01_Chapter_ Introduction 2021-08-15 21.08.47(4).pdf

Chapter 1 (3 Lectures) Fifth Edition Chemistry: The Central Science  Chemistry: The Central CHAPTER 1 Science 1.1 The Study of Chemistry 1.2 Classification of Matter 1.3 Sci...

Chapter 1 (3 Lectures) Fifth Edition Chemistry: The Central Science  Chemistry: The Central CHAPTER 1 Science 1.1 The Study of Chemistry 1.2 Classification of Matter 1.3 Scientific Measurement 1.4 The Properties of Matter 1.5 Uncertainty in Measurement 1.6 Using Units and Solving Problems 2 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.  1.2 Classification of Matter States of Matter Elements Compounds Mixtures 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). Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 3  1.1 The Study of Chemistry Chemistry You May Already Know Chemistry is the study of matter and the changes that matter undergoes. Matter is what makes up our bodies, our belongings, our physical environment, and in fact our universe. Matter is anything that has mass and occupies space. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 4  1.1 The Study of Chemistry Chemistry You May Already Know Although it can take many different forms, all matter consists of various combinations of atoms of only a relatively small number of simple substances called elements. The properties of matter depend on which of these elements it contains and on how the atoms of those elements are arranged. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 5 1.2 Classification of Matter Topics States of Matter Elements Compounds Mixtures 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). Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 6  1.2 Classification of Matter (1) States of Matter Chemists classify matter as either a substance or a mixture of substances. A substance is a form of matter that has a specific composition and distinct properties. Examples are salt (sodium chloride), iron, water, mercury, carbon dioxide, and oxygen. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 7  1.2 Classification of Matter States of Matter All substances can, in principle, exist as a solid, a liquid, and a gas, the three physical states. Solids & liquids :Condensed phase LiquidsCopyright & gases: Fluids © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 8  1.2 Classification of Matter States of Matter A substance is a form of matter that has a definite (constant) composition and distinct properties. Examples are salt (sodium chloride), iron, water, mercury, carbon dioxide, and oxygen. Substances can be either a. elements (such as iron (Fe), mercury (Hg), and oxygen (O2)) b. or compounds (such as salt (NaCl), water (H2O)), and carbon dioxide (CO2)). Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 9  1.2 Classification of Matter (5) Elements An element is a substance that cannot be separated into simpler substances by chemical means. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 10  1.2 Classification of Matter (6) Compounds Most elements can combine with other elements to form compounds. A compound is a substance composed of atoms of two or more elements chemically united in fixed Proportions (NaCl, H2O, CO2). A compound cannot be separated into simpler substances by any physical process. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 11  1.2 Classification of Matter Mixtures A mixture is a combination of two or more substances in which the substances retain their distinct identities. When we dissolve a teaspoon of sugar in a glass of water, we get a homogeneous mixture because the composition of the mixture is uniform throughout. If we mix sand with iron filings, however, the sand and the iron filings remain distinct and discernible from each other. This type of mixture is called a heterogeneous mixture because the composition is not uniform. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 12  1.2 Classification of Matter Mixtures Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 13  1.2 Classification of Matter Mixtures Homogeneous Mixture Heterogeneous Mixture ‘Homo’ means same ‘Hetero’ means different Uniform composition Non-uniform composition Particles are distributed Particles are distributed uniformly non-uniformly It can be separated out It can be separated out physically physically Example: sand with iron filings Example: sugar in water, rainwater Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 14  1.2 Classification of Matter (9) Mixtures Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 15 1.3 Scientific Measurement Topics SI Base Units Mass Temperature Derived Units: Volume and Density 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). Define and provide examples of derived units. Calculate the density of a substance. Use density to relate mass and volume. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 16  1.3 Scientific Measurement SI (international system of units) Base Units Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 17  1.3 Scientific Measurement SI Base Units Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 18  1.3 Scientific Measurement Mass Mass is a measure of the amount of matter in an object or sample. 1 kg = 1000g = 1 × 103 g Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 19  1.3 Scientific Measurement Temperature There are two temperature scales used in chemistry. The Celsius scale was originally defined using the freezing point (0° C) and the boiling point (100° C) of pure water at sea level. The SI base unit of temperature is the kelvin. Kelvin is known as the absolute temperature scale, meaning that the lowest temperature possible is 0 K, a temperature referred to as “absolute zero.” Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 20  1.3 Scientific Measurement Temperature Units of the Celsius and Kelvin scales are equal in magnitude, so a degree Celsius is equivalent to a kelvin. K = °C + 273.15 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 21 SAMPLE PROBLEM 1.1 Strategy Normal human body temperature can range over the course of the day from about 36°C in the early morning to about 37°C in the afternoon. Express these two temperatures and the range that they span using the Kelvin scale. Strategy Use K = °C + 273.15 to convert temperatures from the Celsius scale to the Kelvin scale. Then convert the range of temperatures from degrees Celsius to kelvin, keeping in mind that 1°C is equivalent to 1 K. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 22 SAMPLE PROBLEM 1.1 Solution Solution 36°C + 273 = 309K 37°C + 273 = 310K The range of 1°C is equal to a range of 1 K. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 23  1.3 Scientific Measurement Temperature 5°C temperature in Celsius = (temperature in Fahrenheit − 32°F) × 9°F 9°F temperature in Fahrenheit = × (temperature in degrees Celsius) + 32°F 5°C Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 24  SAMPLE PROBLEM 1.2 Strategy A body temperature above 39°C constitutes a high fever. Convert this temperature to the Fahrenheit scale. Strategy We are given a temperature in Celsius and are asked to convert it to Fahrenheit. Setup 9°F temperature in Fahrenheit = × (temperature in Celsius) + 32°F 5°C Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 25 SAMPLE PROBLEM 1.2 Solution Solution 9°F temperature in Fahrenheit = × 39°C + 32℉ = 102.2°F 5°C Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 26  1.3 Scientific Measurement Derived Units: Volume and Density There are many quantities, such as volume and density, that require units not included in the base SI units. In these cases, we must combine base units to derive appropriate units for the quantity. The derived SI unit for volume, the meter cubed (m3), is a larger volume than is practical in most laboratory settings. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 27  1.3 Scientific Measurement Derived Units: Volume and Density The more commonly used metric unit, the liter (L), is derived by cubing the decimeter (one-tenth of a meter) and is therefore also referred to as the cubic decimeter (dm3). Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 28  1.3 Scientific Measurement Derived Units: Volume and Density Density is the ratio of mass to volume. 𝑚 𝑑= 𝑉 1 gΤcm3 = 1 gΤmL = 1000 kgΤm3 1 gΤL = 0.001 gΤmL Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 29  SAMPLE PROBLEM 1.3 Strategy Ice cubes float in a glass of water because solid water is less dense than liquid water. (a) Calculate the density of ice given that, at 0°C, a cube that is 2.0 cm on each side has a mass of 7.36 g, and (b) determine the volume occupied by 23 g of ice at 0°C. Strategy (a) Determine density by dividing mass by volume. (b) use the calculated density to determine the volume occupied by the given mass. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 30 SAMPLE PROBLEM 1.3 Setup Setup (a) We are given the mass of the ice cube, but we must calculate its volume from the dimensions given. The volume of the ice cube is (2.0 cm)3 , or 8.0 cm3. (b) Rearranging the density equation to solve for volume gives 𝑉 = 𝑚/𝑑. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 31 SAMPLE PROBLEM 1.3 Solution Solution 7.36g (a) 𝑑 = = 0.92 gΤcm3 or 0.92 gΤmL 8.0cm3 (b) 𝑉 = 23 g 0.92 gΤcm3 = 25cm3 or 25 mL Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 32 1.4 The Properties of Matter Topics Physical Properties Chemical Properties Extensive and Intensive Properties Learning Objectives Understand the difference between chemical changes (chemical reactions) and physical changes. Distinguish between chemical properties and physical properties. Understand the difference between extensive properties and intensive properties, and provide examples of both. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 33  1.4 The Properties of Matter Physical Properties Substances are identified by their properties and composition. Properties of a substance may be quantitative (measured and expressed with a number) or qualitative (not requiring explicit measurement). A physical property is one that can be observed and measured without changing the identity of a substance. Melting is a physical change; We can recover the original ice by cooling the water until it freezes. Therefore, the melting point of a substance is a physical property. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 34  1.4 The Properties of Matter Physical Properties Melting is a physical change; one in which the state of matter changes, but the identity of the matter does not change. We can recover the original ice by cooling the water until it freezes. Therefore, the melting point of a substance is a physical property. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 35  1.4 The Properties of Matter Chemical Properties The statement “Hydrogen gas burns in oxygen gas to form water” describes a chemical property of hydrogen, because to observe this property we must carry out a chemical change—burning in oxygen (combustion), in this case. After a chemical change, the original substance (hydrogen gas in this case) will no longer exist. What remains is a different substance (water, in this case). We cannot recover the hydrogen gas from the water by means of a physical process, such as boiling or freezing. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 36 1.4 The Properties of Matter Chemical Properties Chemical properties cannot be determined just by viewing or touching the substance; the substance’s internal structure must be affected for its chemical properties to be investigated. Chemical properties are not reversible Burning of gasoline Souring of milk Combustion of sugar to produce CO2 and H2O Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 37  1.4 The Properties of Matter Extensive and Intensive Properties All properties of matter are either extensive or intensive. The measured value of an extensive property depends on the amount of matter. Mass is an extensive property. The value of an intensive property does not depend on the amount of matter. Density and temperature are intensive properties. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 38 SAMPLE PROBLEM 1.4 The diagram in (a) shows a compound made up of atoms of two elements (represented by the green and red spheres) in the liquid state. Which of the diagrams in (b) to (d) represent a physical change, and which diagrams represent a chemical change? Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 39 SAMPLE PROBLEM 1.4 Strategy Strategy A physical change does not change the identity of a substance, whereas a chemical change does change the identity of a substance. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 40 SAMPLE PROBLEM 1.4 Solution Solution Diagrams (b) and (c) represent chemical changes. Diagram (d) represents a physical change. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 41 1.5 Uncertainty in Measurement Topics Significant Figures Calculations with Measured Numbers Accuracy and Precision Learning Objectives Learn the rules for determining significant figures in 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. Define and use the terms precision and accuracy when describing measured quantities. 42 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.  1.5 Uncertainty in Measurement Significant Figures Chemistry makes use of two types of numbers: exact and inexact. Exact numbers include numbers with defined values, such as 2.54 in the definition 1 inch (in) = 2.54 cm 1000 in the definition 1 kg = 1000 g 12 in the definition 1 dozen = 12 objects. Numbers measured by any method other than counting are inexact. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 43  1.5 Uncertainty in Measurement Significant Figures An inexact number must be reported in such a way as to indicate the uncertainty in its value. This is done using significant figures. Significant figures are the meaningful digits in a reported number. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 44  1.5 Uncertainty in Measurement Significant Figures The number of significant figures in any number can be determined using the following guidelines: 1. Any digit that is not zero is significant (112.1 has four significant figures). 2. Zeros located between nonzero digits are significant (305 has three significant figures, and 50.08 has four significant figures). Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 45  1.5 Uncertainty in Measurement Significant Figures 3. Zeros to the left of the first nonzero digit are not significant (0.0023 has two significant figures, and 0.000001 has one significant figure). 4. Zeros to the right of the last nonzero digit are significant if the number contains a decimal point (1.200 has four significant figures Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 46  1.5 Uncertainty in Measurement Significant Figures 5. Zeros to the right of the last nonzero digit in a number that does not contain a decimal point may or may not be significant (100 may have one, two, or three significant figures—it is impossible to tell without additional information. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 47  1.5 Uncertainty in Measurement Significant Figures To avoid ambiguity in such cases, it is best to express such numbers using scientific notation [Appendix 1]. 1.3 × 102 two significant figures 1.30 × 102 three significant figures Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 48  SAMPLE PROBLEM 1.5 Determine the number of significant figures in the following measurements: (a) 443 cm (b) 15.03 g (c) 0.0356 kg (d) 3.000 3 10−7 L (e) 50 mL (f) 0.9550 m Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 49 SAMPLE PROBLEM 1.5 Strategy Strategy All nonzero digits are significant, so the goal will be to determine which of the zeros is significant. Setup Zeros are significant if they appear between nonzero digits or if they appear after a nonzero digit in a number that contains a decimal point. Zeros may or may not be significant if they appear to the right of the last nonzero digit in a number that does not contain a decimal point. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 50  SAMPLE PROBLEM 1.5 Solution Solution (a) 443 cm (b) 15.03 g (c) 0.0356 kg (d) 3.000 3 10−7 L (e) 50 mL (f) 0.9550 m (a) 3; (b) 4; (c) 3; (d) 4; (e) 1 or 2, an ambiguous case; (f) 4 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 51  1.5 Uncertainty in Measurement Calculations with Measured Numbers 1. In addition and subtraction, the answer cannot have more digits to the right of the decimal point than the original number with the smallest number of digits to the right of the decimal point. 102.50 ← two digits after the decimal point + 0.231 ← three digits after the decimal point 102.731 ← round to 102.73 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 52  1.5 Uncertainty in Measurement Calculations with Measured Numbers 143.29 ← two digits after the decimal point −20.1 ← one digit after the decimal point 123.19 ← round to 123.2 If the leftmost digit to be dropped is less than 5, round down. If the leftmost digit to be dropped is equal to or greater than 5, round up. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 53  1.5 Uncertainty in Measurement Calculations with Measured Numbers 2. In multiplication and division, the number of significant figures in the final product or quotient is determined by the original number that has the smallest number of significant figures. 1.4 × 8.011 = 11.2154 ← round to 11 11.57 = 0.037825290964 ← round to0.03783 305.88 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 54  1.5 Uncertainty in Measurement Calculations with Measured Numbers 3. Exact numbers can be considered to have an infinite number of significant figures and do not limit the number of significant figures in a calculated result. 3 × 2.5g = 7.5g Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 55  1.5 Uncertainty in Measurement Calculations with Measured Numbers 4. In calculations with multiple steps, rounding the result of each step can result in “rounding error.” In general, it is best to retain at least one extra digit until the end of a multistep calculation to minimize rounding error. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 56  SAMPLE PROBLEM 1.6 Strategy Perform the following arithmetic operations and report the result to the proper number of significant figures: (a) 317.5 mL + 0.675 mL (b) 47.80 L – 2.075 L (c) 13.5 g ÷ 45.18 L (d) 6.25 cm × 1.175 cm (e) 5.46 × 102 g + 4.991 × 103 g Strategy Apply the rules for significant figures in calculations, and round each answer to the appropriate number of digits. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 57  SAMPLE PROBLEM 1.6 Setup Setup (a) 317.5 mL + 0.675 mL (b) 47.80 L – 2.075 L (c) 13.5 g ÷ 45.18 L (d) 6.25 cm × 1.175 cm (e) 5.46 × 102 g + 4.991 × 103 g (a) The answer will contain one digit to the right of the decimal point to match 317.5, which has the fewest digits to the right of the decimal point. (b) The answer will contain two digits to the right of the decimal point to match 47.80. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 58 SAMPLE PROBLEM 1.6 Setup, Cont. Setup (a) 317.5 mL + 0.675 mL (b) 47.80 L – 2.075 L (c) 13.5 g ÷ 45.18 L (d) 6.25 cm × 1.175 cm (e) 5.46 × 102 g + 4.991 × 103 g (c) The answer will contain three significant figures to match 13.5, which has the fewest number of significant figures in the calculation. (d) The answer will contain three significant figures to match 6.25. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 59  SAMPLE PROBLEM 1.6 Setup, Cont.1 Setup (a) 317.5 mL + 0.675 mL (b) 47.80 L – 2.075 L (c) 13.5 g ÷ 45.18 L (d) 6.25 cm × 1.175 cm (e) 5.46 × 102 g + 4.991 × 103 g (e) To add numbers expressed in scientific notation, first write both numbers to the same power of 10. That is, 4.991 × 103 = 4.991 × 102 , so the answer will contain two digits to the right of the decimal point (when multiplied by 102 ) to match both 5.46 and 49.91. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 60 SAMPLE PROBLEM 1.6 Solution Solution (a) 317.5 mL +0.675 mL 318.175 mL ← round to 318.2 mL (b) 47.80 L −2.075 L 45.725 L ← round to 45.73 L 13.5 g (c) = 0.298804781 gΤL ← round to 0.299 gΤL 45.18 L (d) 6.25 cm × 1.175 cm = 7.34375 cm2 ← round to 7.34 cm2 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 61  SAMPLE PROBLEM 1.6 Solution, Cont. Solution (e) 5.46 × 102 g +49.91 × 102 g 55.37 × 102 g = 5.537 × 103 g Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 62  SAMPLE PROBLEM 1.7 An empty container with a volume of 9.850 × 102 cm3 is weighed and found to have a mass of 124.6 g. The container is filled with a gas and reweighed. The mass of the container and the gas is 126.5 g. Determine the density of the gas to the appropriate number of significant figures. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 63 SAMPLE PROBLEM 1.7 Strategy Strategy This problem requires two steps: subtraction to determine the mass of the gas, and division to determine its density. Apply the corresponding rule regarding significant figures to each step. Setup 126.5 g – 124.6 g = 1.9 g. Thus, in the division of the mass of the gas by the volume of the container, the result can have only two significant figures. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 64 SAMPLE PROBLEM 1.7 Solution Solution 126.5g −124.6 g mass of gas= 1.9 g 1.9 g 3 ← round to 0.0019 gΤcm3 density = = 0.00193 g Τcm 9.850 × 102 cm3 The density of the gas is 1.9 × 10−3 gΤcm3 65 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.  1.5 Uncertainty in Measurement Accuracy and Precision Accuracy tells us how close a measurement is to the true value. The true value is 0.370 g Precision tells us how close multiple measurements of the same thing are to one another. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 66  1.5 Uncertainty in Measurement Accuracy and Precision Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 67  1.5 Uncertainty in Measurement Accuracy and Precision 68 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1.6 Using Units and Solving Problems Topics Conversion Factors Dimensional Analysis—Tracking Units 69 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.  1.6 Using Units and Solving Problem Conversion Factors A conversion factor is a fraction in which the same quantity is expressed one way in the numerator and another way in the denominator. 1 in = 2.54 cm 1 in 2.54 cm 2.54 cm 1 in Because both forms of this conversion factor are equal to 1, we can multiply a quantity by either form without changing the value of that quantity. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 70  1.6 Using Units and Solving Problems Conversion Factors 2.54cm 12.00 in × = 30.48 cm 1 in Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 71  1.6 Using Units and Solving Problems Dimensional Analysis—Tracking Units The use of conversion factors in problem solving is called dimensional analysis or the factor-label method. 2.54 cm 1m 12.00 in × × = 0.3048 m 1 in 100 cm Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 72 SAMPLE PROBLEM 1.8 Strategy The Food and Drug Administration (FDA) recommends that dietary sodium intake be no more than 2400 mg per day. What is this mass in pounds (lb), if 1 lb = 453.6 g? Strategy This problem requires a two-step dimensional analysis, because we must convert milligrams to grams and then grams to pounds. Assume the number 2400 has four significant figures. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 73 SAMPLE PROBLEM 1.8 Setup Setup The necessary conversion factors are derived from the equalities 1 g = 1000 mg and 1 lb = 453.6 g. 1g 1000 mg 1 lb 453.6 g or and or 1000 mg 1g 453.6 g 1 lb Solution 1g 1 lb 2400 mg × × = 0.005291 lb 1000 mg 453.6 g Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 74 SAMPLE PROBLEM 1.9 Strategy An average adult has 5.2 L of blood. What is the volume of blood in cubic meters? Strategy Convert liters to cubic centimeters and then cubic centimeters to cubic meters. Setup 1 L = 1000 cm3 and 1 cm = 1 × 10−2 m When a unit is raised to a power, the corresponding conversion factor must also be raised to that power in order for the units to cancel appropriately. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 75 SAMPLE PROBLEM 1.9 Solution Solution 3 1000 cm 3 1× −2 10 m 5.2 L × × = 5.2 × 10−3 m3 1L 1 cm Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 76 Dimensional Analysis  Recognize the problem as a dimensional analysis problem. How?  Identify your information given and your information sought. How?  See what conversion factors the question supplies.  Think the problem through focusing on unit to unit conversions, not on numbers Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 77  Extra Problem A motor performs 7.4 × 106 J of work every second. You have 65.2 × 108 grams of copper that need moved. If it takes 9.2 kJ to move one mole of copper, how long (in hours) will it take to move the copper? There are 63.5 grams of copper per mole. Answer: 1mol 9.2 kJ 1000 J sec hr 65.2  10 8 g       35 hours 63.5 g 1mol kJ 7.4  10 J 6 3600 sec Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 78 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 79