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Multiple Choice
1 B
2 B
3 C
4 D
5 A
6 B
7 C
8 C
9 A
10 D
11 C
12 A
13 C
14 D
15 B
16 C
17 C
18 C
19 B
20 B
21 D
22 C
23 D
24 D
25 B
26 A
27 B
28 D
29 D
30 D
31 C
32 D
33 D
34 B
35 A
10 g He x 1 L/.179 g = 55.9 L, 10 g Ne x 1 L/1.78 g = 6.62 L
d = m/V = 20 g/(55.9 L + 6.62 L) x 1000 = 3.25 x 10 -4 g/cm3
E = E4 – E2 = -B/42 - -B/22 = -B/16 – -4B/16 = 3B/16
From low energy to high energy: 1s-2s-2p-3s-3p-4s-3d4p-5s-4d-5p-6s-4f-5d-6p-7s-5f-6d-7p
For n = 3: l = 0, 1, 2 (not 3); sublevels include 3s, 3p and
3d; total electrons is 2n2 = 2(32) = 18.
Radius decreases from bottom to top and from right to
left.
Electron affinity generally decreases from left to right;
except columns 2, 15 and 18.
9 O atoms have a total mass of 144 g, which is more
than half of the 250 g per formula.
2 C5H6NS + 17 O2  10 CO2 + 6 H2O + 2 NO2 + 2 SO2
2 + 17 + 10 + 6 + 2 + 2 = 39
.3 mol P x 2 mol/2 mol = .3 mol PI3 x 412 g/mol = 124 g
150 g I2 x 1 mol/254 g x 2 mol/3 mol = 0.394 mol
3 x 2 mol Al; 3 x 2 mol O x 16 g/mol;
3(2 + 5) mol atoms x 6.02 x 1023 atom/mol
Diffusion rate (r1/r2 = (MM2/MM1)½) depends on MM
(smaller MM = faster rate).
Most polar has the greatest electronegativity difference,
which is predicted by the greatest gap on the PeriodicT.
v1/v2 = (MM2/MM1)½  482/682 = (MM2/32)½  MM2 = 16
CH4 has a MM of 16
Two changes were made; volume, which has no effect
and temperature, which increases pressure  D.
P1V1/T1 = P2V2/T2  2.00 L/373 K = X L/473 K
 X = 2.54 L
Higher pressure produces most compact (most dense)
state  liquid (solid/liquid line has a negative slope).
0.01 atm is below the triple point  sublimation.
Metallic solids conduct in both solid and liquid states;
ionic solids only conduct in the liquid state.
Covalent network and molecular substances are nonconduction in all states, but C.N. has a high melting pt.
Lowest melting point would be for non-polar molecule
with the lowest MM (weakest dispersion forces).
+H because bonds are broken; +S because liquid is
more disordered; -V because water is more dense.
Q = (C + mc)T
Greatest positive S is when there is the greatest
increase in the number of gas molecules.
I is reversed  +286; II is reversed  +414; III is doubled
 2(-425)
Nonspontaneous at higher temperatures  -H and
-S; reaction occurs at 298  -G.
Rate is affected by Ea (smaller = faster), catalyst
(increases rate), and temperature (higher = faster).
Zero order means that the concentration of A does not
affect the reaction rate.
H = Ea – Ea'  Ea' = 159 kJ – (-218 kJ) = 377 kJ
(B) and (C) because 2 NO2 is reactant for slow step.
0.6 mol CS2 x 3 mol Products/1 mol CS2 = 1.8 mol
1.5 mol O2 x 3 mol Products/3 mol O2 = 1.5 mol
10 g H x 1 mol/1 g = 10 mol/7.5 = 1.33 x 3 = 4
90 g C x 1 mol/12 g = 7.5 mol/7.5 = 1 x 3 = 3
The resulting solution must have a concentration
between the two solutions added together.
Catalyst lowers Ea and Ea'.
25.6 g S8 x 256 g/mol x 8 mol H2S/1 mol S8 = 0.8 mol
0.8 mol x 22.4 L/mol = 20 L
0.25 mol N2H4 x 3 mol N2/2 mol x 28 g N2/1 mol = 10.5 g
0.1 mol N2O4 x 3 mol N2/1 mol x 28 g N2/1 mol = 8.4 g
36 D
37 C
38 C
39 A
40 B
41 C
42 C
43 D
44 A
45 D
46 D
47 A
48 A
49 D
50 B
51 D
52 D
53 C
54 B
55 C
56 C
57 B
58 A
59 C
60 D
61 B
62 C
63 A
64 C
65 C
66 D
67 D
68 C
0.0154 mol NaCl x 1 L/0.154 mol = 0.1 L (100 mL)
The most accurate is to use a volumetric flask.
Final volume – Initial volume = Change in volume
35.75 mL – 12.55 mL = 23.20 mL
Phenolphthalein changes the mixture in the flask from
clear (acid) to pink (base) at the endpoint.
0.0464 mol H+ x 1 mol Ba(OH)2/2 mol H+ = 0.0232 mol
0.0232 mol Ba(OH)2/0.0232 L = 1 M
The end point is reached with less Ba(OH)2, which
would decrease the denominator larger [Ba(OH)2].
Ions combining with water (solvation) decreases
disorder (-S).
S < 0 when I2(aq)  I2(s) because solid has more order.
+S when s  l  g (A) and (B) and expansion (D).
Cu2O + H2  2 Cu + H2O
0.80 mol Cu x 63.5 g/mol = 51 g
Tthreshold = H/S  S = H/Tthreshold
S = -20 kJ/400 = -0.05 kJ•mol-1•K-1 (-50 J•mol-1•K-1)
Breaking bonds between CH3OH and H2O molecules is
endothermic, forming bonds is exothermic.
For a first order reaction, the straight line graph is ln[Z]
vs. t.
Mass at the end is for cup + solute + solvent, subtract
from the mass of cup + solvent = mass solute.
Heat needed to dissolve the remaining solute comes
from less water  greater T and H = -q = mcT.
At the normal melting point: G = 0 = H – TS H =
T S
The reaction is spontaneous, but it must not occur at a
fast rate (probably because of large Ea).
Average = (Mass1 x %1) + (Mass2 x %2)
63.5 = (63)(1 – x) + (65)(x) = 63 – 63x + 65 x  25 %
39 Ca  39 K + 0 
20
19
1
It takes 3 half-lives to reduce the radioactivity to 1/8
(1.25/10.0). 195 days/3 = 65 days
From the diagram: E6 = -0.38 eV and E2 = -3.40 eV
E = E2 – E6 = -3.40 eV – (-0.38 eV) = -3.02 eV
The electron can only absorb energy that will move it to
a higher energy level, 9.4 eV is not enough energy.
The orbital diagram for C,1s() 2s() 2p()()( ), has
two unpaired electrons  paramagnetic.
Pauli states that no orbital can contain electrons with
the same spin. Two spins = two electrons.
Heisenberg states that the wave nature of matter limits
what we can know about position and velocity.
# 31: n = 4, l= 1. Being degenerate, then ml = 1, 0 or -1
and ms = +½ or -½  (4, 1, 1, ½) fits requirement.
::S=Se:=S:: has zero formal charge. There are 4 bonds
and 5 pairs of nonbonding electrons.
The 2p electron is in an excited state, otherwise it
would go into 2s.
The 2 2s electrons and 3 2p electrons are valence
(highest energy level)  five.
First ionized electron is from 1s. It takes the most
energy to remove electrons that are close to nucleus.
The biggest jump in energy occurs between 4 and 5,
which means 4 valence electrons  Carbon group.
Radius increases going left and down in periodic T.
Anion is larger than and cation is smaller than atom.
Elements in the same column in the periodic table have
similar chemical properties.
Lattice energy measures ionic bond strength, which is
proportional to charge and 1/radius..
Single bonds are the weakest (CO, O=O, Cl–Cl, NN) 
Cl2.
69 A
70 D
71 B
72 A
73 C
74 D
75 B
76 C
77 D
78 B
79 D
80 D
81 D
82 B
83 D
84 C
85 B
86 D
87 B
88 B
89 A
90 C
91 C
92 B
93 D
94 D
95 B
96 C
97 D
98 A
99 B
100 D
Most polar bond forms between atoms with the greatest
electronegativity difference.
SeCl4 has 34 valence electrons, which require an
expanded octet system (sp3d).
Only SO2 has a single and double bond, which can
exchange places, thus forming resonance forms.
CO: The triple bond between C and O is composed of
one sigma bond and two pi bonds.
Electronegativity difference is greatest between H and F
 most polar = highest dipole moment.
Three H are pushed away from the pair of non-bonding
electrons around P = pyramidal structure.
A bond order of 1.5 means 1 sigma bond and 50% share
of a pi bond, which is the case for O3 (O=O–O).
CCl4 (tetrahedron), CO2 (linear), PCl3 (trigonal pyramid),
PCl5 (trigonal bipyramid), SF6 (octahedron)
Non-bonding electron pairs take up more space than
bonding pairs  H2O (2) < NH3 (1) < CH4 (0).
Acids contain the COOH functional group  (B) (a is a
ketone, c is an alcohol, and d is an ether)
4-methylpentane is the same as 2-methylpentane
because # 4 C = # 2 (left to right vs. right to left).
C1H3–C2C3H: C1 is sp3, C2 is sp, C3 is sp
Both are non-polar, but CCl4 has more electrons
( more polarizable)  stronger dispersion force.
Hydrogen bonding occurs when H is bonded to N, O or
F. Only N2H4 has that arrangement.
Farthest from each other in the gaseous phase, which
is at t5.
Melting occurs along 1st plateau (t2) and boiling along
2nd plateau (t4)  time between is t3.
Heat energy is used to break bonds to form liquid from
solid rather than raise temp  constant T.
At the same temperature, lighter molecules have
greater speed.
40 g of Ar = 1 mol. One mole at STP = 22.4 L. Since P is
2 x standard, then T is 2 x standard (PV = nRT).
PN2 = XN2Ptot
PN2 = (0.25/(0.35 + 0.90 + 0.25))(3.0 atm) = 0.50 atm
Real gases deviated from ideal at low temperatures
(near their boiling point) and high pressure.
(A) metallic bond, (B) ionic bond, (C) covalent bond, and
(D) molecular bond.
Gas solubility increases with greater partial pressure of
the gas in the container or lower temperature.
6 molal = 6 mol ethanol in 1000 g H2O, 1000 g H2O x 1
mol/18 g = 55 mol H2O  fraction: 6/(6 + 55) = 0.1
1 M = 1 mol/L solution. Molality is mol/kg solvent 
density is needed to convert volume to mass.
0.2 mol toluene in 0.8 benzene. 0.8 mole x 80 g = 60 g
benzene  molality = 0.2 mol/0.06 kg = 3 m
The highest boiling point = highest ion concentration.
(A) .2 x 3 = .6, (B) .25 x 3 = .75, (C) .3 x 2 = .6, (D) .4 x 1 = .4
1 CH3OCH3(g) + 3 O2(g)  2 CO2(g) + 3 H2O(g)
0.250 L x 0.400 mol/L x 119 g/mol = 11.9 g
If CO2 is soluble in water, then some gas would remain
in the water and not bubble into the bottle.
CH3OH has the lowest proportion of C (12/32)  given
equal masses; CH3OH would generate the least CO2.
molarity = mol solute/volume solution. Mass to
determine moles solute. Volume to determine volume.
Free Response
1.
a. (1)
S: 1s2 2s22p6 3s23p4
S2-: 1s2 2s22p6 3s23p6
(2)
S2- has greater radius because e increase e-e repulsion
but e-nucleus attraction is the same  larger radius.
(3)
S is attracted to a magnetic field because it has 2 unpaired
electrons in the 3p sublevel. S2- has all paired electrons so
it is not attracted to a magnetic field.
b.
It is easier from S2- because its valence electrons are less
attracted to the nucleus (16) compared to Ar (18).
c. (1)
The forces are stronger in H2S compared to H2O because
S has more electrons than O, which are more polarizable.
(2)
H2O has stronger dipole forces because O has a greater
electronegativity than S, which makes the O-H bond more
polar and strengthens the dipole force between molecules.
2. a.
In both cases the electron removed is from the same
energy level (2p), but F has a greater effective nuclear
charge due to one more proton in its nucleus.
b.
Less because the electron removed from F is from a 2p
orbital, whereas the electron removed from Xe is from a
5p orbital, which is farther from the nucleus.
c.
:::F
..
\ ..
:::O – Xe – O:::
:::F – Xe – F:::
|
.. \
O:::
F:::
d. (1)
Trigonal pyramid
(2)
sp3d2
3. a.
843 mL x 1 L/1000 mL x 1.18 g/1 L = 0.995 g
b.
157.70 g – 0.995 g = 156.71 g
c.
158.08 g – 156.71 g = 1.37 g
d.
PV = nRT
(750/760 atm)(843/1000 L) = n(0.0821 atm•L/mol•K)(296 K)
 n = 0.0342 mol:
MM = m/n = 1.37 g/0.0342 mol = 40.1 g/mol
e.
percent error = |40.1 g – 44.0 g|/44.0 g x 100 = 8.9 %
f. Occurrence 1:
Yes, Dry air (mostly N2 and O2) is less dense than CO2 and
the mass would be less  MM would be less than 44 g.
Occurrence 2:
No, Lower temperature would mean calculated moles at
23.0oC is too small  MM would be greater than 44 g.
g.
Find the mass of the empty flask. Fill the flask with water.
Mass the flask with water. Subtract to find the mass of
water and use the density at 23oC to determine volume.
4. a. (1)
25.0 g CH4 x 1 mol CH4 x 1 mol CH2Cl2 = 1.56 mol CH2Cl2
16.0 g CH4 1 mol CH4
2.58 mol Cl2 x 1 mol CH2Cl2/2 mol Cl2 = 1.29 mol CH2Cl2
(2)
1.29 mol CH2Cl2 x 85.0 g CH2Cl2/1 mol CH2Cl2 = 110. g
b.
242 kJ/1 mol x 103 J/1 kJ x 1 mol/6.02 x 1023 = 4.02 x 10-19 J
c.
E = hc/4.02 x 10-19 J = (6.63 x 10-34 J•s)(3.0 x 108 m/s)/
 = 4.9 x 10-7 m
5. a.
No additional mass was lost during the third heating,  all
the water of hydration has been driven off.
b. (1)
25.825 – 23.977 = 1.848 g H2O x 1 mol/18.02 g = 0.1026 mol
(2)
23.977 – 22.347 = 1.630 g MgCl2 x 1 mol/95.20 g = .01712 mol
0.1026 mol H2O/0.01712 mol MgCl2 = 6 MgCl2•6 H2O
c.
Loss in mass includes solid MgCl2 in addition to the
water, which would make the mass of water loss to large.
6. a.
..
:N  N:
H–N–H
|
H
b.
12.0 g H2 x 1 mol H2/2.0 g H2 x -34 kJ/3 mol H2 = -68 kJ
c.
Reactants because bond energy is released when
reactants turn into products.
d.
The reaction produces less disorder because 4 moles of
gas are changed to 2 moles.
e.
The term TS in G = H –TS is positive and at some
higher temperature will become more positive than H is
negative resulting in a positive G.
7. a.
Initial concentration of S2O32-(aq)
b.
rate = k[S2O32-]n :
trial 2: 0.030 = k(0.075)n  1.50 = 1.50n  n = 1 (1st order)
trial 1: 0.020 = k(0.050)n
c.
0.020 M•s-1 = k(0.050 M)  k = 0.40 s-1
d.
ln(No/Nt) = kt  ln(0.10 M/0.020 M) = (0.40 s-1)t  t = 4.0 s
e.
8.
a. (1)
Fe
[Ar] 4s() 3d()()()()()
Fe3+ [Ar] 3d()()()()()
(2) Fe ____
Fe3+ __X__
b.
NF3
POF3
:O:
•• •• ••
•• || ••
:F–N–F:
:F–P–F:
•• | ••
•• | ••
:F:
:F:
••
••
c.
sp3
107o 109.5o 120o
d.
ArF4
••
:F:
•• \•• ••
: F – Ar – F :
•• ••\ ••
:F:
••
Square planar
Dispersion
Dipole
H-bond
x
x
NH3
x
NF3
9. a.
PV = nRT
(763/760 atm)(843/1000 L) = n(0.0821 atm•L/mol•K)(296 K) 
n = 0.0348 mol
b.
MM = m/n = 1.465 g/0.0348 mol = 42.1 g/mol
c. (1)
percent error = |42.1 g – 40.0 g| x 100 = 5.3 %
40.0 g
(2) Too large _____
Too small __X__
10. a.
m = T/Kf = 2.55oC/1.86oC/m = 1.37 mol/kg
b.
mol solute = (m)(msolvent) = (1.37 mol/kg)(0.100 kg)
mol solute = 0.137 mol
c.
MM = 25.0 g/0.137 mol = 182 g/mol
11. a.
2.00 g x 35.0 g ascorbic acid/100 g table = 0.700 g
b.
0.601 g H2O x 2.02 g H = 0.0674 g H
18.0 g H2O
1.467 g CO2 x 12.0 g = 0.400 g C
44.0 g
1.000 g – 0.0674 g – 0.400 g = 0.533 g O
c.
0.400 g C x 1 mol C = 0.0333 mol C = 1 mol C
12.0 g C
0.0333
0.0674 g H x 1 mol H = 0.0667 mol H = 2 mol H
1.01 g H
0.0333
0.533 g O x 1 mol O = 0.0333 mol O = 1 mol O
16.0 g O0.0333
 empirical formula is CH2O
d.
.0545 L OH- x .102 mol OH- x 1 mol H+ = .00556 mol H+
1L
1 mol OH1.000 g/0.00556 mol = 180. g/mol
e.
mempirical formula = 1(12.0 g) + 2(1.0 g) + 1(16.0 g) = 30 g
180 g/30 g = 6  molecular formula = C6H12O6
f.
C6H12O6 + 6 O2  6 CO2 + 6 H2O
12. a.
T = 21.8 – 25.0 = -3.2oC
b.
The process is endothermic because the temperature
decreased indicating the reaction required energy.
c. (1)
q = mcT = -(5.13 g + 91.95 g)(4.2J g-1 oC-1)(-3.2oC)
q = 1.3 x 103 J
(2)
5.13 g urea x 1 mol/60.0 g = 0.0855 mol
H = q/molurea = 1.3 kJ/0.0855 mol = 15 kJ mol-1
d.
Go = Ho – TSo
-6.9 kJ mol-1 = 14.0 kJ mol-1 – (298 K)So
 So = 0.0701 kJ mol-1 K-1
13. a. (1)
(0.02500 L)(0.250 mol/L) = 6.25 x 10-3 mol
(2)
.00625 mol Cl- x 1 mol Cl2/2 mol Cl- = .003125 mol Cl2
V = nRT/P = (.003125 mol)(0.0821)(295)/0.950 = 0.0797 L
b. (1)
rate = k[Cl-]m[MnO4-]n[H+]p
2.03 x 10-7 = k(0.0312)m(0.00400)n(3.00)p
2.25 x 10-8 = k(0.0104)m(0.00400)n(3.00)p
9 = 3m  m = 2
(2)
rate = k[Cl-]m[MnO4-]n[H+]p
2.03 x 10-7 = k(0.0312)m(0.00400)n(3.00)p
1.02 x 10-7 = k(0.0312)m(0.00200)n(3.00)p
2 = 2n  n = 1
c. (1)
rate = k[Cl-]2[MnO4-][H+]3
(2)
rate = k[Cl-]2[MnO4-][H+]3
2.25 x 10-8 = k(0.0104)2(0.00400)(3.00)3
k = 1.93 x 10-3 M-5s-1