1.
Nitrogen monoxide (NO) is the smallest hormone known, playing a role in numerous
processes in the nervous, immune and cardiovascular systems. Use the phase diagram
of nitrogen monoxide to answer the following. Show work.
a.
Indicate and name the phase change(s) that take place when NO is cooled from
–160 oC to –165 oC at a constant pressure of 250 torr. (3 pts.)
gas
liquid
condensation
b.
freezing
Indicate the phase change(s) that take place when the pressure on NO is
increased from 100 to 1000 torr at a constant temperature of –164 oC. (3 pts.)
gas
solid
deposition
c.
solid
liquid
melting
Estimate the normal boiling point of NO. (2 pts.)
at 760 torr, –156.8 ºC
d.
Estimate the triple point of NO. (2 pts.)
190 torr, –163.5 ºC
f.
Use the phase diagram to explain whether solid NO would float or sink in liquid
NO at its freezing point? (2 pts.)
The melting point line slopes to the left. This means that the most dense
phase is liquid NO. Solid NO would float (like ice in water).
2.
Consider the following pure substances: (28 pts.)
BCl3
SiO2(quartz)
SrCl2
HClO4
a. Classify each as molecular, ionic or other.
molecular
other
ionic
molecular
Xe
POBr3
Cr2O3
ICl3
other
molecular
ionic
molecular
b. Give the valence electron count and the Lewis structure of each substance you classified as
molecular. Do not show shape. For ionic compounds, give charges on the cation and anion.
BCl3
SiO2
24 ve–
SrCl2
Sr2+ Cl–
HClO4
Cl
Cl
B
Xe
32 ve–
O
POBr3
O
Cl
ICl3
Cr3+ O2–
28 ve–
O
H
Cl
Br
Cl
Cr2O3
32 ve–
P
I
Cl
Br
Cl
O
Br
O
c. Draw the VSEPR shape of each substance you classified as molecular (without lone-pair e– or
multiple bonds). Clearly show, using vector notation on the shape, whether the molecular
compounds are polar or nonpolar.
BCl3
SiO2
SrCl2
HClO4
Xe
Cl
POBr3
H
O
P
B
Cl
Cl
Cl
O
nonpolar
polar
Cr2O3
ICl3
O
Br
O
O
Cl
I
Br
Br
Cl
polar
polar
d. Name the dominant intermolecular force in each substance.
BCl3
SiO2
London
56 e–
covalent
bond
SrCl2
HClO4
ion-ion
H-bonding
Xe
POBr3
London
54 e–
2x1
Cr2O3
dipoledipole
128 e–
ICl3
ion-ion
3x2
dipoledipole
104 e–
BCl3
Xe
e. List the substances in order of decreasing (highest to the left) boiling point.
SiO2
Cr2O3
SrCl2
HClO4
POBr3
ICl3
f. Give a specific, quantitative reason for the order if the intermolecular forces are the same.
Product of charges on the ions (for ionic); total electrons for dipole-dipole and London.
See electron totals on d.
Cl
3.
Classify the type of solid formed by each of the substances in question 2 and the
substance in question 4. (5 pts.)
BCl3
POBr3
molecular
SiO2(quartz) covalent (network)
Cr2O3
ionic
SrCl2
ionic
ICl3
molecular
HClO4
molecular
Cs
metallic
Xe
atomic
Cesium is so reactive, not many people have seen a pure sample. Its melting point is
so low it is close to being only the third liquid element on the Periodic Table (after
bromine and mercury).
Use the following data to calculate the heat needed to take 265.8 g of cesium from
25.0 oC to 640.0 oC. (10 pts.)
28.4 oC
641.0 oC
2.092 kJ/mole
67.74 kJ/mole
0.242 J/g∙oC
0.244 J/g∙oC
0.156 J/g∙oC
normal melting point Cs(s)
normal boiling point Cs(ℓ)
ΔHofusion Cs(s)
ΔHovaporization Cs(ℓ)
specific heat capacity Cs(s)
specific heat capacity Cs(ℓ)
specific heat capacity Cs(g)
95
85
Temperature (oC)
4.
molecular
75
65
55
45
35
25
0
1000
2000
3000
4000
Heat (J)
5000
6000
7000
8000
(part of heating curve)
1.
q = msT = 265.8 g x 0.242 J/g∙oC x (28.4 – 25.0 oC) = 218.70 J
2.
Hfusion = 265.8 g x 1 mole/132.905 g x 2092 J/mole = 4183.84 J
3.
q = msT = 265.8 g x 0.244 J/g∙oC x (640.0 – 28.4 oC) = 39665.44 J
qtotal = 218.70 J + 4183.84 J + 39665.44 J = 44.1 kJ
5.
Copper is one of the few elements that can be found in nature as the free element.
Copper crystallizes in a body-centered-cubic lattice (bcc) and has a density of
8.96 g/cm3. Calculate the atomic radius of copper. (10 pts.)
63.546 g Cu
1 mole Cu
2 atoms Cu
1 cm3
x
x
x
 2.3554 x 1023cm3
23
1 mole Cu 6.022 x 10 atoms Cu 1 unit cell 8.96 g Cu
a 
3
V 
3
2.3554 x 1023cm3  2.8665 x 108cm
a 3
2.8665 x 108 cm 3
r 

 1.24 x 10–8 cm = 1.24 Å = 124 pm
4
4
6.
Iron(III) sulfate has been detected by the two martian rovers Spirit and Opportunity
and is indicative of strongly oxidizing conditions of the surface of Mars. In fact, the
Spirit rover became stuck when it drove over a patch of soft iron(III) sulfate that had
been hidden under some normal soil. This meant the end of the mission.
A solution is prepared by dissolving 144.0 g of iron(III) sulfate (FW 399.89 g/mole) in
360.0 mL of water (FW 18.015 g/mole). Calculate:
a.
The molality of the sample. (5 pts.)
1 mole Fe 2 (SO 4 )3
399.89 g Fe 2 (SO 4 )3
1 g H 2O
1 kg
360.0 mL H 2O x
x
1 mL H 2 O 1000 g
144.0 g Fe2 (SO 4 )3 x
m 
= 1.000 m
b.
The normal freezing point of the solution. (5 pts.)
Tf  iKf m  (5)(1.86 o C/m)(1.000 m) = 9.30 oC lower
Tf = 0.0 oC – 9.30 oC = –9.30 oC
c.
The vapor pressure of the solution at 30.0 oC. (5 pts.)
χ water
1 mole H 2O
18.015 g H 2O

= 0.9173
19.9833 mole H 2O  5  0.3601 mole Fe2 (SO4 )3 
Psolution 
360.0 g H 2 O x
χ water Po water =
0.9173 x 31.82 torr = 29.19 torr (0.03841 atm)
7.
Pupfish are a small species of fish noted for being able to live in extreme conditions.
They are found in the very saline, very hot Salt Creek in Death Valley National Park.
Because of the temperature, the amount of oxygen in the water is reduced, but that is
partially offset because of the higher atmospheric pressure below sea level. Fish
generally need an O 2 concentration in water of at least 4 mg/L for survival.
Use the data on the Periodic Table page to determine the concentration of O 2 (in
mg/L) in Salt Creek at 44.4 ºC (112 ºF) and an air pressure of 774.4 torr. (10 pts.)
1 atm
= 0.2134 atm
PO2 = χ O2 Ptotal = 0.209476 x 774.4 torr x
760 torr
CO2  k O2 PO2  9.331 x 104 M/atm O2 x 0.2134 atm O2 = 1.989 x 104 M O2
1.989 x 104 mole O2
31.998 g O2 1000 mg O2
= 6.363 mg O2/L
x
x
L
1 mole O2
1 g O2