Chemistry Practice Exam #1
ANSWERS TO PRACTICE EXAM
1)
Do the following conversions:
a) 54.2 cm3 to in3
(1 in = 2.54 cm)
b) 2.734 x 10-2 nm to cm
 10 9 m  1cm 
  2  
2.734 x10  2 nm
 1nm  10 m 
2.734 x10 9 cm
3
 1in 
54.2cm 3 
 
 2.54cm 
3.31in 3
c) 0.87 g/mL to lb/quart
(1 lb = 453.6 g; 1 quart = 0.946 L)
0.87
g  1lb  1mL  0.946 L 




mL  453.6 g  10 -3 L  1qt 
 lb 
1.8 
 qt 
d) 8.6542 x 107 mg to kg
 10 3 g  1kg 
 3  
8.6542 x10 7 mg
 1mg  10 g 
8.6542 x101 kg  86.542kg
e) 2.867 x 10-9 L to L
 1L 
2.867 x10 9 L 6  
 10 L 
3
2.867 x10 L
-1-
2)
Perform the following mathematical operations and express the results to the correct number of
significant figures:
a) 7.8132 + 0.6 + 18.24
b) (1.34 x 10-3)x(9.1 x 10-6)
= 26.7
= 1.2 x 10-8
3)
Answer the following True or False:
a) The identity of an atom is determined by the number of electrons it has.
F
b) Copper (II) chloride is an ionic compound
T
c) Different isotopes of a given element have different masses.
T
d) The melting of ice is an example of a physical change.
T
e) Density is an example of a physical property of a substance.
T
4)
Co 3 has: Answer is: e)
a) 27 protons, 27 electrons, 59 neutrons
59
27
b) 22 protons, 27 electrons, 37 neutrons
c) 27 protons, 27 electrons, 32 neutrons
d) 27 protons, 30 electrons, 32 neutrons
e) 27 protons, 24 electrons, 32 neutrons
-2-
5)
Give the formulas of the following compounds:
a) Lithium oxide
b) Dinitrogen tetroxide
Li2O
N2O4
c) Iron (III) hydroxide
d) Aluminum sulfate
Fe(OH)3
Al2(SO4)3
e) Chloric acid
HClO3
-3-
6)
Give the names of the following compounds:
a) Na2SO4
b) CuS
Sodium sulfate
Copper (II) sulfide
c) SO2
d) (NH4)2CrO4
Sulfur dioxide
Ammonium chromate
e) Hydrobromic acid
HBr
7)
a) Gallium has two naturally occurring isotopes, 69Ga (with a mass of 68.9256 amu and a natural
abundance of 60.11%) and 71Ga (with a mass of 70.9247 amu and a natural abundance of 39.89%).
Calculate the average atomic mass of gallium, showing all your work.
68.9256 amu·0.6011 + 70.9247 amu·0.3989 = 69.72 amu
-4-
b) If 1 gram is 6.022 x 1023 atomic mass units (amu), calculate the mass in grams of one atom of 69Ga.
1g


 22
68.9256amu
  1.14456 x10 g
23
6
.
022
x
10
amu


8)
Balance the following chemical equations:
a) HI + As2S3  AsI3 + H2S
b) Fe + O2  Fe2O3
6HI + As2S3  2AsI3 + 3H2S
4Fe + 3O2  2Fe2O3
c) HBr + KOH  KBr + H2O
d) Cs + H2O  CsOH + H2
HBr + KOH  KBr + H2O
Balanced as is
2Cs + 2H2O  2CsOH + H2
e) C2H6O + O2  CO2 + H2O
C2H6O + 3O2  2CO2 + 3H2O
-5-