Solutions manual for Chemistry a molecular approach, 4th Edition. 2017. Kathleen Thrush Shaginaw by digitaldownload87

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Contents

piudent Guide to Using: This Solutions antial.

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Chapter 10

Chemical Bonding II: Molecular Shapes, Valence Bond Theory, PTV ee it Eee 1 OO oteretryt ert. ecsexcnancavt ghar connsng cx nmrener cen ess tneo ceils donee ack 381

Chapter 11

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Student Guide to Using This Solutions Manual

struggling student and the advanced The vision of this solutions manual is to provide guidance that is useful for both the

student.

given. This will help in the review of the An important feature of this solutions manual is that answers for the review questions are . major concepts in the chapter.

problem includes Given, Find, The format of the solutions very closely follows the format in the textbook. Each mathematical Conceptual Plan, Solution, and Check sections.

Conceptual Plan: The conceptual plan shows a step-by-step method to solve the problem. In many cases, the given quantities need to be converted to a different unit. Under each of the arrows is the equation, constant, or conversion factor needed to complete this portion of the problem. In the “Problems by Topic” section of the end-of-chapter exercises, the oddnumbered and even-numbered problems are paired. This allows you to use a conceptual plan from an odd-numbered problem in this manual as a starting point to solve the following even-numbered problem. Students should keep in mind that the examples shown are one way to solve the problems. Other mathematically equivalent solutions may be possible.

Given and Find: Many students struggle with taking the written problem, parsing the information into categories, and determining the goal of the problem. It is also important to know which pieces of information in the problem are not necessary to solve the problem and if additional information needs to be gathered from sources such as tables in the textbook.

5.45

Given: m (CO,) = 28.8 g, P = 742 mmHg, and T = 22°C

Find: V

Conceptual Plan: °C > K and mmHg — atm and g > mol then, P, K = "C=

27305

Solution: 7)| —22°@ 1 mol

n = 28.8 ¢ X 44.01

1 atm

1 mol

760 mmHg

2730S) ="295 Ke P =

44.01 g

PV =

T—> V nRT

1 atm

7 - — 0.97631 6316 atm 742 mmaBX << —__—__

= 0.654397 mol PV = nRT

Rearrange to solve for V.

i ° atm

0.654397 mol X 0.08206 xX 295 K a mol: K

0.976316 atm

16.2L

Check: The units (L) are correct. The magnitude of the answer (16 L) makes sense because one mole of an ideal gas under standard conditions (273 K and | atm) occupies 22.4 L. Although these are not standard conditions, they are close

enough for a ballpark check of the answer. Because this gas sample contains 0.65 mol, a volume of 16 L is reasonable.

Solution: The Solution section walks you through solving the problem after the conceptual plan. Equations are rearranged to solve for the appropriate quantity. Intermediate results are shown with additional digits to minimize round-off error. The units are

Check: The Check section confirms that the units in the answer are correct. This section also challenges the student to think about whether the magnitude of the answer makes sense. Thinking about what is a reasonable answer can help uncover errors such as calculation errors.

canceled in each appropriate step.

Matter, Measurement, and Problem Solving Review Questions 1.1

“The properties of the substances around us depend on the atoms, ions, or molecules that compose

them” means that the specific types of atoms and molecules that compose something tell us a great deal about which properties to expect from a substance. A material composed of only sodium and chloride ions will have the properties of table salt. A material composed of molecules with one carbon atom and two oxygen atoms will have the properties of the gas carbon dioxide. If the atoms and molecules change, so do the properties that we expect the material to have. The main goal of chemistry is to seek to understand the behavior of matter by studying the behavior of atoms and molecules. The scientific approach to knowledge is based on observation and experiment. Scientists observe and perform experiments on the physical world to learn about it. Observations often lead scientists to formulate a hypothesis, a tentative interpretation or explanation of their observations. Hypotheses are tested by experiments, highly controlled procedures designed to generate such observations. The results of an experiment may support a hypothesis or prove it wrong—in which case the hypothesis must be modified or discarded. A series of similar observations can lead to the development of scientific law, a brief statement that summarizes past observations and predicts future ones. One or more wellestablished hypotheses may form the basis for a scientific theory. A scientific theory is a model for the way nature is and tries to explain not merely what nature does, but why. The Greek philosopher Plato (427-347 B.c.) took an opposite approach. He thought that the best way to learn about reality was not through the senses, but through reason. He believed that the physical world was an imperfect representation of a perfect and transcendent world (a world beyond space and time). For him, true knowledge came, not through observing the real physical world, but through reasoning and thinking about the ideal one.

1.4

A hypothesis is a tentative interpretation or explanation of the observed phenomena. A law is a concise statement that summarizes observed behaviors and observations and predicts future observations. A theory attempts to explain why the observed behavior is happening.

1.5

Antoine Lavoisier studied combustion and made careful measurements of the mass of objects before and after burning them in closed containers. He noticed that there was no change in the total mass of material within the container during combustion. Lavoisier summarized his observations on combustion with the law of conservation of mass, which states, “In a chemical reaction, matter is neither cre-

ated nor destroyed.”

1.6

John Dalton formulated the atomic theory of matter. Dalton explained the law of conservation of mass, as well as other laws and observations of the time, by proposing that matter was composed of small, indestructible particles called atoms. Because these particles were merely rearranged in chemical changes (and not created or destroyed), the total amount of mass would remain the same.

The statement “that is just a theory” is generally taken to mean that there is no scientific proof behind the statement. This statement is the opposite of the meaning in the context of the scientific theory, where theories are tested again and again.

Chapter 1 Matter, Measurement, and Problem Solving 2 SO 1.8

to its composition. Matter can be classified according to its state—solid, liquid, or gas—and according

Re,

and molecules in a In solid matter, atoms or molecules pack close to each other in fixed locations. Although the atoms and rigid shape. solid vibrate, they do not move around or past each other. Consequently, a solid has a fixed volume relative In liquid matter, atoms or molecules pack about as closely as they do in solid matter, but they are free to move to each other, giving liquids a fixed volume but not a fixed shape. Liquids assume the shape of their container. one another, In gaseous matter, atoms or molecules have a lot of space between them and are free to move relative to

making gases compressible. Gases always assume the shape and volume of their container.

1.10

Solid matter may be crystalline, in which case its atoms or molecules are arranged in patterns with long-range, repeating order, or it may be amorphous, in which case its atoms or molecules do not have any long-range order. A pure substance is composed of only one type of atom or molecule. In contrast, a mixture i$ a substance composed of two or more different types of atoms or molecules that can be combined in variable proportions. A compound is composed of two

An element is a pure substance that cannot be decomposed into simpler substances. or more elements in fixed proportions,

1.13

A homogeneous mixture has the same composition throughout, while a heterogeneous mixture has different compositions in different regions.

1.14

If a mixture is composed of an insoluble solid and a liquid, the two can be separated by filtration, in which the mixture is poured through filter paper (usually held in a funnel). Mixtures of miscible liquids (substances that easily mix) can usually be separated by distillation, a process in which the mixture is heated to boil off the more volatile (easily vaporizable) liquid. The volatile liquid is then recondensed in a condenser and collected in a separate flask.

1.16

A physical property is one that a substance displays without changing its composition, whereas a chemical property is one that a substance displays only by changing its composition via a chemical change.

Changes that alter only state or appearance, but not composition, are called physical changes. The atoms or molecules that compose a substance do not change their identity during a physical change. For example, when water boils, it changes its state from a liquid to a gas, but the gas remains composed of water molecules; so this is a physical change. When sugar dissolves in water, the sugar molecules are separated from each other, but the molecules of sugar and water remain intact. In contrast, changes that alter the composition of matter are called chemical changes. During a chemical change, atoms rearrange, transforming the original substances into different substances. For example, the rusting of iron, the combustion of natural gas to form carbon dioxide and water, and the denaturing of proteins when an egg is cooked are examples of chemical changes. In chemical and physical changes, matter often exchanges energy with its surroundings. In these exchanges, the total energy is always conserved; energy is neither created nor destroyed. Systems with high potential energy tend to change in the direction of lower potential energy, releasing energy into the surroundings.

1.19

Chemical energy is potential energy. It is the energy that is contained in the bonds that hold the molecules together. This energy arises primarily from electrostatic forces between the electrically charged particles (protons and electrons) that compose atoms and molecules. Some of these arrangements—such as the one within the molecules that compose gasoline—have a much higher potential energy than others. When gasoline undergoes combustion, the arrangement of these particles changes, creating molecules with much lower potential energy and transferring a great deal of energy (mostly in the form of heat) to the surroundings. A raised weight has a certain amount of potential energy (dependent on how high the weight is raised) that can be converted to kinetic energy when the weight is released.

1.20

The SI base units include the meter (m) for length, the kilogram (kg) for mass, the second (s) for time, and the Kelvin

(K) for temperature.

The three different temperature scales are Kelvin (K), Celsius (°C), and Fahrenheit (°F). The size of the degree is the same in the Kelvin and the Celsius scales, and they are 1.8 times larger than the degree size for the Fahrenheit scale. Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving a r 1:22

Prefix multipliers are used with the standard units of measurement to change the value of the unit by powers of 10. For example, the kilometer has the prefix “kilo,” meaning 1000 or 10°, Therefore: | kilometer = 1000 meters = 10° meters Similarly, the millimeter has the prefix “milli,” meaning 0.001 or 1Oa%

1 millimeter = 0.001 meters =

107° meters

1.23

A derived unit is a combination of other units. Examples of derived units include speed in meters per second (m/s), volume in meters cubed (m°*), and density in grams per cubic centimeter (g/cm?*).

1.24

The density (d) of a substance is the ratio of its mass (m) to its volume ():

Mas : sei =" Density = ee Volume

The density of a substance is an example of an intensive property, one that is independent of the amount of the substance. Mass is one of the properties used to calculate the density of a substance. Mass, in contrast to density, is an extensive property, one that depends on the amount of the substance. 1.25

An intensive property is a property that is independent of the amount of the substance. An extensive property is a property that depends on the amount of the substance.

1.26

Measured quantities are reported so that the number of digits reflects the uncertainty in the measurement. The non-place-holding digits in a reported number are called significant figures.

1.27

In multiplication or division, the result carries the same number of significant figures as the factor with the fewest significant figures.

1.28

In addition or subtraction, the result carries the same number of decimal places as the quantity with the fewest decimal places.

1.29

When rounding to the correct number of significant figures, round down if the last (or leftmost) digit dropped is four or less and round up if the last (or leftmost) digit dropped is five or more.

1.30

Accuracy refers to how close the measured value is to the actual value. Precision refers to how close a series of measurements are to one another or how reproducible they are. A series of measurements can be precise (close to one another in value and reproducible) but not accurate (not close to the true value).

1.31

Random error is error that has equal probability of being too high or too low. Almost all measurements have some degree of random error. Random error can, with enough trials, average itself out. Systematic error is error that tends toward being either too high or too low. Systematic error does not average out with repeated trials.

L.32

Using units as a guide to solving problems is often called dimensional analysis. Units should always be included in calculations; they are multiplied, divided, and canceled like any other algebraic quantity.

Problems by Topic

The Scientific Approach to Knowledge 1.33

1.34

L355

(a) (b) (c)

This statement is a theory because it attempts to explain why. It is not possible to observe individual atoms. This statement is an observation. This statement is a law because it summarizes many observations and can explain future behavior.

(d)

This statement is an observation.

(a)

This statement is an observation.

(b)

This statement is a law because it summarizes many observations and can explain future behavior.

(c)

This statement is an observation.

(d)

This statement is a theory because it attempts to explain why.

(a) (b)

If we divide the mass of the oxygen by the mass of the carbon, the result is always 4/3. If we divide the mass of the oxygen by the mass of the hydrogen, the result is always 16.

(c)

and ]6:1, These observations suggest that the masses of elements in molecules are ratios of whole numbers [4:3

Chapter 1 Matter, Measurement, and Problem Solving 4 Oi 1.36

galaxies with fragments Many hypotheses may be developed. One hypothesis is that a large explosion generated are still moving away from each other.

that

The Classification and Properties of Matter We)

(a)

(b) (c)

(d)

Vegetable soup is a heterogeneous mixture of broth, chunks of vegetables, and extracts from the vegetables.

(a)

Wine is a generally homogeneous mixture of water, ethyl alcohol, and other components from the grapes. In some cases, there may be sediment present; so it would be a heterogeneous mixture.

(b)

Beef stew is a heterogeneous mixture of thick broth with chunks of beef and vegetables. Iron is a pure substance that is an element (element 26 in the periodic table). Carbon monoxide is a pure substance that is a compound (two or more elements bonded together).

139

1.40

Sweat is a homogeneous mixture of water, sodium chloride, and other components. Carbon dioxide is a pure substance that is a compound (two or more elements bonded together). Aluminum is a pure substance that is an element (element 13 in the periodic table).

Substance

Pure or Mixture

Type (element or compound)

aluminum apple juice hydrogen peroxide chicken soup

pure mixture pure mixture

element neither—mixture compound neither—mixture

Substance

Pure or Mixture

Type (element or compound)

water coffee ice carbon

pure mixture pure pure

compound neither—mixture compound element

(a) (b) (c)

pure substance that is a compound (one type of molecule that contains two different elements) heterogeneous mixture (two different molecules that are segregated into regions) homogeneous mixture (two different molecules that are randomly mixed)

(d)

pure substance that is an element (individual atoms of one type)

1.42

(a) (b) (c) (d)

pure substance that is an element (individual atoms of one type) homogeneous mixture (two different molecules that are randomly mixed) pure substance that is a compound (one type of molecule that contains two different elements) pure substance that is a compound (one type of molecule that contains two different elements)

1.43

(a) (b) (c) (d) (e)

physical property (color can be observed without making or breaking chemical bonds) chemical property (flammability must be observed by making or breaking chemical bonds) physical property (the phase can be observed without making or breaking chemical bonds) physical property (density can be observed without making or breaking chemical bonds) physical property (mixing does not involve making or breaking chemical bonds, so this can be observed without making or breaking chemical bonds)

1.44

(a) (b) (c) (d) (e)

physical property (color can be observed without making or breaking chemical bonds) physical property (odor can be observed without making or breaking chemical bonds) chemical property (reactivity must be observed by making or breaking chemical bonds) chemical property (decomposition involves breaking bonds, so bonds must be broken to observe this property) physical property (the phase of a substance can be observed without making or breaking chemical bonds)

1.45

(a)

chemical property (burning involves breaking and making bonds, so bonds must be broken and made to observe this property) physical property (shininess is a physical property and so can be observed without making or breaking chemical bonds) physical property (odor can be observed without making or breaking chemical bonds) chemical property (burning involves breaking and making bonds, so bonds must be broken and made to observe this property)

1.41

(b) (c) (d)

Chapter 1 Matter, Measurement, and Problem Solving 1.46

(a) (b) (c)

(d) 1.47

physical property (vaporization is a phase change and so can be observed without making or breaking chemical bonds) physical property (sublimation is a phase change and so can be observed without making or breaking chemical bonds) chemical property (rusting involves the reaction of iron with oxygen to form iron oxide; observing this process involves making and breaking chemical bonds) physical property (color can be observed without making or breaking chemical bonds)

(d)

chemical change (new compounds are formed as methane and oxygen react to form carbon dioxide and water) physical change (vaporization is a phase change and does not involve the making or breaking of chemical bonds) chemical change (new compounds are formed as propane and oxygen react to form carbon dioxide and water) chemical change (new compounds are formed as the metal in the frame is converted to oxides)

1.48

(a) (b) (c) (d)

chemical change (new compounds are formed as the sugar burns) physical change (dissolution is a phase change and does not involve the making or breaking of chemical bonds) physical change (this is simply the rearrangement of the atoms) chemical change (new compounds are formed as the silver converts to an oxide)

1.49

(a)

(Cc)

physical change (vaporization is a phase change and does not involve the making or breaking of chemical bonds) chemical change (new compounds are formed) physical change (vaporization is a phase change and does not involve the making or breaking of chemical bonds)

(a)

physical change (vaporization of butane is a phase change and does not involve the making or breaking of

(b)

chemical change (new compounds are formed as the butane combusts) physical change (vaporization of water is a phase change and does not involve the making or breaking of chemical bonds)

(a)

(b) (c)

(b)

chemical bonds)

(c)

Units in Measurement 1.51

(a)

To convert from °F to °C, first find the equation that relates these two quantities. °C =

1.8 answer this in reported digits of number The Note: into the equation and compute the answer. follows significant figure conventions, covered in Section 1.6. °C = = —* = = =a

(b)

;

Now substitute °F

To convert from K to °F, first find the equations that relate these two quantities.

K = °C + 273.15 and°C = a” then Because these equations do not directly express K in terms of °F, you must combine the equations and

solve the equation for °F. Substituting for °C: ; PP ia 2e K = Rai

+ 273.15; rearrange K — 273.15 = mie :

the rearrange 1.8(K — 273.15) = (°F — 32); finally, °F = 1.8(K — 273.15) + 32. Now substitute K into equation and compute the answer.

°F = 1,8(77 — 273.15) + 32 = 1.8(-196) + 32 = —353 + 325 -321°F then convert °C to °F. Alternatively, you could first convert to K to °C, using the formula K = 273.15 + °C, and =—196.15 °C. Then substitute So, 77 K = 273.15 + °C. Rearranging to solve for °C we get °C = 77K — 273.15

. 32 aE sae Multiplying both sides the temperature in °C in the formula °C = z i = getting —196.15 °C = eee 32 to both sides of the equaby 1.8, we get —196.15 °C x 1.8 = °F — 32 and so —353.07 °C = °F — 32. Adding tion we get °F

(c)

=—321.07 °F or —321 °F, with the proper number of significant figures.

two quantities. °C = To convert from °F to °C, first find the equation that relates these

Now substitute °F into the equation and compute the answer.

22109 FP SZ 1.8

ee i8

Chapter 1 Matter, Measurement, and Problem Solving To convert from °F to K, first find the equations that relate these two quantities.

(d)

27415 aod (C=

pr c= 32 18

equations and then Because these equations do not directly express K in terms of °F, you must combine the rRys=152 solve the equation for K. Substituting for °C: K = bana AEM Now substitute °F into the equation and compute the answer.

0; = 32 K= ee

: + 273A5 = oe HQ

= 31026

TS 1S ="37.0' + 273s ‘

Alternatively, you could first convert to °F to °C, using the formula

ini 32

“C = ere

and then convert

0816 °F 25320006616 °F ,. = 37.0 °C. Then ria i8 ; ’ The nha 32 Toconvert from °F to °C, first find the equation that relates these two quantities. “C = oreen

°C to, K.usinig, Ka 273,156 GSO, 6 Ce K = 273.15. 74.37,0-C = 310.26. 1352

(a)

Now substitute °F into the equation and compute the answer. Note: The number of digits reported in this answer follows significant figure conventions, covered in Section 1.6. _ 212 Poet us 100.°C

1°)

1.8

(b)

Begin by finding the equation that relates the quantity that is given (°C) and the quantity you are trying to find (K): K = °C + 273.15. Becatise this equation gives the temperature in K directly, simply substitute the correct

(c)

To convert from K to °F, first find the equations that relate these two quantities:

value for the temperature in °C and compute the answer. C=

K = 22°C + 273.15 = 295K

equations and then solve the equation for °F. Substituting

for °C: K = ade 3

+ 273.15; rearrange

E> 32 , 273:15'= 73 Teattange 1:8(Ko> 273.15) =) (CF ="32); rearrange "F =@(K

— 273.15) +32.

Now substitute K into the equation and compute the answer. °F = 1,8(0.00K — 273.15) + 32 = 1.8(—273.15 K) + 32 = —491.67 + 32 = —459.67°F Begin by finding the equation that relates the quantity that is given (°C) and the quantity you are trying to find (K): K = °C + 273.15. Because this equation does not directly express °C in terms of K, you must solve the equation for °C: °C = K — 273.15. Now substitute K into the equation and compute the answer. =i lad an Oo == 2 O42

(d)

LSS

K = °C + 273.15 and

32 : io Because these equations do not directly express K in terms of °F, you must combine the . Pe rs°32

To convert from °F to °C, first find the equation that relates these two quantities: °C =

°F S 32

. Now substitute °F

into the equation and compute the answer. Note: The number of digits reported in this answer follows significant figure S00 = ope wet BE) conventions, covered in Section 1.6.°C = 8 wei O20 G

Begin by finding the equation that relates the quantity that is given (°C) and the quantity you are trying to find (K). K = °C + 273.15. Because this equation gives the temperature in K directly, simply substitute the correct value for the temperature in °C and compute the answer. K = —62.2°C + 273.15 = 210.9K 1.54

“Eo Se . a To convert from °F to °C, first find the equation that relates these two quantities: °C = Aaa a Now substitute °F into the equation and compute the answer. Note: The number of digits reported in this answer follows significant figure conventions, covered in Section 1.6.

Bel 34ers

oot

102

rung = 56.6667°C! = $6,7°C

Begin by finding the equation that relates the quantity that is given (°C) and the quantity you are trying to find (K): K = °C + 273.15. Because this equation gives the temperature in K directly, simply substitute the correct value for the temperature in °C and compute the answer. K = 56,6667 °C + 273.15 = 329.8K

Chapter 1 Matter, Measurement, and Problem Solvingee

|Wee:

Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions. (a) 10 is equivalent to “nano,” so 1.2 X 10°? m = 1.2 nanometers = 1.2 nm. 10-'° is equivalent to “femto,” so 22 X 107!5s = 22 femtoseconds = 22 fs. (b) (c) 10° is equivalent to “giga,” so 1.5 X 10’?g = 1.5 gigagrams = 1.5 Gg. 10° is equivalent to “mega,” so 3.5 X 10°L = 3.5 megaliters = 3.5 ML. (d)

1.56

Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions.

1.57

(a)

38.8 X 10°g= 3.88 X 10° g; 10° is equivalent to “mega,” so 3.88 X 10°g = 3.88 megagrams = 3.88 Mg.

(bh)

“552%

(c)

23.4 X 10'!'m = 2.34 x 10! m; 10!isequivalent to “tera,” so 2.34 X 10'2m = 2.34 terameters = 2.34 Tm.

(d)

10 “"s = 5.52 101

= 5.52 ns: 10" is equivalent to “nano,” so 5.52 X 10°? s = 5.52 nanoseconds

107

10, ie equivalent to “micro,” so 8.79 X 10° L = 8.79 microliters = 8.79 ules

870 & WT

Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions. (a)

10° is equivalent to “nano,” so 4.5 ns = 4.5 nanoseconds

(b)

10°’ is equivalent to “femto,” so 18 fs = 18 femtoseconds = 18 X 1075s = 1.8 X 10°!*s.

= 4.5 X 10°’s.

Remember that in scientific notation, the first number should be smaller than 10.

(c)

107! is equivalent to “pico,” so 128 pm = 128 X 10°'?m = 1.28 X 107! m. Remember that in scientific notation, the first number should be smaller than 10.

(d)

10° is equivalent to “micro,” so 35 zm = 35 micrometers = 35 X 10 °m = 3.5 X 10° m. Remember that in scientific notation, the first number should be smaller than 10.

1.58

Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions.

(a) (b) (c) (d)

1.59

(b)

“w’ is equivalent to “micro,” or 10~°, so 35 wL = 35 microliters = 35 X 10 °L = 3.5 X 10° L. Remember that in scientific notation, the first number should be smaller than 10. “M” is equivalent to “mega” or 10°, so 225Mm = 225 megameters = 225 X 10°m = 2.25 X 10°m. Remember that in scientific notation, the first number should be smaller than 10. “T” is equivalent to “tera” or 10'*, so 133 Tg = 133 teragrams = 133 X 10's = 1.33 10 *e, Remember that in scientific notation, the first number should be smaller than 10. “c” is equivalent to “centi” or 10°, so 1.5cg = 1.5 centigrams = 1.5 X 10° ¢

Given: 515km_ Find: dm Conceptual Plan: km — m — 1000m

10dm

1 km

1000 ni 10 dm 1 ken x =——— ca

ion: 515 kati x Solution:

5.15

X 10° dm

Check: The units (dm) are correct. The magnitude of the answer (10°) makes physical sense because a decimeter is a much smaller unit than a kilometer. Given: 515km_ Find: cm Conceptual Plan: km —> m— 1000 m

ion: 515 kai x Solution:

100 dm

1000af '100cm 1 kan

= 5.15 X 10’cm

Check: The units (cm) are correct. The magnitude of the answer (10’) makes physical sense because a centimeter is a much smaller unit than a kilometer or a decimeter. (c)

Given: 122.355 s

Find: ms

Conceptual Plan: s — ms

Solution: 122.355 ¥ < =

Gia 1.22355 X 10° ms

a Check: The units (ms) are correct. The magnitude of the answer (10°) makes physical sense because second. a than unit millisecond is a much smaller Given: 122.355s

Find: ks

Conceptual Plan: s — ks Tks 1000s

Chapter 1 Matter, Measurement, andoaProblem Solving

ena Solution: 122.355

KS_ = 1.29355 x 107! ks = 0.122355 ks

1000 s

because a Check: The units (ks) are correct. The magnitude of the answer (10-') makes physical sense second. a kilosecond is a much larger unit than (d)

Given: 3.345 kJ

Find: J

Conceptual Plan: kJ > J 1000J LkJ

Solution: 3.345 kJ x

1000 J

= 3.345 x 10°J

Check: The units (J) are correct. The magnitude of the answer (10°) makes physical sense because a joule is a much smaller unit than a kilojoule. Given: 3.345 X 10°J (from above)

Conceptual Plan:

Find: mJ

J > mJ 1000 mJ 1J

Solution: 3.345 X 10°J X

1000 mJ

= 3,345 X 10° mJ

Check: The units (mJ) are correct. The magnitude of the answer (10°) makes physical sense because a millijoule is a much smaller unit than a joule. 1.60

(a)

Given: 355 km/s_ Find: cm/s Conceptual Plan: km/s m/s — 1000m 1 km

ion: Solution

355 kan

cm/s

100cm 1m

1000 mi

100 cm

= 3.55 x 10’ —s

Check: The units (cm/s) are correct. The magnitude of the answer (10’) makes physical sense because a centimeter is a much smaller unit than a kilometer.

Given: 355 km/s__ Find: m/ms Conceptual Plan: km/s — m/s —

Solution: bate

355kan Les

1000 m

1km

1000 ms

m/ms

= 1080m iLss x x = Lkn 1000 ms 3

— ms

Check: The units (m/ms) are correct. The magnitude of the answer (10°) makes physical sense because the conversion to meters increases the magnitude by a factor of 1000 and the conversion from seconds to milliseconds decreases the magnitude by a factor of 1000.

(b)

Given: 1228 g/L Find: g/mL Conceptual Plan: g/L > g/mL LL 1000 mL

1228 g

Solution: Sree ail Le

eeaOO0dd

= 1.223 = B als

Check: The units (g/mL) are correct. The magnitude of the answer (1) makes physical sense because a milliliter is a much smaller unit than a liter.

Given: 1228 g/L Find: kg/ML Conceptual Plan: g/L > kg/L — l kg

1000g

Solution:

1228

1¥

kg/ML

10°L

1 ML

= 1.228 x 10°——£ © 1000g ~ 1 ML prelate sid

é x

x a

Check: The units (kg/ML) are correct. The magnitude of the answer (10°) makes physical sense because the conversion to kilograms decreases the magnitude by a factor of 1000 and the conversion from liters to megaliters increases the magnitude by a factor of 10°. Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving a (c)

Given: 554 mK/s__

Find: K/s Conceptual Plan: mK/s > K/s lK 1000 mK

Solution:

554 mK

K 2a

IES 1000 mK Ome S Check: The units (K/s) are correct. The magnitude of the answer (107!) makes physical sense because a millikelvin is a much smaller unit than a kelvin.

Given: 554 mK/s__ Find: 4K/ms Conceptual Plan: mK/s + K/s > LK

yK/s > 1s

1000 ms

1000 mK

Solution:

554 mK ls

ss 1000 mK

yK/ms

10 °K 10°

ph 1K

“e = 1000 ms

BUS ms

Check: The units (44K /ms) are correct. The magnitude of the answer (10°) makes physical sense because the conversion to microkelvins increases the magnitude by a factor of 1000 and the conversion from seconds to milliseconds decreases the magnitude by a factor of 1000. (d)

Given: 2.544 mg/mL Find: g/L Conceptual Plan: mg/mL — g/mL —> g/L lg 1000 mL 1000 mg

Solution:

Lids

2.554

1 1000 mg < g ~ ioe = 2.5542 I mL 1000 mg ib It.

Check: The units (g/L) are correct. The magnitude of the answer (3) makes physical sense because the conversion to grams decreases the magnitude by a factor of 1000 and the conversion from milliliters to liters increases the magnitude by a factor of 1000.

Given: 2.544 mg/mL Find: g/mL Conceptual Plan: mg/mL — g/mL —> pg/mL lg 1000 mg

2.554 mg n

10° ng lg

“ 10° ug

Solution ion:

Mg 2.554 xX 10°—

1 mL 1000 mg lg Check: The units (44g/mL) are correct. The magnitude of the answer (10°) makes physical sense because a microgram is a much smaller unit than a milligram.

1.61

(a)

Given: 254,998 m_

Find: km

Conceptual Plan:

m —

1 km

1000 m ion: 2 254,998mi a Solution:

m 1000 nf = 2.54998

x 10*km = 254.998 km

Check: The units (km) are correct. The magnitude of the answer (10) makes physical sense because a kilometer is a much larger unit than a meter. (b)

Given: 254,998 m_ Find: Mm Conceptual Plan: m — Mm 1Mm

10°m-

Solution: 254,998 wi x <= = 2.54998 x 107! Mm = 0.254998 Mm Check: The units (Mm) are correct. The magnitude of the answer (10-') makes physical sense because a megameter is a much larger unit than a meter or a kilometer. (c)

Given: 254,998 m_

Find: mm

Conceptual Plan:

m —

1000 mm 1m

1000 mm = 2.54998 X 10® mm 1 nf sense because a Check: The units (mm) are correct. The magnitude of the answer (10°) makes physical

Solution: 254,998ni x

- millimeter is a much smaller unit than a meter.

Measurement, and Problem Solving

Chapter 1 Matter, 10 MS (d)

Find: cm

Given: 254,998 m_

Conceptual Plan: m —

100 cm Im

100 cm = 2.54998 x 10’cm ia

Solution: 254,998 ni X<

a Check: The units (cm) are correct. The magnitude of the answer (10’) makes physical sense because centimeter is a much smaller unit than a meter, but larger than a millimeter. 1.62

(a)

Given: 556.2 X 10°'*s

Find: ms

Conceptual Plan: s —

1000 ms ls

Solution: 556.2 X 10° ¢ x

1000 ms

= 5.562 X 10’ ms

Check: The units (ms) are correct. The magnitude of the answer Koma) makes physical sense because a millisecond is a much smaller unit than a second. (b)

Given: 556.2 x 10°25

Find: ns

Conceptual Plan: s — ns 10° ns ls

Solution: 556.2 < 10°!?¥ x

10° ns

= 0.5562 ns

Check: The units (ns) are correct. The magnitude of the answer (0.6) makes physical sense because a nanosecond is a much smaller unit than a second.

(c)

Given: 556.2 X 107s

Find: ps

Conceptual Plan: s —

ps 10!ns ls

* 10 12 ps Solution: 556.2 X 10°!?¢ x = 556.2 ps lg

Check: The units (ps) are correct. The magnitude of the answer (10°) makes physical sense because a picosecond is a much smaller unit than a second.

(d)

Given: 556.2 X 10°!2s

Find: fs

Conceptual Plan: s —

10!fs ls

1045 fs

Solution: 556.2 X 107!2 x mite = 5.562 X 10°fe Check: The units (fs) are correct. The magnitude of the answer (10°) makes physical sense because a femtosecond is a much smaller unit than a second.

1.63

Given: | square meter (1 m’)

Find: cm?

Conceptual Plan: 1 m? > cm? 100 cm 1m

Notice that for squared units, the conversion factors must be squared.

Solution: | ai

(100 em)? = | X 10°wae cm*

xX———~—

(1 nf)?

Check: The units of the answer are correct, and the magnitude makes sense. The unit centimeter is smaller than a meter, so the value in square centimeters should be larger than in square meters.

1.64

Given: 4 cm on each edge cube Find: cm* Conceptual Plan: Read the information given carefully. The cube is 4 cm on each side.

lLLweh av V=Il1wh

inacubel=w=h

Chapter 1 Matter, Measurement, and Problem Solving

ea ig ee seems cnc

cep

semaine

orn

Solution: 4cm X 4cm X 4cm = (4cm)? = 64 cm? =60 cm? Check: The units of the answer are correct, and the magnitude makes sense. The unit 4 cm is larger than | cm, so the value in cubic centimeters should be larger.

Density 1.65

Given: m = 2.49 g; V = 0.349cm? Conceptual Plan: m, V > d

Find: d in g/cm? and compare to pure copper.

d=m/V

Compare to the published value: d (pure copper) = 8.96 g/cm*. (This value is in Table 4.) me

Solution: d = Se tia = 7.13 a 0.349 cm cm The density of the penny is much smaller than the density of pure copper (7.13 g/cm’; 8.96 g/cm’), so the penny is not pure copper. Check: The units (g/cm>) are correct. The magnitude of the answer seems correct. Many coins are layers of metals, SO it is not surprising that the penny is not pure copper.

1.66

Given: m = 1.41 kg; V = 0.314L_ Find: din g/cm? and compare to pure titanium. Conceptual Plan: m, V > dthenkg > gthenL > cm’ 1000cm

d=m/ g i Compared to the published value: d (pure titanium)

Solution: d =

1000 1.41 k a KE x

% = 4.51 g/cm’, (This value is in Table 4.)

;= 4.49-2—

Check: The units (g/cm?*) are correct. The magnitude of the answer seems correct. The density of the frame is almost exactly the density of pure titanium (4.49 g/cm? versus 4.51 g/cm?*), so the frame could be titanium.

1.67

Given: m = 4.10 X 10° g;V = 3.25L_

Find: din g/cm"

Conceptual Plan: m, V >

cm‘

dthenL

d=m/V

Solution:

d =

4.10 X 10°g

325K

1000con”

1000cm'

g = 1,26—

cm’

Check: The units (g/cm*) are correct. The magnitude of the answer seems correct.

1.68

Given: m = 371 grams;

V = 19.3mL_

Find: din g/cm? and compare to pure gold.

Conceptual Plan: m, V > dd = m/V Compare to the published value: d (pure gold) = Solution: d =

19.3 g/mL. (This value is in Table 4.)

371 iin heBig igMaa

msm

The density of the nugget is essentially the same as the density of pure gold (19.2 g/mL versus 19.3 g/mL), so the nugget could be gold.

Check: The units (g/cm?) are correct. The magnitude of the answer seems correct and is essentially the same as the density of pure gold. 1.69

(a)

V = 417mL_ d = 1.11 g/cm’; Given:

Conceptual Plan: d, V > m thencm*? d=

Solution: d = m/V_

3 1.1 m= =a.1t—>

m/V

Find: m

+ mL LmL | ome

Rearrange by multiplying both sides of the equation by V.

1 cet’

X >1

417 me = 4.63.X-10 g

im = dxXV |

of the Check: The units (g) are correct. The magnitude of the answer seems correct considering that the value

Chapter 1 Matter, Measurement, and Problem Solving

12 I (b)

Find: VinL

1.11 g/cm*; m =41kg

Given: d =

Conceptual Plan: d, V >

mthenkg

1000 g

ila

> 1s

Tke

1000 em>

Rearrange by multiplying both sides of the equation by V and dividing both sides of the

Solution: d = m/V_ equation by d.

4.1kg

== et

g andcm’

—

1000 g

.LK

LL ont ee 10 1900 ent”

cm" Check: The units (L) are correct. The magnitude of the answer seems correct considering that the value of the density is about 1 g/cm’, 1.70

(a)

Given: d = 0.7857 g/cm’; V = 28.56mL_ Conceptual Plan: d, V

Find: m

— m

d=m/V

Rearrange by multiplying both sides of the equation by V.m = d x V

d = m/V_

Solution:

1 cer’

(07857 “5) x ma

X (28.56 mE) = 22.44¢

Check: The units (g) are correct. The magnitude of the answer seems correct considering that the value of the

(b)

density is less than 1 g/cm’. Given: d = 0.7857 g/cm*; m = 6.54g Conceptual Plan: d,m —

1 mL

d=m/V

Solution: d = m/V__ equation by d.

Find: V

V then cm’? > lem?

Rearrange by multiplying both sides of the equation by V and dividing both sides of the

6.54

im = = 8.320aP x "= = 832mL 0.7857 =~

Ve5s

cm?

Check: The units (mL) are correct. The magnitude of the answer seems correct considering that the value of the density is less than 1 g/cm’, eg

Given: V = 245 L, d = 0.821 g/mL

Conceptual Plan: g/mL —

Find: m

g/L thend,

1000

Solution:

d = m/V_

si= 25K

V > m d=m/V

Rearrange by multiplying both sides of the equation by V.

1000 mE

=———

g )

d X V

X | 0.821— |= 201 2 10°

(

Check: The units (g) are correct. The magnitude of the answer seems correct considering the value of the density is less than | g/mL and the volume is very large. 1.72

Given: d = 0.918 g/cm’, m = 10.0 lbs

Find: V

Conceptual Plan: lb

> g thend,m 453.59g Ib

Solution: 10.0 lbs x

453.59 g

—

d=m/V

= 4.5359 X 10° g; thend = m/V.

Rearrange by multiplying both sides of the

equation by V and dividing both sides of the equation by d. v=

m _ 4.5359 X 10°¢ 0.918

= 4.94 x 10° cm?

Check: The units (cm”) are correct. The magnitude of the answer seems correct considering the value of the density is

less than 1 g/cm’.

Chapter 1 Matter, Measurement, and Problem Solving

The Reliability of a Measurement and Significant Figures Ld

To obtain the readings, look to see where the bottom of the meniscus lies. Estimate the distance between two markings on the device. (a) 73.2 mL—the bottom of the meniscus appears to be sitting just above the 73 mL mark. (b) 88.2 °C—the mercury is between the 88 °C mark and the 89 °C mark, but it is closer to the lower number. (c) 645 mL—the meniscus appears to be just above the 640 mL mark.

1.74

To obtain the readings, look to see where the bottom of the meniscus lies. Estimate the distance between two markings on the device. Use all digits on a digital device. (a) 4.50 mL—the meniscus appears to be on the 4.5 mL mark. (b) 27.43 °C—the mercury is just above the 27.4 °C mark. Note that the tens digit is only labeled every 10 °C. (c) 0.873 g—tead all of the places on the digital display. Remember that

interior zeros (zeros between two numbers) are significant.

leading zeros (zeros to the left of the first nonzero number) are not significant. They only serve to locate the decimal point. trailing zeros (zeros at the end of a number) are categorized as follows: ° Trailing zeros after a decimal point are always significant. Trailing zeros before an implied decimal point are ambiguous and should be avoided by using ° scientific notation or by inserting a decima! point at the end of the number.

(a) (b) (c)

(d)

1.76

1,050,501 km 9.0020 m B.GAVGVGBOGOVVGOO2 s

9.981090 cm

Remember that

uM 2. a

interior zeros (zeros between two numbers) are significant. the leading zeros (zeros to the left of the first nonzero number) are not significant. They only serve to locate decimal point. trailing zeros (zeros at the end of a number) are categorized as follows: Trailing zeros after a decimal point are always significant. ° using Trailing zeros before an implied decimal point are ambiguous and should be avoided by ° scientific notation or by inserting a decimal point at the end of the number.

- (a) (b) (c) (d)

180,701 mi 9.901040 m 9.005710 km 90,201 m

Remember the following rules from Section 1.7. leading zeros only mark the decimal Three significant figures. The 3, 1, and 2 are significant (rule 1). The (a) place and are therefore not significant (rule 3). zeros occur before an implied decimal point (b) | Ambiguous. The 3, 1, and 2 are significant (rule 1). The trailing we would assume three significant figures. on, informati and are therefore ambiguous (rule 4). Without more

or as 3.12000 It is better to write this as 3.12 X 10° to indicate three significant figures

(c)

(d)

(e)

(rule 4). ant (rule iD). Three significant figures. The 3, 1, and 2 are signific ant (rule 1). Five significant figures. The 1s, 3, 2, and 7 are signific

10° to indicate six

occur before an implied decimal point and are thereAmbiguous. The 2 is significant (rule 1). The trailing zeros assume one significant figure. It is better to write would we fore ambiguous (rule 4). Without more information, X 10° to indicate four (rule 4). this as 2 X 10° to indicate one significant figure or as 2.000

Chapter 1 Matter, Measurement, and Problem Solving

1.78

Remember the following rules from Section 1.7. Four significant figures. The ones are significant (rule 1). The leading zeros only mark the decimal place and (a) are therefore not significant (rule 3).

One significant figure. The 7 is significant (rule 1). The leading zeros only mark the decimal place and are therefore not significant (rule 3).

Ambiguous. The 1, 8, and 7 are significant (rule 1). The first 0 is significant because it is an interior 0 (rule 2).

(d) (e)

L729

(a)

(b) (c) (d)

1.80

This is not exact because 7 is an irrational number. The number 3.14 only shows three of the infinite number of significant figures that 7r has. This is an exact conversion because it comes from a definition of the units and so has an unlimited number of significant figures. This is a measured number, so it is not an exact number. There are two significant figures. This is an exact conversion because it comes from a definition of the units and so has an unlimited number of significant figures.

(a) (b) (c) (d)

This is a measured number, so it is not an exact number. There are nine significant figures. This is an exact conversion, and so it has an unlimited number of significant figures. This is a measured number, so it is not an exact number. There are three significant figures. This is an exact conversion because it comes from a definition of the units and so has an unlimited number of significant figures.

(a) (b)

156.9—The 8 is rounded up because the next digit is a 5. 156.8—The last two digits are dropped because 4 is less than 5. 156.8—The last two digits are dropped because 4 is less than S. 156.9—The 8 is rounded up because the next digit is a 9, which is greater than 5.

(c)

(d) 1.82

The trailing zeros occur before an implied decimal point and are therefore ambiguous (rule 4). Without more information, we would assume four significant figures. It is better to write this as 1.087 10° to indicate four significant figures or as 1.08700 X 10° to indicate six (rule 4). Seven significant figures. The 1, 5, 6, and 3s are significant (rule 1). The trailing zeros are significant because they are to the right of the decimal point and nonzero numbers (rule 4). Ambiguous. The 3 and 8 are significant (rule 1). The first 0 is significant because it is an interior zero. The trailing zeros occur before an implied decimal point and are therefore ambiguous (rule 4). Without more information, we would assume three significant figures. It is better to write this as 3.08 X 10 to indicate three significant figures or as 3.0800 X 10* to indicate five (rule 4).

(a)

(b) (c) (d)

7.98 X 10*—The last digits are dropped because 4 is less than 5. 1.55 X 10’—The 8 is rounded up because the next digit is a 9, which is greater than 5. 2.35—The 4 is rounded up because the next digit is a 9, which is greater than 5. 4.54 X 10 °—The 3 is rounded up because the next digit is an 8, which is greater than 5.

Significant Figures in Calculations 1.83

(a)

(b)

(c)

(d)

9.15 + 4.970 = 1.84—Three significant figures are allowed to reflect the three significant figures in the least precisely known quantity (9.15). 1.54 X 0.03060 X 0.69 = 0.033—Two significant figures are allowed to reflect the two significant figures in the least precisely known quantity (0.69). The intermediate answer (0.03251 556) is rounded up because the first nonsignificant digit is a 5. 27.5 X 1.82 + 100.04 = 0.500—Three significant figures are allowed to reflect the three significant figures in the least precisely known quantity (27.5 and 1.82), The intermedi ate answer (0.50029988) is truncated because the first nonsignificant digit is a 2, which is less than 5.

(2.290 X 10°). + (6.7 x 10*) =.34—Two significant figures are allowed to reflect the two significant figures in the least precisely known quantity (6.7 x 10*). The intermediate answer (34.17910448) is truncated because the first nonsignificant digit is a 1, which is less than 5.

Chapter 1 Matter, Measurement, and Problem Solving 1.84

(a)

89.3 X 77.0 X 0.08 = 6 X 10’—One significant figure is allowed to reflect the one significant figure in the least precisely known quantity (0.08). The intermediate answer (5.50088 xX 10) is rounded up because the

(b)

first nonsignificant digit is a 5. (5.01 X 10°) + (7.8 X 10°) = 6.4 X 10°—Two significant figures are allowed to reflect the two significant

(d)

truncated because the first nonsignificant digit is a 2, which is less than 5. 4.005 < 74 x 0.007 = 2—One significant figure is allowed to reflect the one significant figure in the least precisely known quantity (0.007). The intermediate answer (2.07459) is truncated because the first non-significant digit is a 0, which is less than 5. 453 + 2.031 = 223—Three significant figures are allowed to reflect the three significant figures in the

figures in the least precisely known quantity (7.8 X 107); The intermediate answer (6.423076923 x 107) is

least precisely known quantity (453). The intermediate answer (223.042836) is truncated because the first non-significant digit is a 0, which is less than 5.

1.85

(a)

43.7 —2.341 41.359 = 41.4 Round the intermediate answer to one decimal place to reflect the quantity with the fewest decimal places (43.7). Round the last digit up because the first nonsignificant digit is 5. 17.6 +2.838 HBG +110.77 133.508 = 133.5 Round the intermediate answer to one decimal place to reflect the quantity with the fewest decimal places (2.3). Truncate nonsignificant digits because the first nonsignificant digit is 0. 19.6 +58.33 —4.974 72.956 = 73.0 Round the intermediate answer to one decimal place to reflect the quantity with the fewest decimal places (19.6). Round the last digit up because the first nonsignificant digit is 5. 5:99 =SJie 0.418 = 0.42 Round the intermediate answer to two decimal places to reflect the quantity with the fewest decimal places

(5.99). Round the last digit up because the first nonsignificant digit is 8.

1.86

(a)

0.004 +0.09879

0.10279 = 0.103 fewest decimal places Round the intermediate answer to three decimal places to reflect the quantity with the 9. is digit cant nonsignifi first the because up (0.004). Round the last digit

(b)

1239.3 eoah pos

. 1252.45 = 1252.5 places decimal fewest the with quantity the reflect to place decimal one Round the intermediate answer to is 5. (1239.3). Round the last digit up because the first nonsignificant digit

(c)

2.4 Seedy. 0.623 = 0.6 the quantity with the fewest decimal places (2.4). Round the intermediate answer to one decimal place to reflect digit is 2. Truncate nonsignificant digits because the first nonsignificant Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving

(d)

ere |

SSS

532 +73 —48,.523 490.777 = 491 Round the intermediate answer to zero decimal places to reflect the quantity with the fewest decimal places (532). Round the last digit up because the first nonsignificant digit is 7.

Perform operations in parentheses first. Keep track of significant figures in each step by noting the last significant digit in an intermediate result.

(a)

(24.6681

X 2.38) + 332.58 = 58.710078

ano Pats) 391.290078 = 391.3 ae dy The first intermediate answer has one significant digit to the right of the decimal because it is allowed three significant figures [reflecting the quantity with the fewest significant figures (2.38)]. Underline the most significant digit in this answer. Round the next intermediate answer to one decimal place to reflect the quantity with the fewest decimal places (58.7). Round the last digit up because the first nonsignificant digit is 9.

(b)

(c)

(85.3 — 21.489)

63.811

0.0059

~ 0.0059

= 1.081542 * 10%= 1.1 x 10+

The first intermediate answer has one significant digit to the right of the decimal to reflect the quantity with the fewest decimal places (85.3). Underline the most significant digit in this answer. Round the next intermediate answer to two significant figures to reflect the quantity with the fewest significant figures (0.0059). Round the last digit up because the first nonsignificant digit is 8. (512 + 986.7) + 5.44 = 0.5189014

+5.44

(d)

5.9589014 = 5.96 The first intermediate answer has three significant figures and three significant digits to the right of the decimal, reflecting the quantity with the fewest significant figures (512). Underline the most significant digit in this answer. Round the next intermediate answer to two decimal places to reflect the quantity With the fewest decimal places (5.44). Round the last digit up because the first nonsignificant digit is 8. [(28.7 X 10°) + 48.533] + 144.99 = 59135.02 +144.99

59280.01 = 59300 = 5.93 x 104 The first intermediate answer has three significant figures, reflecting the quantity with the fewest significant figures (28.7 X 10°). Underline the most significant digit in this answer. Because the number is so large, when the addition is performed, the most significant digit is the hundreds place. Round the next intermediate answer to the hundreds place and put in scientific notation to remove any ambiguity. Note that the last digit is rounded up because the first nonsignificant digit is 8.

1.88

Perform operations in parentheses first. Keep track of significant figures in each step by noting the last significant digit in an intermediate result.

(a)

(b)

[(1.7 X 10°) + (2.63 x 10°)] + 7.33 = 6.463878 sak fe 13.793878 = 13.8 The first intermediate answer has one significant digit to the right of the decimal because it is allowed two significant figures [reflecting the quantity with the fewest significant figures (1.7 X 10°) ]. Underline the most significant digit in this answer. Round the next intermediate answer to one decimal place to reflect the quantity with the fewest decimal places (6.5). Round the last digit up because the first nonsignificant digit is 9. (568.99 — 232.1) +5.3 = 336.89 + 5.3 = 63.564151 = 64 The first intermediate answer has one significant digit to the right of the decimal to reflect the quantity with the fewest decimal places (232.1). Underline the most significant digit in this answer. Round the next intermediate answer to two significant figures to reflect the quantity with the fewest significan t figures (5.3). Round the last digit up because the first nonsignificant digit is 5. Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving 10° = 9478.1

8.1

(Cc)

(GN4Sr AS = 9.9) XB.

(d)

The first intermediate answer only has significant digits to the left of the decimal, reflecting the quantity with the fewest significant figures (9443 and 45). Underline the most significant digit in this answer. Round the next intermediate answer to two significant figures to reflect the quantity with the fewest significant figures (8.1 X 10°). Round the last digit up because the first nonsignificant digit is 7. (3.14 X 2.4367) — 2.34 = 7.651238

108 = 7.67726 XK Loe 77K 10”

— 2, a 5.311238 = 5.31 The first intermediate answer has three significant figures, reflecting the quantity with the fewest significant figures (3.14). Underline the most significant digit in this answer. This number has two significant digits to the right of the decimal point. Round the next intermediate answer to two significant digits to the right of the decimal point because both numbers have two significant digits to the right of the decimal point. Note that the last digit is truncated because the first nonsignificant digit is 1. 1.89

Given: 11.7 mL liquid; empty flask weight = 124.1 g; flask with liquid weight = 132.8 g Find: d in g/mL Conceptual Plan: Empty flask weight, flask with liquid weight — liquid weight then weight, volume — d (flask with liquid weight) — (flask weight) = liquid weight

d = mass/yolume = m/V

Solution: liquid weight = (flask with liquid weight) — (flask weight) = 132.8 g — 124.1 g = 8.7 g liquid; then ? 8.7 density (d) = re evade St = 0.74358974 g/mL = 0.74g/mL ‘ volume V 1d? ool,

Check: The units (g/mL) are correct. The magnitude of the answer (0.74) makes physical sense because the volume is greater than the mass.

1.90

Given: 9.55 mL liquid; empty flask weight = 157.2 g; flask with liquid weight = 148.4g Find: d in g/mL Conceptual Plan: Empty flask weight, flask with liquid weight — liquid weight then weight, volume — d (flask with liquid weight) — (flask weight) = liquid weight

d = mass/volume = m/V

Solution: liquid weight = (flask with liquid weight) — (flask weight) = 157.2 g — 148.4 g = 8.8 g liquid; then mass m 8.82 density (d) = ol iat beeen = 0.92146597 g/mL = 0.92 g/mL

hae,

.V,

OSE:

Check: The units (g/mL) are correct. The magnitude of the answer (0.92) makes physical sense because the volume is slightly greater than the mass.

Unit Conversions 1.91

(a)

Given: 27.8L

Find: cm?

Conceptual Plan: L > cm? 1000 cm?

‘L

Solution: 27.8Vv

1000 cm?

Fi oe = 2.78 X 10* cm?

S Check: The units (cm*) are correct. The magnitude of the answer (10*) makes physical sense because cm” is much smaller than a liter; so the answer should go up several orders of magnitude. Three significant figures are allowed because of the limitation of 27.8 L (three significant figures).

(b)

Given: 1898 mg _ Find: kg Conceptual Plan: mg > g > kg lg 1000 mg

Solution: 1898 mg Xx

lkg 1000g

ig Lkg = 1,898 X 10 °kg 1000 mg © 1000¢

because a Check: The units (kg) are correct. The magnitude of the answer (10->) makes physical sense mg has 1898 because allowed are figures significant Four kilogram is a much larger unit than a milligram. four significant figures.

Chapter 1 Matter, Measurement, and Problem Solving (c)

Given: 198 km _ Find: cm Conceptual Plan: km > m— cm 1000 m

100 cm

ion: 198 kan x Solution:

1000 ni _ 100 cm 4 x ie? = 1.98 x 10’ cm an

Check: The units (cm) are correct. The magnitude of the answer (10’) makes physical sense because a kilometer is a much larger unit than a centimeter. Three significant figures are allowed because 198 km has three significant figures. 92

(a)

Given: 28.9nm Find: wm Conceptual Plan: nm > m—> wm 102m

10° wm

lnm

10° um

Solution: 28.9 nmi x 10°= nf x

- = 2.89 x 102 um = 0.0289 pm

Check: The units (zm) are correct. The magnitude of the answer (10°?) makes physical sense because a micrometer is a much larger unit than a nanometer. Three significant figures are allowed because 28.9 nm has three significant figures.

(b)

Given: 1432cm°

Find: L

Conceptual Plan: cm? > L 10 5

1000 cm?

Solution: 1432 cm* X

1000 cr?

24320:

Check: The units (L) are correct. The magnitude of the answer (1) makes physical sense because cm® is much smaller than a liter; so the answer should go down several orders of magnitude. Four significant figures are allowed because of the limitation of 1432 cm? (four significant figures).

(c)

Given: 1211 Tm Find: Gm Conceptual Plan: Tm > m > Gm

Solution: Tm x olution 1211

10'2m

10°°Gm

1Tm

10°wf = 10° Gm x = 1211 < 16° Gm 1 BS:

Check: The units (Gm) are correct. The magnitude of the answer (10°) makes physical sense because a terameter is a much larger unit than a gigameter. Four significant figures are allowed because 1211 Tm has four significant figures.

195

(a)

Given: 154cm__ Find: in Conceptual Plan: cm — in Lin 2.54cem

Solution: 154 cm x

1 in

2.54 cm

= 60.62992 in = 60.6 in

Check: The units (in) are correct. The magnitude of the answer (60.6) makes physical sense because an inch is a larger unit than a cm. Three significant figures are allowed because 154 cm has three significant figures.

(b)

Given: 3.14kg Find: g Conceptual Plan: kg > g 1000g Lkg

Solution: 3.14kg x

1000

lkg

: =

3.14 x 10°g

Chapter 1 Matter, Measurement, and Problem Solving i

Check: The units (g) are correct. The magnitude of the answer (10°) makes physical sense because a kilogram is a much larger unit than a gram. Three significant figures are allowed because 3.14 kg has three significant figures.

(c)

Given: 3.5L Find: qt Conceptual Plan: L > qt 1,057 qt TL

, 1.057 qt Solution: 3.5 x cy, = 3.6995 qt = 3.7 qt Check: The units (qt) are correct. The magnitude of the answer (3.7) makes physical sense because a L is a smaller unit than a qt. Two significant figures are allowed because 3.5 L has two significant figures. Round the last digit up because the first nonsignificant digit is a 9.

(d)

Given: 109 mm_ Find: in Conceptual Plan: mm — m —> in Lm 1000 mm

39.37 in Im

Solution: 109 mri x 10001 —*—nam x 30371 2232" = 4.20133 in = 4.29in 1 mt Check: The units (in) are correct. The magnitude of the answer (4) makes physical sense because a mm is a much smaller unit than an inch. Three significant figures are allowed because 109 mm has three significant figures.

1.94

(a)

Given: 1.4in Find: mm Conceptual Plan:in —~ cm

2.54cm lin

Solution: 1.4 in X

m—>

1m 100 cm

2.54 cm 4

1 wf

1000 mm 1m

. 1000 mm

|i 100 cm 1 wf Check: The units (mm) are correct. The magnitude of the answer (36) makes physical sense because a mm is smaller than an inch. Two significant figures are allowed because 1.4 in has two significant figures. Round the last digit up because the first nonsignificant digit is a 5.

(b)

Given: 116 ft

Find: cm

Conceptual Plan: ft

— in in

1 ft

> cm 2.54cem lin

I2id x| 2.54a = 3.5357 X 103cm = 3.54 X 103cm Solution: 116€ x =" Check: The units (cm) are correct. The magnitude of the answer (10°) makes physical sense because a foot is a much larger unit than a cm. Three significant figures are allowed because 116 ft has three significant figures. Round the last digit up because the first nonsignificant digit is a 5. (c)

Given: 1845 kg Find: lb Conceptual Plan: kg > g —

ion: 1845 kg x Solution:

1000 g

lkg

453.6g

1000 g

lig

1 Ib

453.6¢ =

4.0675 ub

3 3 X 10° lb = 4.068 X 10° Ib

Check: The units (Ib) are correct. The magnitude of the answer (10°) makes physical sense because a lb is a smaller unit than a kg. Four significant figures are allowed because 1845 kg and 453.6 g/lb each have four significant figures. Round the last digit up because the first nonsignificant digit is a5.

(d)

Given: 815 yd Find: km Conceptual Plan: yd > m — nee

1.094yd

Solution: 815 yd <

km kore

1000m

Chapter 1 Matter, Measurement, and Problem Solving

eee

Check: The units (km) are correct. The magnitude of the answer (0.7) makes physical sense because a yard is a much smaller unit than a kilometer. Three significant figures are allowed because 815 yd has three significant figures. Round the last digit up because the first nonsignificant digit is a 9. os

Other: running pace = 7.5 miles per hour hr —> min

Find: minutes Given: 10.0km Conceptual Plan: km — mi — 0.6214 mi 1km

1 hr 7.5 mi

60 min 1 hr

0.6214 mi > | bf Y 60 min = 49.712 min = 50. min = 5.0 X 10! min | kan 7.5 mai | hf Check: The units (min) are correct. The magnitude of the answer (50) makes physical sense because she is running almost 7.5 miles (which would take her 60 min = 1 hr). Two significant figures are allowed because of the limitation Solution: 10.0 kan X

of 7.5 mi/hr (two significant figures). Round the last digit up because the first nonsignificant digit is a 7.

1.96

Given: 212 km _ Find: hours Other: riding pace = Conceptual Plan: km — mi — hr 0.6214 mi 1km

18 miles per hour

1 br 18 mi

0.6214 mi 1 hr Xx 18 anf = 7.318711 hr = 7.3h r lan Bi)

ion: 212 kan Xx Solution

Check: The units (hr) are correct. The magnitude of the answer (7) makes physical sense because she is riding over

100 miles (which would take her over 4 hr). Two significant figures are allowed because of the limitation of 24 mi/hr (two significant figures). Truncate the last digit because the first nonsignificant digit is a 1. OT

Given: 17 km/L_

Find: miles per gallon

km mi mi C onceptual tual Plan: — Plan: —— —> —L > zal

Solntlonies 17LVot

0.6214mi

3.785L

1km

1 gallon

: ee 1 kati

: eo = a39 beagggali = agali 1 gal 3

Check: The units (mi/gal) are correct. The magnitude of the answer (40) makes physical sense because the dominating factor is that a liter is much smaller than a gallon; so the answer should go up. Two significant figures are allowed because of the limitation of 17 km/L (two significant figures). Round the last digit up because the first nonsignificant digit is a 9. 1.98

Given: 5.0 gallons

Find: cm*

Conceptual Plan: gal

Solution: 5.0 gal x

> L —> cm? 3.785L

1000cm*

1 gal

3.785% _ 1000cm* x

| gal

aa LY

1.8925 X 10*cm* = 1.9 X 10*cm3

Check: The units (cm*) are correct. The magnitude of the answer (10*) makes physical sense because cm> is much smaller than a gallon; so the answer should go up several orders of magnitude. Two significant figures are allowed because of the limitation of 5.0 gallons (two significant figures). Round the last digit up because the first nonsignificant digit is a 9,

1.99

(a)

Given: 195m?

Find: km?

Conceptual Plan: m’ > km?

(Liam?

(1000 m)?

Notice that for squared units, the conversion factors must be squared.

Solution: 195 mni- x —

(km)?

(1000 nt)?

= 1.95 X 10* km?

Chapter 1 Matter, Measurement, and Problem Solving G r .

5 —4 . Check: The units (km*) are correct. The magnitude of the answer (10 +) makes physical sense because a kilometer is a much larger unit than a meter.

(b)

Given: 195m?

Find: dm?

Conceptual Plan: m? > dm? (10 dm)?

Notice that for squared units, the conversion factors must be squared. Solution: 195 nv? X Boe

mi)

1.95.«

10*dm

Check: The units (dm’) are correct. The magnitude of the answer (10*) makes physical sense because a decimeter is a much smaller unit than a meter. (c)

Given: 195m Find: cm? Conceptual Plan: m? > cm 2 (100 em) ‘ (1m)?

Notice that for squared units, the conversion factors must be squared. Solution: 195 nt? X

(100 cm)?

(1 mt)?

1.95 X 10° cm?

Check: The units (cm*) are correct. The magnitude of the answer (10°) makes physical sense because a centimeter is a much smaller unit than a meter.

1.100

(a)

Given: 115m° Find: km* Conceptual Plan: m*? > km* (1 km)? (1000 m)?

Notice that for cubed units, the conversion factors must be cubed.

Solution: 115 nt x

(1 km)* = 1.15 X 10°’ km? (1000 nt)*

Check: The units (km*) are correct. The magnitude of the answer (10~’) makes physical sense because a kilometer is a much larger unit than a meter.

(b) . Given: 115m°

Find: dm’

Conceptual Plan: m* > dm? (10 dm1)?

(1m) Notice that for cubed units, the conversion factors must be cubed.

Solution: 115 nt x

(10dm)*>

(1 af)?

1.15% 10 dm?

Check: The units (dm’) are correct. The magnitude of the answer (10°) makes physical sense because a decimeter is a much smaller unit than a meter. (c)

Given: 115m°

Find: cm*

Conceptual Plan: m’ > cm* (100 cm)?

(1m)?

Notice that for cubed units, the conversion factors must be cubed.

Solution: 115 nt’ x

(100 cm)?

(1 af)°

= 1.15 X 108 cm?

Measurement, and Problem Solving

Chapter 1 Matter, 22 Be eee 1.101

Given: 435 acres

Other: | acre = 43,560 ft”; 1 mile = 5280 ft

Find: square miles

Conceptual Plan: acres > ft? > mi? 43560 ft2 (1 mi)? (5280 ft)? lacre Notice that for squared units, the conversion factors must be squared.

1 mi)?

x Ces! = = 0.6796875 mi? = 0.680 mi* Solution: 435 aeres X EO | aeres (5280 ff)“ Check: The units (mi’) are correct. The magnitude of the answer (0.7) makes physical sense because an acre is much smaller than a mi2; so the answer should go down several orders of magnitude. Three significant figures are allowed because of the limitation of 435 acres (three significant figures). Round the last digit up because the first nonsignificant digit is a 7. PL O02

(a)

Given: 954 million acres

Find: square miles

Other: 1 acre = 43,560 ft?; 1 mile =

5280 ft

Conceptual Plan: Substitute 10° for million then acres > ft? > mi? 43560 ft?

(1 mi)?

lacre

(5280 ft)?

Notice that for squared units, the conversion must be squared. Solution: 954 million acres = 954 X 10° acres +2

oh Ba = 1.490625 x 10° mi? = 1.49 x 10° mi” 954 X 10° aeres X ae | aeres (5280 f)° 4 Check: The units (mi) are correct. The magnitude of the answer (10°) makes physical sense because an acre

(b)

is much smaller than a mi’; so the answer should go down several orders of magnitude. Three significant figures are allowed because of the limitation of 435 acres (three significant figures). Truncate the last digits because the first nonsignificant digit is a 0. Given: 3.537 million square miles Find: percentage of U.S. land is farmland farmland

Conceptual Plan: Substitute 10° for million then farm = alin otal

lan

X 100%

Note: Units of farmland and total land must be the same.

Solution: 3.537 million mi* = 3.537 X 10° mi?

1.49 x 10° mi?

% farmland = 3.537 X 10°mi? x 100% = 42.12609556%

farmland = 42.1% farmland

Check: The units (%) are correct. The magnitude of the answer (42%) makes physical sense because less and less land is devoted to farmland. Three significant figures are allowed because of the limitation of 435 acres (three significant figures). Truncate the last digit up because the first nonsignificant digit is a 4.

1.103

Given: 141b Find: mL Other: 80 mg/0.80 mL; 15 mg/kg body Conceptual Plan: lb + kg body ~ mg > mL

Solution: 1416 x

1 kg body

15 mg

0.80 mL

2.205 Ib

Lkgbody

80mg_

| kg body rc 15 mg 2.205 Ib

‘ 0.80 mL

| kg-body

80 mg

= 0.9523809524 mL = 0.95 mL

Check: The units (mL) are correct. The magnitude of the answer (1 mL) makes physical sense because it is a reasonable amount of liquid to give to a baby. Two significant figures are allowed because of the statement in the problem. Truncate after the last significant digit because the first nonsignificant digit is a 2.

1.104

Given: 18 lb Find: mL Other: 100 mg/5.0 mL; 10 mg/kg body Conceptual Plan: lb + kg body > mg > mL 1 kg body

10 mg

-2.2051b = L kg body Solution: 18 lb x

5.0mL

100mg

Chapter 1 Matter, Measurement, and Problem Solving oe ag: Sue 0 5 el nel eet ae eats

rrr

rot

Check: The units (mL) are correct. The magnitude of the answer (4 mL) makes physical sense because it is a reason-

able amount of liquid to give to a baby. Two significant figures are allowed because of the statement in the problem. Round the last digit up because the first nonsignificant digit is an 8.

Cumulative Problems 1.105

Given: solar year Find: seconds Other: 60 seconds/minute; 60 minutes/hour; 24 hours/solar day; 365.24 solar days/solar year Conceptual Plan: yr + day > hr > min —> sec 365.24 day

24 hr

60 min

60 sec

| solar yr

1 day

l br

1 min

cnet,

365.24 day

Solution: | solaryr x

| solaryr

24hf

| day

— 60 mim

1 hf

60 sec = 3.1556736 X 10’ sec = 3.1557 X 107 sec | mim

Check: The units (sec) are correct. The magnitude of the answer (107) makes physical sense because each conversion factor increases the value of the answer—a second is many orders of magnitude smaller than a year. Five significant figures are allowed because all conversion factors are assumed to be exact except for the 365.24 days/solar year (five significant figures). Round the last digit up because the first nonsignificant digit is a 7.

1.106

Given: 2.0 hours Find: picoseconds Other: 60 seconds/minute; 60 minutes/hour Conceptual Plan: hr > min > s > ps

Solution: 2.0 hf X

60min

60s

10"ps

1 hr

1 min

Lis

60min | bf

60%

_ 10'ps

x — x | pam

= 7.2 ~ 10 ps

Check: The units (ps) are correct. The magnitude of the answer (10!) makes physical sense because each conversion factor increases the value of the answer—a picosecond is many orders of magnitude smaller than a second. Two significant figures are allowed because all conversion factors are assumed to be exact. There are two significant figures in 2.0 hours.

1.107

(a) (b) (c) (d) (e)

1.108

Extensive—The volume of a material depends on how much is present. Intensive—The boiling point of a material is independent of how much material you have; so these values can be published in reference tables. Intensive—The temperature of a material is independent of how much is present. ~ Intensive—The electrical conductivity of a material is independent of how much material you have; so these values can be published in reference tables. Extensive—The energy contained in material depends on how much is present. If you double the amount of material, you double the amount of energy.

i Meee Given: °C = ¥ eves 1.8

Find: temperature where °F = °C pied,

Conceptual Plan: °C = a

x= 32TE

Solution: x = marr

rine 8x

°C = °F = x and solve forx Se

382 S18xs e392

08s =

‘i

= 32/058. =

—40. — —40. °F = —40.°C Check: The units (°F and °C) are correct. Plugging the result back into the equation confirms that the calculations were done correctly. The magnitude of the answer seems correct because it is known that the result is not between 0 °C and 100 °C. The numbers are getting closer together as the temperature drops. 1.109

Find: temperature where °X = °F. Given: 130°X = 212°F and 10°X = 32°F relating “X and °F. Then set °F = °X equation Conceptual Plan: Use data to derive an temperatures (y = mx + b). two the Solution: Assume a linear relationship between Let y= °Fand let x = °X. Copyright © 2017 Pearson Education, Inc.

z and solve for z.

Chapter 1 Matter, Measurement, and Problem Solving

24 od

The slope of the line (m) is the relative change in the two temperature scales: Ne

PID

32°

eats

ASS MPOSCaON SOK

Solve for intercept (b) by plugging one set of temperatures into the equation: b= 17>F = y) = 15u + b> 32 = (1,5)(10) + b= 32 lI= 1 Fb

°X + 17 (1.5)

Set °F = °X = zand solve for z. z=

152+

17—

-17 = 1.52 — <>

—17 = 0.52 z = —34 —

—34°F = —34°X

Check: The units (°F and °X) are correct. Plugging the result back into the equation confirms that the calculations were done correctly. The magnitude of the answer seems correct because it is known that the result is not between 32 °F and 212 °F. The numbers are getting closer together as the temperature drops. 1.110

Find: temperature where methyl alcohol-boils in “J Given: 17°J = 0°H and 97°J = 120°H Other: methy! alcohol boils at 84 °H Conceptual Plan: Use data to derive an equation relating °J and °H. Then set “H = 84 °H and solve for °J. Solution: Assume a linear relationship between the two temperatures (y = mx + b). Let y = “Jand letx = °H. The slope of the line (m) is the relative change in the two temperature scales. Ea aed we Oya ied _ 80° ~ 0.667 0H 120 = OCIA NO Solve for intercept (b) by plugging one set of temperatures into the equation: y = 0.667x + b— 17 = (0.667)(0) + bb = 17 °J = (0.667) °H + 17 Set °H = 84°H and solve for °J.

°J = (0.667)(84) + 17°F = 56 + 17 = 73°F Check: The units (°J) are correct. Plugging the original data points back into the equation confirms that the calculations were done correctly. The magnitude of the answer seems correct because the result should be between 17 °J and

97 °J and closer to 97 °J than 17 °J. Litt

1G. F = ma = kg(m/s’). Let’s call it N for Newton. Ten tons = 20,000 Ib = (1 kg/2.2 Ib) X 20,000 lb = 4.4 x 10 x 10*kg, deceleration = 55 mi X 0.6km/mi X 10?m/km X 1/3.6 X 10°s*. Exponents = 10* x 10° x 1073 = 10*. So the KN is convenient. For one molecule, the mass is 10°? kg and deceleration is 3 X 10° m/s*. So Exponents = the pN is convenient.

1hi2

107° x 10° =

107”. So

Given: 25 °C and —196°C Find: Why significant figures are 3 and 2, respectively. Conceptual Plan: The problem is stated in units of °C, so it must be converted to another temperature unit to see if the significant figures can change. Try K. Begin by finding the equation that relates the quantity that is given (°C) and the quantity you are trying to find (K). K = °C + 273.15 K = 25°C + 273.15 = 298K —K

= —196°C + 273.15 = 77K

A small positive temperature in °C gains significant figures because of the rules of addition for significant figures— going from 2 to 3. A very negative temperature in °C loses significant figures because of the rules of addition for significant figures—going from 3 to 2.

L113

(a)

L76:%

107 /8.0.%

102

2.2 % 10° Two significant figures are allowed to reflect the quantity with the

fewest significant figures (8.0 < 107). (b)

Write all figures so that the decimal points can be aligned:

0.0187 + 0.0002 — All quantities are known to four places to the right of the decimal place; — 0.0030 so the answer should be reported to four places to the right of the 0.0159

(c)

decimal place, or three significant figures.

[ (136000) (0.000322) /0.082) |(129.2) = 6,899910244 x 10* = 6.9 X 104 Round the intermediate answer

to two significant figures to reflect the quantity with the fewest significant figures (0.082). Round the last digit up because the first nonsignificant digit is 9. Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving ee e 1.114

Given: one gallon of gasoline

Conceptual Plan: gal — awry Solution: | gal x

Find: US dollars

L —

euro >

Other: | euro = $1.38 US and I liter of gasoline in France =

1.35 euro

$US

3.785L

1.35euro

$1.38 US

1 gal

| euro

3.785EK | 1.35¢eu6 . $1.38US | pal x <=oe a

$ 7.051455 US == $7.05 US

Check: The units ($ US) are correct. The magnitude of the answer ($7 US) makes physical sense because the dominating conversion factor is ~4. Three significant figures are allowed because of the limitation of 1.35 euro/L. Truncate the last significant digit because the first nonsignificant digit is a 1.

Fabe:

(a)

Given: cylinder dimensions: length = 22 cm; radius = 3.8 cm; d(gold) = 19.3 g/cm; d(sand) = 3.00g/cm* Find: m(gold) and m(sand) Conceptual Plan: 4 r > Vthend, V > m d=m/V

V=lnr

Solution: V(gold) = V(sand) = (22 cm)(7)(3.8 cm)? = 998.0212 cm? d = m/V Rearrange by multiplying both sides of equation by V. > m = d X V g

m(gold) = (193

cnr )X (998.0212 cat”) = 1.926181 x 10+g = 1.9 x 104g

Check: The units (g) are correct. The magnitude of the answer seems correct considering that the value of the density is ~20 g/cm*. Two significant figures are allowed to reflect the significant figures in 22 cm and 3.8 cm. Truncate the nonsignificant digits because the first nonsignificant digit is a 2.

m(sand) = (3.00=.) (998.0212 cm?) = 2.99406 x 103g = 3.0 x 10°g Check: The units (g) are correct. The magnitude of the answer seems correct considering that the value of the density is 3 g/cm*. This number is much lower than the gold mass. Two significant figures are allowed to reflect the significant figures in 22 cm and 3.8 cm. Round the last digit up because the first nonsignificant digit is a 9.

(b)

LOREO

| Comparing the two values 1.9 x 10*g versus 3.0 * 10° g shows a difference in weight of almost a factor of 10. This difference should be enough to trip the alarm and alert the authorities to the presence of the thief.

Given: r = 1.0 X 10° 3cm;m= 1.7 x 10 ™“g_ Find: density Conceptual Plan:r

—

Vthenm, V >

V = (4/3)nr?

Other: V = (4/3) ar°

d=m/V

Solution: V = (4/3)ar? = (4/3)(a)(1.0 x 1078 cm)? = 4.188790205 x 107°? cm?

d=—= V

1.7 x 10°% 4.188790205

e—__ = 4,058451049 x 10-4, = 4.1 x 104, == cm cm

x 10 ~~" cm

Check: The units (g/cm*) are correct. The magnitude of the answer seems correct considering how small a nucleus 107" cm. Round is compared to an atom. Two significant figures are allowed to reflect the significant figures in 1.0 the last digit up because the first nonsignificant digit is a 5. 1.117

Given: 3.5 lb of titanium Find: volume in in? Other: density of titanium is 4.51 g/cm? Conceptual Plan: lb — gthenm,d — Vthen cm? > in?

453.6 g

ited

tIb a nia Solution: 3.

d= m/V_

7 V a

yeeees ib

dort

(1 in)’ (2.54 cm)?

= 1.5876 X 10°g

Rearrange by multiplying both sides of the equation by V and dividing both sides of the equation by d.

1.5876 X 10°¢ d 4 5:—= cm

= 3.520 < 10° em? cm

(1 in)?

ae = 21 in (2.54 em) = 3.5 < 10’ car % Se

of grams.,Two Check: The units (in*) are correct. The magnitude of the answer seems correct considering the number because the digits ant nonsignific the Truncate lb. 3.5 in significant figures are allowed to reflect the significant figures

26 a 1.118

Chapter 1 Matter, Measurement,A and Problem Solving ae SS Se)

Other: density of iron is 7.86 g/cm?

Given: density (g/cm*) Find: density (Ib/in?) g Ib Ib Conceptual Plan: —,; 7 —_,~7 7,3 cma cm ir 1 lb

7.86 g ion: Solution:

(2.54 cm)?

453.6 g

(1 in)?

1 Ib

(2.54 ent)?

453.6 g

a 3 = (2839557380

(1 in)

a = 0254-5

Check: The units (Ib/in*) are correct. The magnitude of the answer seems correct considering the dominating factor that a gram is smaller than a pound; so the answer should go down. Three significant figures are allowed to reflect the significant figures in 7.86 Ib. Round the last digit up because the first nonsignificant digit is a 9.

Given: cylinder dimensions: length = 2.16 in; radius = 0.22 in;m = 41g Conceptual Plan: in > cmthen/l,r — Vthenm, V > d zones lin

V = Inr*

Find: density.(g/m*)

d=ml/V

Solution: 2.16 itx 2.54om = 5.4864cm = |

0.22 if X 2.54= cm = 0.5588 em = r

V = Imr* = (5.4864 cm)(7)(0.5588 cm)* = 5.3820798 cm?

41 q= "= 2 = 7.617879 =, = 76-2; V_—-5.3820798 cm a cm cm Check: The units (g/cm*) are correct. The magnitude of the answer seems correct considering that the value of the density of iron (a major component in steel) is 7.86 g/cm>. Two significant figures are allowed to reflect the significant figures in 0.22 in and 41 g. Truncate the nonsignificant digits because the first nonsignificant digit is a 1. 1.120

Given: m = 85g _ Find: radius of the sphere (inches) Other: density (aluminum) = 2.7 g/cm? Conceptual Plan: m,d —

V then

V > rthencm

-—> in

V=(4/3)rP

ea 2.54em

d=m/V Solution: d = m/V_ tion by d. 85

yes OTe

Rearrange by multiplying both sides of the equation by V and dividing both sides of the equa-

sdk aeoa

cm?>

V = (4/3)mr?_

Rearrange by dividing both sides of the equation by (4/3). 7° = ~ 7

Take the cube root of both sides of the equation.

eae 1fa —

—ee

(es

cm*

Air

lin

1.95879438cm 6

2.54 cm

1/3

’) = (7.51565009 cm*)'/* = 1.958794386 cm

= 0.7711788923 in = 0.77 in

Check: The units (in) are correct. The magnitude of the answer seems correct. The magnitude of the volume is about

a third of the mass (density is about 3 g/cm*). The radius (in cm) seems correct considering the geometry involved. The magnitude goes down when we convert from centimeters to inches because an inch is bigger than a centimeter. Two significant figures are allowed to reflect the significant figures in 2.7 g/cm? and 85 g. Truncate the nonsignificant digits because the first nonsignificant digit is a 1.

1 A

Given: 185 cubic yards (yd*) of HJO

Find: mass of the HO (pounds)

Other: d(H,O) = 1.00 g/cm? at 4 °C Conceptual Plan: yd? > m? > em’ > (1m)?

(100 cm)*

(1.094yd)>

(1m)

Solution: 185 yd? x

1 nt)?

g—l|b 1.00 g

L Ib

1.00 em?

100

a (1.094 yd)> x Bees (1 af)

453.59 g

cm)? x

ux a 100 cn? ~453.59¢

‘

3.114987377 X 10° lb = 3.11 X 10° Ib

Chapter 1 Matter, Measurement, and Problem Solving

Sree aspire eerste nnn ee cents neers

Check: The units (Ib) are correct. The magnitude of the answer (10° ) makes physical sense because a pool is not a small object. Three significant figures are allowed because the conversion factor with the least precision is the density ‘ . 5. 5 1B [1.00 g/cm” 3 — (3 significant figures) | and the initial size has three significant figures. Truncate after the last digit because the first nonsignificant digit is a 4.

1122

Given: 7655 cubic feet (ft*) of ice

Find: mass of the ice (kg)

Other: d(ice) = 0.917 g/cm? at 0°C

Conceptual Plan: ft* > cm’ > g > kg (30.48 cm)3

Solution: 7655 ft* x

1kg

0.917g

(1 ft)? ~~ 1.00cm?

1000g

(30.48 cm)?

0.917

(1 £03

Lkg

1.00 cm? > 1000¢

= 1.987739274 x 10°kg = 1.99 X 10°k

Gos

Check: The units (kg) are correct. The magnitude of the answer (10°) makes physical sense because an iceberg is a large object. Three significant figures are allowed because the conversion factor with the least precision is the density [(0.917 g/em? — (3 significant figures) |. Round the last digit up because the first nonsignificant digit is a 7.

hizs

Given: 15 liters of gasoline Find: kilometers Other: 52 mi/gal in the city Conceptual Plan: L > gal — mi > km I gal 52 mi 1 km 3.785 L lgal 0.6214 mi Solution: 1S x

1 gal

. 52 mi ’

1 km

3.785 K

legal

0.6214mi

= 3.316327941 x 10°km = 3.3 X 10°km

Check: The units (km) are correct. The magnitude of the answer (10) makes physical sense because the dominating conversion factor is the mileage, which increases the answer. Two significant figures are allowed because the conversion factor with the least precision is 52 mi/gal (two significant figures) and the initial volume (15 L) has two significant figures. Truncate the last digit because the first nonsignificant digit is a 1. It is best to put the answer in scientific notation so that it is clear how many significant figures are expressed.

1.124

Given: 355 mL of gasoline Find: kilometers Other: 57 mi/gal in the city Conceptual Plan: mL — L > gal > mi > km uh,

I gal

57 mi

1 km

1000 mL

3.785 L

lgal

0.6214 mi

1 1k aa x eal Xx Ae! x Sees Be 8.603319984 km = 8.6 km 1000 mE = 3.785v legal 0.6214 mi Check: The units (km) are correct. The magnitude of the answer (8.6) makes physical sense because the dominating conversion factor is the conversion from mL to L, which decreases the answer. Two significant figures are allowed because the conversion factor with the least precision is 57 mi/gal (two significant figures). Truncate the last digit because the first nonsignificant digit is a 0. Solution: 355 mE x

Let 25

Given: radius of nucleus of the hydrogen atom = 1.0 X 10~' cm; radius of the hydrogen atom = 52.9 pm Find: percent of volume occupied by nucleus (%) Conceptual Plan:cm — mthen pm — mthenr — Vthen Vatoms Vaucteus > 7% Vnucteus =

100 cm

ion: 1.0 1.0 X* 10° Vom cm 10 ~ Solution: V = (4/3)ar?

10! pm

(4/3)mr*

% Voueteus =

te = 1.0 x 10° mand 52.9 pm pm x 10!Mm

Vatom

100%

=11 =15.29 X10°" m

Substitute into % V equation.

Voucleus

9%o Vaucleus aae

R007 2

3 ay a r

or Vewcleds

Im x 100

——=

100%

3 (4/3) 7 Fhucleus aS 75, (4/3) ar rom © Ynucleus ere

LOO

‘

Simplify equation.

Substitute numbers and calculate result.

atom

oO,

“

%o Vimcleus —

(10 3< 10-15)? X 100% = (1.890359168 x 107)? x 100% = 6.755118686 x 10°¥ (5.29 x 1071! m)?

and Problem Solving 1 Matter, Measurement, ee ee

Chapter 28 OO Fe a

proton is so small. Check: The units (none) are correct. The magnitude of the answer seems correct (107'?) because a digits up belast the Round cm. 10° x 1.0 in Two significant figures are allowed to reflect the significant figures cause the first nonsignificant digit is a 5. 1.126

Given: radius of neon = 69 pm; 2.69 X 10° atoms per liter Find: percent of volume occupied by neon (%) Conceptual Plan: Assume 1L total volume. pm > m— cmthenr — V then cm? > L then L/atom — L then Vae, Viotai > % Vne 1 ees

100 cm

Verte os= talon

99% Ve= ea 7, Ve

X 10% atoms 2.69 VEE

X 100%%

1 wi 100 cm ig ion: 69 pm x 10! = pirx = 6.9 X 10-7 cm Solution: i

V = (4/3)ar? = (4/3)7(6.9 X 10° cm)? = 1.37605528 x 10° cm* 1.37605528 X 10° cat* x 37605528) % 10571 =

= 137605528

1000 cr®

10 77.L

;

wigs

x 2.69 X 1077 atoms = 3.701588707 x 10° L

Substitute into % V equation.

atom

F _ Vee oy, 32201588707 x 10°K e . be es % Ve = 7 X 100% = iE X 100%s = 3.701588707 X 10°% = 3.7 X 10°% Total

Check: This says that the separation between atoms is very large in the gas phase. The units (%) are correct. The magnitude of the answer seems correct (10); it is known that gases are primarily empty space. Two significant figures are allowed to reflect the significant figures in 69 pm. Truncate the nonsignificant digits because the first nonsignificant digit is a 0.

Given: radius of hydrogen = 212 pm; radius of Ping-Pong ball = 4.0 cm, 6.02 X 107 atoms and balls in a row Find: row length (km) Conceptual Plan: atoms — pm — m — kmand ball ~ cm > m > km 212 pm

1 km

4.0 cm

100 cm

1km

1 atom

10!? pm

1000 m

1 ball

1000 m

212 pm 1 mi 1 km x —— x = 1.28 x 10!! km latom 10pm 1000 nf 4. I oclbatl ee ~~ a a SS oad 10S 100cm ~~ 1000ni

Solution: 6.02 < 107? aterms x

BOD 1 ea

Check: The units (km) are correct. The magnitude of the answers seems correct (10'! and 10!°). The answers are driven

by the large number of atoms or balls. The Ping-Pong ball row is 10° times longer. Three significant figures are allowed to reflect the significant figures in 212 pm. Two significant figures are allowed to reflect the significant figures in 4.0 cm.

1.128

Given: 100 m in 9.69 s; 100 yards in 9.21.s

Find: miles/hr

Conceptual Plan: speed is distance/time; m/s —

yd/s —» m/s > 1m 1.0936 yd

Sandan 100 yd

km/s —

1 km 1000 m

km/s —> miles/s >

miles/hr >

Lkm 1000 m

60s 1 min

100 mt

0.62137 mi 1 km

1 kan >: 0.6213 7m .

+9694 ~ 1000mi 1 wf

si 1 kan

9.21 ~~ 1.0936yd 1000 nf

‘lan

> miles/hr > 60s 1 min

miles/hr and

60 min l hr

60 min 1 hr

ait

¥ 60 min

jap

0.62137 mi by 60

Len

miles/s

0.62137 mi 1km

fii,”

ae:

}

ete 084954 mi/hr = 23.1 mi/hr

60 mim

Libra a ee

ty ae

Check: The units (mi/hr) are correct. The magnitude of the answers seems correct (23 and 22), and they are close to

each other. Assuming that 100 m and 100 yd have three significant figures, three significant figures are allowed to reflect the significant figures in the times.

Chapter 1 Matter, Measurement, and Problem Solving a a Conceptual Plan: g sodium

—

g salt

100 g salt 39.33 g sodium

> g snack mix 100 g snack mix 1.25 g salt

2.40 sedram g ” 100 g-salt ‘- 100 g snack mix Solution: — = 488.1770 g snack mix/day 1 day 39,33 sediam g 1.25 g-sait = 488 g snack mix /day Check: The units (g) are correct. The magnitude of the answer seems correct (500) because salt is less than half

sodium and there is a little over a gram of salt per 100 grams of snack mix. Three significant figures are allowed to reflect the significant figures in the FDA maximum and in the amount of salt in the snack mix.

1.130

Given: 86.6 g lead/100 g galena; 68.5 g galena/100 g ore; 92.5 g lead extracted/100 g lead available; 5.00 cm radius

Other: d(lead) = 11.4g/cm*

sphere

Find: g ore

Conceptual Plan: ‘sphere — Vephere then Vipheres sphere > Msphere(Z lead) — V=

(4/3)9r

d=m/V

g lead available >

g galena — g ore

100 g lead available

100 g galena

100 g ore

92.5 g lead extracted

86.6 g lead

68.5 g galena

Solution: Calculate mshere Vephere = (4/3) 7 rage = (4/3) m (5.00 cm)? = 523.5988 cm? d = m/V Solve for m by multiplying both sides of the equation by V.m = V X d

m = 523.5988 cat’ x 5969.026 glead x

11.4 g lead

1 crt

= 5969.026 glead

100 g lead avattable 2

92.5 gleadextracted

100 g-gatena

86.6 gleadavaitable

100 g ore

68.5 g gatenia

= 1.087811 X 10* gore = 1.09 X 10*g ore Check: The units (g) are correct. The magnitude of the answer seems correct (10*) because lead is so dense and the sphere is not small. Three significant figures are allowed to reflect the significant figures in all of the information given. 1.131

Given: 24.0 kg copper wire; wire is a cylinder of radius = 1.63 mm_ Other: d(copper) = 8.96g/cm*; resistance = 2.061 1/km Conceptual Plan: mm

> m —> cm and kg —> g then 10° m

100 cm

1000g

1 mm

1 kg

d,m— V then V,r>/(cm) “-

cdi

hack

ion: Solution:

shes

Find: resistance (()

~m—>km—>

1 km

2.061 0

100 cm

1000 m

1km

107m _ 100cm |. mm x iar x 1.63 I af

1000 g 24.0 kg x cera = 2.40

= 0.163 ccm

4 eee : 10° g;thend = m/V. Rearrange by multiplying both sides of equation by V to get

240 X 10*g

m = d X V; then divide both sides by d. V = Feed a

aricek

ocaae = 2.6785714 X 10° cm’; then V = Imr’

1 cm?

Rearrange by dividing both sides ofthe equation by mr’ to get

V.__2.6785714 X 10° cm*

—r(0.163 cm)?

= 3.2090623 x 10*cm

then

1 wf lan _ 2.0610 3.2090623 x 10* cri x 100cm x=: 1000 ni - 1 kent = 0.66138779 = 0.6610

Check: The units (1) are correct. The magnitude of the answer seems correct because we expect a small resistance for a material that is commonly used for electrical wiring. Three significant figures are allowed to reflect the signiftcant figures in 1.63 mm and 24.0 kg. Truncate the nonsignificant digits because the first nonsignificant digit is a 3.

Chapter 1 Matter, Measurement, and Problem Solving 1132

Given: aluminum foil; 304 mm wide and 0.016 mm thick; 1.10 kg Find: length of foil Other: d(aluminum) = 2.70 g/cm? Conceptual Plan: for each dimension, mm — m —> cm and kg > g then

d,m—

V then V,w,h

d=m/V

100cm

1000g

1mm

Lkg

1 (cm) > m

V=lrr

Solution: ion: 304 mim x

10>m

10>m_

i ak x

100cm ta

103m

100cm

P; fare x —

30.4 cm and0.016 mam

= 0.0016cm

to get 1.10 kg Xx ea = 1.10 < 10° g; thend = m/V. Rearrange by multiplying both sides of the equation by V

m _ 1.10 X 10°g m = d X V; then divide both sides by d. V =

= 407.40741 cm? then V = 1wh

1 cm?

Rearrange by dividing both sides of equation by (w h) to get

407.40741 cm? . = = §.375974712 X 10°cm (30.4cm) (0.0016 cm) =

then

8.375974712 X 10° cmt Xx

100 cm

= 83.75974712 m = 84m

Check: The units (m) are correct. The magnitude of the answer seems correct because we expect a long length of the foil. Two significant figures are allowed to reflect the significant figures in 0.16 mm. Truncate the nonsignificant digits because the first nonsignificant digit is 0.

Liss

Given: d(liquid nitrogen) = 0.808 g/mL; d(gaseous nitrogen) = 1.15 g/L; 175 L liquid nitrogen; 10.00m x 10.00 m x 2.50mroom_ Find: fraction of room displaced by nitrogen gas Conceptual Plan: L > mL then Viquias Giiquid > Miquia then Set Myguid = Mgas then mas, do, —> Vay, then 1000 mL hee

d=m/v

calculate the V,J5m—> cm’ — L then calculate the fraction displaced

100 cm)*

(| m)-

1000 cm:

= Veoom

; 1000 mL P ; Solution: 175 Vv x RETe = 1.75 X 10° mL. Solve for m by multiplying both sides of the equation by V.

m=VXd=

0.808 g

1.75 X 10° mt x - arnt

1.414 X 10° g nitrogen liquid = 1.414 X 10° g nitrogen gas

d = m/V Rearrange by multiplying both sides of the equation by V and dividing both sides of the equation by d.

Vioomn

1.414 X 10°¢

fiLE

= 1.229565 x 10° L nitrogen gas

= | X w X h = 10.00 m1 X 10.00 nf X 2.50 nf X

Veas

1.229565 X 10°V.

Voom

250 41

(100 em)* IL x = 2.50 X 10° (1m)? 1000 emt* ‘

= 0.491826 = 0.492

Check: The units (none) are correct. The magnitude of the answer seems correct (0.5) because there is a large volume

of liquid and the density of the gas is about a factor of 1000 less than the density of the liquid. Three significant figures are allowed to reflect the significant figures in the densities and the volume of the liquid given.

Chapter 1 Matter, Measurement, and Problem Solving a 1.134

Given: d(mercury at 0.0°C) = 13.596 g/cm*; d(mercury at 25.0 °C) = 13.534 g/cm’; 3.380 g; 0.200 mm diameter capillary Find: distance mercury rises Conceptual Plan: at each temperature m,d + Vthenmm — m — cm then YVroh rs mh

7000 sam

m/V

then calculate the difference between the two heights

Solution:

d = m/V_

Rearrange by multiplying both sides of the equation by V and dividing both sides of the

equation by d. V = *

3.380

at 0.0°C: V = — =

Beste 0.24860253 cm?

13.596 aS cm 5 and at 25.0°C:

3.380 ¢

m — =

= 0.249741392 cm

13.5345 cm: ?-

100 cm =

1 at robin

= 0.200 mm X

= ar“h cm; : then V a

= 0.02 0.0200

Rearrange by dividing both sides of the equation by 7r?.h

at 0.0°C:h =

Le wre

V_0.24860253 cm? mr“

> = 197.8316077 cm = ra a (0.0200 cm)*

—> =

and at 25.0°C:h =

V_0.249741392 cm? ur

—> =

—

a (0.0200 cm)*

= 198.7378852 cm

The difference in height is 198.7378852 cm — 197.8316077 cm = 0.9062775 cm = 1 cm Check: The units (cm) are correct. The magnitude of the answer seems correct (0.7) because there is a relatively

small change in temperature and the two densities are very close to each other. Only one significant figure is allowed because the heights have four significant figures; so the error is in the tenths place.

Challenge Problems 1.135

Force = (mass) X (acceleration), or F = ma, and Pressure = force/area. So if a force of 2.31 X 104 N is applied on an area of 125 cm’, the

Pressure

2.31 x 10*N

125 cat” x

——

= 1.846 x 10°—.

Ses (100 crn)?

Referring to Chapter 5, 1 N/m? = 1 Pa and

{atm 1.136

= 101,325 Pa; so1.846 x 10°. = 1.846 x 10° pa x —“™_ = 18.2atm 101,325P4 maion a= sa ot ae Rp thel

A Newton (N) has units of kg- m/s”. If a dyne has mass measured in units of grams, this will make a dyne a factor of 10° larger than a Newton (N). To get the additional factor of 10? needed to get to a factor of 10°, the length must be in centimeters (cm) 1 dyne

1137

Lkg

g-cm

1000 ¢ ~ 100 om

_,kg:m mF

Referring to the definition of energy in Chapter 6, 1 Joule = 1J = kg-m*/s”. For kinetic energy, if the units of mass are the kilogram (kg) and the units of the velocity are meters/second (m/s), then

kinetic energy units =

m v =kg (") sire S

J. Because a Newton (N) is a unit of force and has units of

Chapter 1 Matter, Measurement, and Problem Solving kg+m/s?, Pressure = force / area and has units of N/m’, and Force = (mass) X (acceleration), or F = ma, then

DRS

3/2 PV units = “5° mt _kgrm = 2 1.138

m=_

kg-m _ %

Given: mass of black hole (BH) = 1 X 10° suns; radius of black hole = one-half the radius of our moon Find: density (g/cm*) Other: radius of our sun = 7.0 X 10° km; average density of our sun = 1.4 X

10° kg/m?; diameter of themoon = 2.16 X 10° miles Conceptual Plan: dgy = mgy/Veu Calculate mpy: Foun > Voun km3,, > m3,, then Veuns dsun > Moun then Moun > Mpy kg > g

(1000 m)?

V = (4/3)9r

Meu

Amn

Calculate Vgy : dnoon > ‘moon 7 RH MI Tmoon =

1/2dmoon

Mao =

1/2 Tmoon

Substitute into dgy

GICET

;

den = Von

km >

1000g

May = (1 X 10°) X mop

m >

cmthenr

“sie

ikg

(4/3)ar?

mpy/ Ven

8 Solution: Calculate mpy :Veun = (4/3)a7rgin = (4/3)7(7.0 X 10° km)? = 1.43675504 x 10'8km

1000 m)?

1.43675504 x 10!8 ken x ee

= 1.43675504 x 1077 m?

Asin = Msun/ Veun Solve for m by multiplying both sides of the equation by V,,,. onivo = Ves oe ee

Mun = (1.43675504 X 10°7m*)(1.4 x 10° kg/m?) = 2.011457056 x 10°? kg mpy = (1 X 10°) X mun = (1 X 10°) X (2.011457056 X 10°° kg) = 2.011457056 x 10% kg

2011457056 x 103kg x 1000 = = 2.011457056 x 10°g I 1 . Calculate Vey :%moon = 74 moon = 3 (2-16 x 10° miles) = 1.08 X 10° miles 1

“habeas

ay = 3/moon = 3 (1.08 x 10° miles) = 540. miles

1 kan

1000 nf | 100cm

Z 40. mites>” x 0.6214mi x lia

1 at

= 8.6900547 x 107cm

V = (4/3)ar> = (4/3)1(8.6900547 x 10’ cm)? = 2.74888227 x 10% cm? _ 2.011457056 x 10°%°g ao 2 Substitute into dgy = mpy =— 3 = 7.31736342 X 10"; = 7.3 x 10!! = Ven —-2.74888227 x 1074 cm =

cm cnr Check: The units (g/cm*) are correct. The magnitude of the answer seems correct (10!) because we expect extremely high numbers for black holes. Two significant figures are allowed to reflect the significant figures in the radius of our sun (7.0 X 10° km) and the average density of the sun (1.4 X 10° kg/m). Truncate the nonsignificant digits because the first nonsignificant digit is a 4.

1.139

Given: 15.0 ppm CO; 8-hour period Find: milligrams of carbon monoxide Other: 0.50 L of air per breath; 20 breaths per minute; carbon monoxid e has a density

means 15.0 L CO per 10° L air

Conceptual plan: hr +

min —> breaths >

60 min | hr

20 breath 1 min

Lair >

0.50 Lair | breath

Leo >

15.0 Leo Tee OM ae

gco > 1.2 8c 1 Leo

of 1.2 g/L; 15.0 ppm CO

M£co 1000 mgco 1 gco

Solution:

8 hf x

60min x | hf

20breath a | min

0.50 Le x L221

15.0L¢« ihe 1000 mgse x aco x cn St Piair le l gO

86.4 mgco = 9 X 10! mgco

Check: The units (mg) are correct. The magnitude of the answer (10°) makes physical sense because there are more than 6 powers of 10 visible in these conversion factors in the numerator and one factor of 10° in the denominator. This means that most of the conversions cancel each other out, but there is still some left in the numerator, Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving

One significant figure is allowed because the conversion factor with the least precision is 20 breaths/minute (one significant figure); the starting time (8 hours) also has one significant figure. Round the last digit up because the first nonsignificant digit is a 6. 1.140

Given: cubic nanocontainers with an edge length = 25 nanometers Find: (a) volume of one nanocontainer; (b) grams of oxygen could be contained by each nanocontainer; (c) grams of oxygen inhaled per hour; (d) minimum number of nanocontainers per hour; (e) minimum volume of nanocontainers. Other: (pressurized oxygen) = 85 g/L; 0.28 g of oxygen per liter; average human inhales about 0.50 L of air per breath and takes about 20 breaths per minute; adult total blood volume = ~5 L

Conceptual Plan: (a)

(b)

—> m=

emthenl

100 cm

10° nm

L —

Vthencm’

v=B

1000 cm?

g pressurized oxygen

85 g oxygen 1 L nanocontainers

(c)

(d)

hr —

min

—

breaths >

60 min

20 breath

1 hr

1 min

grams oxygen

—

L,j;, >

go,

0.50L,i,

0.28 gco

1 breath

esis

number nanocontainers

1 nanocontainer

part (b) grams of oxygen

(e)

number nanocontainers

—

volume nanocontainers

part (a) volume of 1 nanocontainer

Solution:

(a)

1 nf

25am x 10° —2— om x Ver

(b)

100 ——" = 1 nf

25 x 10% em

TL = (2.5% 10cm)? = 15625 X 10° cat” X —___ = 1000 ce*

85 g oxygen 1.5625 X 10° x 1 V nanocontainers = 13281 25ex 10)

1.5625 X 10 -

= 16 x Helodls

ig & Pressurized O,

nanocontainer

g pressurized O>

ye 19!8 ==——______—_

nanocontainer

60 main

20 breath

(c)

Ilhf x

(d)

1.68 X 10° goxygen x =

(ec)

1,3

| bf

| ant

(0.28

~ ~*O.50LG

“x

Bie

lege

| breath

1 nanocontainer

1.3 x 10°? goxygen

1.68 X 10° g oxygen = 1.7 X 10° g oxygen

1.292307692 x 107° nanocontainers

10” nanocontainers

1.292307692 < 10°? nanoeontainers x

156252

107°

nanoeeentainer

= 2,019230769 L= 2.0L

average human is 5 L. This volume is much too large to be feasible because the volume of blood in the Check:

(a)

(b).

sense because these The units (L) are correct. The magnitude of the answer (10—20) makes physical the significant figures in the are very, very tiny containers. Two significant figures are allowed, reflecting up because the first nonsignificant starting dimension [25 nm (2 significant figures)]. Round the last digit ve digit is a 6. very, are these because physical sense The units (g) are correct. The magnitude of the answer GP) makes figures are allowed, reflecting the significant Two inside. fit can molecules few very very tiny containers and Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving

(c)

(d)

(e)

1.141

significant figures in the starting dimension (25 nm) and the given concentration (85 g/L)—2 significant figures in each. Truncate the nonsignificant digits because the first nonsignificant digit is a 2. The units (g oxygen) are correct. The magnitude of the answer (107) makes physical sense because of the conversion factors involved and the fact that air is not very dense. Two significant figures are allowed because it is stated in the problem. Round the last digit up because the first nonsignificant digit is an 8. The units (nanocontainers) are correct. The magnitude of the answer (107°) makes physical sense because these are very, very tiny containers and we need a macroscopic quantity of oxygen in them. Two significant figures are allowed, reflecting the significant figures in both of the quantities in the calculation—2 significant figures. Round the last digit up because the first nonsignificant digit is a 9. The units (L) are correct. The magnitude of the answer (2) makes physical sense because of the magnitudes of the numbers in this step. Two significant figures are allowed, reflecting the significant figures in both of the quantities in the calculation—2 significant figures. Truncate the nonsignificant digits because the first nonsignificant digit is a 1.

Because the person weighs 155 Ib and has a density of 1.0 g/cm*, the volume of the person can be calculated as

453.59¢

155 bh x ———

rier

Icom

x 10* cm’. i oF = 7.030645 s ae

Approximating the volume of a person as a cylinder 4.0 feet tall, V = Jar. Rearranging the equation, solve for r. 1

V \2 = pr (+)

oe | 7.030645 x 10fem®

4.0: 5¢ 248TG

= 13 s4g31om

The circumference is 27r = 277(13.54831 cm) = 85.12655 cm. When the person gains 40.0 Ib of fat, the volume

increaseis40.06 X

453.59¢

Wie

1cm

= 1.97643 X 10*

OOlk es

cm’,

Thus, the new volume is 7.030645 x 10* cm? + 1.97643 x 10* cm? = 9.00708 x 10* cm*. So the new 1

A V \2 radius is r = (+) = a

AROS 9.00708 x 10" cm = eS 30.48 cm : kk T

15.33485 cm, and the new circumference is

2mr = 2m(15.33485 cm) = 96.35170 cm. ; a 96.35170 cm — 85.12655 cm The percent increase in circumference — = X 100% 85.12655 cm

1.142

13.1864% _

13%.

Assume that all of the spheres are the same size. Let x = the percentage of spheres that are copper (expressed as a fraction); so the volume of copper = (427 cm*)x, and the volume of lead = (427 cm*) (1 — x).

Because the density of copper is; 8.96 g/cm’, the mass of copper = (427 cm*)x x 8.9 —

Because the density of lead is 11.4 g/cm’, the mass of lead = (427 cet’)(1 — x) x —

= 3825.92(x) g.

GN 9342 (1.x) o,

Because the total mass is 4.36 kg = 4360 g, 4360 g = 3825.92(x) g + 4893.42 (1 — x) g. Solving for x, 1067.50(x) g = 533.42 g—>x = 0.499691, or 50.% of the spheres are copper. Check: This answer makes sense because the average density of the spheres = 4360 g/427 cm* = 10.2 g/cm? and the average of the density of copper and the density of lead = (8.96 + 11.4)/2 g/cm? = 10.2 g/cm’.

Conceptual Problems 1.143

No. Because the container is sealed, the atoms and molecules can move around but they cannot leave. If no atoms or molecules can leave, the mass must be constant.

1.144

(c) is the best representation. When solid carbon dioxide (dry ice) sublimes, it changes phase from a solid to a gas. Phase changes are physical changes, so no molecular bonds are broken. This diagram shows molecules with one Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving a

carbon atom and two oxygen atoms bonded together in every molecule. The other diagrams have no carbon dioxide molecules.

1.145

This problem is similar to Problem 1.64 except that the dimension is changed to 7 cm on each edge. Given: 7 cm on each edge cube Find: cm®* Conceptual plan: Read the information carefully. The cube is 7 cm on each side. Lweh—ov V=Iwh

inacubel=w=h

Solution:

7cm X 7cm X 7cm = (7cm)? = 343 cm’, or 343 cubes

1.146

To determine which number is large, the units need to be compared. There is a factor of 1000 between grams and kilograms in the numerator. There is a factor of (100)°, or 1,000,000, between cm* and m*, This second factor more than compensates for the first factor. Thus, Substance A with a density of 1.7 g/cm? is denser than Substance B with a density of 1.7 kg/m’,

1.147

Remember that density = mass /volume. (a)

The darker-colored box has a heavier mass but a smaller volume, so it is denser than the lighter-colored box.

(b)

The lighter-colored box is heavier than the darker-colored box, and both boxes have the same volume; so the

(c)

lighter-colored box is denser. The larger box is the heavier box, so it cannot be determined with this information which box is denser.

1.148

(b) .

1.149

(d)

Remember that an observation is the information collected when studying phenomena. A law is a concise statement that summarizes observed behaviors and observations and predicts future observations. A theory attempts to explain . | why the observed behavior is happening. behavior— future explain can and observations many summarizes it because law a like most is This statement (a) many places and many days. (b) ~ This statement is a theory because it attempts to explain why (gravitational forces). This statement is most like an observation because it is information collected to understand tidal behavior. (c) This statement is most like a law because it summarizes many observations and can explain future behavior— (d) many places and many days and months. Copyright © 2017 Pearson Education, Inc.

Chapter 1 Matter, Measurement, and Problem Solving

Questions for Group Work SO

heterogeneous mixture

liquid compound

solid element

liquid compound physical change:

before = liquid Solid element chemical change:

DOH

—>

Ss Ci

before = solid element

ASH

(a)

The

after = solid compound

average

100 wm xX

)

1 <t0*m

(Cc)

0.0001 m

thickness

10°°m pm

(d)

100 microns or 100 wm

(e)

The

distance

1.133

er X10

between

149,600,000 kan X

PlSZ

of a human

hair

is about

Earth

149,600,000,000 m.

1 kan

(f)

1.496 x 10''m

(g) (h)

149,600,000,000 m 149,600,000 km or 149.6 Mm

1 pm

10°

149,600,000

| km

10°

the

10°m

100. microns.¢(jm)..Since.

and

the

Sun

km.

Since

All of these values can be true because when we round 4.73297 km to three significant figures we get 4.73 km; and if we round this value to one significant figure we get 5 km. The number of digits reported indicates how accurately the distance was measured. The odometer on a car could be used for the first measurement. A trip meter on a car could be used for the second measurement. A long tape measure could be used for the last measurement. (hae

Chapter 1 Matter, Measurement, and Problem Solving 2.54cm

; ————

Since | in = 2.54 cm, there are

1 in

1.154

2.54cm

2.54cem

1 in

lin

16.397064 cm?/ in’,

To convert the height of a student from feet and inches into meters: first convert the feet to inches; then add this to the number of inches, and finally convert the inches to centimeters and to meters. For example, if the

height of a student is 5’8.5": 68.5 ui X

2.54 cm

100. em

5f€

12 in ik

177399 m =

’

60 in (exactly); then 60 in + 8.5 in = 68.5 in, and finally 1.74m.

Add all of the heights together to get the sum for your group.

Data Interpretation and Analysis 1.155

(a)

(6))

Using the information presented in Figure b, the density of solid ice at 0 °C is 918 kg/m>. Using the information presented in Figure a, the density of liquid water at 4 °C is 999.97 kg/m’. Since the density of the solid ice is less than the density of the liquid water, the ice will float. When the air temperature drops, the surface of the pond or lake will begin to freeze. This ice will float because the ice is less dense and fish can survive in 4 °C liquid water as long as there is enough dissolved oxygen in this water. 2G) Converting the density of Pluto from g/cm? to kg/m? can be done as follows: S

cas

1k = x «10° ¢

10? ent)? a Tul (1 m)

k m:

re The mean density of Pluto 1.8 — 2.1 g/cm? becomes

1,800 — 2,100 kg/m, which is closer to the density of ice (920 — 940 g/cm?) than the density of Earth’s Moon (3,346.4 kg/m*). This is consistent with the idea that Pluto contains a large proportion of water.

(ii)

(c)

The density of solid ice in Antarctica (at 65.1 °C) is ~925 kg/m/ and the density of ice on Pluto (—233 °C to —223 °C) is extrapolated to be ~938 kg/m>. The density of ice increases as the temperature decreases, so the ice is | — 2 % denser on Pluto. The density of solid ice at —20 °C is 919 kg/m? and the density of liquid water at 4 °C is 999.97 kg/m*. The

volume of the water from Antarctica will be 26.5 X 10°km* x

244 X 107 kim’.

(d)

in Pluto Pluto isis estimated estimated as 1.2 as 1.27 The mass of water in volume of the water will be 3.81 < 107! kg X

919 keAni”

999.97 keAni?

=12,A354230% 10) ke

ice. ee |es 10° g elie: Bh 100 X 10” kg Ptuto Re aerate

1a xy ot km) == 3.81011 X10" km? = 4 X10" kin’, 999.97 kg (10° kant?

Taking all of the assumptions into account, we find that there is more water on Pluto than there is in Antarctica.

Atoms and Elements

Review Questions val

First described by Scottish botanist Robert Brown, Brownian motion is the random motion of particles suspended in a fluid (a liquid or a gas). This motion is the result of the collision of these particles with the moving atoms or molecules (i.e., the particles) in the fluid.

2,2

The first people to propose that matter was composed of small, indestructible particles were Leucippus and Democritus. These Greek philosophers theorized that matter was ultimately composed of small, indivisible particles called atomos. In the sixteenth century, modern science began to emerge. A greater emphasis on observation brought rapid advancement as the scientific method became the established way to learn about the physical world. By the early 1800s, certain observations led the English chemist John Dalton to offer convincing evidence that supported the early atomic ideas of Leucippus and Democritus. The theory that all matter is composed of atoms grew out of observations and laws. The three most important laws that led to the development and acceptance of the atomic theory were the law of conservation of mass, the law of definite proportions, and the law of multiple proportions. John Dalton explained the laws with his atomic theory.

The law of conservation of mass states the following: In a chemical reaction, matter is neither created nor destroyed. In other words, when you carry out any chemical reaction, the total mass of the substances involved in the reaction does not change.

2.4

The law of definite proportions states the following: All samples of a given compound, regardless of their source or how they were prepared, have the same proportions of their constituent elements. This means that elements composing a given compound always occur in fixed (or definite) proportions in all samples of the compound.

oe:

The law of multiple proportions states the following: When two elements (call them A and B) form two different compounds, the masses of element B that combine with | g of element A can be expressed as a ratio of small whole numbers. This means that when two atoms (A and B) combine to form more than one compound, the ratio of B in one compound to B in the second compound will be a small whole number. The law of definite proportions refers to the composition of a particular compound, not to a comparison of different compounds. The law of definite proportions states that a particular compound is always made of the same elements in the same ratio.

2.6

The main ideas of John Dalton’s atomic theory are as follows: (1) Each element is composed of tiny, indestructible particles called atoms. (2) All atoms of a given element have the same mass and other properties that distinguish them from the atoms of other elements. (3) Atoms combine in simple wholenumber ratios to form compounds. (4) Atoms of one element cannot change into atoms of another element. In a chemical reaction, atoms change the way they are bound together with other atoms to form a new substance. The law of conservation of mass is explained by the fourth idea. Because the atoms cannot change into another element, and just change how they are bound together, the total mass will remain constant. The law of constant composition is supported by ideas 2 and 3. Because the atoms of a given element always have the same mass and other distinguishing properties, and they combine in simple whole-number ratios, different samples of the same compound will have the same properties and the same composition. The law of multiple proportions is also supported by ideas 2 and 3 because

Chapter 2 Atoms and Elements

the atoms can combine in simple whole-number ratios; the ratio of the mass of B in one compound to the mass of B in a second compound will also be a small whole number.

Pe |

In the late 1800s, an English physicist named J. J. Thomson performed experiments to probe the properties of cathode rays. Thomson found that these rays were actually streams of particles with the following properties: They traveled in straight lines, they were independent of the composition of the material from which they originated, and they carried a negative electrical charge. He measured the charge-to-mass ratio of the particles and found that the cathode ray particle was about 2000 times lighter than hydrogen. In Millikan’s oil drop experiment, oil was sprayed into fine droplets using an atomizer. The droplets were allowed to fall under the influence of gravity through a small hole into the lower portion of the apparatus where they could be viewed. During their fall, the drops would acquire electrons that had been produced by the interaction of high-energy radiation with air. These charged drops interacted with two electrically charged plates within the apparatus. The negatively charged plate at the bottom of the apparatus repelled the negatively charged drops. By varying the voltage on the plates, the fall of the charged drops could be slowed, stopped, or even reversed. From the voltage required to halt the free fall of the drops and from the masses of the drops themselves, Millikan calculated the charge of each drop. He then reasoned that because each drop must contain an integral number of electrons, the charge of each drop must be a whole-number multiple of the electron’s charge. The magnitude of the charge of the electron is of tremendous importance because it determines how strongly an atom holds its electrons. The plum-pudding model of the atom proposed by J. J. Thomson hypothesized that the negatively charged electrons were small particles electrostatically held within a positively charged sphere. Rutherford’s gold foil experiment directed positively charged a particles at an ultrathin sheet of gold foil. These particles were to act as probes of the gold atoms’ structures. If the gold atoms were indeed like plum pudding—with their mass and charge spread throughout the entire volume of the atom—these speeding probes should pass right through the gold foil with minimum deflection. A majority of the particles did pass directly through the foil, but some particles were deflected, and some even bounced back. Rutherford realized that to account for the deflections, the mass and positive charge of an atom must all be concentrated in a space much smaller than the size of the atom itself.

Rutherford’s nuclear model of the atom has three basic parts: (1) Most of the atom’s mass and all of its positive charge are contained in a small core called the nucleus. (2) Most of the volume of the atom is empty space, throughout which tiny negatively charged electrons are dispersed. (3) There are as many negatively charged electrons outside the nucleus as there are positively charged particles within the nucleus, so that the atom is electrically neutral. The revolutionary part of this theory is the idea that matter, at its core, is much less uniform than it appears. Matter appears solid because the variation in its density is on such a small scale that our eyes cannot see it.

The three subatomic particles that compose atoms are as follows: Protons, which have a mass of 1.67262

Neutrons, which have a mass of 1.67493 Electrons, which have a mass of 0.00091

lie kg or 1.00727 amu and a relative charge of +1

10-7’ kg or 1.00866 amu and a relative charge of 0 10°°’ kg or 0.00055 amu and a relative charge of —1

2.14

The number of protons in the nucleus defines the identity of an element.

aahe

number A is the sum of the The atomic number, Z, is the number of protons in an atom’s nucleus. The atomic mass

neutrons and protons in an atom.

2.16

originates from the Greek The names of the elements were often given to describe their properties. For example, argon mythology or astroRoman or Greek word argos, meaning “inactive.” Other elements were named after figures from was born. discoverer their where or nomical bodies. Still others were named for the places where they were discovered More recently, elements have been named after scientists.

PANE

neutrons. The percent natural abundance Isotopes are atoms with the same number of protons but different numbers of of a given element. sample occurring naturally a in is the relative amount of each different isotope

Oe 2.18

Chapter 2 Atoms and Elements

is the atomic number, and X is the chemical symbol. Isotopes can be symbolized as 4X, where A is the mass number, Z by a hyphen and the mass number of the isoA second notation is the chemical symbol (or chemical name) followed is the mass number. The carbon isotope with a mass tope, such as X-A, where X is the chemical symbol or name and A

of 12 would have the symbol !2C or C-12 or carbon-12.

2419 2.20

ly charged ions are called anions.

An ion is a charged particle. Positively charged ions are called cations. Negative

mass, certain sets of properThe periodic law states the following: When elements are arranged in order of increasing of a series of rows in which ties recur periodically. Mendeleev organized all the known elements in a table consisting were aligned in the mass increased from left to right. The rows were arranged so that elements with similar properties same vertical column.

2a k

of heat and electricMetals are found on the left side and in the middle of the periodic table. They are good conductors often shiny, and are they (ductile), wires into ity; they can be pounded into flat sheets (malleable), they can be drawn they tend to lose electrons when they undergo chemical changes. are solids at Nonmetals are found on the upper-right side of the periodic table. Their properties are more varied: Some room temperature, while others are liquids or gases. As a whole, they tend to be poor conductors of heat and electricity and to gain electrons when they undergo chemical changes. Metalloids lie along the zigzag diagonal line that divides metals and nonmetals. They show mixed properties. Several metalloids are also classified as semiconductors because of their intermediate and temperature-dependent electrical conductivity.

453)

(a)

Noble gases are in group 8A and are mostly unreactive. As the name implies, they are all gases in their natural state.

(b)

Alkali metals are in group 1A and are all reactive metals.

Alkaline earth metals are in group 2A and are also fairly reactive. Halogens are in group 7A and are very reactive nonmetals.

PIB}

Main-group metals tend to lose electrons, forming cations with the same number of electrons as the nearest noble gas. Main-group nonmetals tend to gain electrons, forming anions with the same number of electrons as the nearest following noble gas.

2.24

Atomic mass represents the average mass of the isotopes that compose the element. The average calculated atomic mass is weighted according to the natural abundance of each isotope.

Atomic mass = >’ (fraction of isotope n) X (mass of isotope n) n

PEE:

In a mass spectrometer, the sample is injected into the instrument and vaporized. The vaporized atoms are then ionized by an electron beam. The electrons in the beam collide with the vaporized atoms, removing electrons from the atoms and creating positively charged ions. Charged plates with slits in them accelerate the positively charged ions into a magnetic field, which deflects them. The amount of deflection depends on the mass of the ions—lighter ions are deflected more than heavier ones are. Finally, the ions strike a detector and produce an electrical signal that is recorded.

2.26

The result of the mass spectrometer is the separation of the atoms in the sample according to their mass, producing a mass spectrum. The position of each peak on the x-axis gives the mass of the isotope, and the intensity (indicated by the height of the peak) gives the relative abundance of that isotope.

Zsa

A mole is an amount of material. It is defined as the amount of material containing 6.0221421 < 10” particles (Avogadro’s number). The numerical value of the mole is defined as being equal to the number of atoms in exactly 12 grams of pure carbon-12. It is useful for converting number of atoms to moles of atoms and moles of atoms to number of atoms.

2.28

The mass corresponding to a mole of one element is different from the mass corresponding to a mole of another element because the mass of an atom of each element is different. A mole is a specific number of atoms; so the heavier the mass of each atom, the heavier the mass of one mole of atoms.

Chapter 2 Atoms and Elements

Problems by Topic The Laws of Conservation of Mass, Definite Proportions, and Multiple Proportions 229

Given: 1.50 g hydrogen; 12.0 g oxygen Find: grams of water vapor Conceptual Plan: total mass reactants = total mass products Solution: Mass of reactants = 1.50 g hydrogen + 12.0 g oxygen = 13.5 grams Mass of products = mass of reactants = 13.5 grams water vapor Check: According to the law of conservation of mass, matter is neither created nor destroyed in a chemical reaction. Since water vapor is the only product, the masses of hydrogen and oxygen must combine to form the mass of water vapor.

i)bo o

Given: 21 kg gasoline; 84 kg oxygen Find: mass of carbon dioxide and water Conceptual Plan: total mass reactants = total mass products Solution: Mass of reactants = 21 kg gasoline + 84kg oxygen = 105kg Mass of products = mass of reactants = 105 kg of carbon dioxide and water Check: According to the law of conservation of mass, matter is neither created nor destroyed in a chemical reaction. Since carbon dioxide and water are the only products, the masses of gasoline and oxygen must combine to form the mass of carbon dioxide and water.

twoe) ps

Given: sample 1: 38.9 g carbon, 448 g chlorine; sample 2: 14.8 g carbon, 134 g chlorine Find: consistent with definite proportions Conceptual Plan: Determine mass ratio of samples 1 and 2 and compare. mass of chlorine

mass of carbon

Soluti tal 448 g chorine ution: SAL en ene 38.9 g carbon

bes = 134 g chlorine S “» Bat Fed: Roca (4.8 g carbon

Loe

Results are not consistent with the law of definite proportions because the ratio of chlorine to carbon is not the same.

Check: According to the law of definite proportions, the mass ratio of one element to another is the same for all samples of the compound. tN Wwbo

Given: sample 1: 6.98 grams sodium, 10.7 grams chlorine; sample 2: 11.2 g sodium, 17.3 grams chlorine Find; consistent with definite proportions Conceptual Plan: Determine mass ratio of samples 1 and 2 and compare. mass of sodium

ane

PAG

10.7 g chorine 34 33

eccdiin,

gchlorine _ Ge 17.3

Thos

lo gscdue

Results are consistent with the law of definite proportions.

Check: According to the law of definite proportions, the mass ratio of one element to another is the same for all samples of the compound.

253

Given: mass ratio sodium to fluorine = 1.21:1; sample = 28.8 g sodium Conceptual Plan: g sodium — g fluorine

Find: g fluorine

mass of fluorine mass of sodium

| g fluorine flu g fluorine ain = 23.8 g x 1 i 28. gsediam ion: 28.8 Solution:

is reasonable because it is less Check: The units of the answer (g fluorine) are correct. The magnitude of the answer than the grams of sodium.

Chapter 2 Atoms and Elements 2.34

kg magnesium Given: sample 1: 1.65 kg magnesium, 2.57 kg fluorine; sample 2: 1.32 2 Find: g fluorine in sample fluorine (kg) —> mass fluorine (g) Conceptual Plan: mass magnesium and mass fluorine > mass ratio > mass 1000 g kg

mass of fluorine mass of magnesium

2.57kg fluorine Solution: mass ratio =

—_—‘1.56 kg fluorine

1.65 kg magnesium ~~ 1,00 kg magnesium : 1.56 kg fluorine ” 1000 g

= 2.06 X 10° g fluorine

kg Sr og maenestuna ke [dz megnestunl

because it is Check: The units of the answer (g fluorine) are correct. The magnitude of the answer is reasonable greater than the mass of magnesium and the ratio is greater than 1.

Given: 1g osmium: sample 1 = 0.168 g oxygen; sample 2 = 0.3369 g oxygen Find: consistent with multiple proportions Conceptual Plan: Determine mass ratio of oxygen. mass of oxygen sample 2 mass of oxygen sample 1

The ratio is a small whole number. Results are consistent with multiple proportions.

= 2,00

ee Solution: 2 0.168 g oxygen

Check: According to the law of multiple proportions, when two elements form two different compounds, the masses of element B that combine with 1 g of element A can be expressed as a ratio of small whole numbers.

2.30

Given: | g palladium: compound A: 0.603 g S; compound B: 0.301 g S; compound C: 0.151 gS Find: consistent with multiple proportions Conceptual Plan: Determine mass ratio of sulfur in the three compounds. mass of sulfur sample A — mass of sulfursample A mass of sulfur sample B mass of sulfur sample C

Solution:

0.603 g S in compound A

0.301 gSincompoundB

0.603 g S in compound A : = 3.99~4 0.151 g S in compound C

= 2.00

0.301 g S in compound B

i50

_—smass of sulfur sample B mass of sulfur sample C

dee

0.151gSincompoundC The ratio of each is a small whole number. Results are consistent with multiple proportions. Check: According to the law of multiple proportions, when two elements form two different compounds, the masses of element B that combine with | g of element A can be expressed as a ratio of small whole numbers.

Given: sulfur dioxide = 3.49 g oxygen and 3.50 g sulfur; sulfur trioxide = 6.75 g oxygen and 4.50 g sulfur Find: mass oxygen per g S for each compound and then determine the mass ratio of oxygen mass of oxygen in sulfur dioxide mass of sulfur in sulfur dioxide

mass of oxygen in sulfur trioxide mass of sulfur in sulfur trioxide

3.49 goxygen 0.997 g oxygen — = 3.50 g sulfur 1 g sulfur 1.50 g oxygen in sulfurtrioxide 1.50 3

Solution: sulfur dioxide =

(997s oxyeen Waultur dioxide 5S

— 5

mass of oxyen in sulfur trioxide mass of oxyen in sulfur dioxide

sulfur trioxide =

6.75

goxygen eae 4.50 g sulfur

_1.50 g oxygen Slee 1 g sulfur

Thee ratio ratio is i converted from ; 50: to 3:2:2 because bec 1.50:1 the law

of multiple proportions states that the ratio is in small whole numbers. Ratio is in small whole numbers and is consistent with multiple proportions. Check: According to the law of multiple proportions, when two elements form two different compounds, the masses of element B that combine with | g of element A can be expressed as a ratio of small whole numbers.

2.38

Given: sulfur hexafluoride = 4.45 g fluorine and 1.25 g sulfur; sulfur tetrafluoride = 4.43 g fluorine and 1.87 g sulfur Find: mass fluorine per g S for each compound and then determine the mass ratio of fluorine mass of fluorine in sulfur hexafluoride

mass of fluorine in sulfur tetrafluoride

mass of oxyen in sulfur hexafluoride

mass of sulfur in sulfur hexafluoride

mass of sulfur in sulfur tetrafluoride

mass of oxyen in sulfur tetrafluoride

Solution: sulfur hexafluoride =

sulfur tetrafluoride =

4.45 gfluorine 1.25 g sulfur

4.43 g fluorine 1.87 g sulfur

3.56 g fluorine ry

g sulfur

2.369 g fluorine = ; 1 g sulfur

Chapter 2 Atoms and Elements F ee 3.56 g fluorine in sulfur hexafluoride 2.369 g fluorine in sulfur tetrafluoride

1.50 3 1 2

The ratio is converted from 1.50:1 to 3:2 because

the law of multiple proportions states that the ratio is in small whole numbers. The ratio is in small whole numbers and is consistent with multiple proportions. Check: According to the law of multiple proportions, when two elements form two different compounds, the masses of element B that combine with | g of element A can be expressed as a ratio of small whole numbers.

Atomic Theory, Nuclear Theory, and Subatomic Particles (a)

Sulfur and oxygen atoms have the same mass. /nconsistent with Dalton’s atomic theory because only atoms of the same element have the same mass. All cobalt atoms are identical. Consistent with Dalton’s atomic theory because all atoms of a given element have the same mass and other properties that distinguish them from atoms of other elements. Potassium and chlorine atoms combine in a 1:1 ratio to form potassium chloride. Consistent with Dalton’s atomic theory because atorns combine in simple whole-number ratios to form compounds. Lead atoms can be converted into gold. Inconsistent with Dalton’s atomic theory because atoms of one element cannot change into atoms of another element.

(a)

All carbon atoms are identical. Consistent with Dalton’s atomic theory because all atoms of a given element have the same mass and other properties that distinguish them from atoms of other elements. An oxygen atom combines with 1.5 hydrogen atoms to form a water molecule. /nconsistent with Dalton’s atomic theory because atoms combine in simple whole-number ratios to form compounds. An oxygen atom actually combines with two hydrogen atoms to form a water molecule. Two oxygen atoms combine with a carbon atom to form a carbon dioxide molecule. Consistent with Dalton’s atomic theory because atoms combine in simple whole-number ratios to form compounds. The formation of a compound often involves the destruction of one or more atoms. /nconsistent with Dalton’s atomic theory. Atoms change the way they are bound together with other atoms when they form a new substance, but they are neither created nor destroyed.

(b)

(a)

(b)

(d)

2.42

(a)

(b)

(c)

(d)

The volume of an atom is mostly empty space. Consistent with Rutherford’s nuclear theory because most of the volume of the atom is empty space, throughout which tiny, negatively charged electrons are dispersed. The nucleus of an atom is small compared to the size of the atom. Consistent with Rutherford’s nuclear theory because most of the atom’s mass and all of its positive charge are contained in a small core called the nucleus. ‘Neutral lithium atoms contain more neutrons than protons. /nconsistent with Rutherford’s nuclear theory because it did not distinguish where the mass of the nucleus came from other than from the protons. Neutral lithium atoms contain more protons than electrons. /nconsistent with Rutherford’s nuclear theory because there are as many negatively charged particles outside the nucleus as there are positively charged particles in the nucleus.

Since electrons are smaller than protons and since a hydrogen atom contains only one proton and one electron, it must follow that the volume of a hydrogen atom is mostly due to the proton. Inconsistent with Rutherford’s nuclear theory because most of the volume of the atom is empty space, throughout which tiny, negatively charged electrons are dispersed. A nitrogen atom has seven protons in its nucleus and seven electrons outside its nucleus. Cc rene with Rutherford’s nuclear theory because there are as many negatively charged particles outside the nucleus as there are positively charged particles in the nucleus. A phosphorus atom has 15 protons in its nucleus and 150 electrons outside its nucleus. Inconsistent with Rutherford’s nuclear theory because there are as many negatively charged particles outside the nucleus as there are positively charged particles in the nucleus. The majority of the mass of a fluorine atom is due to its nine electrons. Inconsistent with Rutherford’s nuclear theory because most of the atom’s mass and all of its positive charge are contained in a small core called the nucleus.