ࡱ> IKH'` LAbjbj"9"9 ;~@S@S8fl l l l l l l B ^;^;^;8;l<F<X>(>>>???cFeFeFeFeFeFeF$GhFJ|Fl ?????Fl l >>{FBBB?l >l >cFB?cFBBl l B>< O~^;@pBFTF0FBJ@lJBJl B ??B?????FFA???F????$!:;$:; t J> l l l l l l  CHAPTER 1 | Matter, Energy, and the Origins of the Universe 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 containers 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 H2 or Br2, or even more highly associated as in S8 or P4. 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 (redred) condenses to a slightly disordered phase while the other element (blueblue) 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. 1.10. 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. 1.13. 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. Solve (a) Distillation purifies a substancenot a chemical change. (b) Combustion transforms the fuel (such as wood) into carbon dioxide and watera chemical change. (c) Filtration separates substances from each othernot a chemical change. (d) Condensation changes a vapor into a liquidnot a chemical change. Think about It Distillation, filtration, and condensation all involve physical changes, not chemical changes. 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. 1.22. 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. 1.40. 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. 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:  EMBED Equation.DSMT4  For the soft drink already priced per half-liter, the price per liter will be double. Solve  EMBED Equation.DSMT4   EMBED Equation.DSMT4  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. 1.44. 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). Analyze The number of gallons used in the trip can be computed by  EMBED Equation.DSMT4  Solve  EMBED Equation.DSMT4  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. 1.52. Collect and Organize In this problem we use the density to find the mass of ethanol in 65.0 mL. This uses the definition density = mass/volume. Analyze We can easily solve this problem by rearranging the density equation:  EMBED Equation.DSMT4  Solve  EMBED Equation.DSMT4  Think about It With a density of less than a gram per milliliter, we expect that we would have a mass lower than 65 g for the 65 mL sample of ethanol. 1.63. Collect and Organize In this problem we use the mass of a carat (the unit of weight for diamonds) to find the mass of a large diamond and then use the density to calculate the volume of that large diamond. Analyze We need the fact that 1 carat = 0.200 g and that the density is defined as mass per volume. To find the volume of the diamond we can rearrange the density equation to read  EMBED Equation.DSMT4  Solve The mass of the 5.0 carat diamond is  EMBED Equation.DSMT4  The volume of the diamond is then  EMBED Equation.DSMT4  Think about It For this relatively large diamond in terms of carats, the mass is fairly small (one gram is about a fifth of the mass of a nickel) and so even though the density is relatively low, the volume is also quite small. 1.70 Collect and Organize For each of the quantities given, we choose those that contain three significant figures. Analyze Writing all the quantities in scientific notation will help determine the number of significant figures in each. 7.02 6.452 302 = 3.02 ( 102 6.02 ( 1023 12.77 = 1.277 ( 101 3.43 Solve The quantities that have three significant figures are (a) 7.02, (c) 302, (d) 6.02 ( 1023, and (f ) 3.43. Think about It Remember that a zero between two other digits is always significant. 1.71. Collect and Organize We are to express the result of each calculation to the correct number of significant figures. Analyze The rules regarding the significant figures that carry over in calculations are given in Section 1.8 in the textbook. Remember to operate on the weak-link principle. Solve (a) The least well-known value has three significant figures so the calculator result of 17.363 is reported as 17.4 with rounding up the tenths place. (b) The least well-known value has only one significant figure so the calculator result of 1.044 ( 1013 is reported as 1 ( 1013. (c) The least well-known value has three significant figures so the calculator result of 5.701 ( 1023 is reported as 5.70 ( 1023. (d) The least well-known value has three significant figures so the calculator result of 3.5837 ( 103 is reported as 3.58 ( 103 with rounding. Think about It Indicating the correct number of significant figures for a calculated value indicates the level of confidence we have in our calculated value. Reporting too many significant figures would indicate a higher level of precision in our number than we actually have. 1.72. Collect and Organize We are to express the result of each calculation to the correct number of significant figures. Analyze The rules regarding the significant figures that carry over in calculations are given in Section 1.8 in the textbook. Remember to operate on the weak-link principle. Solve (a) The least well-known value has two significant figures so the calculator result of 1.5506 ( 101 is reported as 1.5 ( 101. (b) The least well-known value has three significant figures so the calculator result of 146.3988 is reported as 146. (c) The least well-known value has three significant figures so the calculator result of 2.25857 ( 102 is reported as 2.26 ( 102. (d) The least well-known value has three significant figures so the calculator result of 3.5700 ( 103 is reported as 3.57 ( 103. Think about It Indicating the correct number of significant figures for a calculated value indicates the level of confidence we have in our calculated value. Reporting too many significant figures would indicate a higher level of precision in our number than we actually have. 1.75. Collect and Organize We are asked in this problem to convert from kelvins to Celsius degrees. Analyze The relationship between the Kelvin temperature scale and the Celsius temperature scale is given by  EMBED Equation.DSMT4  Rearranging this gives the equation to convert Kelvin to Celsius temperatures:  EMBED Equation.DSMT4  Solve  EMBED Equation.DSMT4  Think about It Because 4.2 K is very cold, we would expect that the Celsius temperature would be very negative. It should not, however, be lower than 273.15 K, since that is the lowest temperature possible. 1.87. Collect and Organize This question considers the runoff of nitrogen every year into a stream caused by a farmers application of fertilizer. We must consider that not all of the fertilizer contains nitrogen and not all of the fertilizer runs off into the stream. We must also account for the flow of the stream in taking up the nitrogen runoff. Analyze First, we have to determine the amount of nitrogen that is in the fertilizer (10% of 1500 kg). Then, we need to find how much of that nitrogen gets washed into the stream (15% of the mass of N in the fertilizer). Our final answer must be in milligrams of N, so we can convert the mass of N that gets washed into the stream from kilograms to milligrams.  EMBED Equation.DSMT4  Next, we need to know how much water flows through the farm each year via the stream. To find this, we must convert the rate of flow in cubic meters per minute to liters per year. We can convert this through one line using dimensional analysis with the following conversions:  EMBED Equation.DSMT4  Solve The amount of N washed into the stream each year is  EMBED Equation.DSMT4  The amount of stream water flowing through the field each year is  EMBED Equation.DSMT4  The additional concentration of N that is added to the stream by the fertilizer is  EMBED Equation.DSMT4  Think about It The calculated amount of nitrogen that is added to the stream seems reasonable. The concentration is relatively low because the stream is moving fairly swiftly and the total amount of nitrogen that washes into the stream over the course of the year is not too great. The problem, however, does not tell us if this amount would cause harm to the plant and animal life in the stream. 1.88. Collect and Organize For this problem we try to identify which cylinder is made of aluminum and which is made of titanium by comparing experimentally determined densities with the actual known densities. Analyze (a) To calculate the volume of each cylinder from its dimensions, we will have to use the equation for volume of a cylinder:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (b) To calculate the volume from the water displacement method, we need only find the difference in water volume for each cylinder from the diagram in Figure P1.88. (c) To determine the method with the most significant figures we will compare the answers in parts a and b. (d) To compute the density for each cylinder we use the equation for density:  EMBED Equation.DSMT4  Solve (a) Volumes of the cylinders from their measured dimensions:  EMBED Equation.DSMT4  (b) Volume of cylinders from water displacement measurements:  EMBED Equation.DSMT4  (c) Neither. As seen in the above calculations, both the volume measurement from the water displacement method and from the cylinders dimension have two significant figures for the volume calculation. (d) From part a:  EMBED Equation.DSMT4  From part b we obtain answers with two significant figures for the density:  EMBED Equation.DSMT4  Think about It How we make measurements is very important to the values we can report for those measurements. In this problem, neither method provided more significant figures for the calculation.      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massvolumeor mass==densityvolumeCompObj#%HiObjInfo&JEquation Native K_1352192792J)F BO~O~ FMathType 5.0 Equation MathType EFEquation.DSMT49qDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Mass of ethanol==65.0 mL 0.789 gmL==51.3 g FMathType 5.0 Equation MathType EFEquation.DSMT49q:lDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!EDOle RCompObj(*SiObjInfo+UEquation Native VV/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Volume =  massdensity FMathType 5.0 Equation MathType EFEquation.DSMT49q_1352192794.FO~O~Ole \CompObj-/]iObjInfo0_Equation Native `_1352192796,E3FO~O~Ole gCompObj24hisDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  5.0 carat 0.200 g1 carat ==1.0 g FMathType 5.0 Equation MathType EFEquation.DSMT49qhDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   1.0 g3.51 g/cm ObjInfo5jEquation Native k_1352194355|8FO~O~Ole r3 ==0.28 cm 3 FMathType 5.0 Equation MathType EFEquation.DSMT49q# \DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!EDCompObj79siObjInfo:uEquation Native v?_1352194363mO=FO~O~/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  K =  "o C + 273.15 FMathType 5.0 Equation MathType EFEquation.DSMT49q( \DSMT5WinAllBasicCodePagesOle {CompObj<>|iObjInfo?~Equation Native DTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   "o C =  K   273.15 FMathType 5.0 Equation MathType EFEquation.DSMT49q_1352194370BFO~O~Ole CompObjACiObjInfoD \DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   "o C =  4.2 K   273.15 =  269.0 "o CEquation Native _1352192904GFO~O~Ole CompObjFHi FMathType 5.0 Equation MathType EFEquation.DSMT49q<DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Mass ObjInfoIEquation Native _13521929061;LFO~O~Ole of fertilizer in kg0.10 = mass of N in fertilizer in kgMass of N in fertilizer in kg0.15 = mass of N washed into the stream in kgMass of N that washes into the stream in kg 1000 g1 kg 1000 mg1 g==mass of N that washes into the stream in mg FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjKMiObjInfoNEquation Native _1357464506@TQFO~O~ DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   1000 L1 m 3 , 1 hr60 min, 1 d24 hr, 1 yr365.25 d FMathType 5.0 Equation MathType EFEquation.DSMT49qk\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_DOle CompObjPRiObjInfoSEquation Native _E_E_A  1500 kg  0.10 = 150 kg N in the fertilizer 150 kg 0.15 = 22.5 kg N washed into the stream in one year22.5 kg  1000 g1 kg 1000 mg1 g==2.2510 7  mg of N washed into the stream in one year_1357464549cVFO~O~Ole CompObjUWiObjInfoX FMathType 5.0 Equation MathType EFEquation.DSMT49qk\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   1.4 Equation Native _1352192913[FO~O~Ole CompObjZ\im 3 1 min 1000 L1 m 3  60 min1 hr 24 hr1 d  365.25 d1 yr==7.3610 8  L/yr FMathType 5.0 Equation MathType EFEquation.DSMT49qDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   2.2510 7  mg NObjInfo]Equation Native _1352192915Yh`FO~O~Ole /yr7.3610 8  L/yr==0.031 mg/L FMathType 5.0 Equation MathType EFEquation.DSMT49qDDSMT5WinAllBasicCodePagesCompObj_aiObjInfobEquation Native _1357466968eFO~O~Times New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Volume of cylinder = height of cylinderp(radius) 2Ole CompObjdfiObjInfogEquation Native z FMathType 5.0 Equation MathType EFEquation.DSMT49qk^\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  where radius =  1   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FMathType 5.0 Equation MathType EFEquation.DSMT49q: DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_1352192919jFO~O~Ole CompObjikiObjInfolEquation Native V_1352192922^woFO~O~Ole  CompObjnp i_E_E_A  Density =  massvolume FMathType 5.0 Equation MathType EFEquation.DSMT49qlDSMT5WinAllBasicCodePagesObjInfoqEquation Native _1352192924tFO~O~Ole Times New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Volume of Cylinder A = 5.1 cmp(0.60 cm) 2 ==5.8 cm 3 Volume of Cylinder B = 5.9 cmp(0.65 cm) 2 ==7.8 cm 3 FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjsuiObjInfovEquation Native  _1352192926r6yFO~O~DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Volume of Cylinder A = 30.8 mL   25.0 mL = 5.8 mLVolume of Cylinder B = 32.8 mL   25.0 mL = 7.8 mL FMathType 5.0 Equation MathType EFEquation.DSMT49qOle ,CompObjxz-iObjInfo{/Equation Native 0DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Density of Cylinder A =  15.560 g5.8 mL==2.7 g/mLDensity of Cylinder B =  35.536 g7.8 mL==4.6 g/mL FMathType 5.0 Equation MathType EFEquation.DSMT49q_1352192928~FO~O~Ole <CompObj}=iObjInfo?Equation Native @1TableJSummaryInformation(LDocumentSummaryInformation8TDDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourierPSTimesMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Density of Cylinder A =  15.560 g5.8 mL==2.7 g/mLDensity of Cylinder B =  35.536 g7.8 mL==4.6 g/mLOh+'0 $0 P \ h tChapter X | Chapter TitleQ7 l!?h,5W*ZVS OLԻ|祋<w.TEwvd'Hեj/2,`^ aX(1qcE`M&ؐM.A4גKa&7rd La "J'DxaɌI{t 62N'tx > _X嶳eXkXn}al~χd0fDd 0  # A2؜C=56O`!؜C=56@`+:xڥT=Q6A-DQ%krɭf@6+r)Ncc!SIe/YpF8oz6yvv|fg |t88c.c|28;_HdI3,C-[_}/*+ sܿ ֟/h!e"zo28jXxz-18['LDHa薺k= .%n: |!Dbh5 X˰? 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