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CHEM 110
Page 1 of 14
EXAM 2 PRACTICE PROBLEMS PACKET – ANSWER KEY
1) Write the conversion factor between milliliters and liters (use symbols): 1000 mL = 1 L
2) Write the conversion factor between meters and centimeters (use symbols): 1 m = 100 cm
3) Write the conversion factor between grams and kilograms (use symbols): 1000 g = 1 kg
Metric Prefix (name) Metric Prefix (symbol) Exponent
Question 4) micro− μ 10−6
Question 5) giga− G 109
Question 6) nano− n 10−9
Question 7) kilo− k 103
CHEM 110
Page 2 of 14
Questions 8 – 17: Write the name Questions 18 – 27: Write the chemical symbol
Each name is worth 2 points. Each chemical symbol is worth 2 points.
Every three spelling errors = 3% deduction Write capital and lower case letters clearly
8) Ne neon 18) potassium K
9) N nitrogen 19) silver Ag
10) Ca calcium 20) sulfur S
11) U uranium 21) copper Cu
12) Ga gallium 22) mercury Hg
13) Li lithium 23) boron B
14) Sr strontium 24) krypton Kr
15) Co cobalt 25) bromine Br
16) W tungsten 26) plutonium Pu
17) Mg magnesium 27) carbon C
Note: For Astatine (At), I will not take points for including or omitting it. Different sources conflict
on whether astatine is a metalloid or not.
Question 28) In the table below, shade in the metalloids.
CHEM 110
Page 3 of 14
Question 29) In the table below, circle the alkali metals. Question 30) In the table below, circle the halogens.
Some notes
Alkali Metals: Hydrogen is not considered to be an alkali metal. No points will be deducted from me if you included. However, CHEM 120
you will be expected to know this.
Halogens: Astatine and Element 117: Please include these in the halogens.
CHEM 110
Page 4 of 14
PART 2 (NON-CALCULATOR SECTION); SUGGESTED TIME ALLOTMENT: 15 MIN – 25 MIN
Approximately 25 to 35 questions.
31) Rewrite the following types of radiation in order of increasing (shortest to longest) wavelength.
blue light gamma radiation infrared yellow light
gamma blue yellow infrared
32) P: 1s2 2s2 2p6 3s2 3p3 How many valence electrons does phosphorus have? 5
33) Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 How many valence electrons does bromine have? 7
C 34) WOTF (which of the following) is NOT an example of an intensive property?
D 35) WOTF is an NOT an example of a chemical change?
A 36) WOTF is an NOT a metalloid?
C 37) WOTF is an NOT considered a metal?
38) Plum Pudding Model
Question 39 (Answers will vary):
Category 1 (any one of the phase changes): (list a substance) with freezing, melting, vaporization,
condensation, sublimation, or deposition
Category 2 (solubility/miscibility): dissolving salt in water, water and alcohol mixing
Category 3 (physical mutilation): bending, breaking, tearing, flattening (malleability),
stretching (ductility)
So, appropriate answers to Question 39:
1) Water freezing
2) Sugar dissolving in water
3) Cutting a piece of paper in half.
CHEM 110
Page 5 of 14
40) deposition
41) gas
Answer to Question 42
Malleability is the ability to flatten a substance (1 pt). Malleability is a physical property
because a chemical change is not necessary to observe this property (1 pt). Malleability is
an intensive property because this property is not dependent upon the amount of
substance present (1 pt).
CHEM 110
Page 6 of 14
Questions 43 through 48 are True/False questions.
If the statement is true, write a T in the answer box.
If the statement is false:
Write an F in the answer box
In the box, write a word or phrase to replace the bold/italicized word to make the statement true.
F 43) “Atoms of the same element are identical with the same physical properties.”
varying (or different)
F 44) Rutherford conducted the cathode ray tube experiment.
Gold Foil Experiment
F 45) The symbol for atomic mass is Z.
atomic number
T 46) A heterogeneous mixture has inconsistent composition throughout the sample.
T 47) Elements cannot be broken down into simpler substances through chemical reactions.
T 48) An electron has a relative mass of 1/1840 when compared to the mas of a proton.
T 49) The SI base unit for mass is the kilogram.
CHEM 110
Page 7 of 14
D 50) Iodine gas cooling down to become iodine solid. A) chemical change
G 51) Water heating up to become steam. B) chemical property
E 52) Red dye added to water turns the water red. C) condensation
A 53) Two clear colorless liquids turn red when added together. D) deposition
B 54) Iron does not react with nitrogen gas. E) dissolving
H 55) Alcohol is miscible with water. F) kinetic energy
F 56) The energy associated with a ball brick falling to the ground. G) physical change
I 57) Energy stored within the chemical bonds of an explosive. H) physical property
I) potential energy
J) sublimation
CHEM 110
Page 8 of 14
Answer to Question 58
If the Plum Pudding Model were correct (diffuse positive cloud with electrons scattered
throughout), then 100% of the alpha particles should have passed straight through the gold foil (3
pt).
Answer to Question 59
A cathode ray results only when the gas tube is filled with gaseous vapor of any substance (3 pt).
The beam was found to be negatively charged because it was attracted to the positive plate of the
external circuit (2 pt).
alpha particle
source () Au
POWER
SOURCE
− cathode
+ anode
+
−
external positive plate
external negative plate
EXTERNAL
POWER
SOURCE
Partially evacuated vacuum
tube, containing Hg vapor
CHEM 110
Page 9 of 14
61) B
62) A
63) D
64) F
65) C
66) 3 Hz
Answer to Question 67
The Unknown contains Element A and Element E.
Element A: All of the lines in Element A are present in the same position as the lines in the
Unknown.
Element B contains two green lines that are not in the Unknown in the same position. Therefore,
Element B can be eliminated.
Element C contains two orange lines that is completely absent in the Unknown. There are also
two red lines that are not in the same position as the red lines in the Unknown. Therefore,
Element C can be eliminated.
Element D contains a teal (aqua) line that is completely absent in the Unknown. Therefore
Element D cannot be in the unknown. There is also a purple line in Element D that is not in the
same position as any of the purple lines that are in the Unknown.
Element E: All of the lines in Element E are present in the same position as the lines in the
Unknown.
CHEM 110
Page 10 of 14
PART 3 (CALCULATOR SECTION): SUGGESTED TIME ALLOTMENT: 15 MIN – 25 MIN
Question 68) Equation 2: hydrogen(g) + nitrogen(g) ammonia(g)
6.05 g 14.01 g 20.06 g
Use the Law of Conservation of Mass to determine the mass of ammonia. You do not have to show your work.
Question 69) Equation 1: methanol(l) + salicylic acid(g) methyl salicylate(g) + water(l)
32.04 g 138.13 g 152.15 g 18.02
CHEM 110
Page 11 of 14
Composition of Various Isotopes – Table 1
#no Z Atomic Symbol #p+ A #e− Isotope Name
70) 33 27 Co2760
27 60 27 Cobalt-60
71) 115 77 Ir77192 77 192 77 Iridium-192
72) 32 30 Zn3062 30 62 30 Zinc-62
73) Al: 1s2 2s2 2p6 3s2 3p1
74) Ba: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2
75) Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
CHEM 110
Page 12 of 14
76) The table below contains information about the four naturally occurring isotopes of chromium metal. Use
the information to determine the atomic mass of chromium. Remember: for these problems, I want you
to ignore significant figure rules. The final answer should have a total of four significant digits. Show your work.
Isotope Atomic Mass of Isotope Natural Abundance of Isotope
Chromium-50 50 amu 4.34%
Chromium-52 52 amu 83.79%
Chromium-53 53 amu 9.50%
Chromium-54 54 amu 2.37%
Chromium-50: (4.34
100) × 50 amu = 2.17 amu 2.1700 amu
Chromium-52: (83.79
100) × 52 amu = 43.5708 amu 43.5708 amu
Chromium-53: (9.50
100) × 53 amu = 5.035 amu 5.0350 amu
Chromium-54: (2.37
100) × 54 amu = 1.2798 amu + 1.2798 amu
52.0556 amu
Remember the instruction: regardless of significant figure rules, I always want four significant digits only
for atomic mass values.
So: x52.06 amux
You should have confidence that your answer is correct if it is very close to the atomic mass value on the
periodic table.
CHEM 110
Page 13 of 14
77a) Nitroglycerin is a powerful contact explosive. It is 18.53% nitrogen by mass. Write the conversion factor
that can be derived from the percent-by-mass value of nitrogen in nitroglycerin. (The chemical formula
for nitroglycerin is C3H5N3O9. Assume mass values are measured in grams.
18.53 g N = 100 g C3H5N3O9
77b) If a sample of nitroglycerin contains 0.400 grams of nitrogen. Calculate the mass (in grams) of the
entire sample. Show your work.
(0.400 g N
1) (
100 g C3H5N3O9
18.53 g N) = x2.16 g C3H5N3O9x
77c) Nitroglycerin has a density of 1.6 g/mL. Write the conversion factor that can be derived from this
density value.
1.6 g C3H5N3O9 = 1 mL C3H5N3O9
77d) Calculate the mass of nitrogen (in grams) of a 50.00-mL sample of nitroglycerin. Hint: You may need to
use both conversion factors (from Question 77a and Question 77c).
(50.00 mL C3H5N3O9
1) (
1.6 g C3H5N3O9
1 mL C3H5N3O9) (
18.53 g N
100 g C3H5N3O9) = x14.82 g Nx
CHEM 110
Page 14 of 14
77e) A sample of nitroglycerin is poured into a spherical delivery apparatus with a radius of 2.50 cm.
Determine the mass of nitrogen (in grams) of the amount of nitroglycerin that will fill the delivery
apparatus.
Vsphere = (4
3) π(r)3
Helpful Hints
Solve this in pieces
First, determine the volume of the sphere.
The volume of sphere will be equal to the volume of nitroglycerin. Assume 1 cm3 = 1 mL
The volume of nitroglycerin can be converted to mass of nitroglycerin (through density conversion factor)
The mass of nitrogen can then be determined through the %-by-mass conversion factor.
Vsphere = (4
3) π(r)3 = (
4
3) π(2.50 cm)3 = (
4
3) π (15.625 cm3) = 65.4 cm3
(65.4 cm3 sphere
1) (
1 mL sphere
1 cm3 sphere) (
1 mL C3H5N3O9
1 mL sphere) = 65.4 mL C3H5N3O9
(65.4 mL C3H5N3O9
1) (
1.6 g C3H5N3O9
1 mL C3H5N3O9) (
18.53 g N
100 g C3H5N3O9) = x19.4 g Nx
EXTRA CREDIT
I am leaning against extra credit questions for this Exam, but I’m not 100% decided. And the test will be shorter than this practice packet.