CHEMISTRY CP
Name:________________________________
Per. ____
REVIEW FOR UNIT ONE TEST- ANSWER KEY
Introductory Material: Scientific Method, Scientific Notation, Density
1. You will need to know the Metric Conversion Chart for this test. As a sample problem, give the numerical value
for each of the prefixes:
A. Mega 1,000,000
E. deci 0.1
B. Kilo 1,000
F. centi 0.01
C. Hecto 100
G. milli 0.001
D. Deka 10
2. Metric Conversions: Give the value of the following in the units indicated.
a. 6.92 mm = 0.00692 m
b. 8 Kg = 80,000 dg
c. 35 dam = 0.35 Km
d. 690 ml = 0.690 L
e. .068 m = 68 mm
f. 934 cm = 0.0934 hm
g. .55 L = 550 mL
h. 6.45 Kg = 6,450,000 mg
i. 75 mg = 0.075g
3. Be able to recognize which step of the scientific method is being described from different examples. Remember
the basic steps are the following:
a) observation, b) hypothesis, c) experimentation or testing, d) collecting or recording data, e) drawing
conclusions.
As an example, which step of the scientific method would each of the following be?
(1) After years of testing and analyzing data a scientist determines that chemicals in a city’s drinking water
are causing an increase in the intelligence of children in the city. E- Drawing Conclusions
(2) A biology student notices many frog tadpoles in a particular pond have two tails. A- Observation
(3) The biology student prepares tables and graphs of all the information she has collected. D - Collecting or
recording data
(4) The biology student guesses that waste water from a sewage treatment plant which empties into the pond
is causing the two tailed tadpoles. B - Hypothesis
(5) The biology student obtains some freshly laid frog eggs and keeps some in the pond water and others in
distilled water while waiting for them to hatch. C – Experimentation or testing
Problems: Calculate the following using correct units and significant digits.
4. A metal has a volume of 4.2 cm3 and a mass of 16.0 g. What is its density?
D = M/V = (16.0 g/ 4.2 cm3) = 3.8 g/cm3
5. An object has a mass of 75 g and a volume of 80.0 cm3. What is its density?
D = M/V = 75 g / 80.0 cm3) = .94 g/cm3
6. An object has a density of 4.5 g/cm3 and a mass of 20.6 g. What is its volume?
V = M/D = 20.6 g/ 4.5 g/cm3 = 4.6 cm3
7. Do the following multiplication and division problems and round off to the proper number of significant
digits in the final answer.
A. 342÷16 = 21.375 = 21
B. 2300÷110 = 20.90909091 = 21
C. (23.002) (.0560) = 1.288122 =1.29
D. (84.00) (9001) = 756084 =756100
Perform the following operations and express your answer in proper scientific notation with the proper
number of significant digits and proper units.
8. (4.1×103cm) (3.6×10-4cm) = 14.76 × 10-1 cm2 = 1.5 × 100 cm2 OR 1.5 cm2
9. 1.6×10-4g = 0.2 × 10-2 g/dm3 = 2 × 10-3 g/dm3
8×10-2dm3
10. (6.1×10-4Kg) (4.4×10-2m) = 26.84 × 10-6 = 2.3 × 10-4 kg•m
(4.2×103s) (2.79×10-5s)
11.718 × 10-2
s2
11. 7.0×106m3 = 4.7 × 101 m2
1.5×105m
12. Terms to Know:
A. Conservation of Mass: Mass can neither be created nor destroyed. Their total amounts remain constant
B. Chemistry (the four part definition): Chemistry is the study of matter, its structure, reactions with other
matter and the energy involved
C. Volume: Amount of space that an object occupies
D. Density: Comparison of the amount of matter in an object to the space it occupies; mass per unit
volume
E. Weight: Pull of gravity on an object (can change depending on where you are)
F. Mass: Amount of matter in an object (will not change depending on where you are)
G. Accuracy: How closely individual measurements agree with the true value
H. Precision: How close individual measurements agree with each other
I. Mole: Number of basic particles of a substance in a gram molecular mass of that substance 6.02 × 1023
particles
VOLUME VS MASS READINGS FOR SAMPLES OF
SUBSTANCE X
a. What is the volume 20 grams of
substance X would have? 10 mL
20
30
b. What is the mass 15ml of substance X
would have? 30 g
c. Calculate the approximate slope of the
line. y2-y1/x2-x1 = 40-30/20-15 = 2
10
Mass in grams
40
50
13. Be able to read and interpret data off a line graph.
5
10
15
20
25
30
35
d. Calculate the approximate density of
the substance X. 2 g/mL
Volume in mL
Moles and Mole Conversions
1. What is Avogadro’s Number? 6.02 × 1023
2. How many particles are in 1 Mole of carbon? 6.02 × 1023 particles
3. What is the mass that is equal to 1 Mole of Argon (Ar)? 39.40 g
4. Complete the following Mass to Mole Conversions: DIVIDE BY MOLAR MASS
a. 56 g of carbon (C)
b. .36 grams of helium (He)
c. 225 grams of gold (Au)
𝒈
𝒈
𝒈
a. 56 g/12.01 𝑴𝒐𝒍𝒆 = 4.7 M
b. .36 g/4.00 𝑴𝒐𝒍𝒆 = 0.090 M
c. 225 g/197 𝑴𝒐𝒍𝒆 = 1.14 M
5. Complete the following Mole to Mass Conversions: MULTIPLY BY MOLAR MASS
a. 5 moles oxygen (O)
b. 11.5 moles of neon (Ne)
c. .35 moles of potassium (K)
𝒈
𝒈
𝒈
a. 5 Mole × 16.01 𝑴𝒐𝒍𝒆 = 80 g
b. 11.5 Mole × 20.18 𝑴𝒐𝒍𝒆 = 232 g c. .35 Mole × 39.10 𝑴𝒐𝒍𝒆 = 14 g
6. Complete the following Molecules to Mole Conversions: DIVIDE BY AVOGADRO’S NUMBER
a. 1.01 × 1024 molecules of hydrogen
b. 5.85 × 1025 molecules of helium
23
6.02 × 10 moleules
6.02 × 1023 moleules
1
0
a. 0.168 × 10 Moles  1.68 × 10 Moles
b. 0.972 × 102 Moles 9.72× 101 Moles
7. Complete the following Mole to Molecules Conversions: MULTIPLY BY AVOGADRO’S NUMBER
a. 25.4 moles of carbon dioxide
b. 6.85 moles of krypton
a. 25.4 Moles × (6.02 × 1023molecules)
b. 6.85 Moles × (6.02 × 1023molecules)
1 Mole
1 Mole
152.91 × 1023molecules = 1.53 × 1025molecules
41.237 × 1023molecules = 4.12 × 1024 molecules
Structure of Matter, Physical/Chemical Properties and States of Matter
1. Which of the following are chemical changes and which are physical changes?
a. melting iron physical
e. receiving a tattoo chemical
b. burning hydrogen chemical
f. tearing paper physical
c. cutting a piece of wood physical
g. exploding dynamite chemical
d. digesting a Big Mac chemical
h. evaporating salt water physical
2. Identify each of the following as being elements, compounds, homogeneous mixtures (solutions) or
heterogeneous mixtures. ***Nonfat milk is homogeneous and Full fat milk is heterogeneous***
a. iron element
h. ocean water homogeneous mixture
b. dirt heterogeneous mixture
i. lemonade homogeneous mixture
c. rubbing alcohol compound
j. water compound
d. salt water homogeneous mixture
k. corn syrup homogeneous mixture
e. Italian salad dressing heterogeneous mixture
l. granite heterogeneous mixture
f. Nonfat milk homogeneous mixture
g. mercury element
3. Identify each of the following as either chemical properties or physical properties. If they are physical properties
identify them as either intensive or extensive.
a. burning in air chemical
f. volume physical – extensive
b. boiling point physical - intensive
g. density physical - intensive
c. malleability physical - intensive
h. shape physical - intensive
d. color physical - intensive
i. odor physical - intensive
e. melting point physical - intensive
j. reactivity/stability chemical
4. Know the difference between the three states of matter in terms of shape, volumes, compressibility, and particle
spacing.
Solid
Liquid
Gas
Description
-Definite shape
-Takes shape of container
-Fills container
-Definite volume
-Constant volume
-Changeable volume
-No compressibility
-No compressibility
-Easily compressed
-Very little motion/
-Particles slide/flow past
-Particles move freely
Vibrate in place
each other
5. Know the four different methods for separating mixtures mentioned in the book: filtration, distillation,
crystallization and chromatography.
♦ Filtration: a separation technique that uses a porous barrier to separate a solid from a liquid
(example is sand and water mixture through filter paper and funnel)
♦ Crystallization: a separation technique that results from the formation of pure solid particles of a
substance from a solution contained a dissolved substance (example is rock candy forming from sugar
water)
♦
♦
Chromotography: a technique that separates components from a mixture, the mobile phase, as they
travel across the surface of another material, stationary phase (example is black ink separates into
different components as it spreads across filter paper)
Distillation: a separation technique that is based on the differences in boiling points of the
substances involved (example is distilling water to remove impurities)
6. Conservation of mass problems:
a. From a process that splits water to form hydrogen and oxygen, a scientist collected 20.0 grams of
hydrogen and 159 grams of oxygen. How much water was originally involved in the process?
Water  hydrogen + oxygen
x grams =
20.0 grams + 159 grams
x grams = 179 grams
b. In a flask, 20.3 grams of aluminum reacted with 200.0 grams of bromine to form aluminum bromide.
After the reaction, no aluminum remained and 17 grams of bromine remained un-reacted. How many
grams of bromine reacted? How many grams of aluminum bromide were formed?
Aluminum + bromine
 aluminum bromide
20.3 grams
+ 200.0 grams =
x grams
20.3 grams + 200.0 grams = 220.3 grams
Bromine(reacted) = Bromine(used) – Bromine(remained) = 200.0 g – 17 g = 183 grams
Aluminum bromide(formed) = Aluminum + Bromine(reacted) = 20.3 g + 183 g = 203.3 grams
7. Percent by mass problems:
a. A 156 gram sample of an unknown compound contains 25 grams of hydrogen. What is the percent
by mass of hydrogen in the compound?
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
25 𝑔𝑟𝑎𝑚𝑠
% by mass = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 × 100 = 156 𝑔𝑟𝑎𝑚𝑠 × 100 = 16%
b. If 7.0 grams of substance X reacts with 21 grams of substance Y to form the compound XY, what is
the percent by mass of each element in the compound?
X + Y = XY
7.0 g + 21 g = 28 g
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
% by mass X = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 × 100 =
% by mass Y =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
× 100 =
7.0 𝑔𝑟𝑎𝑚𝑠
28 𝑔𝑟𝑎𝑚𝑠
× 100 = 25%
21 𝑔𝑟𝑎𝑚𝑠
28 𝑔𝑟𝑎𝑚𝑠
× 100 = 75% OR (100% - 25% = 75%
c. Two unknown compounds are tested. Compound 1 contains 90.0 grams of hydrogen and 180 grams
of oxygen. Compound 2 contains 45 grams of hydrogen and 90.0 grams of oxygen. Are the
compounds the same?
Compound 1: 90.0 grams + 180 grams = 270 grams
90.0 𝑔𝑟𝑎𝑚𝑠
× 100 = 33% hydrogen
100% - 33% = 67% oxygen
270 𝑔𝑟𝑎𝑚𝑠
Compound 2: 45 grams + 90.0 grams = 135 grams
45 𝑔𝑟𝑎𝑚𝑠
× 100 = 33% hydrogen
100% - 33% = 67% oxygen
135 𝑔𝑟𝑎𝑚𝑠
YES, THE COMPOUNDS ARE THE SAME
8. Law of Definite and Multiple Proportions
a. A sample of a certain lead compound contains 6.46 g of lead for each gram of oxygen. A second
sample has a mass of 68.54 g and contains 28.76 g of oxygen. Are the two samples the same?
Compound 1: Set up a lead to oxygen ratio
6.46 𝑔𝑟𝑎𝑚𝑠
= 6.46
1 𝑔𝑟𝑎𝑚𝑠
Compound 2: 68.54 grams – 28.76 grams = 39.78 grams lead
Set up a lead to oxygen ratio
39.78 𝑔𝑟𝑎𝑚𝑠
=1.38 Because the ratios are NOT equal, these are NOT the same compound
28.76 𝑔𝑟𝑎𝑚𝑠
b.
Two different compounds are formed by the elements carbon and oxygen. The first compound
contains 42.9% by mass carbon and 57.1% by mass oxygen. The second compound contains 27.3%
by mass carbon and 72.7% by mass oxygen. Show that the data are consistent with the Law of
Multiple Proportions.
Compound 1: Set up a carbon to oxygen ratio
42.9 𝑔𝑟𝑎𝑚𝑠
57.1 𝑔𝑟𝑎𝑚𝑠
= 0.751
Compound 2: Set up a lead to oxygen ratio
27.3 𝑔𝑟𝑎𝑚𝑠
72.7 𝑔𝑟𝑎𝑚𝑠
= 0.376
Set up a ratio of Compound 1 to Compound 2:
𝐶𝑂𝑀𝑃 1
𝐶𝑂𝑀𝑃 2
=
0.751
0.376
= 𝟐: 𝟏