Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before
submitting your answer.
ANSWER:
Reset
n­pentane
Help
You would expect the viscosity of n­hexane to be larger mainly due to one more carbon atom in the chain .
one less carbon atom in the
chain
hydrogen bonding
Correct
Viscosity tends to increase with the strength of intermolecular interactions. Since n­hexane is longer and has a greater molecular weight than n­
pentane, you would expect the dispersion forces of n­hexane to be stronger, which would make it a more viscous liquid.
Part B
Compare the viscosity of neopentane, (CH3 )4 C, and n­pentane.
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Homework 11 Chapter 11
Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before
submitting your answer.
ANSWER:
Reset
the linear shape means the
molecule is more polar
Help
You would expect the viscosity of n­pentane to be larger mainly because the cylindrical shape results in stronger dispersion forces when compared to the other
hydrocarbon.
the spherical shape results in
weaker dispersion forces
neopentane
the symmetrical shape means
the molecule is less polar
Correct
Viscosity tends to increase with the strength of intermolecular interactions. Since n­pentane is longer and more cylindrical than neopentane (at the
same molecular weight), you would expect the dispersion forces of n­pentane to be stronger, which would make it a more viscous liquid.
Sample Exercise 11.3 Practice Exercise 1 with feedback
Part A ­ Calculating ΔH for Temperature and Phase Changes
What information about water is needed to calculate the enthalpy change for converting 1 mol H2 O(g) at 100 ∘ C to H2 O(l) at 80 ∘ C?
ANSWER:
heat of vaporization and specific heat of H2 O(g)
heat of fusion and specific heat of H2 O(l)
heat of vaporization and specific heat of H2 O(l)
heat of fusion
heat of vaporization
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Homework 11 Chapter 11
Correct
Two calculations are required to obtain the the enthalpy change for converting 1 mol H2 O(g) at 100 ∘ C to H2 O(l) at 80 ∘ C. The two calculations
are the enthalpy change associated with converting vapor to liquid (the heat of vaporization), and the enthalpy change associated with cooling the
liquid from 100 ∘ C to 80 ∘ C, (the specific heat of the liquid).
Sample Exercise 11.4 Practice Exercise 1 with feedback
Part A ­ Relating Boiling Point to Vapor Pressure
In the mountains, water in an open container will boil when,
ANSWER:
its temperature is 100 ∘ C
its vapor pressure equals atmospheric pressure
its critical temperature exceeds room temperature
enough energy is supplied to break covalent bonds
none of these is correct
Correct
Problem 11.50
Part A
Acetone, H3 CCOCH3 , has a boiling point of 56∘ C. Based on the data given in the figure, would you expect acetone to have a higher or lower vapor
pressure than ethanol at 25∘ C?
ANSWER:
Acetone has a higher vapor pressure.
Acetone has a lower vapor pressure.
Acetone has the same vapor pressure.
Correct
Problem 11.61
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Homework 11 Chapter 11
Use the phase diagram to answer the following questions:
Part A
What is the approximate value of the normal melting point?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
T
= 24 K
Correct
Part B
Below what pressure value will solid neon sublime?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
P
= 0.43 atm
Correct
Part C
At temperature T =25 ∘ C can neon be liquefied by compressing it?
ANSWER:
cannot be liquefied
can be liquefied
Correct
Sample Exercise 11.6 Practice Exercise 1 with feedback
Part A ­ Properties of Liquid Crystals
Liquid crystalline phases are produced by which of the following?
ANSWER:
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Homework 11 Chapter 11
three dimensional order among molecules
short, flexible molecules
rod­shaped molecules
complete lack of order among molecules
highly branched molecules
Correct
Liquid crystals are formed when the molecules have order but not in all three dimensions.
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