Type:
Solution Manual
Chemistry A Molecular Resource: Approach Edition:
6th Edition
Author(s):
Nivaldo J. Tro
1 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. 1.2 The main goal of chemistry is to seek to understand the behavior of matter by studying the behavior of atoms and molecules. 1.3 The scientific approach to knowledge is based on observation and experimentation. 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 a scientific law: a brief statement that summarizes past observations and predicts future ones. One or more well-established 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 carefully measured 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 created 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. 1.7 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 repeatedly. 1.8 Matter can be classified according to its state—solid, liquid, or gas—and according to its composition. 1.9 In solid matter, atoms or molecules pack close to each other in fixed locations. Although the atoms and molecules in a solid vibrate, they do not move around or past each other. Consequently, a solid has a fixed volume and a rigid shape. In liquid matter, atoms or molecules pack about as closely as they do in solid matter, but they are free to move relative to each other, giving liquids a fixed volume but not a fixed shape. Liquids assume the shape of their container. In gaseous matter, atoms or molecules have a lot of space between them and are free to move relative to one another, 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. 1.11 A pure substance is composed of only one type of atom or molecule. In contrast, a mixture is a substance composed of two or more different types of atoms or molecules that can be combined in variable proportions. 1.12 An element is a pure substance that cannot be decomposed into simpler substances. A compound is composed of two 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). 1.15 Mixtures of miscible liquids (substances that easily mix in all proportions) 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. 1.17 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. 1.18 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. 1.21 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 the degree size is 1.8 times larger than the degree size for the Fahrenheit scale. 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 103 . Therefore
1 kilometer = 1000 meters = 103 meters Similarly, the millimeter has the prefix “milli,” meaning 0.001 or 10−3 .
1 millimeter = 0.001 meters = 10−3 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 (m3 ), and density in grams per cubic centimeter (g/cm3 ). 1.24 The density (d) of a substance is the ratio of its mass (m) to its volume (V):
Density =
Mass m or d = Volume V
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 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 tends toward being either too high or too low. Systematic error does not average out with repeated trials. 1.32 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 a. This statement is a theory because it attempts to explain why. It is not possible to observe individual atoms. b. This statement is an observation. c. This statement is a law because it summarizes many observations and can explain future behavior. d. This statement is an observation. 1.34 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. 1.35 a. If we divide the mass of the oxygen by the mass of the carbon, the result is always 4/3. b. If we divide the mass of the oxygen by the mass of the hydrogen, the result is always 16. c. These observations suggest that the masses of elements in molecules are ratios of whole numbers [4:3 and 16:1, respectively, for parts (a) and (b)]. d. Atoms combine in small whole-number ratios and not as random weight ratios. 1.36 Many hypotheses may be developed. One hypothesis is that a large explosion generated galaxies with fragments that are still moving away from each other.
The Classification and Properties of Matter 1.37 a. Sweat is a homogeneous mixture of water, sodium chloride, and other components. b. Carbon dioxide is a pure substance that is a compound (two or more elements bonded together). c. Aluminum is a pure substance that is an element (element 13 in the periodic table). d. Vegetable soup is a heterogeneous mixture of broth, chunks of vegetables, and extracts from the vegetables. 1.38 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. c. Iron is a pure substance that is an element (element 26 in the periodic table). d. Carbon monoxide is a pure substance that is a compound (two or more elements bonded together). 1.39 Substance
Pure or Mixture
Type (element or compound)
aluminum
pure
element
apple juice
mixture
neither—mixture
hydrogen peroxide
pure
compound
chicken soup
mixture
neither—mixture
1.40 Substance
Pure or Mixture
Type (element or compound)
water
pure
compound
coffee
mixture
neither—mixture
ice
pure
compound
carbon
pure
element
1.41 a. pure substance that is a compound (one type of molecule that contains two different elements) b. heterogeneous mixture (two different molecules that are segregated into regions) c. homogeneous mixture (two different molecules that are randomly mixed) d. pure substance that is an element (individual atoms of one type) 1.42 a. pure substance that is an element (individual atoms of one type) b. homogeneous mixture (two different molecules that are randomly mixed) c. pure substance that is a compound (one type of molecule that contains two different elements) d. pure substance that is a compound (one type of molecule that contains two different elements) 1.43 a. physical property (color can be observed without making or breaking chemical bonds) b. chemical property (flammability must be observed by making or breaking chemical bonds) c. physical property (the phase can be observed without making or breaking chemical bonds) d. physical property (density can be observed without making or breaking chemical bonds) e. 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. physical property (color can be observed without making or breaking chemical bonds) b. physical property (odor can be observed without making or breaking chemical bonds) c. chemical property (reactivity must be observed by making or breaking chemical bonds) d. chemical property (decomposition involves breaking bonds, so bonds must be broken to observe this property) e. 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) b. physical property (shininess is a physical property, so it can be observed without making or breaking chemical bonds) c. physical property (odor can be observed without making or breaking chemical bonds) d. chemical property (burning involves breaking and making bonds, so bonds must be broken and made to observe this property) 1.46 a. physical property (vaporization is a phase change, so it can be observed without making or breaking chemical bonds) b. physical property (sublimation is a phase change, so it can be observed without making or breaking chemical bonds) c. chemical property (rusting involves the reaction of iron with oxygen to form iron oxide; observing this process involves making and breaking chemical bonds) d. physical property (color can be observed without making or breaking chemical bonds) 1.47 a. chemical change (new compounds are formed as methane and oxygen react to form carbon dioxide and water) b. physical change (vaporization is a phase change and does not involve the making or breaking of chemical bonds) c. chemical change (new compounds are formed as propane and oxygen react to form carbon dioxide and water) d. chemical change (new compounds are formed as the metal in the frame is converted to oxides) 1.48 a. chemical change (new compounds are formed as the sugar burns) b. physical change (dissolution is a phase change and does not involve the making or breaking of chemical bonds) c. physical change (this is simply a rearrangement of the atoms) d. chemical change (new compounds are formed as the silver converts to an oxide) 1.49 The metabolism of carbohydrates, such as sugar in living organisms, is a chemical change because sugars are broken down to form other compounds and generate energy. 1.50 The act of swallowing water is a physical change, but water is an important reactant in many biochemical reactions in the body where it is used to convert various molecules. 1.51 a. physical change (vaporization is a phase change and does not involve the making or breaking of chemical bonds) b. chemical change (new compounds are formed) c. physical change (vaporization is a phase change and does not involve the making or breaking of chemical bonds) 1.52 a. physical change (vaporization of butane is a phase change and does not involve the making or breaking of chemical bonds) b. chemical change (new compounds are formed as the butane combusts) c. physical change (vaporization of water is a phase change and does not involve the making or breaking of chemical bonds)
Units in Measurement 1.53 a. To convert from °F to °C, first find the equation that relates these two quantities. °C =
°F − 32 . Now substitute °F into the equation and compute the 1.8
answer. Note: The number of digits reported in this answer follows significant figure conventions covered in Section 1.6.
°F − 32 0. = = 0. °C 1.8 1.8 b. To convert from K to °F, first find the equations that relate these two quantities. °F − 32 K = °C + 273.15 and °C = 1.8 Because these equations do not directly express K in terms of °F, you must combine the equations and then solve the equation for °F. Substituting for °C °C =
results in the following:
°F − 32 °F − 32 + 273.15; rearrange K − 273.15 = ; 1.8 1.8 rearrange 1.8(K − 273.15) = (°F − 32); finally, °F = 1.8(K − 273.15) + 32. Now substitute K into the equation and compute the answer. °F = 1.8(77 − 273.15) + 32 = 1.8(−196) + 32 = −353 + 32 = −321 °F Alternatively, you could first convert to K to °C, using the formula K = 273.15 + °C, and then convert °C to °F. So, 77 K = 273.15 + °C. Rearranging °F − 32 to solve for °C we get °C = 77K − 273.15 = −196.15 °C. Then substitute the temperature in °C in the formula °C = , getting 1.8 °F − 32 −196.15 °C = . Multiplying both sides by 1.8, we get −196.15 °C × 1.8 = °F − 32, so −353.07 °C = °F − 32. Adding 32 to both sides of – – 1.8 the equation we get °F = −321 –.07 °F or −321 °F, with the appropriate number of significant figures. °F − 32 c. To convert from °F to °C, first find the equation that relates these two quantities. °C = . Now substitute °F into the equation and compute the 1.8 K=
answer.
−109 °F − 32 °F − 141 = = −78.3 °C 1.8 1.8 d. To convert from °F to K, first find the equations that relate these two quantities. °C =
K = °C + 273.15 and °C =
°F − 32 1.8
Because these equations do not directly express K in terms of °F, you must combine the equations and then solve the equation for K. Substituting for
°F − 32 + 273.15. 1.8 Now substitute °F into the equation and compute the answer. (98.6 − 32) 66.6 K= + 273.15 = + 273.15 = 37.0 + 273.15 = 310.2 K 1.8 1.8 °F − 32 Alternatively, you could first convert to °F to °C, using the formula °C = , and then convert °C to K, using K = 273.15 + °C. So, 1.8 98.6 °F − 32 66.6 °F °C = = = 37.0 °C. Then K = 273.15 + 37.0 °C = 310.2 K. 1.8 1.8 °C, K =
1.54
a. To convert from °F to °C, first find the equation that relates these two quantities. °C =
°F − 32 1.8
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. Remember that the 1.8 is exact because it is part of the definition of the temperature scales.
°C =
212 °F − 32 °F 1.8
=
180. = 100. °C 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. Because this equation gives the temperature in K directly, simply substitute the correct value for the temperature in °C and compute the answer.
K = 22 °C + 273.15 = 295 K
°F − 32 . Because these equations do not 1.8 °F − 32 + 273.15; directly express K in terms of °F, you must combine the equations and then solve the equation for °F. Substituting for °C, K = 1.8 °F − 32 rearrange K − 273.15 = ; rearrange 1.8(K − 273.15) = (°F − 32); rearrange °F = 1.8(K − 273.15) + 32. Now substitute K into the 1.8
c. To convert from K to °F, first find the equations that relate these two quantities: K = °C + 273.15 and °C =
equation and compute the answer.
°F = 1.8(0.00 K − 273.15) + 32 = 1.8(−273.15 K) + 32 = −491.67 + 32 = −459.67 °F d. 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. °C = 2.735 − 273.15 = −270.42 °C
1.55 To convert from °F to °C, first find the equation that relates these two quantities. °C =
°F − 32 . Now substitute °F into the equation and compute the 1.8
answer. Note: The number of digits reported in this answer follows significant figure conventions covered in Section 1.6.
°C =
−128.6 °F − 32 °F 1.8
=
−160.6 = −89.22 °C 1.8
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 = −89.22 °C + 273.15 = 193.93 K 1.56 To convert from °F to °C, first find the equation that relates these two quantities. °C =
°F − 32 . Now substitute °F into the equation and compute the 1.8
answer. Note: The number of digits reported in this answer follows significant figure conventions covered in Section 1.6.
°C =
134 °F − 32 °F 1.8
=
102 = 56. 6667 °C = 56.7 °C – 1.8
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 = 329.8 K
–
1.57 Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions. a. 10−6 is equivalent to “micro,” so 3.4 × 10−6 m = 3.4 micrometers = 3.4 μm. b. 10−12 is equivalent to “pico,” so 11 × 10−12 s = 11 picoseconds = 11 ps. c. 103 is equivalent to “kilo,” so 7.4 × 103 g = 7.4 kilograms = 7.4 kg. d. 106 is equivalent to “mega,” so 5.2 × 106 L = 5.2 megaliters = 5.2 ML. 1.58 Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions. a. 12.1 × 108 g = 1.21 × 109 g; 109 is equivalent to “giga,” so 1.21 × 109 g = 1.21 gigagrams = 1.21 Gg. b. 45.9 × 10−7 s = 4.59 × 10−6 s; 10−6 is equivalent to “micro,” so 4.59 × 10−6 s = 4.59 microseconds = 4.59 μs. 4 3 3 3 c. 11.2 × 10 m = 112 × 10 m; 10 is equivalent to “kilo,” so 112 × 10 m = 112 kilometers = 112 km. −7 −6 −6 −6 d. 21.3 × 10 L = 2.13 × 10 L; 10 is equivalent to “micro,” so 2.13 × 10 L = 2.13 microliters = 2.13 μL. 1.59 Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions. a. 10−9 is equivalent to “nano,” so 4.5 ns = 4.5 nanoseconds = 4.5 × 10−9 s. b. 10−15 is equivalent to “femto,” so 18 fs = 18 femtoseconds = 18 × 10−15 s = 1.8 × 10−14 s. Remember that in scientific notation, the first number should be smaller than 10.
−12
−12
−10
is equivalent to “pico,” so 128 pm = 128 × 10 c. 10 m = 1.28 × 10 m. Remember that in scientific notation, the first number should be smaller than 10. −6 −6 −5 d. 10 is equivalent to “micro,” so 35 μm = 35 micrometers = 35 × 10 m = 3.5 × 10 m. Remember that in scientific notation, the first number should be smaller than 10. 1.60 Use Table 2 to determine the appropriate prefix multiplier and substitute the meaning into the expressions. −6 −6 −5 a. “μ” is equivalent to “micro,” or 10 , so 35 μL = 35 microliters = 35 × 10 L = 3.5 × 10 L. Remember that in scientific notation, the first number should be smaller than 10. 6 6 8 b. “M” is equivalent to “mega” or 10 , so 225 Mm = 225 megameters = 225 × 10 m = 2.25 × 10 m. Remember that in scientific notation, the first number should be smaller than 10. c. “T” is equivalent to “tera” or 1012 , so 133 Tg = 133 teragrams = 133 × 1012 g = 1.33 × 1014 g. Remember that in scientific notation, the first number should be smaller than 10. d. “c” is equivalent to “centi” or 10−2 , so 1.5 cg = 1.5 centigrams = 1.5 × 10−2 g
1.61 b. Given: 515 km
Find: dm
Conceptual Plan:
km → m → dm 1000 m 1 km
10 dm 1m
1000 m
Solution: 515 km ×
1 km
×
10 dm = 5.15 × 106 dm 1 m
Check: The units (dm) are correct. The magnitude of the answer (106 ) makes physical sense because a decimeter is a much smaller unit than a kilometer. Given: 515 km Find: cm Conceptual Plan:
km → m → cm 1000 m 1 km
100 dm 1m
100 cm = 5.15 × 107 cm 1 m 1 km Check: The units (cm) are correct. The magnitude of the answer (107 ) makes physical sense because a centimeter is a much smaller unit than a kilometer Solution: 515 km ×
or a decimeter. c. Given: 122.355 s Conceptual Plan:
1000 m
×
Find: ms
s → ms 1000 ms 1s
Solution: 122.355 s ×
1000 ms = 1.22355 × 105 ms 1 s
Check: The units (ms) are correct. The magnitude of the answer (105 ) makes physical sense because a millisecond is a much smaller unit than a second. Given: 122.355 s Find: ks Conceptual Plan:
s → ks 1 ks 1000 s
1 ks = 1.22355 × 10−1 ks = 0.122355 ks 1000 s Check: The units (ks) are correct. The magnitude of the answer (10−1 ) makes physical sense because a kilosecond is a much larger unit than a second. Solution: 122.355 s × d. Given: 3.345 kJ Conceptual Plan:
Find: J
kJ → J 1000 J 1 kJ
Solution: 3.345 kJ ×
1000 J 1 kJ
= 3.345 × 103 J
Check: The units (J) are correct. The magnitude of the answer (103 ) makes physical sense because a joule is a much smaller unit than a kilojoule. Given: 3.345 × 103 J (from above) Find: mJ Conceptual Plan:
J → mJ 1000 mJ 1J
Solution: 3.345 × 103 J ×
1000 mJ 1 J
= 3.345 × 106 mJ
Check: The units (mJ) are correct. The magnitude of the answer (106 ) makes physical sense because a millijoule is a much smaller unit than a joule. 1.62 a. Given: 355 km/s Conceptual Plan:
Find: cm/s
km/ s → m / s → cm/ s 1000 m 1 km
100 cm 1m
355 km 1000 m 100 cm cm × × = 3.55 × 107 1 m 1s s 1 km Check: The units (cm/s) are correct. The magnitude of the answer (107 ) makes physical sense because a centimeter is a much smaller unit than a Solution:
kilometer. Given: 355 km/s
Find: m/ms