ࡱ> u @ wbjbj)) ,KzKz;;;;;;;2=8RF2=\+6.*dpԯԯFg)i)i)i)i)i)i)$.Rd0););;ԯ+z;ԯ;g)g)#! ;;''ԯ* pmTn!e/#:K',+0\+i#x1+bx1t''";Df<;;;;x1;''$r |c))2=2=B2=2=CHAPTER  DOCPROPERTY ChapterNumber \* MERGEFORMAT 1  DOCPROPERTY ChapterTitle \* MERGEFORMAT Chemistry and Measurement Chapter Terms and Definitions Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred terms are italicized in the text. Where a term does not fall directly under a text section heading, additional information is given for you to locate it. experiment  observation of natural phenomena carried out in a controlled manner so that the results can be duplicated and rational conclusions obtained (1.2) law  concise statement or mathematical equation about a fundamental relationship or regularity of nature (1.2) hypothesis  tentative explanation of some regularity of nature (1.2) theory  tested explanation of basic natural phenomena (1.2) scientific method*  the creative process of understanding the physical world that involves hypothesis formation, experimentation, and modification of theory followed by more experimentation (1.2) mass  quantity of matter in a material (1.3) matter  whatever occupies space and can be perceived by our senses (1.3) law of conservation of mass  total mass remains constant during a chemical change (chemical reaction) (1.3) weight*  force of gravity exerted on an object (1.3) fluids*  liquids and gases: kinds of matter that flow easily and change their shapes in response to slight outside forces (1.4) compressibility*  property of a gas enabling it to be pressed into a smaller space((1.4) expansibility*  property of a gas enabling it to spread out to fill a given space (1.4) solid  form of matter characterized by rigidity; has fixed volume and shape (1.4) liquid  form of matter that is a relatively incompressible fluid; has fixed volume but no fixed shape (1.4) gas  form of matter that is an easily compressible fluid; has neither fixed volume nor fixed shape (1.4) vapor*  the gaseous state of any matter that normally exists as a liquid or solid (1.4) states of matter  the three forms in which matter exists: solid, liquid, and gas (1.4) physical change  change in the form of matter but not in its chemical identity (1.4) distillation*  process in which a liquid is vaporized and then condensed; used to separate substances that differ in volatility (1.4) condenser*  cooled tube in which vapor changes back into liquid (1.4) receiver*  container in which distilled material is collected (1.4) chemical change (chemical reaction)  change in which one or more kinds of matter are transformed into a new kind of matter or several new kinds of matter (1.4) physical property  characteristic of a material that can be observed without changing its chemical identity (1.4) chemical property  characteristic of a material involving its ability to undergo chemical change (1.4) substance  kind of matter that cannot be separated into other kinds of matter by any physical process (1.4) element  kind of matter composed of only one chemically distinct type of atom; substance that cannot be chemically decomposed into simpler substances (1.4) compound  kind of matter composed of atoms of two or more elements chemically combined in fixed proportions; substance that can be chemically decomposed into two or more elements (1.4) law of definite proportions (law of constant composition)  a pure compound, whatever its source, always contains definite or constant proportions of the elements by mass (1.4) mixture  material that can be separated by physical means into two or more substances (1.4) heterogeneous mixture  mixture that consists of physically distinct parts, each with different properties (1.4) homogeneous mixture (solution)  mixture that is uniform in its properties throughout given samples (1.4) phase  one of several different homogeneous materials present in the portion of matter under study (1.4) chromatography*  technique for separating a mixture that is based on partitioning of the components of the mixture between stationary and mobile phases (Instrumental Methods: Separation of Mixtures by Chromatography) paper chromatography*  chromatographic procedure in which components of a mixture are dissolved in a solution and carried up a paper strip at different rates (Instrumental Methods: Separation of Mixtures by Chromatography) column chromatography*  chromatographic procedure in which a mixture is placed on a stationary column and washed down with solvent (Instrumental Methods: Separation of Mixtures by Chromatography) chromato-*  color (Instrumental Methods: Separation of Mixtures by Chromatography) gas chromatography (GC)*  chromatographic procedure in which a mixture of gases or highly volatile liquids is separated using a carrier gas passing through a solid, or viscous liquid on a solid support, packed in a column (Instrumental Methods: Separation of Mixtures by Chromatography) carrier*  unreactive gas that flows through the solid column in a gas chromatograph (Instrumental Methods: Separation of Mixtures by Chromatography) retention time*  time it takes for a substance to travel through the column to the detector in a gas chromatograph (Instrumental Methods: Separation of Mixtures by Chromatography) chromatogram*  chart recording of peaks corresponding to the passage of different substances by the detector in a gas chromatograph (Instrumental Methods: Separation of Mixtures by Chromatography) unit  fixed standard of measurement (1.5) precision  agreement among measured values of a quantity (1.5) accuracy  agreement of a measured value with the true value (1.5) significant figures  the digits in a measured number (or result of a calculation with measured numbers) that include all certain digits plus a final one having some uncertainty (1.5) number of significant figures  number of digits reported for the value of a measured or calculated quantity, indicating the precision of the value (1.5) scientific notation  method of writing numbers in the form A ( 10n, where A is a number with a single nonzero digit to the left of the decimal point, and n is an integer, or whole number (1.5) exact number  number that arises when we count items or sometimes when we define a unit (1.5) rounding  dropping nonsignificant digits in a calculation result and adjusting the last digit reported (1.5) livre*  eighteenth-century unit of measurement for the pound, equivalent to 9216 grains (1.6, margin note) metric system*  decimal set of weights and measures (1.6) International System (SI)  group of metric units internationally accepted in 1960 as the standard units of scientific measurement (1.6) SI base units  SI units from which all others can be derived (1.6) SI prefix  prefix used in the International System to indicate a power of 10 (1.6) meter (m)  SI base unit of length (about 39 inches) (1.6) angstrom ()  traditional non-SI unit of length (1 = 10 10 m) (1.6) kilogram (kg)  SI base unit of mass (about 2.2 lb) (1.6) second (s)  SI base unit of time; unit on absolute temperature scale (1.6) Celsius scale  temperature scale in general scientific use, in which there are exactly 100 units between the freezing and boiling points of water (1.6) kelvin (K)  SI base unit of temperature (1.6) absolute temperature*  temperature reported using a scale on which the lowest point that can be attained theoretically is zero (1.6) area*  length times width (1.7) speed*  rate of change of distance with time (1.7) SI derived unit  unit obtained by combining SI base units (1.7) volume*  length times width times height (1.7) liter (L)  traditional unit of volume (1 L = 1 dm3) (1.7) density  mass per unit volume (1.7) dimensional analysis (factor-label method)  method of calculation that includes the units for quantities and treats them as numbers (1.8) conversion factor  ratio equivalent to 1 that converts a quantity expressed in one unit to a quantity expressed in another unit (1.8) Chapter Diagnostic Test For questions 13, write the letter of the one best answer.  SEQ NL_a \r 0 \h  SEQ NL1 1. Chemistry is classified as a scientific endeavor because it  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . investigates physical changes in matter.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . is based on a close relationship between theory and experiment.  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . is founded on a molecular classification of matter.  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . depends on a quantitative relationship between atoms and molecules.  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . is continually expanding into new areas in an attempt to explain natural phenomena.  SEQ NL_a \r 0 \h  SEQ NL1 2. The derived unit of pressure in SI units would be  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h .  EMBED Equation.3 .  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h .  EMBED Equation.3 .  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h .  EMBED Equation.3 .  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h .  EMBED Equation.3 .  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . either (a) or (d).  SEQ NL_a \r 0 \h  SEQ NL1 3. An irregularly shaped object has a mass of 2.3 ( 10(2 g and a density of 3.4 ( 103 kg/m3. The volume of this object would be calculated to be  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h .  EMBED Equation.3  (  EMBED Equation.3  ( 2.3 ( 10(2 g.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h .  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3 .  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h .  EMBED Equation.3  (  EMBED Equation.3  ( 2.3 ( 10(2 g.  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h .  EMBED Equation.3  (  EMBED Equation.3  ( 2.3 ( 10(2 g.  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . none of the above.  SEQ NL_a \r 0 \h  SEQ NL1 4. The number of significant figures in a numerical quantity tells us the _________________ of the measurement.  SEQ NL_a \r 0 \h  SEQ NL1 5. Absolute temperature is measured in units called ____________________.  SEQ NL_a \r 0 \h  SEQ NL1 6. Match each value in the left-hand column with its expression in scientific notation in the right-hand column. _____(1)201(a)1.4916 ( 104_____(2)0.0000750(b)2.01 ( 10(2_____(3)14.916 ( 10-5(c)7.50 ( 10(5_____(4)0.750 ( 104(d)2.01 ( 102_____(5)0.0201(e)1.4916 ( 10(4_____(6)14,916(f)7.50 ( 103  SEQ NL_a \r 0 \h  SEQ NL1 7. Give the number of significant figures in each of the following numbers.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 1745.0  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 0.0156  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 2.89 ( 104  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 1.000 ( 10(3  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 6200  SEQ NL_a \r 0 \h  SEQ NL1 8. Which of the following is (are) correctly rounded off?  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 74.63 to two significant figures equals 75.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 2.4501 to two significant figures equals 2.4.  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 0.093374 to four significant figures equals 0.09337.  SEQ NL_a \r 0 \h  SEQ NL1 9. Write the correct values on the lines.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 1 pg = _______ g.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 1 g = _______ ng.  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 1 kg = _______ g.  SEQ NL_a \r 0 \h  SEQ NL1 10. Perform the following arithmetic operations and report the answers in scientific notation with the correct number of significant figures.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h .  EMBED Equation.3   seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 0.0076 + 24.6 (0.000064)  SEQ NL_a \r 0 \h  SEQ NL1 11. Make the following conversions. Observe significant figures.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 6.45 ( 104 g to pounds  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 3.2 ( 102 m to inches  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 4.5 gal to L  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 91.2(F to (C  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 14.7 lb/in2 to g/m2  SEQ NL_a \r 0 \h  SEQ NL1 12. In an electrolysis experiment, 0.0004 cm of copper was plated onto an electrode. What was the thickness of the copper in nanometers?  SEQ NL_a \r 0 \h  SEQ NL1 13. The mass of the moon is 7.345 ( 1022 kg. Its volume is 2.1991 ( 1025 cm3. What is the density of the moon in grams per cubic centimeter?  SEQ NL_a \r 0 \h  SEQ NL1 14. There is growing concern today over depletion of the ozone layer in our atmosphere. The mass of this protective layer has been estimated at 2 ( 109 metric tons. If the density of ozone is 2.144 ( 10-3 g/cm3, what is the volume corresponding to the estimated mass? (1 metric ton = 1000 kg.) Answers to Chapter Diagnostic Test If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer. SEQ NL1 \r 0 \h   SEQ NL_a \r 0 \h  SEQ NL1 1. b (1.2)  SEQ NL_a \r 0 \h  SEQ NL1 2. d (1.6, 1.7)  SEQ NL_a \r 0 \h  SEQ NL1 3. a (1.7, 1.8, PS Sk. 5)  SEQ NL_a \r 0 \h  SEQ NL1 4. precision (1.5)  SEQ NL_a \r 0 \h  SEQ NL1 5. kelvins (1.6)  SEQ NL_a \r 0 \h  SEQ NL1 6. ( seq NL_1_ 1) d, ( seq NL_1_ 2) c, ( seq NL_1_ 3) e, ( seq NL_1_ 4) f, ( seq NL_1_ 5) b, ( seq NL_1_ 6) a (Appendix A)  SEQ NL_a \r 0 \h  SEQ NL1 7.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 5  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 3  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 3  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 4  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . uncertain (1.5)  SEQ NL_a \r 0 \h  SEQ NL1 8. a, c (1.5, PS Sk. 2)  SEQ NL_a \r 0 \h  SEQ NL1 9.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 1012  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 109  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 103 (1.6)  SEQ NL_a \r 0 \h  SEQ NL1 10.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 2.02 ( 1010  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 9.2 ( 10-3 (1.5, PS Sk. 2)  SEQ NL_a \r 0 \h  SEQ NL1 11.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 142 lb (1.8, PS Sk. 2, 6)  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 1.3 ( 104 in (1.8, PS Sk. 2, 6)  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 17 L (1.8, PS Sk. 2, 6)  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 32.9(C (1.6, PS Sk. 2, 3)  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 1.03 ( 107 g/m2 (1.8, PS Sk. 2, 6)  SEQ NL_a \r 0 \h  SEQ NL1 12. 4 ( 103 nm (1.6, PS Sk. 2, 6)  SEQ NL_a \r 0 \h  SEQ NL1 13. 3.340 g/cm3 (1.7, 1.8, PS Sk. 4, 6)  SEQ NL_a \r 0 \h  SEQ NL1 14. 9 ( 1017 cm3 (1.7, 1.8, PS Sk. 5, 6) Summary of Chapter Topics 1.1 Modern Chemistry: A Brief Glimpse  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Learning Objective Provide examples of the contributions of chemistry to humanity. 1.2 Experiment and Explanation  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Learning Objectives Describe how chemistry is an experimental science. Understand how the scientific method is an approach to performing science. In Sections 1.1 and 1.2 you are introduced to modern chemistrya quantitative science where the final test of truth is repeatable experimental observation. It is important to understand the difference between laws made by governments, which any person may choose to break, and observed natural laws. You cannot choose to break the law of gravity! Can you think of any theories? Examples are the theory of evolution, from biology, and Einsteins theory of relativity, from physics. You will learn about chemistrys atomic theory in Chapter 2. 1.3 Law of Conservation of Mass Learning Objectives Explain the law of conservation of mass. Apply the law of the conservation of mass. (Example 1.1) seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h   seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Problem-Solving Skill SEQ NL1 \r 0 \h   seq NL1 1 seq NL_a \r 0 \h . Using the law of conservation of mass. Given the masses of all substances in a chemical reaction except one, calculate the mass of this one substance (Example 1.1). Exercise 1.1 You place 1.85 grams of wood in a vessel with 9.45 grams of air and seal it. Then you heat the vessel strongly so that the wood burns. After the experiment, you weigh the ash that remains after the wood burns and find that its mass is 0.28 gram. What is the mass of wood converted to gas by the end of the experiment? Wanted: Mass of gases (in grams) that were formed when the wood burns Given: 1.85 grams of wood, 0.28 gram of ash. The gases are lost when the wood burns to give the ash. Solution: 1.85 grams wood ( 0.28 gram ash 1.57 grams gases  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h 1.4 Matter: Physical State and Chemical Constitution  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Learning Objectives Compare and contrast the three common states of matter: solid, liquid, and gas. Describe the classifications of matter: elements, compounds, and mixtures (heterogeneous and homogeneous). Understand the difference between chemical changes (chemical reactions) and physical changes. Distinguish between chemical properties, and physical properties. Be sure that you understand the difference between a material and a substance. A material is any sample of matter that has mass and occupies space. A substance is a type of material that is unique. It has its own chemical formula and composition and its own specific properties. There are two types of substances: elements and compounds. Exercise 1.2 Potassium is a soft, silvery-colored metal melting at 64(C. It reacts vigorously with water, with oxygen, and with chlorine. Identify all the physical properties given in this description. Identify all the chemical properties given. Known: Definitions of physical properties and chemical properties Solution: Three physical propertiessoft, silvery-colored, melting point 64(C; four chemical propertiesit is a metal (its chemical identity), it reacts vigorously with water, it reacts vigorously with oxygen, and it reacts vigorously with chlorine.  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h 1.5 Measurement and Significant Figures  seq NL1 \r 0 \h Learning Objectives Define and use the terms precision and accuracy when describing measured quantities. Learn the rules for determining significant figures in reported measurements. Know how to represent numbers using scientific notation. Apply the rules of significant figures to reporting calculated values. Be able to recognize exact numbers. Know when and how to apply the rules for rounding. Use significant figures in calculations. (Example 1.2)  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Problem-Solving Skill SEQ NL1 \r 1\h   seq NL1 2 seq NL_a \r 0 \h . Using significant figures in calculations. Given an arithmetic setup, report the answer to the correct number of significant figures and round it properly (Example(1.2). Only since the wide use of calculators has the subject of significant figures become important. Previously, when a slide rule was used for calculations, there was little chance to report too many figures. Remembering the rules for reporting significant figures is helpful, but the best way to learn the process is to practice it as much as possible. Note that exact numbers have no effect on the number of significant figures you should report in the answer to a calculation. The number of significant figures in a calculation result will depend only on the numbers of significant figures in quantities having uncertainties. Scientific notation greatly simplifies chemical calculation. Remember that in converting a number to scientific notation, we are not changing its value. We are merely changing the expression of its value. Students often have trouble determining the sign of the power of 10 when writing numbers in scientific notation. When we move the decimal point to the left or right, we are in effect dividing or multiplying the original number by 10 for each place we move it. We must then indicate the opposite operation with the power of 10. For example, 49,012 equals 4.9012 ( 104. We moved the decimal four places to the left, indicating division by 104. Therefore, we must multiply the smaller number by 104. As another example, putting 0.0036024 in scientific notation, we get 3.6024 ( 10(3. This time we moved the decimal to the right, indicating multiplication by 103. Therefore, we must indicate division by 103. We do this by multiplying by 10(3. Get used to writing any numbers less than 0.01 and greater than 999 in scientific notation. Remember that any number written in scientific notation shows the precision in measurement of the number (significant figures). Only the digits in the number, not the exponent, are significant. As an exercise, write the following numbers in scientific notation, referring to Appendix A for help:  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 6093  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 4218  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 0.00413  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 0.00006987 The answers are SEQ NL_a \r 0 \h   seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 6.093 ( 103  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 4.218 ( 103  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 4.13 ( 10(3  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 6.987 ( 10(5 A point worth remembering is that any number to the zero power is 1, so 100 = 1. You will see this again in later chapters. Exercise 1.3 Give answers to the following arithmetic setups. Round to the correct number of significant figures. SEQ NL_a \r 0 \h   seq NL_a \* alphabetic a seq NL_1_ \r 0 \h .  EMBED Equation.3   seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 8.91 6.435  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 6.81 6.730  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 38.91 ( (6.81 6.730) Solution: (a) 4.9. The calculator result is 4.864671429, but we must report only two digits because 9.1 has but two. Following rule 1 of the rounding procedure, we increase 8 to 9 because the leftmost digit to be dropped is 6. (b) 2.48. The calculator result is 2.475, but we report only two decimal places because 8.91 has but two. Following rule 1 of the rounding procedure, we increase 7 to 8 and drop the 5. (c) 0.08. The calculator value is correct, so we must report only two decimal places because 6.81 has but two. (d) 3. We do the computation in parentheses first: 38.91 ( (0.08) = 3.1128 Since 0.08 has but one significant figure, we round off according to rule 2 of the rounding procedure. Thus our answer has one digit.  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h 1.6 SI Units Learning Objectives Become familiar with the SI (metric) system of units including the SI prefixes. Convert from one temperature scale to another. (Example 1.3)  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Problem-Solving Skill  seq NL1 3 seq NL_a \r 0 \h . Converting from one temperature scale to another. Given a temperature reading on one scale, convert it to another scaleCelsius, Kelvin, or Fahrenheit (Example 1.3). Can you imagine what it would be like to go to a party where everybody spoke to you in a different language and you had to have an interpreter to understand what was said? This is what it was like in the scientific world before the use of SI units. It is crucial that you memorize the metric prefixes mega-, kilo-, deci-, centi-, milli-, micro-, nano-, and pico- and their symbols. Also memorize the SI base units for the base quantities mass, length, time, and temperature. Exercise 1.4 Express the following quantities using an SI prefix and a base unit. For instance, 1.6 ( 10(6 m = 1.6 (m. A quantity such as 0.000168(g could be written 0.168 mg or 168 (g.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 1.84 ( 10(9 m  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 5.67 ( 10(12 s  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 7.85 ( 10(3 g  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 9.7 ( 103 m  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 0.000732 s  seq NL_a \* alphabetic f seq NL_1_ \r 0 \h . 0.000000000154 m Solution: SEQ NL_a \r 0 \h   seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 1.84 nm (nanometer)  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 5.67 ps (picosecond)  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 7.85 mg (milligram)  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 9.7 km (kilometer)  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 0.732 ms (millisecond) or 732 (s (microsecond)  seq NL_a \* alphabetic f seq NL_1_ \r 0 \h . 0.154 nm (nanometer) or 154 pm (picometer) Exercise 1.5 SEQ NL_a \r 0 \h   seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . A person with a fever has a temperature of 102.5(F. What is this temperature in degrees Celsius?  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . A cooling mixture of dry ice and isopropyl alcohol has a temperature of (78(C. What is this temperature in kelvins? Solution: SEQ NL_a \r 0 \h   seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . (C =  EMBED Equation.3  =  EMBED Equation.3  =  EMBED Equation.3  = 39.2(C  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . K = (C + 273.15 = 78 + 273.15 = 195 K  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h 1.7 Derived Units  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Learning Objectives Define and provide examples of derived units. Calculate the density of a substance. (Example 1.4) Use density to relate mass and volume. (Example 1.5)  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Problem-Solving Skills SEQ NL1 \r 3\h   seq NL1 4 seq NL_a \r 0 \h . Calculating the density of a substance. Given the mass and volume of a substance, calculate the density (Example 1.4).  seq NL1 5 seq NL_a \r 0 \h . Using the density to relate mass and volume. Given the mass and density of a substance, calculate the volume, or given the volume and density, calculate the mass (Example 1.5). Exercise 1.6 A piece of metal wire has a volume of 20.2 cm3 and a mass of 159 g. What is the density of the metal? The metal is either manganese, iron, or nickel, which have densities of 7.21, 7.87, and 8.90 g/cm3, respectively. From which metal is the wire made? Solution: d =  EMBED Equation.DSMT4  =  EMBED Equation.3  = 7.87 g/cm3 The object is made of iron. Exercise 1.7 Ethanol (grain alcohol) has a density of 0.789 g/cm3. What volume of ethanol must be poured into a graduated cylinder to equal 30.3 g? Solution: Since d =  EMBED Equation.DSMT4 , V =  EMBED Equation.DSMT4  = 30.3 g (  EMBED Equation.3  = 38.4 cm3  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h 1.8 Units and Dimensional Analysis (Factor-Label Method)  seq NL1 \r 0 \h  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Learning Objectives Apply dimensional analysis to solving numerical problems. Convert from one metric unit to another metric unit. (Example 1.6) Convert from one metric volume to another metric volume. (Example 1.7) Convert from any unit to another unit. (Example 1.8)  seq NL_EVEN \r 0 \h  seq NL_ODD \r 0 \h  seq NL_Eqn \r 0 \h  seq NL_Sec \r 1 \h Problem-Solving Skill SEQ NL1 \r 5\h   seq NL1 6 seq NL_a \r 0 \h . Converting units. Given an equation relating one unit to another (or a series of such equations), convert a measurement expressed in one unit to a new unit. (Examples 1.6, 1.7, and 1.8) One of the most useful methods for solving problems in all fields of science is dimensional analysis. This problem-solving technique cannot be overstressed. If you are not totally familiar with this technique, review Section 1.8 of your textbook. The time you now spend learning and mastering this method will reward you tenfold during your science education. As you review this section in your textbook, pay particular attention to the fact that in every problem and/or example the unit is always written with the number. These units are used to set up the problem correctly and to assure you of getting the correct answer. Get into the habit now of using units as part of the problem as well as part of the answer. It is a good idea to memorize the relationships in text Table 1.4. You will find that this wont take much effort if you give yourself lots of practice working the conversion problems. You will eventually remember the factors just from using them. Exercise 1.8 The oxygen molecule (the smallest particle of oxygen gas) consists of two oxygen atoms a distance of 121 pm apart. How many millimeters is this distance? Solution: We have been given the distance in picometers and must convert to millimeters: 121 pm (  EMBED Equation.3  (  EMBED Equation.3 = 121 pm (  EMBED Equation.3  (  EMBED Equation.3 converts pm to mconverts m to mm = 1.21 ( 10(7 mm Note that we report the answer in scientific notation. Exercise 1.9 A crystal is constructed by stacking small, identical pieces of crystal, much as you construct a brick wall by stacking bricks. A unit cell is the smallest such piece from which a crystal can be made. The unit cell of a crystal of gold metal has a volume of 67.6(3. What is this volume in cubic decimeters? Solution: We have been given the volume in 3 and must convert to dm3: 67.6 3 (  EMBED Equation.3  (  EMBED Equation.3 = 67.6 3 (  EMBED Equation.3  (  EMBED Equation.3 converts to m3 converts m3 to dm3 = 6.76 ( 10(26 dm3 Exercise 1.10 Using the definitions 1 in = 2.54 cm and 1 yd = 36 in (both exact), obtain the conversion factor for yards to meters. How many meters are there in 3.54 yd? Solution:  EMBED Equation.3  =  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3  = 1.093613298 yd/mconverts to yd/cmconverts to yd/m Because all conversion factors are exact, this number is exact to an infinite number of figures. We will use a conversion factor of 1.094 in the next calculation: 3.54 yd = 3.54 yd (  EMBED Equation.3  = 3.24 m converts yd to m Additional Problems SEQ NL1 \r 0 \h   SEQ NL_a \r 0 \h  SEQ NL1 1. A 5.00-g sample of magnesium ribbon burns in oxygen. When the bright blue-white flame subsides, a white ash is left. If the ash weighs 8.29 g, what mass of oxygen combined with the magnesium during the reaction?  SEQ NL_a \r 0 \h  SEQ NL1 2. Perform the following calculations, and report the answers to the correct number of significant figures.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h .  EMBED Equation.3   seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 8.941 + 2.11  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 4.785 2.1003  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 10.56 17.8 ( 0.04  SEQ NL_a \r 0 \h  SEQ NL1 3. Round the following numbers to the designated number of significant figures, and express the answer in scientific notation.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 0.004977 to three significant figures  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 13.955 to four significant figures  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 200,143,000 to three significant figures  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . (0.19223 ( 106 to one significant figure  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 0.00000088852932956 to eight significant figures  SEQ NL_a \r 0 \h  SEQ NL1 4. Express each of the following quantities using the most appropriate SI prefix and base unit. Give the abbreviation, such as dm for decimeter, for each.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 4.3 ( 10(6 g  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 6.8 ( 10(3 L  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 8.7 ( 10(9 s  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 3.82 ( 103 m  SEQ NL_a \r 0 \h  SEQ NL1 5. Express each of the following as an SI base unit. (Use scientific notation.)  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 2.5 g  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 9.8 ps  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 4.7 mg  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 2.54 cm  SEQ NL_a \r 0 \h  SEQ NL1 6. Convert  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 5.89 kg to mg.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 4.01 (L to mL.  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 6.28 to cm.  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . (40(C (exact) to (F.  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 98.6(F to kelvins.  SEQ NL_a \r 0 \h  SEQ NL1 7. How many milliliters of soft drink are in a 12-oz can of Classic Coke? (32 fluid(oz = 1(qt.)  SEQ NL_a \r 0 \h  SEQ NL1 8. The density of iron is 7.86 g/cm3. Express this in units of ounces per cubic inch.  SEQ NL_a \r 0 \h  SEQ NL1 9. Suppose that your car averages 45 miles per gallon of gasoline when traveling on the interstate highway. If your tank contains 12 gallons of gasoline, can you make the trip from Atlanta, Georgia, to New Orleans, Louisiana, a distance of 772 km, without stopping for gas?  SEQ NL_a \r 0 \h  SEQ NL1 10. A thermos bottle contains liquid nitrogen, which boils at (323(F. Is the bottle cold enough to contain liquid oxygen, solid oxygen, or both? The melting point of oxygen is (219(C; the boiling point of oxygen is (183(C.  SEQ NL_a \r 0 \h  SEQ NL1 11. The Hope diamond, probably the most famous diamond in the world, weighs 45.52 carats. Its density is 3.51 g/cm3. What is the volume of this blue gem? [1(carat((c) = 0.200 g.]  SEQ NL_a \r 0 \h  SEQ NL1 12. A soft plastic material weighing 1.0 lb is shaped into a cylinder 13.5 cm in height and 6.8 cm in diameter. Will this cylinder sink or float when dropped into water? Assume that the density of water is 1.0 g/mL.  SEQ NL_a \r 0 \h  SEQ NL1 13. The densest known form of matter is the metal osmium, with a density of 22.48 g/cm3. A grapefruit with a diameter of 10.50 cm weighs about lb. How many pounds does a sphere of osmium of this size weigh? Answers to Additional Problems If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer. SEQ NL1 \r 0 \h   SEQ NL_a \r 0 \h  SEQ NL1 1. 8.29 g residue 5.00 g magnesium ribbon = 3.29 g oxygen (1.3, PS Sk. 1)  SEQ NL_a \r 0 \h  SEQ NL1 2.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 7.9  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 11.05  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 2.685  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 10.56 (.7) = 9.9 (1.5, PS Sk. 2)  SEQ NL_a \r 0 \h  SEQ NL1 3.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 4.98 ( 10(3  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 1.396 ( 101  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 2.00 ( 108  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . (2 ( 105  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 8.8852933 ( 10(7 (1.5)  SEQ NL_a \r 0 \h  SEQ NL1 4.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 4.3 micrograms ((g)  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 6.8 milliliters (mL)  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 8.7 nanoseconds (ns)  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 3.82 kilometers (km) (1.6)  SEQ NL_a \r 0 \h  SEQ NL1 5.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 2.5 g (  EMBED Equation.3  = 2.5 ( 103 kg  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 9.8 ps (  EMBED Equation.3  = 9.8 ( 1012 s  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 4.7 mg (  EMBED Equation.3  = 4.7 ( 103 kg  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 2.54 cm (  EMBED Equation.3  = 2.54 ( 10(2 m (1.6)  SEQ NL_a \r 0 \h  SEQ NL1 6.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 5.89 kg (  EMBED Equation.3  (  EMBED Equation.3  = 5.89 ( 106 mg  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 4.01 (L (  EMBED Equation.3  (  EMBED Equation.3  = 4.01 ( 103 mL  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 6.28 (  EMBED Equation.3  (  EMBED Equation.3  = 6.28 ( 10-8 cm (1.6, 1.8, PS Sk. 6)  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h .  EMBED Equation.3  + 32(F = 40(F (exact)  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . (98.6 (F 32 (F) (  EMBED Equation.3  = 37.0(C 37.0(C + 273.15 = 310.2 K (1.6, PS Sk. 3)  SEQ NL_a \r 0 \h  SEQ NL1 7. Calculate the number of milliliters (mL) represented by 12 fluid oz. 12 oz (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3  = 3.5 ( 102 mL (1.8, PS Sk. 6)  SEQ NL_a \r 0 \h  SEQ NL1 8.  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3  = 4.54 oz/in3 (1.8, PS Sk. 6)  SEQ NL_a \r 0 \h  SEQ NL1 9. Yes. Calculate the number of kilometers the car will travel on 12 gallons of gasoline. 12 gal (  EMBED Equation.3  (  EMBED Equation.3  = 8.7 ( 102 km The distance from Atlanta to New Orleans is 772 km; therefore, you can make the trip without stopping for gas. (1.8, PS Sk. 6)  SEQ NL_a \r 0 \h  SEQ NL1 10. The bottle is cold enough to contain liquid oxygen but not solid oxygen. (C = (323 (F 32 (F) (  EMBED Equation.3  = 197 (C (1.6, PS Sk. 3)  SEQ NL_a \r 0 \h  SEQ NL1 11. Since d =  EMBED Equation.3 , V =  EMBED Equation.3 : V = 45.52 c (  EMBED Equation.3  (  EMBED Equation.3  V = 2.59 cm3 (1.7, 1.8, PS Sk. 5, 6)  SEQ NL_a \r 0 \h  SEQ NL1 12. The cylinder will float because its density is less than that of water; V = h(r2, r =  EMBED Equation.3 . Density =  EMBED Equation.3  =  EMBED Equation.3  (  EMBED Equation.3  = 0.93 g/cm3 (1.7, PS Sk. 4)  SEQ NL_a \r 0 \h  SEQ NL1 13. Since d =  EMBED Equation.3 , m = dV; V =  EMBED Equation.3 (r3, r =  EMBED Equation.3  m =  EMBED Equation.3  (  EMBED Equation.3  ( ( (  EMBED Equation.3  cm3 (  EMBED Equation.3  = 30.04 lb (1.7, 1.8, PS Sk. 5, 6) Chapter Post-Test SEQ NL1 \r 0 \h   SEQ NL_a \r 0 \h  SEQ NL1 1. In the scientific method, one experiment can disprove a theory if the experiment is repeatable. True/False  SEQ NL_a \r 0 \h  SEQ NL1 2. The law of conservation of mass states that ________________________________________ ___________________________________________________________________________.  SEQ NL_a \r 0 \h  SEQ NL1 3. Which of the following could be classified as a possible theory? (There may be more than one answer.)  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . Animals have a special sense to warn them of upcoming natural disasters such as earthquakes.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . Water boils at 100(C at 1 atm pressure.  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . Sodium reacts with chlorine to form sodium chloride.  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . One inch equals 2.54 centimeters.  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . Molecules are composed of atoms bound together.  SEQ NL_a \r 0 \h  SEQ NL1 4. The derived unit of volume in SI units is  seq NL_a \* alphabetic a seq N 345678ab{|  & ( . 2       , 0 z|RTbfbdnp *.lnz~>B j h5 h5jhU\ h\jhUhU7} (   |Tp nv##t##s#s#s#s#s#s#s#s#s#s#s##s#s#s#s#s#s#s#s#s#s#s#s#sNO52uwtv  $^`nr | h!j!x!|! 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L.  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . m3.  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . none of the above.  SEQ NL_a \r 0 \h  SEQ NL1 5. Which of the following equals the volume of 469 ft3 of gas in liters?  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 469 ft3 (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3   seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 469 ft3 (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3   seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 469 ft3 (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3   seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 469 ft3 (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3  (  EMBED Equation.3   SEQ NL_a \r 0 \h  SEQ NL1 6. Express the following numbers in scientific notation.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . (0.00042  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 37.4 ( 10-6  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 19,758  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 250,000,000  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 0.1603 ( 1010  SEQ NL_a \r 0 \h  SEQ NL1 7. Which of the following give(s) the correct number of significant figures for the respective number?  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 104.710 has six significant figures.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 0.000305 has six significant figures.  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . (91.20 ( 104 has four significant figures.  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 5.62 has three significant figures.  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 0.00000001 has eight significant figures.  SEQ NL_a \r 0 \h  SEQ NL1 8. Round off the following numbers to the designated number of significant figures.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 46,574 to three significant figures  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 6.5535 to four significant figures  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 0.009374 to three significant figures  SEQ NL_a \r 0 \h  SEQ NL1 9. Which of the following answers give(s) the correct number of significant figures for the indicated arithmetic calculation?  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h .  EMBED Equation.3  = 2.935 ( 103  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . (64.4)1(0.00342) + 1.43 ( 105 = 6.74 ( 105  SEQ NL_a \r 0 \h  SEQ NL1 10. Complete the following conversions. Observe significant figures.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 17.3 in = ___ cm.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 185 lb = ___ kg.  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 1000 mL = ___ gal.  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 145(C = ___ (F.  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 55 mi/h = ___ m/s.  SEQ NL_a \r 0 \h  SEQ NL1 11. The freezing point of ethyl alcohol is (117.6(C. Using this information, indicate whether each of the following statements is true or false.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . This is colder than (179.7(F.  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . This is colder than 175 K.  SEQ NL_a \r 0 \h  SEQ NL1 12. Light travels through space at the speed of 1.86 ( 105 mi/s. Using this information, answer the following questions.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . What is the speed of light in centimeters per second?  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . A light-year is the distance light travels in 1 year. How many miles are there in 1 light-year?  SEQ NL_a \r 0 \h  SEQ NL1 13. A jeweler was asked to determine the identity of the metal in a gold-colored lapel pin. The volume of the pin was determined to be 0.8721 cm3, and its mass was determined to be 14.790 g. Calculate the density of the metal, and determine if the pin was made from brass (density 8.5 g/cm3), 14-karat gold (density 15 g/cm3), or 18karat gold (density 17 g/cm3).  SEQ NL_a \r 0 \h  SEQ NL1 14. The planet Jupiter is often referred to as a failed star because its density, 1.33 g/cm3, is about that of our sun. If the volume of Jupiter is 1.44 ( 1030 cm3, by what factor is its mass less than the mass of the sun (approximately 2 ( 1030 kg)? Answers to Chapter Post-Test If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer. SEQ NL1 \r 0 \h   SEQ NL_a \r 0 \h  SEQ NL1 1. True. (1.2)  SEQ NL_a \r 0 \h  SEQ NL1 2. Mass remains constant during a chemical change. (1.3)  SEQ NL_a \r 0 \h  SEQ NL1 3. a, e (1.2)  SEQ NL_a \r 0 \h  SEQ NL1 4. d (1.7)  SEQ NL_a \r 0 \h  SEQ NL1 5. c (1.7, 1.8, PS Sk. 5, 6)  SEQ NL_a \r 0 \h  SEQ NL1 6.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . (4.2 ( 104  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 3.74 ( 105  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 1.9758 ( 104  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 2.5 ( 108  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 1.603 ( 109 (Appendix A)  SEQ NL_a \r 0 \h  SEQ NL1 7. a, c, d (1.5)  SEQ NL_a \r 0 \h  SEQ NL1 8.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 4.66 ( 104  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 6.554  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 0.00937 (1.5)  SEQ NL_a \r 0 \h  SEQ NL1 9. b (1.5, PS Sk. 2)  SEQ NL_a \r 0 \h  SEQ NL1 10.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 43.9 cm  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 83.9 kg  seq NL_a \* alphabetic c seq NL_1_ \r 0 \h . 0.2642 gal  seq NL_a \* alphabetic d seq NL_1_ \r 0 \h . 293(F  seq NL_a \* alphabetic e seq NL_1_ \r 0 \h . 25 m/s (1.6, 1.8, PS Sk. 3, 6)  SEQ NL_a \r 0 \h  SEQ NL1 11.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . False  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . True (1.6, PS Sk. 3)  SEQ NL_a \r 0 \h  SEQ NL1 12.  seq NL_a \* alphabetic a seq NL_1_ \r 0 \h . 2.99 ( 1010 cm/s  seq NL_a \* alphabetic b seq NL_1_ \r 0 \h . 5.87 ( 1012 mi (1.7, 1.8, PS Sk. 6)  SEQ NL_a \r 0 \h  SEQ NL1 13. d = 16.96 g/cm3; the metal is 18-karat gold (1.7, PS Sk. 4)  SEQ NL_a \r 0 \h  SEQ NL1 14. m = 1.92 ( 1027 kg; 1000 times less massive than our sun (1.7, 1.8, PS Sk. 5, 6)      PAGE 18 Chapter  DOCPROPERTY "ChapterNumber" \* MERGEFORMAT 1:  DOCPROPERTY "ChapterTitle" \* MERGEFORMAT Chemistry and Measurement Chapter  DOCPROPERTY "ChapterNumber" \* MERGEFORMAT 1:  DOCPROPERTY "ChapterTitle" \* MERGEFORMAT Chemistry and Measurement  PAGE 19 Copyright Houghton Mifflin Company. All rights reserved. Copyright Houghton Mifflin Company. All rights reserved. Copyright Houghton Mifflin Company. 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