Chemistry 4th Edition Solution Manual

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Solve your textbook questions with ease using Chemistry 4th Edition Solution Manual, a comprehensive and easy-to-follow guide.

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Chapter 1
Chemistry: The Central Science
Practice Problems C
1.1 (iii)
1.2 (i) and (i)
1.3 pink liquid = grey solid < blue solid < yellow liquid < blue liquid < green solid
1.4 physical: (iii), chemical: (i), (ii) is neither
1.5 12 blue cubes, infinite number of significant figures;
2 × 10 1 red spheres, one significant figure
1.6 2.4 × 10 2 lbs.
1.7 2.67 g/cm3
1.8 (a) 4 red blocks/1 object
(b) 1 object/1 yellow block
(c) 2 white blocks/1 yellow block
(d) 1 yellow block/6 grey connectors
1.9 375 red bars; 3500 yellow balls
Applying What You’ve Learned
a) The recommended storage-temperature range for cidofovir is 20°C−25°C.
b) The density of the fluid in a vial is 1.18 g/mL. (The density should be reported to three significant figures.)
c) The recommended dosage of cidofovir for a 177-lb man is 4 × 102 mg or 0.4 g.
d) 1.18 × 10 3 g/L, 1.18 × 10 3 kg/m3 .

Chapter 1 -- Chemistry: The Central Science 2
Questions and Problems
1.1 Chemistry is the study of matter and the changes that matter undergoes.
Matter is anything that has mass and occupies space.
1.2 The scientific method is a set of guidelines used by scientists to add their experimental results to the
larger body of knowledge in a given field. The process involves observation, hypothesis,
experimentation, theory development, and further experimentation.
1.3 A hypothesis explains observations. A theory explains data from accumulated experiments and
predicts related phenomena.
1.4 a. Hypothesis – This statement is an opinion.
b. Law – Newton’s Law of Gravitation.
c. Theory – Atomic Theory.
1.5 a. Law – Newton’s 2nd Law of Motion.
b. Theory – Big Bang Theory.
c. Hypothesis – It may be possible but we have no data to support this statement.
1.6 a. O and H b. C and H c. H and Cl d. N
1.7 a. C and O b. F and H c. N and H d. O
1.8 a. Matter is anything that has mass and occupies space. Examples include air, seawater, concrete, an
automobile, or a dog.
b. A substance is a form of matter that has definite (constant) composition and distinct properties.
Examples include iron, silver, water, or sugar.

Chapter 1 -- Chemistry: The Central Science 3
c. A mixture is a combination of two or more substances in which the substances retain their distinct
identities. Examples include milk, salt water, air, or steel.
1.9 Examples of homogeneous mixtures: apple juice or root beer.
Examples of heterogeneous mixtures: chocolate chip cookie or vinaigrette salad dressing.
1.10 Examples of elements (see the front cover for a complete list): oxygen, platinum, sodium, cobalt.
Examples of compounds: sugar, salt, hemoglobin, citric acid.
An element cannot be separated into simpler substances by chemical means. A compound can be separated
into its constituent elements by a chemical reaction.
1.11 There are 118 known elements.
1.12 Li: Lithium
F: Fluorine
P: Phosphorus
Cu: Copper
As: Arsenic
Zn: Zinc
Cl: Chlorine
Pt: Platinum
Mg: Magnesium
U: Uranium
Al: Aluminum
Si: Silicon
Ne: Neon
1.13 a. K (potassium)
b. Sn (tin)
d. B (boron)
e. Ba (barium)
g. S (sulfur)
h. Ar (argon)

Chapter 1 -- Chemistry: The Central Science 4
c. Cr (chromium) f. Pu (plutonium) i. Hg (mercury)
1.14 a. hydrogen: element
b. water: compound
c. gold: element
d. sugar: compound
1.15 a. The sea is a heterogeneous mixture of seawater and biological matter, but seawater, with the biomass
filtered out, is a homogeneous mixture.
b. element
c. compound
d. homogeneous mixture
e. heterogeneous mixture
f. homogeneous mixture
g. heterogeneous mixture
1.16 a. liquid b. gas c. mixture d. solid
1.17 a. element b. compound c. compound d. element
1.18 a. Chemistry Units: meter (m), centimeter (cm), millimeter (mm)
SI Base Unit: meter (m)
b. Chemistry Units: cubic decimeter (dm 3 ) or liter (L), milliliter (mL), cubic centimeter (cm3 )
SI Base Unit: cubic meter (m3
)

Chapter 1 -- Chemistry: The Central Science 5
c. Chemistry Units: gram (g)
SI Base Unit: kilogram (kg)
d. Chemistry Units: second (s)
SI Base Unit: second (s)
e. Chemistry Units: kelvin (K) or degrees Celsius (°C)
SI Base Unit: kelvin (K)
1.19 a. 1 × 10 6
b. 1 × 10 3
c. 1 × 10−1
d. 1 × 10−2
e. 1 × 10−3
f. 1 × 10−6
g. 1 × 10−9
h. 1 × 10−12
1.20 For liquids and solids, chemists normally use g/mL or g/cm3 as units for density.
For gases, chemists normally use g/L as units for density. Gas densities are generally very low, so the
smaller unit of g/L is typically used. 1 g/L = 0.001 g/mL.
1.21 Weight is the force exerted by an object or sample due to gravity. It depends on the gravitational force
where the weight is measured.
Mass is a measure of the amount of matter in an object or sample. It remains constant regardless of where it
is measured.
Since gravity on the moon is about one sixth that on Earth,
  1
weight on the moon 168 lbs on Earth 6
 
  
  28 lbs
1.22 Kelvin is known as the absolute temperature scale, meaning the lowest possible temperature is 0 K.
The units of the Celsius and Kelvin scales are the same, so conversion between units is a matter of addition:
K C 273.15  
The freezing point of water is defined as 0°C. The boiling point of water is defined as 100°C.
In the Fahrenheit scale, the freezing point of water is 32°F and the boiling point of water is 212°F. Since the
difference is 180°F, compared to 100°C between the freezing and boiling points of water, one degree
Fahrenheit represents a smaller change in temperature than one degree Celsius. To convert between these

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Chapter 1 -- Chemistry: The Central Science 6
two temperature scales, use:
  5 C
temperature in Celsius temperature in F 32 F 9 F

   
1.23 Strategy: Use the density equation:
m
d V

Solution: 586 g
188 mL
m
V
   3.12 g / mLd
1.24 mass of ethanol ൌ ଴.଻ଽ଼ ௚
ଵ ௠௅ ൈ 205 ܮ ݉ൌ ૚૟૝ ࢍ
1.25 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between
Celsius and Fahrenheit given in Section 1.3 of the text. Substitute the temperature values given
in the problem into the appropriate equation.
Setup: Conversion from Fahrenheit to Celsius:
5 C
C ( F 32 F) 9 F

      
Conversion from Celsius to Fahrenheit:
9 F
F C 32 F
5 C
 
      
 
Solution: a. 5 C
C (95 F 32 F) 9 F

      
 35 C
b. 5 C
C (12 F 32 F) 9 F

       
 11 C
c. 5 C
C (102 F 32 F) 9 F

      
 39 C

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Chapter 1 -- Chemistry: The Central Science 7
d. 5 C
C (1852 F 32 F) 9 F

      
 1011 C
e. 9 F
F 273.15 C 32 F
5 C
 
        
  459.67 F 
1.26 a.   5 C
C 105 F 32 F 9 F

      
 41°C
b. 9 F
F 11.5 C 32 F
5 C
 
        
  11.3°F
c. 3 9 F
F 6.3 10 C 32 F
5 C
 
        
 
4
1.1×10 °F
1.27 Strategy: Use the density equation.
Solution: volume of water =ܸ ൌ ௠
ௗ ൌ ଶ଻.଴ ௚
଴.ଽଽଶ ௚ ௠௅⁄ ൌ 27.2ܮ݉
1.28
3
87.6 g
volume of platinum 21.5 g/cm
  3
4.07 cm
1.29 Strategy: Use the equation for converting °C to K.
Setup: Conversion from Celsius to Kelvin:
K = °C + 273.15
Solution: a. K = 115.21C + 273.15 = 388.36 K
b. K = 37°C + 273 = 3.10 × 10 2 K
AMPS Solution Co

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Chapter 1 -- Chemistry: The Central Science 8
c. K = 357°C + 273 = 6.30 × 102 K
Note that when there are no digits to the right of the decimal point in the original temperature,
we use 273 instead of 273.15.
1.30 a. °C = K – 273 = 77 K – 273 = –196°C
b. C = 4.22 K  273.15 = 268.93C
c. C = 600.61 K  273.15 = 327.46C
AMPS Solution Co
1.31 The picture on the right best illustrates the measurement of the boiling point of water using the
Celsius and Kelvin scales. A temperature on the Kelvin scale is numerically equal to the temperature
in Celsius plus 273.15.
1.32 The relative densities of the aluminum differ based upon the volume the shapes occupy. The larger
and flatter ball spreads its mass over a larger area, creating a larger volume and lowering the overall
density relative to the same volume of water causing the material to float. The smaller rolled up ball
minimizes its mass over a small volume creating a larger overall density relative to the same volume of
water, causing it to sink. The density of the aluminum remains constant.
1.33 Qualitative data does not require explicit measurement. Quantitative data requires measurement and
is expressed with a number.
1.34 Physical properties can be observed and measured without changing the identity of a substance. For
example, the boiling point of water can be determined by heating a container of water and measuring the
temperature at which the liquid water turns to steam. The water vapor (steam) is still H2 O, so the identity of
the substance has not changed. Liquid water can be recovered by allowing the water vapor to contact a cool
surface, on which it condenses to liquid water.
Chemical properties can only be observed by carrying out a chemical change. During the
measurement, the identity of the substance changes. The original substance cannot be recovered by any
physical means. For example, when iron is exposed to water and oxygen, it undergoes a chemical change to
produce rust. The iron cannot be recovered by any physical means.
1.35 An extensive property depends on the amount of substance present. An intensive property is
independent of the amount of substance present.

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Chapter 1 -- Chemistry: The Central Science 9
1.36 a. extensive
b. extensive
c. intensive
d. extensive
1.37 a. Quantitative. This statement involves a measurable distance.
b. Qualitative. This is a value judgment. There is no numerical scale of measurement for artistic
excellence.
c. Qualitative. If the numerical values for the densities of ice and water were given, it would be a
quantitative statement.
d. Qualitative. The statement is a value judgment.
e. Qualitative. Even though numbers are involved, they are not the result of measurement.
1.38 a. Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are
changed.
b. Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter
(different composition).
c. Physical property. The measurement of the boiling point of water does not change its identity or
composition.
d. Physical property. The measurement of the densities of lead and aluminum does not change their
composition.
e. Chemical property. When uranium undergoes nuclear decay, the products are chemically different
substances.
1.39 a. Physical Change. The material is helium regardless of whether it is located inside or outside the balloon.
b. Chemical change in the battery.

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Chapter 1 -- Chemistry: The Central Science 10
c. Physical Change. The orange juice concentrate can be regenerated by evaporation of the water.
d. Chemical Change. Photosynthesis changes water, carbon dioxide, etc., into complex organic matter.
e. Physical Change. The salt can be recovered unchanged by evaporation.
1.40 Mass is extensive and additive: 44.3 + 115.2 = 159.5 g
Temperature is intensive: 10°C
Density is intensive: 1.00 g/mL
1.41 Mass is extensive and additive: 37.2 + 62.7 = 99.9 g
Temperature is intensive: 20°C
Density is intensive: 11.35 g/cm3
1.42 a. Exact. The number of tickets is determined by counting.
b. Inexact. The volume must be measured.
c. Exact. The number of eggs is determined by counting.
d. Inexact. The mass of oxygen must be measured.
e. Exact. The number of days is a defined value.
1.43 Using scientific notation avoids the ambiguity associated with trailing zeros.
1.44 Significant figures are the meaningful digits in a reported number. They indicate the level of
uncertainty in a measurement. Using too many significant figures implies a greater certainty in a
measured or calculated number than is realistic.
1.45 Accuracy tells us how close a measurement is to the true value. Precision tells us how close multiple
measurements are to one another. Having precise measurements does not always guarantee an
accurate result, because there may be an error made that is common g to all the measurements.

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Chapter 1 -- Chemistry: The Central Science 11
1.46 a. The decimal point must be moved eight places to the right, making the exponent –8.
0.000000027 = 2.7 × 10 –8
b. The decimal point must be moved two places to the left, making the exponent 2.
356 = 3.56 × 102
c. The decimal point must be moved four places to the left, making the exponent 4.
47,764 = 4.7764 × 104
d. The decimal point must be moved two places to the right, making the exponent –2.
0.096 = 9.6 × 10–2
1.47 Strategy: To convert an exponential number N × 10n to a decimal number, move the decimal n places to
the left if n < 0, or move it n places to the right if n > 0. While shifting the decimal, add place-
holding zeros as needed.
Solution: a. 1.52  10 –2 = 0.0152
b. 7.78  10 –8 = 0.0000000778
c. 1 × 10 –6 = 0.000001
d. 1.6001 × 103 = 1600.1
1.48 a.  1
145.75 2.3 10 145.75 0.23
     2
1.4598×10
b. 4
2 2
79500 7.95×10
2.5×10 2.5×10
  2
3.2×10
c.        3 4 3 3
7.0 10 8.0 10 7.0 10 0.80 10   
        3
6.2×10

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Chapter 1 -- Chemistry: The Central Science 12
d.    4 6
1.0×10 9.9×10  10
9.9×10
1.49 a. Addition using scientific notation.
Strategy: A measurement is in scientific notation when it is written in the form N × 10n, where 0 ≤ N <
10 and n is an integer. When adding measurements that are written in scientific notation,
rewrite the quantities so that they share a common exponent. To get the “N part” of the
result, we simply add the “N parts” of the rewritten numbers. To get the exponent of the
result, we simply set it equal to the common exponent. Finally, if need be, we rewrite the
result so that its value of N satisfies 0 ≤ N < 10.
Solution: Rewrite the quantities so that they have a common exponent. In this case, choose the
common exponent n = –3.
0.0095 = 9.5×10 –3
Add the “N parts” of the rewritten numbers and set the exponent of the result equal to the
common exponent.
3
3
3
9.5 10
8.5 10
18.0 10




 

Rewrite the number so that it is in scientific notation (so that 0 10N  ).
18.0  10 –3 = 1.8  10 –2
b. Division using scientific notation.
Strategy: When dividing two numbers using scientific notation, divide the “N parts” of the numbers in
the usual way. To find the exponent of the result, subtract the exponent of the devisor from
that of the dividend.
Solution: Make sure that all numbers are expressed in scientific notation.
653 = 6.53  10 2
Divide the “N parts” of the numbers in the usual way.
6.53  5.75 = 1.14
Subtract the exponents.

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Chapter 1 -- Chemistry: The Central Science 13
1.14  10 +2 – (–8) = 1.14  10 +2 + 8 = 1.14  1010
c. Subtraction using scientific notation.
Strategy: When subtracting two measurements that are written in scientific notation, rewrite the
quantities so that they share a common exponent. To get the “N part” of the result, we
simply subtract the “N parts” of the rewritten numbers. To get the exponent of the result, we
simply set it equal to the common exponent. Finally, if need be, we rewrite the result so that
its value of N satisfies 0 10N  .
Solution: Rewrite the quantities sot that they have a common exponent. Rewrite 850,000 in such a way
that n = 5.
850,000 = 8.5  10 5
Subtract the “N parts” of the numbers and set the exponent of the result equal to the common
exponent.
5
5
5
8.5 10
9.0 10
0.5 10

 
 
Rewrite the number so that 0 10N  (ignore the sign of N when it is negative).
0.5  10 5 = 5  10 4
d. Multiplication using scientific notation.
Strategy: When multiplying two numbers using scientific notation, multiply the “N parts” of the
numbers in the usual way. To find the exponent of the result, add the exponents of the two
measurements.
Solution: Multiply the “N parts” of the numbers in the usual way.
3.6  3.6 = 13
Add the exponents.
13  10 –4 + (+6) = 13  10 2
Rewrite the number so that it is in scientific notation (so that 0 10N  ).

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Chapter 1 -- Chemistry: The Central Science 14
13  10 2 = 1.3  10 3
1.50 a. four
b. two
c. five
d. two, three, or four
e. three
f. one
g. one
h. two
1.51 a. one
b. three
c. three
d. four
e. three
f. one
g. one or two
1.52 a. 10.6 m b. 0.79 g c. 16.5 cm
1.53 a. Division
Strategy: The number of significant figures in the answer is determined by the original number having
the smallest number of significant figures.
Solution: 7.310 km
5.70 km  1.283
The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three
significant digits. Therefore, the answer has only three significant digits.
The correct answer rounded off to the correct number of significant figures is:
1.28
Think
About It: Why are there no units?

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Chapter 1 -- Chemistry: The Central Science 15
b. Subtraction
Strategy: The number of significant figures to the right of the decimal point in the answer is
determined by the lowest number of digits to the right of the decimal point in any of the
original numbers.
Solution: Writing both numbers in the decimal notation, we have
0.00326 mg
0.0000788 mg
0.0031812 mg
The bold numbers are nonsignificant digits because the number 0.00326 has five digits to the
right of the decimal point. Therefore, we carry five digits to the right of the decimal point in
our answer.
The correct answer rounded off to the correct number of significant figures is:
0.00318 mg = 3.18×10 –3 mg
c. Addition
Strategy: The number of significant figures to the right of the decimal point in the answer is
determined by the lowest number of digits to the right of the decimal point in any of the
original numbers.
Solution: Writing both numbers with exponents = +7, we have
(0.402  10 7 dm) + (7.74  10 7 dm) = 8.14  10 7 dm
Since 7.74  10 7 has only two digits to the right of the decimal point, two digits are carried to
the right of the decimal point in the final answer.
1.54 Student A’s results are neither precise nor accurate. Student B’s results are both precise and
accurate. Student C’s results are precise but not accurate.
1.55 Tailor Z’s measurements are the most accurate. Tailor Y’s measurements are the least accurate.
Tailor X’s measurements are the most precise. Tailor Y’s measurements are the least precise.

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