Unformatted text preview: 1 Chapter 2 Exercise Key Chapter 2 Exercise Key
Exercise 2.1 – Periodic Table: Complete the following table. Name Symbol Group
Number aluminum Al 13, 3A
or IIIA metal representative element 3 solid silicon Si 14, 4A
or IVA metalloid representative element 3 solid nickel Ni 10, 8B or
VIIIB metal transition metal 4 solid sulfur S 16, 6A
or VIA nonmetal representative element 3 solid fluorine F 17, 7A
or VIIA nonmetal representative element 2 gas cesium Cs 1, 1A
or IA metal representative element 6 solid mercury Hg 12, 2B or
IIB metal transition metal 6 liquid uranium U (No
group
number) metal inner transition metal 7 solid manganese Mn 7, 7B
or VIIB metal transition metal 4 solid strontium Sr 2, 2A
or IIA metal representative element 5 solid bromine Br 17 nonmetal representative element 4 liquid silver Ag 1B metal transition metal 5 solid bismuth Bi VA metal representative element 6 solid carbon C 14 nonmetal representative element 2 solid Copyright 2004 Mark Bishop Metal,
Nonmetal or
Metalloid? Representative Element,
Transition Metal or Inner
Transition Metal? Number
for Period Solid,
Liquid,
or Gas? 2 Chapter 2 Exercise Key Exercise 2.2 – Group Names: Write the name of the group to which each of the following elements belongs.
a. helium
b. Cl noble gases c. magnesium halogens d. Na Exercise 2.3 – Isotope Symbolism:
Symbol alkaline earth metals alkali metals Complete the following table. Atomic
number Mass
number Number
protons Number
neutrons Number
electrons Charge 59
28 Ni 28 59 28 31 28 0 32
16 S2 16 32 16 16 18 2− Pb 2+ 82 207 82 125 80 2+ As 33 75 33 42 33 0 Br − 35 79 35 44 36 − 207
82 75
33
79
35 Exercise 2.4 – MetricMetric Conversion Factors:
include the following metric units.
103 J
1 kJ a. joule and kilojoule b. meter and centimeter
c. liter and gigaliter 102 cm
1m or 1 cm
10−2 m or 1 μg
10−6 g 109 L
1 GL d. gram and microgram 106 μg
1g e. gram and megagram 106 g
1 Mg f. pascal and millipascal Copyright 2004 Mark Bishop Write conversion factors that 103 mPa
1 Pa 3 Chapter 2 Exercise Key Exercise 2.5 – Significant Figures: Identify whether each of the following values is
exact or not. If it is not exact, write the number of significant figures it has.
a. 8.0 in 8.0 mL (derived from a measurement)
Measurements never lead to exact values. Zeros to the right of nonzero
digits and to the right of the decimal point are significant. 8.0 is not exact
and is two significant figures.
b. 80 from 80 desks in a classroom (determined by counting them)
Counting leads to exact values. 80 is exact.
c. 2000 in 2000 lb
1 ton EnglishEnglish conversion factors with units of the same type of
measurement top and bottom come from definitions, so the values in them
are exact. 2000 is exact.
d. 453.6 in 453.6 g
1 lb 2.54 cm
, the values in Englishmetric conversion factors that
1 in.
we will see are calculated and rounded off. They are not exact. 453.6 is
four significant figures.
Except for e. 103 in 103 mg
1g Metricmetric conversion factors with units of the same type of
measurement top and bottom come from definitions, so the values in them
are exact. 103 is exact.
f. 0.1067 in 0.1067 oz (calculated from its measured mass of 3.023 g)
Measurements never lead to exact values. Thus, values calculated from
measurements are not exact. Zeros to the left of nonzero digits are not
significant, and zeros between nonzero digits are significant. 0.1067 is not
exact and is four significant figures.
g. 0.006665 in 0.006665 lb (calculated from the 0.1067 oz described in part f.)
Values calculated from nonexact values are not exact. 0.006665 is not
exact and has four significant figures.
h. 10 in 10% of the tablet desks in a room are for left handed people (determined
by counting 8 lefthanded desks and counting 80 desks total and then
calculating the percent)
Values calculated from exact values and not rounded off are exact. The 10
is exact. Copyright 2004 Mark Bishop 4 Chapter 2 Exercise Key i. 21 from 21% of the desks have initials carved in them (determined by
counting 17 desks with initials and counting 80 desks total and then
calculating the percent)
Even if a value is calculated from exact values, if the answer is rounded
off, the rounded answer is not exact. 17 divided by 80 and multiplied by
100 yields 21.25. Thus, the 21% includes a value that was rounded off.
The 21 is not exact and has two significant figures.
j. 6.00 x 103 from the temperature of the surface of the sun, 6.00 x 103 °C.
The value 6.00 x 103 must have come from a measurement, an estimate, or
a calculation, so it is not exact and has three significant figures.
Exercise 2.6 – Rounding Off Answers Derived from Multiplication and Division: A
first class stamp allows you to send a letter with a mass up to 1/2 ounce. You weigh a
letter and find it has a mass of 10.5 grams. What is its mass in ounces? Report your
answer to the correct significant figures. Can you mail this letter with one stamp? ⎛ lb ⎞ ⎛ 16 oz ⎞
? oz = 10.5 g ⎜
⎟ = 0.370 oz
⎟⎜
⎝ 453.6 g ⎠ ⎝ lb ⎠
Exercise 2.7  Rounding Off Answers Derived from Multiplication and Division:
The reentry speed of the Apollo 10 space capsule was 11.0 km/s. How many hours
would it have taken for the capsule to fall through 25.0 miles of the stratosphere?
5280 ft ⎞ ⎛ 12 in. ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ ⎛ 1 km ⎞ ⎛ 1 s ⎞ ⎛ 1 min ⎞ ⎛ 1 hr ⎞
? hr = 25.0 mi ⎛
⎟⎜
⎟⎜
⎟⎜
⎜
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 1 mi ⎠ ⎝ 1 ft ⎠ ⎝ 1 in. ⎠ ⎝ 102 cm ⎠ ⎝ 103 m ⎠ ⎝ 11.0 km ⎠ ⎝ 60 s ⎠ ⎝ 60 min ⎠ = 1.02 x 10−3 hr
Exercise 2.8  Rounding Off Answers Derived from Addition and Subtraction:
Report the answers to the following calculations to the correct number of decimal
positions. Assume that each number is ±1 in the last decimal position reported.
a. 684 − 595.325 = 89 b. 92.771 + 9.3 = 102.1 Exercise 2.9  Rounding Off Answers: The mass of a liquid can be found by first
weighing a container, adding the liquid to the container, weighing the container and the
liquid, and finding the mass of the liquid by subtracting the mass of the container from
the total mass of container and liquid. A container is found to have a mass of 42.6 g.
When 10.2 mL of a liquid is added to the container, the mass increases to 50.7 g. What is
the density of this substance? The setup for this problem is below. Do the calculation
and report your answer to the correct significant figures.
? g 50.7 g − 42.6 g
=
= 0.79 g/mL
mL
10.2 mL Copyright 2004 Mark Bishop Chapter 2 Exercise Key 5 Exercise 2.10 – Unit Analysis: The average human body contains 13 gallons of water.
What is this volume in quarts? ⎛ 4 qt ⎞
? qt = 13 gal ⎜
⎟ = 52 qt
⎝ 1 gal ⎠
Exercise 2.11 – Unit Analysis: The diameter of a proton is 2 x 10−15 meters. What is
this diameter in nanometers? ⎛ 109 nm ⎞
−6
? nm = 2 x 10−15 m ⎜
⎟ = 2 x 10 nm
⎝ 1m ⎠
Exercise 2.12 – Unit Analysis: The mass of an electron is 9.1093897 x 10−31 kg. What
is this mass in nanograms? ⎛ 103 g ⎞ ⎛ 109 ng ⎞
−19
? lb = 9.1093897 x 10−31 kg ⎜
⎟⎜
⎟ = 9.1093897 x 10 ng
⎝ 1 kg ⎠ ⎝ 1 g ⎠
Exercise 2.13 – Unit Analysis: There are 2035 tons of sulfuric acid used to make JellO
each year. Convert this to kilograms.
2000 lb ⎞ ⎛ 453.6 g ⎞ ⎛ 1 kg ⎞
6
? kg = 2035 tons ⎛
⎟ ⎜ 3 ⎟ = 1.846 x 10 kg
⎜
⎟⎜
⎝ 1 ton ⎠ ⎝ 1 lb ⎠ ⎝ 10 g ⎠
Exercise 2.14 – Unit Analysis: A piece of Styrofoam has a mass of 88.978 g and a
volume of 2.9659 L. What is its density in g/mL?
? g 88.978 g ⎛ 1 L ⎞
=
⎜
⎟ = 0.030000 g/mL
mL 2.9659 L ⎝ 103 mL ⎠ Exercise 2.15 – Unit Analysis: The density of blood plasma is 1.03 g/mL. A typical
adult has about 2.5 L of blood plasma. What is the mass in kilograms of the blood plasma
in this person? ⎛ 103 mL ⎞ ⎛ 1.03 g ⎞ ⎛ 1 kg ⎞
?kg = 2.5 L ⎜
⎟⎜ 3 ⎟
⎟⎜
⎝ 1 L ⎠ ⎝ 1 mL ⎠ ⎝ 10 g ⎠
or 1.03 kg ⎞
?kg = 2.5 L ⎛
⎜
⎟ = 2.6 kg
⎝ 1L ⎠ Exercise 2.16 – Unit Analysis: Pain information is transferred through the nervous
system between 12 and 30 meters per second. If a student drops a textbook on her toe,
how long will it take for the pain information with a velocity of 18 meters per second to
travel 6.0 feet to reach the brain?
12 in. ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ ⎛ 1 s ⎞
? s = 6.0 ft ⎛
⎟⎜
⎟⎜
⎟ = 0.10 s
⎜
⎟⎜
⎝ 1 ft ⎠ ⎝ 1 in. ⎠ ⎝ 102 cm ⎠ ⎝ 18 m ⎠ Copyright 2004 Mark Bishop 6 Chapter 2 Exercise Key Exercise 2.17 – Unit Analysis: An electron takes 6.2 x 10−9 seconds to travel across a
TV set that is 22 inches wide. What is the velocity of the electron in km/hr?
? km
22 in. ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ ⎛ 1 km ⎞ ⎛ 60 s ⎞ ⎛ 60 min ⎞
=
⎜
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎟
hr
6.2 x 10−9 s ⎝ 1 in. ⎠ ⎝ 102 cm ⎠ ⎝ 103 m ⎠ ⎝ 1 min ⎠ ⎝ 1 hr ⎠ = 3.2 x 108 km/hr
Exercise 2.18 – Unit Analysis: The mass of the ocean is about 1.8 x 1021 kg. If the
ocean contains 0.041% by mass calcium ions, Ca2+, how many tons of Ca2+ are in the
ocean? ⎛ 0.041 kg Ca 2+ ⎞ ⎛ 2.205 lb ⎞ ⎛ 1 ton ⎞
? ton Ca 2+ = 1.8 x 1021 kg ocean ⎜
⎟
⎟⎜
⎟⎜
⎝ 100 kg ocean ⎠ ⎝ 1 kg ⎠ ⎝ 2000 lb ⎠
= 8.1 x 1014 ton Ca2+
Exercise 2.19 – Unit Analysis: While you are at rest, your heart pumps about 5.0 liters
of blood per minute. Your brain gets about 15% by volume of your blood. What volume
of blood in liters is pumped through your brain in 1.0 hour of rest?
60 min ⎞ ⎛ 5.0 L total ⎞ ⎛ 15 L to brain ⎞
? L to brain = 1.0 hr ⎛
⎜
⎟⎜
⎟⎜
⎟ = 45 L
⎝ 1 hr ⎠ ⎝ 1 min ⎠ ⎝ 100 L total ⎠ Exercise 2.20 – Unit Analysis: A normal adult has from 4 to 6 million red blood cells
per mm3 of blood. Consider a person with 5.5 L of blood and 5 x 106 red blood cells per
mm3 of blood. How many red blood cells does this person have?
3 ⎛ 103 mL ⎞ ⎛ 1 cm3 ⎞ ⎛ 1 m ⎞ ⎛ 103 mm ⎞ ⎛ 5 x 106 rbc ⎞
? rbc = 5.5 L blood ⎜
⎟⎜
⎟⎜
⎟⎜ 2
⎟⎜
⎟
3
⎝ 1 L ⎠ ⎝ 1 mL ⎠ ⎝ 10 cm ⎠ ⎝ 1 m ⎠ ⎝ 1 mm blood ⎠
3 or ⎛ 103 cm3
? rbc = 5.5 L blood ⎜
⎝ 1L ⎞ ⎛ 10 mm ⎞ ⎛ 5 x 106 rbc ⎞
⎟ ⎜ 1 cm ⎟ ⎜ 1 mm3 blood ⎟
⎠⎝
⎠⎝
⎠ = 3 x 1013 red blood cells Copyright 2004 Mark Bishop 3 7 Chapter 2 Exercise Key Exercise 2.21 – Temperature Conversions:
a. N,Ndimethylaniline, C6H5N(CH3)2, melts at 2.5 °C. What is
N,Ndimethylaniline’s melting point in °F and K?
1.8 °F ⎞
°F = 2.5 °C ⎛
⎜
⎟ + 32 °F = 36.5 °F
⎝ 1 °C ⎠ K = 2.5 °C + 273.15 = 275.7 K
b. Benzenethiol, C6H5SH, melts at 5.4 °F. What is benzenethiol’s melting point
in °C and K?
°C = (5.4 °F − 32 °F) 1 °C
= −14.8 °C
1.8 °F K = −14.8 °C + 273.15 = 258.4 K
c. The hottest part of the flame for a Bunsen burner is found to be 2.15 x 103 K.
What is this temperature in °C and °F? °C = 2.15 x 103 K − 273.15 = 1.88 x 103 °C
1.8 °F ⎞
3
°F = 1.88 x 103 °C ⎛
⎜
⎟ + 32 °F = 3.42 x 10 °F
⎝ 1 °C ⎠ or 3.41 x 103 °F if the unrounded answer to the first calculation is used in
the second calculation. Copyright 2004 Mark Bishop ...
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This note was uploaded on 03/03/2012 for the course CHEM 100 taught by Professor Mark during the Fall '06 term at Monterey Peninsula College.
 Fall '06
 Mark
 Periodic Table

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