Semester 1 Review Problems
A.P. Chemistry
DIRECTIONS: This sheet is not meant to be turned in; it is merely a reminder of some of the topics that could be on
the final exam. (NOTE: Conversions that are required on the exam will be included, except for metric conversions.)
1.
Which of the following are chemical changes versus the physical changes: burning paper, subliming dry ice, sticking a
tack in a cork board, the result of mixing baking soda with vinegar.
Answer:
burning paper – C
subliming dry ice – P
sticking a tack in a cork board – P
the result of mixing baking soda with vinegar – C
2.
Give an example of a sample of data that is precise but not accurate, knowing that the theoretical value is 12.90 g.
Answer:
Theoretical = 12.90 g
Precise by not accurate = All of the values are close to each other but not close to the theoretical value.
3.
How many significant figures would result in the following calculation?
(2.06–1.2+900.) x 2.9156 ÷ 0.10
Answer:
(2.06–1.2+900.) x 2.9156 ÷ 0.10 = 900.86 x 2.9156 ÷ 0.10 = 26265.474
2 significant figures
4.
Convert 1.961 kg to ounces.
Answer:
1.961kg 2.20lb 16oz
 69.03oz
1kg 1lb
5.

The separation technique of distillation is based on what physical property.
Answer:
Differing boiling points
6.
Give an example of a heterogeneous mixture and a homogeneous mixture.
Answer:
Heterogeneous – chocolate mix
Homogeneous - brass
7.
Name N2O3.
Answer:
Dinitrogen trioxide
8.
What would be the formulas of the following ions (in order): hypoarsenite, arsenite, arsenate, and perarsenate.
Answer:
Hypoarsenite = AsO23–
Arsenite = AsO33–
Arsenate = AsO43–
Perarsenite = AsO53–
9.
If the initial volume of a gas was 500 mL at a pressure of 101.3 kPa, when the volume is compressed to 450 mL, what is
the pressure in torr?
Answer:
101.3 kPa = 760 torr
PV = PV
(760 torr)(500 mL) = (P)(450 mL)
P = 844 torr
10.
If the initial volume of a gas is double, the initial pressure tripled, and the amount of moles halved, what would happen
to the temperature?
Answer:
P1V1 P2V2

n1T1 n2T2
PV (3P)(2V )
 1
nT1
( 2 n)T2

11.

½ T1 = 6 T2
So the final temperature (T 2) would be 12x that of the initial temperature (T 1)
A piece of copper wire at 97.0˚C is dropped into 52.0 mL of H2O at room temperature, which is 23.0˚C. What is the
equilibrium temperature of the system if the wire has a mass of 4.85 grams? (NEEDED: specific heat of copper = 0.385
J/g·˚C; specific heat of water = 4.18 J/g·˚C)
Answer:
q = mc∆t
qH2O = –qice
1 g water = 1 mL water
J
(52.0 g)(4.184 g C )(T – 23.0˚C) = –(4.85 g)(0.385
217.36T – 4999.28 = –1.86725T + 181.12325
219.22725T = 5180
T = 23.6˚C

J
g C
)(T – 97.0˚C)

12.
Write a net ionic equation for the reaction between solutions of nickel(II) chloride and sodium sulfite.
Answer:
NiCl2(aq) + Na2SO3(aq) → NiSO3(s) + 2 NaCl(aq)
Ni2+(aq) + SO32–(aq) →NiSO3(s)
13.
Write the formula of ferric borate.
Answer:
FeBO3
14.
An electron in the hydrogen atom jumps from the energy level n = 7 to the energy level n = 3. Answer the following
questions:
a. How much energy is being absorbed or emitted?
b. What are the frequency and wavelength of the photon of light produced?
c. Which portion of the electromagnetic spectrum is this energy change associated with? (i.e., UV, visible, etc.)
Answer:



 1
1 
ERH  2  2 
n f ni , c , Eh
1 1 
E  2.18 x1018  2  2   1.98 x1019 J
3 7 
–19
1.98x10 J 
were emitted
–34
–19
1.98x10 J = (6.63x10 J·s)(  )
 = 2.99x1014 Hz
3.0x108 m/s =  (2.99x1014 Hz)
 = 1.01x10–6 m

IR spectrum
2 ways of knowing: #1 = wavelength is  1000 nm, above the red 700 nm wavelength
#2 = the electrons falls to the 3rd E. L., which signifies the Paschen series, all of which emit

IR radiation

15.
A certain reaction has the following general form:
a A(g) → b B(g)
At a particular temperature and [A]o = 2.00x10–2 M, concentration versus time data were collected for this reaction, and
a plot of ln[A] versus time resulted in a straight line with a slope value of –2.97x10–2 min–1.
a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.
b. Calculate the half-life for this reaction.
c. How much time is required for the concentration of A to decrease to 2.50x10 –3M?
Answer:
A straight line of ln[A] versus time means that the reaction is a first order process.
Rate law: Rate = k [A]1
[A] 
ln  t  kt
[A]
Integrated rate law:  o 
Rate constant: k = – slope = –(–2.97x10–2 min–1) = 2.97x10–2 min–1
The half-life can be found by plugging into the integrated rate law with half of the concentration as [A] t:
2
1.00x10
M 
2
1
ln 
 2.97x10 min  t
2
2.00x10 M 
t 12  23.3min
 
1


16.
17.
2
The the two concentrations can be plugged into the integrated rate law to solve for time:
2.50x103 M 
2
1
ln 
 2.97x10 min t 
2
2.00x10 M 
t  70.0min
What is the electron configuration for manganese and the manganese (II) ion?

Answer:
 Mn = 1s2 2s2 2p6 3s2 3p6 4s2 3d5
Mn2+ = 1s2 2s2 2p6 3s2 3p6 3d5
9.60 grams of oxygen gas and 44.90 grams of sulfur dioxide gas exert a total pressure of 950 mmHg in a closed
container. Find the partial pressure of the sulfur dioxide.
Answer:
9.60gO2 1mol
 0.300molO2
32g
44.90gSO2 1mol
 0.0.701molSO2
64.06g
 SO2 
0.0.701molSO2
 0.700
0.0.701molSO2  0.300molO2
Ptotal  SO2 = PSO2
(950 mmHg)(0.700) = PSO2
PSO2 = 670 mmHg

18. Write the formula of sulfurous acid.
Answer:
H2SO3
19.
23.10 grams of aluminum metal are dissolved in 500. mL of 3.76 M hydrochloric acid, HCl, to yield aluminum chloride
and hydrogen gas. How many liters of hydrogen gas are produced, assuming STP?
Answer:
23.10gAl 1mol
 0.856molAl
26.98g
(0.500L)(3.76M)  1.88molHCl
2 Al(s)
+ 6 HCl(aq)
0.856 mol
1.88 mol
– 2x
– 6x
0.856 – 2(0.313) =
1.88 – 6(0.313) = 0
0.229 mol
First determine which will act as the limiting reactant:
I
C
E

→
2 AlCl3(aq) +
0
+ 2x
2(0.313) =
0.626 mol
3 H2(g)
0
+ 3x
3(0.313) =
0.939 mol
0.856molAl
 0.428
2
1.88molHCl
 0.313
6
The calculated amount from HCl is lower, so HCl is the limiting reactant AND 0.313 mol is the value of x.
Plug this value in, see above.

20.

21.

22.
0.939molH 2 22.4L
 21.0LH2
1mol
How many atoms of chromium, Cr, are in 12.9 gram of sodium dichromate?
Answer:
12.9gNa2Cr2O7
1mol
2Cr
6.02x10 23 atoms
 5.93x10 22 atomsCr
2611.98g 1Na2Cr2O7
1mol
Name H2S.
Answer:
Hydrosulfuric acid
Bromine gas is diatomic and Argon gas is monoatomic. Which gas is faster and what is the ratio of the velocities?
Answer:
M Br
 Ar
159.8


 Br
M Ar
39.9
 Ar
2
 Br
2
2
2
Argon is 2x faster than Bromine.
23.
The following apparatus is set up in a laboratory where A and B are columns of mercury, x represents the pressure of an
unknown gas, and y represents the pressure of the atmosphere. The top gas changer has a volume of 1 unit. The bottom
chamber has a volume of 3 units. The valves are closed off from each other by a stopper.
Find X and Y given these initial conditions: the closed-ended manometer has a mercury height difference of 400 mmHg
at point A and the open-ended manometer has a mercury height difference of 430 mmHg at point B.
Answer:
@ A – closed manometer
the difference in height of a closed manometer equals the pressure of the container being measured  400
mmHg = x
@ B – open manometer
the level of mercury at point B is higher than the level of mercury on the open end, meaning that the
atmospheric pressure, y, is greater than the pressure of the container; the pressure of the 
container is 400 mmHg
and the atmospheric pressure is 430 greater  y = 400 + 430 = 830 mmHg
24.
Octyl acetate is an ether compound that smells like oranges. A sample of octyl acetate consists of 0.227 grams
hydrogen, 0.362 grams oxygen, and 1.34 grams carbon. Also, 0.0256 mole of octyl acetate has a mass of 4.40 grams.
Find the empirical and molecular formulas
of octyl acetate.
Answer:
0.227gH 1mol
 0.225molH
1.01g
0.362gO 1mol
 0.0226molO
16g

1.34gC 1mol
 0.112molC
12.01g
0.225molH 0.0226molO 0.112molC
:
:
0.0226mol 0.0226mol 0.0226mol
10H:1O:5C
Empirical Formula = H10OC5

4.40g
171.875g / mol
0.0256mol
= Molar mass of octyl acetate
171.875g / mol
2
(10(1.01)  1(16)  5(12.01))

Molecular Formula = H20O2C10
25.

Write the formula of potassium cyanide.
Answer:
KCN
26.
How much energy is lost when 450. grams of liquid water at 65.0˚C cools and freezes at 0˚C?
Answer:
∆H = qwater + qfreezing
∆H = mc∆t + ∆Hfus
∆H = (450. g)(4.184
J
g C
)(65 – 0˚C) +
 450.g 1mol 6.01kJ 1000 J 





18.02 g 1mol
1kJ 

∆H = 122265 Joules + 150083 Joules = +272348 J = +272 kJ
of the solids Cs, LiCl, and CHI3 with one of the following sets of properties:
Associate each
a. A white solid melting at 613˚C; the liquid conducts, while the solid does not.
b. A white solid which is lustrous when cut. Both the solid and liquid are electrical conductors.
c. A yellow solid with a characteristic odor having a melting point of 120˚C.
Answer:
a. LiCl
b. Cs
c. CHI3
27.
28.
Draw the Lewis dot structure of formaldehyde, CH2O. Determine its geometry, the approximate bond angles, the
number of sigma and pi bonds, whether the molecule exhibits resonance, the type of bond(s), whether the molecule is
polar, and which intermolecular forces the molecule has.
Answer:
H
| ..
H–C=O:
geometry: trig. planar
bond angles: 120˚
sigma bonds: 3
pi bonds: 1
resonance: no
type of bonds: C–H = polar covalent
C=O = polar covalent
polar: yes (higher electronegativity on oxygen)
intermolecular forces: London (dispersion/van der Waals) and Dipole, but not Hydrogen Bonding
29.
You are given three liquids: HI, NH3, and CH4. Which of these will be the most volatile? Which will have the most
surface tension?
Answer:
HI has London and Dipole forces, NH3 has London, Dipole and H-bonding, and CH4 has only London.
CH4 will be the most volatile
NH3 will have the greatest surface tension
30.
An aqueous solution is 22.0% lithium chloride, LiCl, by mass. The density of the solution is 1.127 g/mL. What are the
molality, mole fraction, and molarity of LiCl in the solution?
Answer:
22.0% LiCl = 22.0 grams LiCl and 78.0 grams H2O
density = 1.127 g/mL
Molality:
moleSolute
m

kgSolvent
22.0gLiCl
42.39g / mol
0.0780kgH2O
m = 6.65 m
Molarity:


dsolution  1.127g / mL
Vsolution 
22.0gLiCl  78.0gH 2O
 88.7mL
1.127g / mL
moleSolute
M

LSolution
22.0gLiCl
42.39g / mol
0.0887L
M = 5.85 M
Mole Fraction:

 LiCl
22.0gLiCl
moleSolute
42.39g / mol






22.0gLiCl
moleSolute  moleSolvent 
 78.0gH 2O

42.39g
/
mol
18.02g
/
mol

 

XLiCl = 0.107
31.
32.
Write a balanced net ionic equation which represents what will happen when a solution of mercury(II) chlorate is added
to solid sodium phosphate.
Answer:
Hg(ClO3)2(aq) + Na3PO4(s) → Hg3(PO4)2(s) + NaClO3(aq)
3 Hg2+(aq) + 2 Na3PO4(s) → 1 Hg3(PO4)2(s) + 6 Na+(aq)
Write a balanced equation showing what happens when potassium chlorate is heated .
Answer:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)