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Solution Manual for Chemistry: The Science in Context, 3rd Edition

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    1 CHAPTER 1 | Matter, Energy, and the Origins of the Universe 1.1. Collect and Organize In Figure P1.1(a) we are shown “molecules” each consisting of one red sphere and one blue sphere and in Figure P1.1(b) we have separate blue spheres and red spheres. In each figure we are to identify whether the substance(s) depicted is a solid, liquid, or gas and if the figures show pure elements or compounds. Analyze A pure substance (whether element or compound) is composed of all the same type of molecule or atom, not a mixture of two kinds. An element is composed of all the same type of atom and a compound is composed of two or more types of atoms. Solids have a definite volume and a highly ordered arrangement where the particles are close together, liquids also have a definite volume but have a disordered arrangement of particles that are close together, and gases have disordered particles that fill the volume of the container and are far apart from each other. Solve (a) Because the particles each consist of one red sphere and one blue sphere, all the particles are the same — this is a pure compound. The particles fill the container and are disordered, so these particles are in the gas phase. (b) Because Figure P1.1(b) shows a mixture of red and blue spheres, this is depicting a mixture of blue element atoms and red element atoms. The blue spheres fill the container and are disordered, so these particles are in the gas phase. The red spheres have a definite volume and are slightly disordered, so these particles are in the liquid phase. Think about It Remember that both elements and compounds may be either pure or present in a mixture. 1.2. Collect and Organize In Figure P1.2(a) we are shown “atoms” of only red spheres and in Figure P1.2(b) we have “molecules” consisting of two red spheres or two blue spheres. In each figure we are to identify whether the substance(s) depicted is a solid, liquid, or gas and if the figures show pure elements or compounds. Analyze A pure substance (whether element or compound) is composed of all the same type of molecule or atom, not a mixture of two kinds. An element is composed of all the same type of atom and a compound is composed of two or more types of atoms. Solids have a definite volume and a highly ordered arrangement where the particles are close together, liquids also have a definite volume but have a disordered arrangement of particles that are close together, and gases have disordered particles that fill the volume of the container and are far apart from each other. Solve (a) Because all the atoms are of the same type, Figure P1.2(a) depicts a pure element. The particles take up a definite volume and are ordered, so this element is in the solid phase. (b) Because there is a mixture of blue diatomic molecules and red diatomic molecules, Figure P1.2(b) depicts a mixture of two elements. Both the blue and red diatomic particles fill the container’s volume and are highly disordered; the mixture depicted is in the gas phase. Think about It Elements do not need to be present as single atoms. They may be diatomic as in H 2 or Br 2 , or even more highly associated as in S 8 or P 4 . 1.3. Collect and Organize In this question we are to consider whether the reactants as depicted undergo a chemical reaction and/or a phase change.

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    2 | Chapter 1 Analyze Chemical reactions involve the breaking and making of bonds in which atoms are combined differently in the products compared to that of the reactants. In considering a possible phase change, solids have a definite volume and a highly ordered arrangement where the particles are close together, liquids also have a definite volume but have a disordered arrangement of particles that are close together, and gases have disordered particles that fill the volume of the container and are far apart from each other. Solve In Figure P1.3 two pure elements (red – red and blue – blue) in the gas phase recombine to form a compound (red – blue) in the solid phase (ordered array of molecules). Therefore , answer b describes the reaction shown. Think about It A phase change does not necessarily accompany a chemical reaction. We will learn later that the polarity of the product will determine whether or not a substance will be in the solid, liquid, or gaseous state at a given temperature. 1.4. Collect and Organize In this question we are to consider whether the reactants as depicted undergo a chemical reaction (either recombination or decomposition) and/or a phase change. Analyze Chemical reactions involve the breaking and making of bonds in which atoms are combined differently in the products compared to that of the reactants. In considering a possible phase change, solids have a definite volume and a highly ordered arrangement where the particles are close together, liquids also have a definite volume but have a disordered arrangement of particles that are close together, and gases have disordered particles that fill the volume of the container and are far apart from each other. Solve In Figure P1.4 we see that no recombination of the diatomic molecules occurs. The pure element (red – red) condenses to a slightly disordered phase while the other element (blue – blue) remains in the gas phase. Therefore , answer a describes the reaction pictured. Think about It Cooling of air in this fashion to different temperatures separates the components of air. 1.5 Collect and Organize From the space - filling model shown , we are to write the formula for the chemical represented. Analyze From the Atomic Color Palette shown , on the inside back cover of the textbook, we see that this model contains three hydrogen atoms bonded to a carbon ato m which in turn is bonded to an O – H unit. Solve H 3 COH or CH 4 O Think about It This model represent s methanol , an d the presence of the O – H unit classifies it as an alcohol. Sometimes methanol is named “wood alcohol.” 1.6. Collect and Organize From the ball and stick model of acetone shown , we are to write the chemical formula. Analyze From the Atomic Color Palette shown on the inside back cover of the textbook, we see that this model contains two CH 3 units bonded to a C – O unit.

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    Matter, Energy, and the Origins of the Universe | 3 Solve H 3 C(CO)CH 3 or C 3 H 6 O Think about It Acetone is classified as a ketone and is a useful solvent. It is the main component in nail - polish remover. 1.7. Collect and Organize This question considers if and how matter and energy are related. In particular, we consider whether the sun is all mass or all energy. Analyze Einstein showed that matter and energy are interconvertible through E = mc 2 . Solve The sun is an example where matter is being changed into energy through nuclear fusion reactions. Therefore, both students are correct. Think about It Through Einstein’s equation we see that a little bit of mass contains a great deal of energy locked into the nuclei of the atoms. 1.8. Collect and Organize In this question, we consider how elements and compounds compare. Analyze Compounds can be made from the elements and have different types of atoms in them. Elements are composed of atoms all of the same kind. Solve Compounds are different from elements in that they are made up of two or more elements, these elements can be separated from each other (but elements cannot be separated further), and compounds have different chemical and physical properties from the elements that compose them. Elements are also rarely found in nature. Compounds are similar to elements in that they are composed of atoms, have definite physical and chemical properties, and can be isolated in pure form. Think about It By combining the different elements with each other, we can arrive at many, many compounds which are used as fuels, medicines, plastics, etc. 1.9. Collect and Organize For this question we are to list some chemical and physical properties of gold. Analyze A chemical property is seen when a substance undergoes a chemical reaction thereby becoming a different substance. A physical property can be seen without any transformation of one substance into another. Solve One chemical property of gold is its resistance to corrosion (oxidation). Gold’s physical properties include its density, color, melting temperature, and electrical and thermal conductivity. Think about It Another metal that does not corrode (or rust) is platinum. Platinum and gold, along with palladium, are often called “noble metals.”

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    4 | Chapter 1 1.10. Collect and Organize For this question we are to compare the physical properties of gold and silver. Analyze Physical properties include color, metallic luster, malleability, ductility, melting point, boiling point, density, electrical conductivity, and thermal conductivity. Solve Both gold and silver have metallic luster, are malleable, and conduct electricity. However, gold and silver have different densities, different melting temperatures, and different colors. Think about It The yellow color of pure gold compared to most metals, which are silvery, is due to relativistic effects in the atom. 1.11. Collect and Organize This question asks us to use differences in physical or chemical properties to separate a mixture of substances. Using filtration, we are to propose a method to separate salt from sand in a mixture. Analyze The difference between sand and salt is salt’s solubility in water versus sand’s insolubility in water. Solve By adding water to the salt – sand mixture, the salt will dissolve. Passing the sand – solution mixture through a filter will leave the sand on the filter. The salt can be recovered by evaporating the water from the solution that passed through the filter. Think about It The filtration method works very well for a mixture of two substances, one of which is soluble and the other of which is insoluble in a solvent, and it is used often to separate and purify a desired product in a chemical reaction. 1.12. Collect and Organize We are to describe how we can use distillation to remove the salt from seawater. Analyze Distillation involves the boiling of an impure liquid until the vaporization of the pure liquid occurs. The pure liquid, now a pure compound, is isolated by cooling the vapor through a condenser. Solve When seawater is boiled, only the water enters the vapor phase, leaving the salts and other impurities behind. Condensing the vaporized water gives pure water. Think about It By distilling seawater, we can render the undrinkable seawater safe to drink. 1.13. Collect and Organize For the four processes named, we are to determine which involve a chemical change. Analyze Chemical changes involve transforming one substance into another to give that substance different physical and chemical properties.

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    Matter, Energy, and the Origins of the Universe | 5 Solve (a) Distillation purifies a substance — not a chemical change. (b) Combustion transforms the fuel (such as wood) into carbon dioxide and water — a chemical change. (c) Filtration separates substances from each other — not a chemical change. (d) Condensation changes a vapor into a liquid — not a chemical change. Think about It Distillation, filtration, and condensation all involve physical changes, not chemical changes. 1.14. Collect and Organize We are to determine whether gasohol (mixture of gasoline and ethanol) is a heterogeneous or a homogeneous mixture. Analyze In a homogeneous mixture the components are evenly distributed throughout the mixture , giving a uniform appearance to the eye. A heterogeneous mixture contains distinct, observable, individual components. Solve Because the ethanol is dissolved in the gasoline, we do not see regions of ethanol distinct from regions of gasoline. Gasohol, therefore, is homogeneous. Think about It Most cars on the road today can run well on a 10% ethanol to 90% gasoline fuel mixture. There is widespread interest in using ethanol as a fuel, despite its expense, because of lower CO and NO x emissions and as a means to replace reliance on petroleum, a fossil fuel. 1.15. Collect and Organize For the foods listed, we are to determine which are heterogeneous. Analyze A heterogeneous mixture has visible regions of different compositions. Solve There are clear regions of differing composition in a Snickers bar (b) and in an uncooked hamburger (d), but not in solid butter (a) or in grape juice (c). Think about It When butter is melted, you notice that there are milk solids and clear regions that are definitely discernible. Therefore, homogeneous solid butter becomes heterogeneous when heated. 1.16. Collect and Organize For the foods listed, we are to determine which are homogeneous. Analyze Homogeneous mixtures have the same composition throughout. Solve Freshly brewed coffee and vinegar (a, b) are homogeneous mixtures. A slice of white bread and a slice of ham (c, d) are heterogeneous mixtures. Think about It A slice of white bread is considered to be heterogeneous because its crust is different from the interior bread , and the bread contains gas bubbles that are clearly seen as tiny holes in the bread.

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    6 | Chapter 1 1.17. Collect and Organize For the foods listed, we are to determine which are heterogeneous. Analyze A heterogeneous mixture has visible regions of different compositions. Solve There are clear regions of differing composition in orange juice (with pulp) (d), but not in apple juice, cooking oil, solid butter, or tomato juice (a – c, e). Think about It When butter is melted, you notice that there are milk solids and clear regions that are definitely discernible. Therefore, homogeneous solid butter becomes heterogeneous when heated. 1.18. Collect and Organize For the substances listed, we are to determine which are homogeneous. Analyze Homogeneous mixtures have the same composition throughout. Solve A wedding ring, sweat, and compressed air in a scuba tank (a, b, e) are homogeneous. Nile River water and human blood (c, d) are heterogeneous. Think about It A gold wedding ring is made up of an alloy (a solid solution of one metal dissolved in another) of gold with another metal such as palladium or copper to give the soft gold metal strength and make it less expensive than 100% gold. 1.19. Collect and Organize We are asked in this question to name three properties to distinguish between table sugar, water, and oxygen. Analyze We can distinguish between substances using either physical properties (color, melting point, density, etc.) or chemical properties (chemical reactions, corrosion, flammability, etc.). Solve We can distinguish between table sugar, water, and oxygen by examining their physical states (sugar is a solid, water is a liquid, and oxygen is a gas) and by their densities, melting points, and boiling points. Think about It These three substances are also very different at the atomic level. Oxygen is a pure element made up of diatomic molecules, water is a liquid compound made up of discrete molecules of hydrogen and oxygen (H 2 O), and table sugar is a solid compound made up of carbon, hydrogen, and oxygen atoms. 1.20. Collect and Organize We are asked in this question to name three properties to distinguish between table salt, sand, and copper. Analyze We can distinguish between substances using either physical properties (color, melting point, density, etc.) or chemical properties (chemical reactions, corrosion, flammability, etc.). Solve We can distinguish between table salt, sand, and copper by examining their color (salt is composed of small cubic white crystals, sand is irregularly shaped and many - colored, and copper is reddish). Salt will dissolve in

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    Matter, Energy, and the Origins of the Universe | 7 water while sand and copper will not. Copper conducts electricity while solid table salt and sand do not. The densities of these substances will also differ. Think about It These three substances are also very different at the atomic level. Table salt is a crystalline ionic compound composed of sodium cations and chloride anions. Sand is a compound most commonly composed of silica, a compound of silicon and oxygen. Copper is a pure element and a metal. 1.21. Collect and Organize From the list of properties of sodium, we are to determine which are physical and which are chemical properties. Analyze Physical properties are those that can be observed without transforming the substance into another substance. Chemical properties are only observed when one substance reacts with another and therefore is transformed into another substance. Solve Density, melting point, thermal and electrical conductivity, and softness (a – d) are all physical properties while tarnishing and reaction with water (e and f) are both chemical properties. Think about It Because the density of sodium is less than that of water, a piece of sodium will float on water as it reacts. 1.22. Collect and Organize From the list of properties of hydrogen gas, we are to determine which are physical and which are chemical properties. Analyze Physical properties are those that can be observed without transforming the substance into another substance. Chemical properties are only observed when one substance reacts with another and therefore is transformed into another substance. Solve Density, boiling point, and electrical conductivity (a, c, and d) are all physical properties while the reaction of hydrogen with oxygen (b) is a chemical property. Think about It Because the density of hydrogen gas is lower than that of any other gas, a lightweight balloon filled with hydrogen will float in air like the more familiar helium balloon. 1.23. Collect and Organize In this question we consider whether distillation or filtration would be a suitable method to separate a protein (enzyme) in egg whites, for example, from water in a solution. Analyze Distillation vaporizes a liquid in a mixture and then condenses it in pure form. Filtration separates suspended solids from a liquid. Solve Distillation will separate the water and the dissolved proteins that have formed a homogeneous solution with the water. This would have to be accomplished, however, at low temperature, because heating the solution of the enzyme might cause a chemical change. Think about It Filtration is not appropriate because the enzyme in water forms a homogeneous solution.

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    8 | Chapter 1 1.24. Collect and Organize For this question we consider how to separate two phases of water: ice and liquid. Analyze We have two physical methods available to separate mixtures: distillation and filtration. Distillation vaporizes a liquid in a mixture and then condenses it in pure form. Filtration separates suspended solids from a liquid. Solve Water could be separated from ice by simply pouring the water off of the heterogeneous mixture — a crude filtration. Think about It Distillation will not work for the obvious reason that heating the ice – water mixture for the distillation will result in melting of the ice. 1.25. Collect and Organize For each of the elements listed, we are to determine the state (gas, liquid, or solid) at room temperature and pressure. Analyze Some periodic tables indicate which elements are liquids, which are solids, and which are gases under ordinary conditions. Solve At ordinary temperatures Fe is a solid, O 2 is a gas, and Hg is a liquid. Think about It Mercury is unique in that it is a liquid metal. This arises from relativistic effects that cause the mercury atoms to not be as chemically reactive and to be less willing to share their electrons with other mercury atoms , and so the metallic bond between mercury atoms is weak. 1.26. Collect and Organize From the list of consumer products , we are to decide which are solid at room temperature and pressure. Analyze Solids have a definite volume and shape of their own. Liquids have a definite volume but take on the shape of the container. Gases have no definite volume but rather fill their container. Solve At ordinary temperatures sea salt and ready - to - eat Jell - O are solids , but ketchup is not. Think about It Ketchup would be classified as a liquid since it takes on the shape of its container. 1.27. Collect and Organize We are to explain if an extensive property can be used to identify a substance. Analyze An extensive property is one that, like mass, length, and volume, is determined by size or amount. Solve Extensive properties will change with the size of the sample and therefore cannot be used to identify a substance.

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    Matter, Energy, and the Origins of the Universe | 9 Think about It We could, for example, have the same mass of feathers and lead, but their mass alone will not tell us which mass measurement belongs to which — the feathers or the lead. 1.28. Collect and Organize Of the properties listed, we are to choose which are intensive properties. Analyze An intensive property is not dependent on the size or amount of the sample. Solve Of the properties on the list, freezing point and temperature are intensive properties. Heat content depends on sample size. Think about It Intensive properties are related to chemical interactions between atoms and molecules in the substance. 1.29. Collect and Organize In this question we think about the information needed to formulate a hypothesis. Analyze A hypothesis is a tentative explanation for an observation. Solve To form a hypothesis we need at least one observation, experiment, or idea (from examining nature). Think about It A hypothesis that is tested and shown to be valid can become a theory. 1.30. Collect and Organize / Analyze In this question we consider how a hypothesis becomes a theory. Solve A theory is formed from a hypothesis when the hypothesis has been extensively tested by many observations and experiments. A theory is the best (current) possible explanation that is extensively supported by experimentation. Think about It A theory, further tested over time, may be elevated further to become a scientific law. 1.31. Collect and Organize We are to consider whether we can disprove a hypothesis. Analyze A hypothesis is a tentative explanation for an observation. Solve It is possible to disprove a scientific hypothesis. In fact, many experiments are designed to do just that as the best test of the hypothesis’s validity. Think about It It is even possible to disprove a theory (albeit harder to do so) or cause a theory to be modified when new evidence, a new experimental technique, or new data from a new instrument give observations that are counter to the explanation stated by the theory.

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    10 | Chapter 1 1.32. Collect and Organize In this question we consider why the atomic theory is universally accepted. Analyze Atomic theory states that all matter is composed of atoms. This theory is accepted despite the fact that scientists have not seen atoms directly. Solve The atomic theory is universally accepted because of the sheer volume of evidence from various experiments and observations that support the hypothesis that matter is composed of atoms. Think about It In the 1980s, the techniques of scanning tunneling microscopy and atomic force microscopy were developed to give us indirect “pictures” of atoms and molecules. 1.33. Collect and Organize We are to define theory as used in conversation. Analyze Theory in everyday conversation has a different meaning than it does in science. Solve Theory in normal conversation is someone’s idea or opinion or speculation that can easily be changed and may not have much evidence or many arguments to support it. Think about It A theory in science is a generally accepted and highly tested explanation of observed facts. 1.34. Collect and Organize We consider in this question whether a theory can be proven. Analyze Theory in science is the best (current) possible explanation that is extensively supported by experimentation and observations. Solve Theory is nearly equivalent to fact in science, without being the absolute truth. A theory is hard to prove absolutely, but has many, many supporting experiments whose observations strongly support the theory. Think about It One experiment that is counter to the explanation for a phenomenon explained by the theory could disprove a theory, so theories may be toppled and replaced with new explanations and theories. 1.35. Collect and Organize We are to compare SI units to English units. Analyze SI units are based on a decimal system to describe basic units of mass, length, temperature, energy, etc. English units vary. Solve SI units, which were based on the original metric system, can be easily converted into a larger or smaller unit by multiplying or dividing by multiples of 10. English units are more complicated to manipulate. For example, to convert miles to feet you have to know that there are 5280 feet in 1 mile and to convert gallons to quarts you have to know that 4 quarts are in 1 gallon.

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    Matter, Energy, and the Origins of the Universe | 11 Think about It Once you can visualize a meter, a gram, and a liter, using the SI system is quite convenient. 1.36. Collect and Organize / Analyze In this question we are to suggest two reasons why SI units are not more widely used in the United States. Solve English units instead of SI units are used everywhere in the United States because many of our manufacturing facilities have been built to make parts in inches or to bottle liquids in gallons. It has also been difficult for people used to buying and measuring in the English units to convert their thinking so as to visualize a kilometer instead of a mile or a liter instead of a quart. Think about It The only widespread everyday use of an SI unit in the United States is the 2 L soda bottle. 1.37. Collect and Organize This problem asks for the conversion of the speed of light in meters per second into kilometers per hour. We have to convert both distance and time into other units. We have to convert from meters (a short unit of distance) to kilometers (a longer unit of distance) and from seconds (a short unit of time) to an hour (a longer unit of time). There are several ways to convert these units; the most direct would be from meters to kilometers and from seconds to minutes to hours. We will need, therefore, the following equivalencies: 1000 meters = 1 kilometer 60 seconds = 1 minute 60 minutes = 1 hour Analyze The conversion of units for this problem can be expressed as ratios. Since we are converting from meters to kilometers the following ratio for unit conversions is appropriate: 1 km 1000 m To convert seconds to hours the following conversion ratios are appropriate: 1 min 1 hr and 60 s 60 min Solve 8 9 2.9979 10 m 1 km 60 s 60 min 1.0792 10 km/hr s 1000 m 1 min 1 hr       Think about It The speed of light in kilometers per hour is nearly the same order of magnitude as the speed in meters per second. This makes sense in that to convert from meters to kilometers we divided by 1000 and to convert from seconds to hours we multiplied by 3.6  10 3 . The two conversions, therefore, nearly offset each other. 1.38. Collect and Organize To compute the runner’s speed we have to use the definition: speed = distance/time. In the marathon runner’s case we are given distance in miles and time in hours plus additional minutes. The first calculation of speed, therefore, in miles per hour will not require any unit conversion. That result, however, will be used to compute the runner’s speed in meters per second using conversions for miles to meters and hours to seconds. Analyze The equation to compute speed is given by distance Speed = time

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    12 | Chapter 1 Because the time is given as 3 hours 40 minutes, we will have to convert the 40 minutes into a part of an hour using the fact that 1 hour = 60 minutes. We can then divide the marathon distance by this time in hours to get the speed in miles per hour. To convert this speed to meters per second, we can use the following conversions: 1 km 0.6214 mi and 1000 m 1 km 1 min 1 hr and 60 s 60 min Solve First, the number of hours for the runner to complete the marathon is 1 hr 3 hr + 40 min 3.67 hr 60 min   (a) The speed in miles per hour is 26.2 mi speed = 7.14 mi/hr 3.67 hr  (b) Converting this speed to meters per second gives mi 1 km 1000 m 1 hr 1 min 7.14 3.19 m/s hr 0.6214 mi 1 km 60 min 60 s      Think about It Both of these values seem reasonable. A walking pace is about 3 miles per hour, so running could be imagined to be twice that fast. Also, 3 meters per second easily can be run by a fast runner. 1.39. Collect and Organize For this problem we need to convert the distance of the Olympic mile (1500 m) in meters to miles and then to feet to compare that distance to an actual mile using a ratio and then converting that into a percentage. Analyze For converting the distance we can make use of these conversions: 1 km 0.6214 mi 5280 ft , , and 1000 m 1 km 1 mi To determine the percentage the Olympic mile distance is compared to the actual mile we will use Olympic mile distance in feet % distance = 100 5280 ft  Solve 1 km 0.6214 mi 5280 ft 1500 m 4921 ft 1000 m 1 km 1 mi     4920 ft % distance = 100 93.20% 5280 ft   Think about It This calculation shows that the Olympic mile is just a little bit shorter than the actual mile. 1.40. Collect and Organize To compare the prices of the soft drinks, we will have to convert to a common unit of volume, either ounces or liters.

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    Matter, Energy, and the Origins of the Universe | 13 Analyze We can use the following conversions to convert ounces to quarts to liters for the soft drink priced at $1.00 for 24 oz: 1 qt 0.9464 L and 32 oz 1 qt For the soft drink already priced per half - liter , the price per liter will be double. Solve $1.00 32 oz 1 qt $1.41/L 24 oz 1 qt 0.9464 L    $0.75 $1.50/L 0.50 L  Comparing the two prices in common units, it is better to buy the soft drink in the 24 oz bottles, not the half - liter bottles. Think about It This problem can also be solved by converting the half - liter volume to ounces. In that case, the price per ounce in the 24 oz bottle would be 4.2¢ while the price per ounce in the half - liter bottle would be 4.4¢. The conclusion is the same in either case. 1.41. Collect and Organize A light - year is the distance light travels in 1 year. To determine the distance of 4.3 light - years in kilometers, we will first have to convert 4.3 years into seconds, then use the speed of light to determine the distance the light travels over that amount of time. Analyze The length of time of 4.3 years in seconds can be found using the conversions 1 yr 1 d 1 hr 1 min , , , 365.25 d 24 hr 60 min 60 s The distance of 4.3 light - years in meters can be found from the speed of light: distance of 4.3 light - years = speed of light (in m/s)  4.3 yr (in seconds) This can be converted into kilometers using 1 km 1000 m Solve 8 365.25 d 24 hr 60 min 60 s 4.3 yr 1.36 10 s 1 yr 1 d 1 hr 1 min       8 8 13 2.9979 10 m 1 km Distance of Proxima Centauri = (1.36 10 s) 4.1 10 km s 1000 m       Think about It This is a very large distance since light travels so fast. The light - year, being such a large distance, is an ideal unit for expressing astronomical distances. 1.42. Collect and Organize To solve this problem, we need to know the volume of water in liters that are to be removed from the swimming pool. Using that volume and the rate at which the water can be siphoned, we can find the time it will take to remove the water. Analyze The volume of water to be removed in cubic meters can be found from Length of pool (in m) width of pool (in m) depth of water to be removed (in m)  

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    14 | Chapter 1 This volume will have to be converted to liters through the conversion 3 3 1 L 1 10 m   The time it would take to siphon the water is determined by the rate at which the siphon pump operates volume of water to be siphoned in liters Time to siphon the water = rate at which the water can be siphoned in liters per second Solve The volume of the water to be siphoned out of the pool is (using 6 in = 0.50 ft) 3 1 m 50.0 m 25.0 m 3.0 cm 37.5 m 100 cm           Converting this into gallons 3 4 3 3 1 L 37.5 m 3.75 10 L 1 10 m      The amount of time it would take to siphon this water is 4 3.75 10 L Time to siphon the water = 7210 s 5.2 L/s 1 min 7210 s 120 min or 2.0 h 60 s     Think about It This may be a surprisingly long time to siphon only 3 cm of water from the pool, but the total volume to be siphoned is quite large due to the size of the pool. 1.43. Collect and Organize To find the Calories burned by the wheelchair marathoner in a race , we can first find the number of hours the race will be for the marathoner at the pace of 13.1 miles per hour. The Calories burned can then be calculated from that value and the rate at which the marathoner burns Calories. Analyze The time it takes for the marathoner to complete the race will be given by distance of the marathon Time to complete the marathon = pace of the marathoner The Calories burned will be computed by Calories burned Calories burned = length of the marathon race hour  Solve 26.2 mi Time to complete the marathon = 2.00 hr 13.1 mi/hr  665 Cal Calories burned = 2.00 hr = 1330 Cal hr  Think about It This problem could be solved without touching a calculator! Because it takes the marathoner 2.00 hr to complete the race, the Calories she burns is simply twice the number of Calories she burns in one hour. 1.44. Collect and Organize To find how much gas is needed , we need only to convert the miles traveled (389 mi) into the gallons that would be used by using the known value of the gas mileage for the SUV (18 mi/gal).

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    Matter, Energy, and the Origins of the Universe | 15 Analyze The number of gallons used in the trip can be computed by 1 gallon Number of gallons used = number of miles traveled number of miles  Solve 1 gal Number of gallons used = 389 mi 22 gal 18 mi   Think about It This is a reasonable value. As a quick estimate you can round the trip to 400 miles and the gas mileage to 20 miles per gallon. A quick mental computation gives this to be about 20 gallons. The above answer is, therefore, in the correct range. 1.45. Collect and Organize This problem asks for a simple conversion of length: from meters to miles. Analyze The conversions that we need include meters to kilometers and kilometers to miles. 1 km 0.6214 mi , 1000 m 1 km Solve 3 1 km 0.6214 mi 4.0 10 m 2.5 mi 1000 m 1 km     Think about It The answer is reasonable because 4000 meters would be a little over 2 miles when estimated. It is surprising, though, for a natural piece of silk to be that long. 1.46. Collect and Organize This problem involves the conversion of one volume unit into another. We have to convert from cubic inches into liters via the cubic centimeter and cubic liter. We have to be careful to “cube” the units of length in the conversions to get the volume. Analyze The conversions needed are 3 1 in 1 m 1 m , , 2.54 cm 100 cm 1000 L Solve 3 3 3 3 2.54 cm 1 m 1000 L 292 in 4.79 L 1 in 100 cm 1 m                       Compared to the 2005 Thunderbird with a 3.9 L engine, the 1955 Thunderbird had a bigger engine. Think about It This might be a difficult conversion to visualize but the value of 4.79 L is reasonable because the volume of an engine is on the liter scale, not a tenth or tens of liters. 1.47. Collect and Organize We can solve this by converting the time it takes the runner to complete the race into seconds, then converting the distance of the race from kilometers into meters. The runner’s speed is the ratio of distance (meters) to time (seconds).

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