CHEMISTRY WKST KEY: SEMESTER 2 REVIEW
1)
a) 7
b) 5
c) 9
2)
a) 6
b) 2
c) 50
3)
red, orange, yellow, green, blue, indigo, violet
longest wavelength
4)
A) crest
highest energy
highest frequency
B) wavelength
C) node
D) trough
5)
6)
p. 1
a) [Ne] 3s2 3p3
d) [Kr] 5s2 4d10 5p6
b) [Ar] 4s1 3d5
e) [Ar] 4s2 3d10 4p6
c) [Xe] 6s2 4f14 5d8
7)
a) 3s2 3p6 3d10
b) 5s1
8)
λ = 3.2 x 10−17 m
c = 3.00 x 108 m/s
c) 3s2 3p6
c = λν
ν=
3.00 x 108 ms
c
=
= 9.375 x 1024 s-1
λ
3.2 x 10-17 m
ν = 9.4 x 1024 s−1
9)
Atoms form ions to become like noble gases and be more stable.
10)
a) ↑→
b) ↑→
c) ↓←
d) ↓←
11)
a) 6
b) 3
c) 6
d) 8
12)
a) polar covalent
c) nonpolar covalent
b) nonpolar covalent
d) ionic
E) amplitude
CHEMISTRY WKST KEY: SEMESTER 2 REVIEW
p. 2
13)
Ionic bonds involve a transfer of an electron from one atom to another and the resulting oppositely charged ions
attract each other. Covalent bonds form when two half-filled orbitals overlap, allowing the atoms to share the two
electrons.
14)
Ionic compounds have ionic bonds between its particles (ions). They form a lattice structure and have high
melting and boiling points because of the strong bonding. Ionic compounds are electrolytes (conduct electricity)
when melted or dissolved in water.
Molecular compounds have covalent bonds between the atoms within the molecules but have weak forces
(dipole-dipole, hydrogen bonding, or London Forces) between the molecules. Therefore they have low melting
and boiling points because of the weak forces between the molecules. They are also nonelectrolytes (don’t
conduct electricity).
15)
16)
17)
a) bent
b) trigonal pyramidal
c) tetrahedral
ELECTRONIC
SHAPE
MOLECULAR
SHAPE
SF4
trigonal
bipyramidal
SO3
FORMULA
LEWIS STRUCTURE
DRAWING
POLARITY
IMF
teeter
totter
polar
DD
trigonal
planar
trigonal
planar
nonpolar
LF
NH4+
tetrahedral
tetrahedral
nonpolar
LF
CH2O
trigonal
planar
trigonal
planar
polar
DD
ClF5
octahedral
square
pyramidal
polar
DD
Atoms get smaller going across the period left-to-right because the nuclear charge is increasing as the atoms gain
more protons. The stronger force pulls the electrons, which are all in the same valence shell, closer to the
nucleus.
CHEMISTRY WKST KEY: SEMESTER 2 REVIEW
p. 3
18)
Atoms get larger going down the groups primarily because larger energy levels are being filled with electrons.
Also, the core electrons (the electrons inside the valence shell) shield the valence electrons from the nucleus so
they move farther away from the nucleus.
19)
Chemical reactions take place in the valence shell.
20)
a)
H2S → H2S is polar with dipole-dipole forces while CCl4 is nonpolar with London Forces. H2S has stronger
forces and therefore high boiling point.
b)
H2O → H2O is polar with hydrogen bonding while SF4 is polar with dipole-dipole forces. H2O has stronger
forces and therefore a lower vapor pressure.
c)
NaCl → NaCl is ionic with ionic bonds between its particles. I2 is nonpolar with London Forces between its
particles. NaCl has the stronger forces, therefore has the higher melting point.
21)
a) C6H6
stuff
b) H2O
c) H2O
litmus
phenolphthalein
acid
base
water
acid
red
clear
pH
<7
>7
=7
base
blue
pink
pOH
>7
<7
=7
24)
HCl → hydrochloric acid
HBr → hydrobromic acid
HI → hydroiodic acid
25)
a) HCl → H+ + Cl−
26)
a) HCl + NaOH → H2O(l) + NaCl
(H2O is polar and dissolves polar stuff—ionic
(like “c” and “d” acts like it is polar. C6H6 is
nonpolar and will dissolve nonpolar stuff.)
d) H2O
22)
23)
e) C6H6
HNO3 → nitric acid
HClO3 → chloric acid
HClO4 → perchloric acid
H2SO4 → sulfuric acid
b)
H2Cr2O7 ⇄ H+ + HCr2O7−
HCr2O7− ⇄ H+ + Cr2O72−
b) H2SO4 + 2NH4OH → 2H2O(l) + (NH4)2SO4
CHEMISTRY WKST KEY: SEMESTER 2 REVIEW
27)
a)
b)
28)
a)
molecular
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
ionic
Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3−(aq)
net ionic
Ag+(aq) + Cl−(aq) → AgCl(s)
molecular
3Pb(ClO3)2(aq) + 2K3PO4(aq) → Pb3(PO4)2(s) + 6KClO3(aq)
ionic
3Pb2+(aq) + 6ClO3−(aq) + 6K+(aq) + 2PO43−(aq) → Pb3(PO4)2(s) + 6K+(aq) + 6ClO3−(aq)
net ionic
3Pb2+(aq) + 2PO43−(aq) → Pb3(PO4)2(s)
pH = 14 – pOH
14 = pH + pOH
pH = 14 – 11.25 = 2.75
b)
pH = −log[H+]
pH = −log(0.0445 M) = 1.352
c)
pOH = −log[OH−]
pOH = −log(5.25 x 10−9 M) = 8.280
14 = pH + pOH
pH = 14 – pOH = 14 – 8.280 = 5.720
−OR−
29)
p. 4
10-14
10-14
=
= 1.90 x 10-6 M
[OH-]
5.25 x 10-9 M
10−14 = [H+][OH−]
[H+]=
pH = −log[H+]
pH = −log(1.90 x 10−6 M) = 5.721
Na = 1.25 N
Va = 42.55 mL
Nb = ?
Vb = 25.00 mL
NaVa = NbVb
Nb =
(1.25 N)(42.55 mL)
Na Va
=
= 2.1275 N
Vb
25.00 mL
Nb = 2.13 N
30)
a) HI
b) H2SO3
c) H3PO4
31)
a) carbonic acid
b) acetic acid
c) nitric acid
32)
a)
b)
c)
NaCl
I2
ethanol
33)
M=
g solute
(MM solute)(L soln)
M=
15.0 g NaCl
= 0.205339 M = 0.205 M
g
(58.44 mol
)(1.25 L)
CHEMISTRY WKST KEY: SEMESTER 2 REVIEW
g solute
(MM solute)(kg solvent)
72.55 g Ba(NO3)2
= 0.5551942 m = 0.5552 m
g
(261.35 mol
)(0.5000 kg)
34)
m=
35)
% solute
mass solute
=
100
mass solution
x
100.0 g BaCl2
=
100 300.0 g solution
mass soln = mass solute + mass solvent
density H2O = 1.00 g/mL
300.0 x = 10000.0
x = 33.33% BaCl2
m=
p. 5
36)
N = nM
N = (3)(2.50 M) = 7.50 N
37)
∆Tb = Kbmi
∆Tb = (0.512 °C
)(1.25 m)(3) = 1.92°C
m
BP = 100°C + 1.92°C = 101.92°C
∆Tf = Kfmi
∆Tf = (1.86
°C
m
)(1.25 m)(3) = 6.975°C = 6.98°C
FP = 0°C – 6.98°C = −6.98°C
38)
NaCl = K3PO4 < Ba(NO3)2 < C6H12O6
39)
unsaturated
40)
stir or shake; heat; increase surface area by grinding up into a powder
41)





42)
a)
Catalysts speed up reactions by lowering the activation energy so more particles can meet that requirement.
b)
Adding heat will increase the KE of the particles. Therefore, more particles will end up with at least the
activation energy and be able to react.
nature of reactants and products
concentration (surface area)
temperature
energy (activation energy)
outside influences (catalysts and inhibitors)
43)
a) 3
44)
A
45)
a)
exothermic → products have less energy than reactants so energy must have been lost by the reaction.
b)
products → have less energy than the reactants; less energy = more stable
46)
b) 1
c) 6
(Orders are the exponents in the rate law. Overall order is the sum.)
 frequency of collisions → gotta have a lot of collisions.
 correct orientation → particles must line up right in order to react
 energy → particles need to have enough energy to react (need at least the activation energy)
CHEMISTRY WKST KEY: SEMESTER 2 REVIEW
47)
48)
49)
a)
shifts to the left (towards reactants)
b)
amount of CO2 decreases
a)
shifts to the right (towards products)
b)
amount of C2H6 decreases
a)
shifts to the right (towards products)
b)
amount of O2 decreases
50)
Cl2(g) + CHCl3(g) → HCl(g) + CCl4(g)
51)
step 2
52)
Cl, CCl3
53)
rate = k[Cl2][CHCl3]
54)
a)
 [A] doubles between trials 1 and 2 while [B] remains the same. Rate goes up 4 times for these trials.
Therefore, the order for [A] = 2.
 [B] doubles between trials 2 and 3 while [A] remains the same. Rate goes up 2 times for these trials.
Therefore, the order for [B] = 1.
rate = k[A]2[B]
b)
k=
rate
[A]2 [B]
Since k is the same for all 3 trials, it doesn’t matter which trial you use. I chose to use trial 1.
k=
c)
3.0 x 10-4 Ms-1
2
(0.20 M) (0.20 M)
= 0.038 M-2 s-1
rate = k[A]2[B] = (0.038 M−2s−1)(0.30 M)2(0.10 M) = 3.4 x 10−4 Ms−1
2
55)
a)
[SO3 ]
2
[SO2 ] [O2 ]
2
b)
p. 6
Keq =
[SO3 ]
2
[SO2 ] [O2 ]
=
(0.00462 M)2
2
(0.0960 M) (0.0425 M)
= 0.0545
CHEMISTRY WKST KEY: SEMESTER 2 REVIEW
THIS POINT FORWARD IS FOR THE UNDERCLASSMEN’S WKST.
56)
4
57)
a)
b)
58)
a) alkyne
b) alkane
c) alkene
59)
a) ester
b) carboxylic acid
c) alcohol
d) alkene
e) alkane
f) alkane
d) carboxylic acid
e) ester
f) alcohol
p. 7