1. Chapter 2 Scientific Measurements 1 Vanessa N. Prasad-Permaul CHM 1025 Valencia Community College

2. Uncertainty in Measurements  A measurement is a number with a unit attached.  It is not possible to make exact measurements, thus all measurements have uncertainty.  We will generally use metric system units. These include:  The meter, m, for length measurements  The gram, g, for mass measurements  The liter, L, for volume measurements 2

3. Length Measurements  Let’s measure the length of a candy cane.  Ruler A has 1 cm divisions, so we can estimate the length to ± 0.1 cm. The length is 4.2 ± 0.1 cm.  Ruler B has 0.1 cm divisions, so we can estimate the length to ± 0.05 cm. The length is 4.25 ± 0.05 cm. 3

4. Uncertainty in Length  Ruler A: 4.2 ± 0.1 cm; Ruler B: 4.25 ± 0.05 cm.  Ruler A has more uncertainty than Ruler B.  Ruler B gives a more precise measurement. 4

5. EXAMPLE 2.1 Uncertainty in Measurement Ruler A has an uncertainty of ±0.1 cm, and Ruler B has an uncertainty of ±0.05 cm. Thus, (a)Ruler A can give the measurements 2.0 cm and 2.5 cm. (b)Ruler B can give the measurements 3.35 cm and 3.50 cm. Solution Which measurements are consistent with the metric rulers shown in Figure 2.2? (a)Ruler A: 2 cm, 2.0 cm, 2.05 cm, 2.5 cm, 2.50 cm (b)Ruler B: 3.0 cm, 3.3 cm, 3.33 cm, 3.35 cm, 3.50 cm Figure 2.2 Metric Rulers for Measuring Length On Ruler A, each division is 1 cm. On Ruler B, each division is 1 cm and each subdivision is 0.1 cm.

6. EXERCISE 2.1 Uncertainty in Measurement Which measurements are consistent with the metric rulers shown in Figure 2.2? (a)Ruler A: 1.5 cm, 1.50 cm, 1.55 cm, 1.6 cm, 2.00 cm (b)Ruler B: 0.5 cm, 0.50 cm, 0.055 cm, 0.75 cm, 0.100 cm Practice Exercise Figure 2.2 Metric Rulers for Measuring Length On Ruler A, each division is 1 cm. On Ruler B, each division is 1 cm and each subdivision is 0.1 cm.

7. Mass Measurements  The mass of an object is a measure of the amount of matter it possesses.  Mass is measured with a balance and is not affected by gravity.  Mass and weight are not interchangeable. 7

8. Mass Versus Weight  Mass and weight are not the same.  Weight is the force exerted by gravity on an object. 8

9. Volume Measurements  Volume is the amount of space occupied by a solid, a liquid, or a gas.  There are several instruments for measuring volume, including:  Graduated cylinder  Syringe  Buret  Pipet  Volumetric flask 9

10. Significant Digits  Each number in a properly recorded measurement is a significant digit (or significant figure).  Significant digits express the uncertainty in the measurement.  When you count significant digits, start counting with the first nonzero number.  Let’s look at a reaction measured by three stopwatches. 10

11. Significant Digits, Continued  Stopwatch A is calibrated to seconds (±1 s); Stopwatch B to tenths of a second (±0.1 s); and Stopwatch C to hundredths of a second (±0.01 s). 11 Stopwatch A reads 35 s; B reads 35.1 s; and C reads 35.08 s. –35 s has two significant figure. –35.1 s has three significant figures. –35.08 has four significant figures.

12. EXAMPLE 2.2 Significant Digits In each example, we simply count the number of digits. Thus, (a)5(b)3 (c)1(d)4 Notice that the leading zero in (b) and (c) is not part of the measurement but is inserted to call attention to the decimal point that follows. Solution State the number of significant digits in the following measurements: (a)12,345 cm(b)0.123 g (c)0.5 mL(d)102.0 s

13. EXERCISE 2.2 Significant Digits State the number of significant digits in the following measurements: (a)2005 cm(b)25.000 g (c)25.0 mL(d)0.25 s Practice Exercise

14. Significant Digits and Placeholders  If a number is less than 1, a placeholder zero is never significant.  Therefore, 0.5 cm, 0.05 cm, and 0.005 cm all have one significant digit.  If a number is greater than 1, a placeholder zero is usually not significant.  Therefore, 50 cm, 500 cm, and 5000 cm all have one significant digit. 14

15. EXAMPLE 2.3 Significant Digits In each example, we count the number of significant digits and disregard placeholder zeros. Thus, (a)2(b)4 (c)3(d)2 Solution State the number of significant digits in the following measurements: (a)0.025 cm(b)0.2050 g (c)25.0 mL(d)2500 s

16. EXERCISE 2.3 Significant Digits State the number of significant digits in the following measurements: (a)0.050 cm (b)0.0250 g (c)50.00 mL(d)1000 s Practice Exercise

17. Exact Numbers  When we count something, it is an exact number.  Significant digit rules do not apply to exact numbers.  An example of an exact number: There are seven coins on this slide. 17

18. Rounding Off Nonsignificant Digits  All numbers from a measurement are significant. However, we often generate nonsignificant digits when performing calculations.  We get rid of nonsignificant digits by rounding off numbers.  There are three rules for rounding off numbers. 18

19. Rules for Rounding Numbers 1. If the first nonsignificant digit is less than 5, drop all nonsignificant digits. 2. If the first nonsignificant digit is greater than or equal to 5, increase the last significant digit by 1 and drop all nonsignificant digits. 3. If a calculation has two or more operations, retain all the nonsignificant digits until the final operation and then round off the answer. 19

20. Rounding Examples  A calculator displays 12.846239 and 3 significant digits are justified.  The first nonsignificant digit is a 4, so we drop all nonsignificant digits and get 12.8 as the answer.  A calculator displays 12.856239 and 3 significant digits are justified.  The first nonsignificant digit is a 5, so the last significant digit is increased by one to 9. All the nonsignificant digits are dropped, and we get 12.9 as the answer. 20

21. Rounding Off and Placeholder Zeros  Round the measurement 151 mL to two significant digits.  If we keep two digits, we have 15 mL, which is only about 10% of the original measurement.  Therefore, we must use a placeholder zero: 150 mL  Recall that placeholder zeros are not significant.  Round the measurement 2788 g to two significant digits.  We get 2800 g. Remember, the placeholder zeros are not significant, and 28 grams is significantly less than 2800 grams. 21

22. EXAMPLE 2.4 Rounding Off To locate the first nonsignificant digit, count three digits from left to right. If the first nonsignificant digit is less than 5, drop all nonsignificant digits. If the first nonsignificant digit is 5 or greater, add 1 to the last significant digit. (a)22.3 (Rule 2)(b)0.345 (Rule 1) (c)0.0720 (Rule 1)(d)12,300 (Rule 2) In (d), notice that two placeholder zeros must be added to 123 to obtain the correct decimal place. Solution Round off the following numbers to three significant digits: (a)22.250(b)0.34548 (c)0.072038(d)12,267

23. EXERCISE 2.4 Rounding Off Round off the following numbers to three significant digits: (a)12.514748(b)0.6015261 (c)192.49032(d)14652.832 Practice Exercise

24. Adding and Subtracting Measurements  When adding or subtracting measurements, the answer is limited by the value with the most uncertainty. 24 Let’s add three mass measurements. The measurement 106.7 g has the greatest uncertainty (± 0.1 g). The correct answer is 107.1 g. 106.7g 0.25g + 0.195g 107.145g

25. EXAMPLE 2.5 Addition/Subtraction and Rounding Off In addition or subtraction operations, the answer is limited by the measurement with the most uncertainty. Since 30.5 mL has the most uncertainty (±0.1 mL), we round off to one decimal place. The answer is 5.0 mL and is read “five point zero milliliters.” Solution Add or subtract the following measurements and round off your answer: 35.45 mL – 30.5 mL

26. EXERCISE 2.5 Addition/Subtraction and Rounding Off Add or subtract the following measurements and round off your answer: (a)8.6 cm + 50.05 cm(b)34.1 s – 0.55 s Practice Exercise

27. Multiplying and Dividing Measurements  When multiplying or dividing measurements, the answer is limited by the value with the fewest significant figures.  Let’s multiply two length measurements: (5.15 cm)(2.3 cm) = 11.845 cm 2  The measurement 2.3 cm has the fewest significant digits—two.  The correct answer is 12 cm 2. 27

28. EXAMPLE 2.6 Multiplication/Division and Rounding Off In multiplication and division operations, the answer is limited by the measurement with the least number of significant digits. (a)In this example, 50.5 cm has three significant digits and 12 cm has two. (50.5 cm) (12 cm) = 606 cm 2 The answer is limited to two significant digits and rounds off to 610 cm 2 after inserting a placeholder zero. The answer is read “six hundred and ten square centimeters.” Solution Multiply or divide the following measurements and round off your answer: (a)50.5 cm  12 cm(b)103.37 g/20.5 mL

29. EXAMPLE 2.6 Multiplication/Division and Rounding Off (b)In this example, 103.37 g has five significant digits and 20.5 mL has three. The answer is limited to three significant digits and rounds off to 5.04 g/mL. Notice that the unit is a ratio; the answer is read as “five point zero four grams per milliliter.”

30. EXERCISE 2.6 Multiplication/Division and Rounding Off Multiply or divide the following measurements and round off your answer. (a)(359 cm) (0.20 cm) (b)73.950 g/25.5 mL Practice Exercise

31. Exponential Numbers  Exponents are used to indicate that a number has been multiplied by itself.  Exponents are written using a superscript; thus, (2)(2)(2) = 2 3.  The number 3 is an exponent and indicates that the number 2 is multiplied by itself 3 times. It is read “2 to the third power” or “2 cubed”.  (2)(2)(2) = 2 3 = 8 31

32. Powers of 10  A power of 10 is a number that results when 10 is raised to an exponential power.  The power can be positive (number greater than 1) or negative (number less than 1). 32

33. EXAMPLE 2.7 Converting to Powers of 10 The power of 10 indicates the number of places the decimal point has been moved. (a)We must move the decimal five places to the left; thus, 1  10 5. (b)We must move the decimal eight places to the right; thus, 1  10 –8. Solution Express each of the following ordinary numbers as a power of 10: (a)100,000 (b)0.000 000 01

34. EXERCISE 2.7 Converting to Powers of 10 Express each of the following ordinary numbers as a power of 10: (a)10,000,000 (b)0.000 000 000 001 Practice Exercise Which of the following lengths is less: 1  10 3 cm or 1  10 –3 cm? Concept Exercise

35. EXAMPLE EXERCISE 2.8 Converting to Ordinary Numbers The power of 10 indicates the number of places the decimal point has been moved. (a)The exponent in 1  10 4 is positive 4, and so we must move the decimal point four places to the right of 1, thus, 10,000. (b)The exponent in 1  10 –9 is negative 9, and so we must move the decimal point nine places to the left of 1, thus, 0.000 000 001. Solution Express each of the following powers of 10 as an ordinary number: (a)1  10 4 (b)1  10 –9 s

36. EXAMPLE EXERCISE 2.8 Converting to Ordinary Numbers Express each of the following powers of 10 as an ordinary number: (a)1  10 10 (b)1  10 –5 Practice Exercise Which of the following masses is less: 0.000 001 g or 0.000 01 g? Concept Exercise

37. Scientific Notation  Numbers in science are often very large or very small. To avoid confusion, we use scientific notation.  Scientific notation utilizes the significant digits in a measurement followed by a power of 10. The significant digits are expressed as a number between 1 and 10. 37 D.DD x 10 n power of 10 significant digits

38. Applying Scientific Notation  To use scientific notation, first place a decimal after the first nonzero digit in the number followed by the remaining significant digits.  Indicate how many places the decimal is moved by the power of 10.  A positive power of 10 indicates that the decimal moves to the left.  A negative power of 10 indicates that the decimal moves to the right. 38

39. Scientific Notation, Continued There are 26,800,000,000,000,000,000,000 helium atoms in 1.00 L of helium gas. Express the number in scientific notation.  Place the decimal after the 2, followed by the other significant digits.  Count the number of places the decimal has moved to the left (22). Add the power of 10 to complete the scientific notation. 39 2.68 x 10 22 atoms

40. Another Example The typical length between two carbon atoms in a molecule of benzene is 0.000000140 m. What is the length expressed in scientific notation?  Place the decimal after the 1, followed by the other significant digits.  Count the number of places the decimal has moved to the right (7). Add the power of 10 to complete the scientific notation. 40 1.40 x 10 -7 m

41. EXAMPLE EXERCISE 2.9 Scientific Notation We can write each value in scientific notation as follows: (a)Place the decimal after the 2, followed by the other significant digits (2.68). Next, count the number of places the decimal has moved. The decimal is moved to the left 22 places, so the exponent is +22. Finally, we have the number of helium atoms in 1.00 L of gas: 2.68  10 22 atoms. (b)Place the decimal after the 6, followed by the other significant digits (6.65). Next, count the number of places the decimal has shifted. The decimal has shifted 24 places to the right, so the exponent is –24. Finally, we have the mass of a helium atom: 6.65  10 –24 g. Solution Express each of the following values in scientific notation: (a)There are 26,800,000,000,000,000,000,000 helium atoms in a one liter balloon filled with helium gas. (b)The mass of one helium atom is 0.000 000 000 000 000 000 000 006 65 g.

42. EXAMPLE EXERCISE 2.9 Scientific Notation Express each of the following values as ordinary numbers: (a)The mass of one mercury atom is 3.33  10 –22 g. (b)The number of atoms in 1 mL of liquid mercury is 4.08  10 22. Practice Exercise Which of the following masses is greater: 1  10 –6 g or 0.000 01 g? Concept Exercise

43. Scientific Calculators  A scientific calculator has an exponent key (often “EXP” or “EE”) for expressing powers of 10.  If your calculator reads 7.45 E-17, the proper way to write the answer in scientific notation is 7.45 x 10 -17.  To enter the number in your calculator, type 7.45, then press the exponent button (“EXP” or “EE”), and type in the exponent (17 followed by the +/– key). 43

44. Unit Equations  A unit equation is a simple statement of two equivalent quantities.  For example:  1 hour = 60 minutes  1 minute = 60 seconds  Also, we can write:  1 minute = 1/60 of an hour  1 second = 1/60 of a minute 44

45. Unit Factors  A unit conversion factor, or unit factor, is a ratio of two equivalent quantities.  For the unit equation 1 hour = 60 minutes, we can write two unit factors: 1 hour or 60 minutes 60 minutes 1 hour 45

46. EXAMPLE EXERCISE 2.10 Unit Conversion Factors We first write the unit equation and then the corresponding unit factors. (a)There are 16 ounces in a pound, and so the unit equation is 1 pound = 16 ounces. The two associated unit factors are (b)There are 4 quarts in a gallon, and so the unit equation is 1 gallon = 4 quarts. The two unit factors are Solution Write the unit equation and the two corresponding unit factors for each of the following: (a)pounds and ounces(b)quarts and gallons Write the unit equation and the two corresponding unit factors for each of the following: (a)hours and days(b)hours and minutes Practice Exercise

47. EXAMPLE EXERCISE 2.10 Unit Conversion Factors Continued How many significant digits are in the following unit equation? 1 hour = 3600 seconds Concept Exercise

48. Unit Analysis Problem Solving  An effective method for solving problems in science is the unit analysis method.  It is also often called dimensional analysis or the factor-label method.  There are three steps to solving problems using the unit analysis method. 48

49. Steps in the Unit Analysis Method 1. Write down the unit asked for in the answer. 2. Write down the given value related to the answer. 3. Apply a unit factor to convert the unit in the given value to the unit in the answer. 49

50. Unit Analysis Problem How many days are in 2.5 years?  Step 1: We want days.  Step 2: We write down the given: 2.5 years.  Step 3: We apply a unit factor (1 year = 365 days) and round to two significant figures. 50

51. Another Unit Analysis Problem A can of soda contains 12 fluid ounces. What is the volume in quarts (1 qt = 32 fl oz)?  Step 1: We want quarts.  Step 2: We write down the given: 12 fl oz.  Step 3: We apply a unit factor (1 qt = 12 fl oz) and round to two significant figures. 51

52. EXAMPLE EXERCISE 2.11 Unit Analysis Problem Solving A can of soda contains 12 fluid ounces (fl oz). What is the volume in quarts (given that 1 qt = 32 fl oz)? Dr. Pepper A 12 fl oz can of soda contains 355 mL

53. EXAMPLE EXERCISE 2.11 Unit Analysis Problem Solving Strategy Plan Step 1: What unit is asked for in the answer? Step 2: What given value is related to the answer? Step 3: What unit factor should we apply? Since the unit equation is 1 qt = 32 fl oz, the two unit factors are 1 qt/32 fl oz, and its reciprocal 32 fl oz/1 qt. Continued Unit Analysis Map

54. EXAMPLE EXERCISE 2.11 Unit Analysis Problem Solving We should apply the unit factor 1 qt/32 fl oz to cancel fluid ounces, which appears in the denominator. The given value, 12 fl oz, limits the answer to two significant digits. Since the unit factor 1 qt/32 fl oz is derived from an exact equivalent, 1 qt = 32 fl oz, it does not affect the significant digits in the answer. Solution A can of soda contains 355 mL. What is the volume in liters (given that 1 L = 1000 mL)? Practice Exercise How many significant digits are in the following unit equation? 1 L = 1000 mL Concept Exercise Continued

55. Another Unit Analysis Problem, Continued A marathon is 26.2 miles. What is the distance in kilometers (1 km = 0.62 mi)?  Step 1: We want km.  Step 2: We write down the given: 26.2 mi.  Step 3: We apply a unit factor (1 km = 0.62 mi) and round to two significant figures. 55

56. EXAMPLE EXERCISE 2.12 Uncertainty in Measurement Strategy Plan Step 1: What unit is asked for in the answer? Step 2: What given value is related to the answer? Step 3: What unit factor should we apply? Since the unit equation is 1 mi = 1760 yd, the two unit factors are 1 mi/1760 yd, and its reciprocal 1760 yd/1 mi. A marathon covers a distance of 26.2 miles (mi). If 1 mile is exactly equal to 1760 yards, what is the distance of the race in yards? Boston Marathon Marathon athletes run a distance of 26.2 miles.

57. EXAMPLE EXERCISE 2.12 Uncertainty in Measurement Continued Unit Analysis Map We should apply the unit factor 1760 yd/1 mi to cancel miles, which appears in the denominator. The given value, 26.2 mi, limits the answer to three significant digits. Since the unit factor 1760 yd/1 mi is derived from an exact equivalent, 1 mi = 1760 yd, it does not affect the significant digits in the answer. Solution

58. EXAMPLE EXERCISE 2.12 Uncertainty in Measurement How many significant digits are in the following unit equation? 1 km = 0.62 mi Concept Exercise Continued Answer: 42 km Given that a marathon is 26.2 miles, what is the distance in kilometers (given that 1 km = 0.62 mi)? Practice Exercise

59. Critical Thinking: Units  When discussing measurements, it is critical that we use the proper units.  NASA engineers mixed metric and English units when designing software for the Mars Climate Orbiter.  The engineers used kilometers rather than miles.  1 kilometer is 0.62 mile.  The spacecraft approached too close to the Martian surface and burned up in the atmosphere. 59

60. The Percent Concept  A percent, %, expresses the amount of a single quantity compared to an entire sample.  A percent is a ratio of parts per 100 parts.  The formula for calculating percent is shown below: 60

61. Calculating Percentages  Sterling silver contains silver and copper. If a sterling silver chain contains 18.5 g of silver and 1.5 g of copper, what is the percent of silver in sterling silver? 61

62. Percent Unit Factors  A percent can be expressed as parts per 100 parts.  25% can be expressed as 25/100 and 10% can be expressed as 10/100.  We can use a percent expressed as a ratio as a unit factor.  A rock is 4.70% iron, so 62

63. EXAMPLE EXERCISE 2.13 The Percent Concept Sterling Silver Sterling silver has a high luster and is found in fine utensils and jewelry. Sterling silver contains silver and copper metals. If a sterling silver chain contains 18.5 g of silver and 1.5 g of copper, what is the percent of silver?

64. EXAMPLE EXERCISE 2.13 The Percent Concept To find percent, we compare the mass of silver metal to the total mass of the silver and copper in the chain, and multiply by 100%. Genuine sterling silver is cast from 92.5% silver and 7.5% copper. If you carefully examine a piece of sterling silver, you may see the jeweler’s notation.925, which indicates the item is genuine sterling silver. Solution Strategy Plan Step 1: What is asked for in the answer? Step 2: What given value is related to the answer? Step 3: What unit factor should we apply? No unit factor is required. Continued

65. EXAMPLE EXERCISE 2.13 The Percent Concept A 14-karat gold ring contains 7.45 g of gold, 2.66 g of silver, and 2.66 g of copper. Calculate the percent of gold in the 14- karat ring. Practice Exercise If a gold alloy contains 20% silver and 5% copper, what is the percent of gold in the alloy? Concept Exercise Continued

66. Percent Unit Factor Calculation The Earth and Moon have a similar composition; each contains 4.70% iron. What is the mass of iron in a lunar sample that weighs 235 g?  Step 1: We want g iron.  Step 2: We write down the given: 235 g sample.  Step 3: We apply a unit factor (4.70 g iron = 100 g sample) and round to three significant figures. 66

67. EXAMPLE EXERCISE 2.14 Percent as a Unit Factor Strategy Plan Step 1: What unit is asked for in the answer? Step 2: What given value is related to the answer? Step 3: What unit factor should we apply? From the definition of percent, 4.70 g iron = 100 g sample; the two unit factors are 4.70 g iron/100 g sample, and its reciprocal 100 g sample/4.70 g iron. The Moon and Earth have a similar composition and each contains 4.70% iron, which supports the theory that the Moon and Earth were originally a single planet. What is the mass of iron in a lunar sample that weighs 235 g? Unit Analysis Map

68. EXAMPLE EXERCISE 2.14 Percent as a Unit Factor We should apply the unit factor 4.70 g iron/100 g sample to cancel grams sample, which appears in the denominator. The given value and unit factor each limits the answer to three significant digits. Solution A Moon sample is found to contain 7.50% aluminum. What is the mass of the lunar sample if the amount of aluminum is 5.25 g? Practice Exercise Water is 11.2% hydrogen by mass. What two unit factors express the percent hydrogen in water? Concept Exercise Continued

69. Chemistry Connection: Coins  A nickel coin contains 75.0 % copper metal and 25.0 % nickel metal, and has a mass of 5.00 grams.  What is the mass of nickel metal in a nickel coin? 69

70. Chapter Summary  A measurement is a number with an attached unit.  All measurements have uncertainty.  The uncertainty in a measurement is dictated by the calibration of the instrument used to make the measurement.  Every number in a recorded measurement is a significant digit. 70

71. Chapter Summary, Continued  Placeholding zeros are not significant digits.  If a number does not have a decimal point, all nonzero numbers and all zeros between nonzero numbers are significant.  If a number has a decimal place, significant digits start with the first nonzero number and all digits to the right are also significant. 71

72. Chapter Summary, Continued  When adding and subtracting numbers, the answer is limited by the value with the most uncertainty.  When multiplying and dividing numbers, the answer is limited by the number with the fewest significant figures.  When rounding numbers, if the first nonsignificant digit is less than 5, drop the nonsignificant figures. If the number is 5 or more, raise the first significant number by 1, and drop all of the nonsignificant digits. 72

73. Chapter Summary, Continued  Exponents are used to indicate that a number is multiplied by itself n times.  Scientific notation is used to express very large or very small numbers in a more convenient fashion.  Scientific notation has the form D.DD x 10 n, where D.DD are the significant figures (and is between 1 and 10) and n is the power of ten. 73

74. Chapter Summary, Continued  A unit equation is a statement of two equivalent quantities.  A unit factor is a ratio of two equivalent quantities.  Unit factors can be used to convert measurements between different units.  A percent is the ratio of parts per 100 parts. 74

75. © 2011 Pearson Education, Inc. State the number of significant digits in the measured quantity 0.8020 mL. a. two b. three c. four d. five

76. © 2011 Pearson Education, Inc. State the number of significant digits in the measured quantity 0.8020 mL. a. two b. three c. four d. five

77. © 2011 Pearson Education, Inc. To the correct number of significant digits, what is the sum of 22.9898 g + 79.904 g? a. 102.89 g b. 102.893 g c. 102.894 g d. 102.8938 g

78. © 2011 Pearson Education, Inc. To the correct number of significant digits, what is the sum of 22.9898 g + 79.904 g? a. 102.89 g b. 102.893 g c. 102.894 g d. 102.8938 g

79. © 2011 Pearson Education, Inc. To the correct number of significant digits, what is the product of 2.1 cm × 1.45 cm × 3.654 cm? a. 11 cm 3 b. 11.1 cm 3 c. 11.12 cm 3 d. 11.13 cm 3

80. © 2011 Pearson Education, Inc. To the correct number of significant digits, what is the product of 2.1 cm × 1.45 cm × 3.654 cm? a. 11 cm 3 b. 11.1 cm 3 c. 11.12 cm 3 d. 11.13 cm 3

81. © 2011 Pearson Education, Inc. Which is the largest mass? a. 0.000392 × 10 2 g b. 0.0453 g c. 4.73 × 10 –2 g d. 5.27 × 10 –3 g

82. © 2011 Pearson Education, Inc. Which is the largest mass? a. 0.000392 × 10 2 g b. 0.0453 g c. 4.73 × 10 –2 g d. 5.27 × 10 –3 g

83. © 2011 Pearson Education, Inc. Water is 11.9% hydrogen and 88.1% oxygen. What is the mass of hydrogen and oxygen in 100.0 g of water? a. 0.119 g hydrogen and 0.881 g oxygen b. 1.19 g hydrogen and 8.81 g oxygen c. 11.9 g hydrogen and 88.1 g oxygen d. 119 g hydrogen and 881 g oxygen

84. © 2011 Pearson Education, Inc. Water is 11.9% hydrogen and 88.1% oxygen. What is the mass of hydrogen and oxygen in 100.0 g of water? a. 0.119 g hydrogen and 0.881 g oxygen b. 1.19 g hydrogen and 8.81 g oxygen c. 11.9 g hydrogen and 88.1 g oxygen d. 119 g hydrogen and 881 g oxygen