CONCEPTUAL EXAMPLE 1.1
Risk–Benefit Analysis
Some medical doctors think heroin is more effective for the relief of severe pain than other medicines.
However, heroin is highly addictive and often renders the user unable to function in society. Do a risk–benefit
analysis for the use of heroin in treating the pain of (a) a young athlete’s broken leg and (b) a terminally ill
cancer patient.
Solution
a.
b.
The heroin would provide the benefit of pain relief, but its use for such purposes has been judged to be
too risky by the U.S. Food and Drug Administration. The DQ is low.
The heroin would provide the benefit of pain relief. The risk of addiction in a dying person is irrelevant.
Heroin is used this way in Great Britain, but it is banned for any purpose in the United States. The DQ is
uncertain. (Both answers involve judgments that are not clearly scientific; people can differ in their
assessments of each.)
Exercise 1.1A
Chloramphenicol is a powerful antibacterial drug that often destroys bacteria unaffected by other drugs. It is
highly dangerous to some individuals, however, causing fatal aplastic anemia in about 1 in 30,000 people. Do a
risk–benefit analysis for the use of chloramphenicol in (a) sick farm animals, from which people might
consume milk or meat with residues of the drug, and (b) a person with Rocky Mountain spotted fever faced
with a high probability of death or permanent disability.
Exercise 1.1B
A four-year clinical trial (6800 participants) of the drug tamoxifen in healthy women at high risk showed a
49% lower rate of breast cancer than a group of 6800 women taking a placebo. Tamoxifen has a low rate of
serious side effects, including potentially fatal blood clots, uterine cancer, hot flashes, loss of libido, and
cataracts. Do a risk–benefit analysis for the use of tamoxifen in (a) all women and (b) women at high risk.
CONCEPTUAL EXAMPLE 1.1
Risk–Benefit Analysis
continued
Exercise 1.1C
Rotavirus is the most common cause of severe diarrhea among children. This virus causes hospitalization of
approximately 55,000 children each year in the United States and the death of over 600,000 children annually
worldwide. A vaccine can prevent most cases. There is a strong association between the vaccine and bowel
obstruction in some infants. Do a risk–benefit analysis for the use of the vaccine in (a) the United States and
(b) worldwide.
CONCEPTUAL EXAMPLE 1.2
Mass and Weight
On the planet Mercury gravity is 0.376 times that on Earth. (a) What would be the mass on Mercury of a person
who has a mass of 62.5 kilograms (kg) on Earth? (b) What would be the weight on Mercury of a person who
weighs 124 pounds (lb) on Earth?
Solution
a.
b.
The person’s mass would be the same (62.5 kg) as on Earth; the quantity of matter has not changed.
The person would weigh only 0.376 x 124 lb = 46.6 lb; the force of attraction between planet and person
is only 0.376 times that on Earth.
Exercise 1.2A
At the surface of Venus, the force of gravity is 0.903 times that on Earth’s surface. (a) What would be the mass
of a standard 1.00-kg object on Venus? (b) A man who weighs 198 lb on Earth would weigh how much on the
surface of Venus?
Exercise 1.2B
On Jupiter, at the boundary between the gaseous atmosphere and the liquid that makes up the bulk of the
planet, the force of gravity is 2.34 times that on Earth. (a) What would be the mass of a 52.5-kg woman at that
location on Jupiter? (b) A man who weighs 212 lb on Earth would weigh how much on Jupiter?
CONCEPTUAL EXAMPLE 1.3
Chemical Change and Physical
Change
Which of the following events involve chemical changes and which involve physical changes?
a.
Your hair is cut.
b.
Lemon juice converts milk to curds and whey.
c.
Water boils.
d.
Water is broken down into hydrogen gas and oxygen gas.
Solution
We examine each change and determine whether there has been a change in composition or structure. In other
words, we ask “Have new substances that are chemically different been created?” If so, the change is chemical;
if not, it is physical.
a.
Physical change: The composition of the hair is not changed by cutting.
b.
Chemical change: The compositions of curds and whey are different from the composition of the
milk.
c.
Physical change: Liquid water and invisible water vapor formed when liquid water boils have the
same composition; the water merely changes from a liquid to a gas.
d.
Chemical change: New substances, hydrogen and oxygen, are formed.
Exercise 1.3A
Which of the following events involve chemical changes and which involve physical changes?
a.
Gasoline vaporizes from an open container.
b.
A piece of magnesium metal burns in air to form a white powder called magnesium oxide.
c.
A dull knife is sharpened with a whetstone.
CONCEPTUAL EXAMPLE 1.3
Chemical Change and Physical
Change continued
Exercise 1.3B
Which of the following events involve chemical changes and which involve physical changes?
a.
A steel wrench left out in the rain becomes rusty.
b.
A stick of butter melts.
c.
A wooden log is burned.
d.
A piece of wood is ground up into sawdust.
CONCEPTUAL EXAMPLE 1.4
Elements and Compounds
Which of the following represent elements and which represent compounds?
C Ca HI BN In HBr
Solution
C, Ca, and In represent elements (each is a single symbol). HI, BN, and HBr are composed of two symbols
each and represent compounds.
Exercise 1.4A
Which of the following represent elements and which represent compounds?
He CuO No NO KI Os
Exercise 1.4B
How many different elements are represented in the entire list of Exercise 1.4A?
EXAMPLE 1.5
Prefixes and Powers of Ten
Convert each of the following measurements to a unit that replaces the power of ten by a prefix.
a.
2.89 x 10–3 g
b.
4.30 x 103 m
Solution
Our goal is to replace each power of ten with the appropriate prefix from Table 1.5. For example, 10 –3 = 0.001,
corresponding to milli(unit). (It doesn’t matter what the unit is; here we are dealing only with the prefixes.)
a.
10–3 corresponds to the prefix milli-; 2.89 mg; that is, 10–3 x 9 (unit) = milli(unit).
b.
103 corresponds to the prefix kilo-; 4.30 km; that is, 103 x (unit) = kilo(unit).
Exercise 1.5
Convert each of the following measurements to a unit that replaces the power of ten by a prefix.
a.
7.24 x 103 g
b.
4.29 x 10–6 m
c.
7.91 x 10-3 s
d.
2.29 x 10–2 g
e.
7.90 x 106 m
EXAMPLE 1.6
Prefixes and Powers of Ten
5 Use exponential notation to express each of the following measurements in terms of an SI base unit.
a.
4.12 cm
b.
947 ms
c.
3.17 nm
Solution
a.
b.
c.
Our goal is to find the power of ten that relates the given unit to the SI base
unit. That is, centi(base unit) = 10-2 x (base unit)
4.12 centimeter = 4.12 x 10-2 m
To change microsecond to the base unit second, we replace the prefix micro- by 10-6.
To obtain an answer in the conventional exponential form, we also need to replace the coefficient
947 by 9.47 x 102. The result of these two changes is
947 ms = 947 x 10-6 s = 9.47 x 102 x 10-6 s = 9.47 x 10-4 s
To change nanometer to the base unit meter, we replace the prefix nano- by 10–9. The answer in
exponential form is 3.17 x 10-9 m.
Exercise 1.6A
Use exponential notation to express each of the following measurements in terms of an SI base unit.
a.
7.45 nm
b.
5.25 ms
d.
1.415 km
e.
2.06 mm
Exercise 1.6B
Use exponential notation to express each of the following measurements in terms of an SI base unit.
a.
284 nm
b.
119 ms
d.
754 km
e.
6.19 x 106 mm
EXAMPLE 1.7
Math, Length, Area, Volume
5 Without doing a detailed calculation, determine which of the following is a reasonable
(a) mass (weight) and (b) height for a typical two-year-old child.
Mass:
10 mg 10 g 10 kg
100 g
Height: 85 mm 85 cm 850 cm 8.5 m
Solution
a.
In customary units, a two-year-old child should weigh about 20–25 lb. Knowing that 1 kg is about 2.2 lb,
the only reasonable answer is 10 kg.
b.
In customary units, a two-year-old child should be about 30–36 in. tall. Knowing that 1 cm is a little less
than 0.5 in., the answer must be a little more than twice 30–36, or a bit more than 70–72 cm. Thus the only
reasonable answer 85 cm.
Exercise 1.7A
Without doing a detailed calculation, determine which of the following is a reasonable area for the front cover
of your textbook.
500 mm2
50 cm2
500 cm2
50 m2
Exercise 1.7B
Without doing a detailed calculation, determine which of the following is a reasonable volume for your
textbook.
1600 mm3
16 cm3
1600 cm3
1.6 m3
EXAMPLE 1.8
Unit Conversions
5 Convert (a) 1.83 kg to grams and (b) 729 mL to milliliters.
Solution
a.
We start with the given quantity 1.83 kg and use the equivalence 1 kg = 1000 g to form a conversion factor
(Appendix A) that allows us to cancel the unit kg and end with the unit g.
1.83 kg x
b.
1000 g
1 kg
x = 1830 g
Here we start with the given quantity 729 mL and use the equivalences 106 mL = 1 L and 103 mL = 1 L to
form conversion factors that allows us to cancel the unit mL and end with the unit mL.
729 mL x
1L
106 mL
x
1000 mL
1L
= 0.729 mL
Exercise 1.8
Convert (a) 7.5 m to millimeters, (b) 2056 mL to liters, (c) 2.06 g to micrograms, and (d) 0.738 cm to
millimeters.
EXAMPLE 1.9
Mass, Volume, and Density
5 Answer the following without doing a detailed calculation. (a) Which has the greater volume, a 50.0-g block
of copper or a 50.0-g block of gold? (b) Which has the greater mass, 225 mL of ethyl alcohol or 225 mL
of hexane?
Solution
a.
From Table 1.6 we see that the density of gold (19.3 g/cm3) is greater than that of copper (8.94 g/cm3). It
takes a larger block of copper to have a mass of 50.00 g than of gold. A 50.0-g block of copper has a
greater volume than a 50.0-g block of gold.
b.
From Table 1.6 we see that the density of ethyl alcohol (0.789 g/mL) is greater than that of hexane (0.660
g/mL). Because it is more dense (that is, it has more mass in each unit of volume), 225 mL of ethyl alcohol
has a greater mass than does 225 mL of hexane.
Exercise 1.9A
Without doing a detailed calculation, determine which has the greater volume, a 500.0-g block of ice or a
500.0-g block of magnesium.
Exercise 1.9B
Wood does not dissolve in water and will float on water if its density is less than that of water. Padouk
(d = 0.86 g/cm3) and ebony (d = 1.2g/cm3) are tropical woods. Which will float on water and which will sink
in water?
EXAMPLE 1.10
Density from Mass and Volume
5 What is the density of iron if 156 g of iron occupies a volume of 20.0 cm3?
Solution
The given quantities are
m = 156 g
We can use the equation that defines density.
V = 20.0 cm3
and
mass (m)
d=
m
V
volume (V)
=
156 g
= 7.80 g/cm3
20.0 cm3
density (d)
Exercise 1.10A
What is the density, in grams per milliliter, of salt solution if 52.5 mL has a mass of 58.5 g?
Exercise 1.10B
What is the density, in grams per cubic centimeter, of a metal alloy if a cube that measures 2.00 cm on an edge
has a mass of 80.2g?
EXAMPLE 1.11
Mass from Density and Volume
5 What is the mass of 1.00 L of gasoline if its density is 0.703 g/mL?
Solution
We can express density as a ratio, 0.703 g/1.00 mL, and use it as a conversion factor. We also need the factor
1000 mL/1 L to convert to milliliters.
m=dxV=
0.703 g
1 mL
x1Lx
1000 mL
1L
= 703 g
Exercise 1.11A
What is the mass of 200.0 mL of glycerol, which has a density of 1.264 g/mL at 20 °C?
Exercise 1.11B
What is the mass of a lead brick that is 2.50 cm x 3.50 cm x 8.25 cm on an edge? (The density of lead at 20 °C
is 11.34 g/cm3.)
EXAMPLE 1.12
Volume from Mass and Density
5 What volume is occupied by 461 g of mercury? (See Table 1.6)
Solution
Here we can express the inverse of density as a ratio, 1.00 mL/13.6 g, and use it as a conversion factor.
V = 461 g x
1 mL
13.534 g
= 34.1 mL
Exercise 1.12A
What volume is occupied by a 845-g piece of magnesium? (See Table 1.6.)
Exercise 1.12B
What volume is occupied by 225 g of hexane? (See Table 1.6.)
EXAMPLE 1.13
Temperature Conversions
5 Ether boils at 36 °C. What is the boiling point of ether on the Kelvin scale?
Solution
K = °C + 273.15
K = 36 + 273.15 = 309 K
Exercise 1.13A
What is the boiling point of water (100 °C) expressed in kelvins?
Exercise 1.13B
Express a temperature of –78 °C in kelvins.
EXAMPLE 1.14
Energy Conversions
5 When 1.00 g of gasoline burns, it yields about 10.3 kcal of energy. What is the quantity of energy in
kilojoules (kJ)?
Solution
10.3 kcal x
1000 cal
1 kcal
x
4.184 J
1 cal
x
1 kJ
1000 J
= 43.1 kJ
Exercise 1.14A
A European woman on average consumes food with an energy content of 7525 kJ per day. What is her daily
intake in kilocalories (food calories)?
Exercise 1.14B
It takes about 12 kJ of energy to melt a single ice cube. How many kilocalories does it take to melt three
ice cubes?