CHM 134 General Chemistry I
Exam I Review – Dr. Steel
Name
1. Complete the table below.
Part A
Part B
Type
Formula
Name
×××
×××
ionic
K3PO4
potassium phosphate
Zn2+
Br1+
ionic
ZnBr2
zinc bromide
×××
×××
molecular
SeI6
selenium hexaiodide
×××
×××
molecular
P3N5
triphosphorus pentanitride
NH41+
S2-
ionic
(NH4)2S
ammonium sulfide
Fe3+
CO32-
ionic
Fe2(CO3)3
iron (III) carbonate
×××
×××
acid
H2SO4
sulfuric acid
×××
×××
molecular
N2O5
dinitrogen pentaoxide
Cu2+
O2-
ionic
CuO
copper (II) oxide
×××
×××
molecular
S2F10
disulfur decafluoride
Pb4+
O2-
ionic
PbO2
lead (IV) oxide
H
Cl
acid
HCl
hydrochloric acid
2. Playing a carnival game you swing a hammer so that it hits a target with 108.3 pounds of
force. The area of the hammer head is 25.8 in2. Determine the pressure (Pressure =
Force/Area) of the hit in units of N/m2. You may find these conversions useful:
1 in. = 2.54 cm
1 N = 101.97 pounds
F 108.3 lb
=
= 4.20 inlb2
A 25.8 in 2
1N
1 in 2
(100 cm) 2
lb
4.20 in 2 ×
×
×
= 63.8 mN2
101.97 lb (2.54 cm) 2
1m2
P=
3. A compound forms between magnesium and the dichromate ion (Cr2O72-). Answer the
following questions.
What are the name and chemical formula of the compound?
magnesium dichromate
MgCr2O7
What is the mass percentage of chromium in the compound?
Mass percent of Cr =
mass Cr
2(52.00 g)
=
× 100% = 43.28%
total mass of compound 24.31 g + 2(52.00 g) + 7(16.00 g)
How many moles of chromium are there in a 15.753-g sample of the compound?
15.753 g MgCr2 O 7 ×
1 mole MgCr2 O 7
2 mol Cr
×
= 0.13111 molCr
240.31 g MgCr2 O 7 1 mol MgCr2 O 7
4. Determine the empirical formula of an unknown compound if you are able to collect the
following information about a 2.581-g sample of the compound:
It is 19.36% calcium.
It contains 1.501×1022 atoms of chlorine.
The remainder of the sample is oxygen.
Ca:
Cl:
O:
(Using the first two amounts, there are
1.198 g of oxygen atoms.)
1 mol Ca
= 0.012467 mol Ca ÷ 0.012467 = 1
40.08 g Ca
1 mole Cl
1.501 × 10 22 atoms Cl ×
= 0.024925 mol Cl ÷ 0.012467 = 2
6.022 × 10 23 atoms
0.1936 × 2.581 g compound ×
1.197 g O ×
1 mol O
= 0.074813 mol O ÷ 0.012467 = 6
16.00 g O
Formula: CaCl2O6 (or more likely Ca(ClO3)2)
34 2−
5. Identify each quantity for this atomic symbol: 16 S
Number of protons
Number of electrons
16
Number of neutrons 18
18
6. Complete the calculation at the right, expressing
your answer with the appropriate units and
number of significant figures.
186.7 kg − 132.3 kg + 17.66 kg
=
(2.051 m)2
72.06 kg
= 17.1302 kg 2 = 17.1 kg 2
2
m
m
4.20 66 m