2.  Scientific Notation  Units of Measurement  Significant Figures  Conversion Calculations  Density CH104 1

3. Scientific Notation CH1042

4.  If a number is larger than 1 X left Move decimal point X places left to get a number between 1 and 10. 1 2 3, 0 0 0, 0 0 0. XThe resulting number is multiplied by 10 X. = 1.23 x 10 8 CH104 3

5.  If a number is smaller than 1 X right Move decimal point X places right to get a number between 1 and 10. 0. 0 0 0 0 0 0 1 2 3 -7 = 1.23 x 10 -7 -XThe resulting number is multiplied by 10 -X. CH104 4

6. Write in Scientific Notation: 25 = 8931.5 = 0.000593 = 0.0000004 = 3,210. = 2.5 x 10 1 8.9315 x 10 3 5.93 x 10 - 4 4 x 10 - 7 3.210 x 10 3 CH104 5

7. x 10 1.44939 x 10 -2 = 1.44939 E-2 0.0144939 On Calculator 1.44939 (-) 2 EE Means x 10 ChangeSign CH104 6

8.  Write 4.2 x 10 3 in regular numerical format. 4200  4200  Write 35,500 in scientific notation. 3.55 x 10 4  3.55 x 10 4  Write 6.830 x 10 -2 in regular format. 0.06830  0.06830  Write 0.00096 in scientific notation. 9.6 x 10 -4  9.6 x 10 -4 CH104 7

9. Select the correct scientific notation for each. A. 0.000 000 08 m 1) 8 x 10 8 m 2) 8 x 10 -8 m 3) 0.8 x 10 -7 m B. 72 000 L 1) 7.2 x 10 4 L 2) 72 x 10 3 L 3) 7.2 x 10 -4 8 CH104

10. Select the correct scientific notation for each. A. 0.000 000 08 m 1) 8 x 10 8 m 2) 8 x 10 -8 m 3) 0.8 x 10 -7 m B. 72 000 L 1) 7.2 x 10 4 L 2) 72 x 10 3 L 3) 7.2 x 10 -4 9 CH104

11. Write each as a standard number. A. 2.0 x 10 -2 s 1) 200 s2) 0.0020 s 3) 0.020 s B. 1.8 x 10 5 g 1) 180 000 g2) 0.000 018 g3) 18 000 g 10 CH104

12. Write each as a standard number. A. 2.0 x 10 -2 s 1) 200 s2) 0.0020 s 3) 0.020 s B. 1.8 x 10 5 g 1) 180 000 g2) 0.000 018 g3) 18 000 g 11 CH104

13. Multiplication XYXY Add Exponents (10 X )( 10 Y )= 10 X+Y (10 2 ) ( 10 3 ) = 10 2+3 = 10 5 (10)(10) (10)(10)(10)= 100,000 1 - 47 © 1997, West Educational Publishing. CH104 12

14. Division Subtract Exponents XX-Y (10 X )= 10 X-Y Y (10 Y ) (10 2 ) = 10 2-(3) = 10 -1 (10 3 ) -1 (10)(10) = 1 = 10 -1 = 0.1 (10)(10)(10) (10) CH10413

15. 2.1 Units of Measurement CH10414

16.  Units are important has little meaning, just a number has some meaning - money more meaning - person’s salary CH104 15

17.  English units.  English units. Still commonly used in daily life. ◦ For Example:  Common English measures of volume  1 tablespoon=3 teaspoons  1 cup= 16 tablespoons  1 pint=2 cups  1 quart=2 pints  1 gallon=4 quarts  1 peck=2 gallons  1 bushel= 4 pecks  Not often used in scientific work CH104 16

18. MetricSICommon Conversions Length Units of Measurement meter (m) meter (m) 1 m = 1.09 yd 2.54 cm = 1 in 1 m = 100 cm = 1000 mm CH10417

19. MetricSICommon Conversions Length Volume Units of Measurement meter (m) meter (m) 1 m = 1.09 yd 2.54 cm = 1 in 1 m = 100 cm = 1000 mm liter (L) cubic meter (m 3 ) 1 L = 1.06 qt 946 mL = 1 qt 1 L = 1000 mL CH10418

20. MetricSICommon Conversions Length Volume Mass Units of Measurement meter (m) meter (m) 1 m = 1.09 yd 2.54 cm = 1 in 1 m = 100 cm = 1000 mm liter (L) cubic meter (m 3 ) 1 L = 1.06 qt 946 mL = 1 qt 1 L = 1000 mL gram (g) Kilogram (kg) 1 kg = 2.20 lb 1 kg = 1000 g 454 g = 1 lb CH10419

21. material in an object The amount of material in an object  Mass: Mass (in g’s) of a 1L Bowling Ball > a 1 L Balloon Weight: Mass Vs. Weight Pull of Gravity on an object. Weight of Person on Earth > Person on Moon CH104 20

22. How much would you weigh on another planet? http://www.exploratorium.edu/ronh/weight/ CH10421

23. MetricSICommon Conversions Length Volume Mass Time Temp Units of Measurement meter (m) meter (m) 1 m = 1.09 yd 2.54 cm = 1 in 2.54 cm = 1 in liter (L) cubic meter (m 3 ) gram (g) Kilogram (kg) 1 kg = 2.20 lb 1 L = 1.06 qt 946 mL = 1 qt Celsius ( o C) Kelvin (K) o C = ( o F-32)/1.8 K = o C + 273 second (s) second (s) 60 s = 1 min CH10422

24. For each of the following, indicate whether the unit describes 1) length, 2) mass, or 3) volume. ____ A. A bag of tomatoes is 4.5 kg. ____ B. A person is 2.0 m tall. ____ C. A medication contains 0.50 g of aspirin. ____ D. A bottle contains 1.5 L of water. 23 CH104

25. For each of the following, indicate whether the unit describes 1) length, 2) mass, or 3) volume. ____ A. A bag of tomatoes is 4.5 kg. ____ B. A person is 2.0 m tall. ____ C. A medication contains 0.50 g of aspirin. ____ D. A bottle contains 1.5 L of water. 24 1 2 2 3 CH104

26. Identify the measurement that has an SI unit. A. John’s height is _____. 1) 1.5 yd2) 6 ft 3) 2.1 m B. The race was won in _____. 1) 19.6 s2) 14.2 min3) 3.5 h C. The mass of a lemon is _____. 1) 12 oz2) 0.145 kg3) 0.62 lb D. The temperature is _____. 1) 85 °C2) 255 K3) 45 °F 25 CH104

27. 26 CH104 Identify the measurement that has an SI unit. A. John’s height is _____. 1) 1.5 yd2) 6 ft 3) 2.1 m B. The race was won in _____. 1) 19.6 s2) 14.2 min3) 3.5 h C. The mass of a lemon is _____. 1) 12 oz2) 0.145 kg3) 0.62 lb D. The temperature is _____. 1) 85 °C2) 255 K3) 45 °F

28. 2.2 Measured Numbers and Significant Figures 27 CH104

29. Exact Numbers = from counting or by definition 12 coins per package 12 coins 1 package 12 coins 1 package 12 coins 1 package 12 coins = 1 dozen coins 12 coins 1 dozen coins 12 coins 1 dozen coins 12 coins = CH10428

30. 29 from numbers in a defined relationship when objects are counted CH104

31. Measured Numbers = estimated using a tool uncertainty All measurements contain some uncertainty. We make errors Tools have limits CH10430

32. How close are we to the true value?Truth Precision How well do our values agree?Consistency CH104 31

33. Our goal! Truth and Consistency Values we can trust. CH104 32

34. 6.76.8 Length of object is between 6.7 and 6.8 The next digit would be a guess. 6.76 3 significant figures If use 6.76 then have error of + 0.01cm and have 3 significant figures. CH10433

35. accuracy & precision. Expresses accuracy & precision. You can’t report values more accurate than the methods of measurement used. 6.76 3 6.76 units = 3 significant figures Certain Digits Uncertain Digit CH104 34

36. . l 8.... l.... l 9.... l.... l 10.. cm 35 What is the length of the red line? 1) 9.0 cm 2) 9.03 cm 3) 9.04 cm CH104

37. The length of the red line could be reported as 2) 9.03 cm or 3) 9.04 cm The estimated digit may be slightly different. Both readings are acceptable. 36. l 8.... l.... l 9.... l.... l 10.. cm CH104

38. 8.45 mL We can estimate the value to be 8.45 mL but cannot be more accurate than that. 8.45 has 3 sig figs. Significant figures 8.48.5 Meniscus is between 8.4 and 8.5 The next digit would be a guess.

39. Sig Figs don’t depend on the decimal point. millimeters  255 millimeters centimeters  25.5 centimeters decimeters  2.55 decimeters meters  0.255 meters decameters  0.0255 decameters CH104 38

40. 0.00421 Leading zero are Captive zeros are significant. 4012 are Trailing zeros behind decimal are significant. 114.20 Captive zero Trailing zero are not Leading zeros are not significant. 3 3 sig figs 4 4 sig figs 5 5 sig figs CH104 39

41. 32,000 Are the 0’s significant? 2 2 sig figs = 3 3 sig figs = 4 4 sig figs = 5 5 sig figs = 3.2 x 10 4 3.20 x 10 4 3.200 x 10 4 3.2000 x 10 4 32,000. CH104 40

42. 1025 km 2.00 mg 0.00570 520 Significant figures: Rules for zeros Three Three (only trailing zero behind decimal is significant, leading zeros are not) Four Four (Captive zeros are significant) Three Three (trailing zeros behind decimal are significant) Two Two (No decimal, zero assumed insignif) CH10441

43. In scientific notation: All digits, including zeros in the coefficient, are significant. Scientific NotationNumber of Significant Figures___________ 8 x 10 4 m 8.0 x 10 4 m 8.00 x 10 4 m 42 1 2 3 CH104

44. State the number of significant figures in each of the following measurements: A. 0.030 m B. 4.050 L C. 0.0008 g D. 2.80 m 43 CH104

45. State the number of significant figures in each of the following measurements: A. 0.030 m B. 4.050 L C. 0.0008 g D. 2.80 m 44 2 4 1 3 CH104

46. 45 CH104 A. Which answer(s) contains 3 significant figures? 1) 0.47602) 0.00476 3) 4.76 x 10 3 B. All the zeros are significant in 1) 0.00307 2) 25.300 3) 2.050 x 10 3 C. The number of significant figures in 5.80 x 10 2 is 1) one3) two3) three

47. 46 CH104 A. Which answer(s) contains 3 significant figures? 1) 0.47602) 0.00476 3) 4.76 x 10 3 B. All the zeros are significant in 1) 0.00307 2) 25.300 3) 2.050 x 10 3 C. The number of significant figures in 5.80 x 10 2 is 1) one3) two3) three

48. In which set(s) do both numbers contain the same number of significant figures? 1) 22.0 and 22.00 2) 400.0 and 4.00 x 10 2 3) 0.000015 and 150 000 47 CH104

49. 48 Both numbers contain two (2) significant figures. CH104 In which set(s) do both numbers contain the same number of significant figures? 1) 22.0 and 22.00 2) 400.0 and 4.00 x 10 2 3) 0.000015 and 150 000

50. A. Exact numbers are obtained by 1. using a measuring tool 2. counting 3. definition B. Measured numbers are obtained by 1. using a measuring tool 2. counting 3. definition 49 CH104

51. A. Exact numbers are obtained by 1. using a measuring tool 2. counting 3. definition B. Measured numbers are obtained by 1. using a measuring tool 2. counting 3. definition 50 CH104

52. Classify each of the following as exact (E) or measured (M) numbers. Explain your answer. A. __ Gold melts at 1064 °C. B. __ 1 yard = 3 feet C. __ The diameter of a red blood cell is 6 x 10 -4 cm. D. __ There are 6 hats on the shelf. E. __ A can of soda contains 355 mL of soda. 51 CH104

53. 52 M M M E E CH104 Classify each of the following as exact (E) or measured (M) numbers. Explain your answer. A. __ Gold melts at 1064 °C. B. __ 1 yard = 3 feet C. __ The diameter of a red blood cell is 6 x 10 -4 cm. D. __ There are 6 hats on the shelf. E. __ A can of soda contains 355 mL of soda.

54. 2.3 Significant Figures in Calculations 53 CH104

55. Write with 4 Significant Figures : 2.579 035 1st insignificant digit 34.20 221 34.20 becomes 34.20 2.580 becomes 2.580 5 4 > 5 > 5 round up < 5 < 5 round down. > 5 > 5 round up < 5 < 5 round down. Sometimes a calculated answer shows too many significant digits so we need to round. CH104 54

56. 4 1.5 0.2 12 more Sometimes a calculated answer requires more significant digits so we need to add zeros. Calculated answer Zeros added to give 3 significant figures 4.00 1.50 0.200 12.0 CH104 55

57. Adjust the following calculated answers to give answers with three significant figures. A. 824.75 cm B. 0.112486 g C. 8.2 L 56 CH104

58. Adjust the following calculated answers to give answers with three significant figures. A. 824.75 cm B. 0.112486 g C. 8.2 L First digit dropped is greater than 4. 825 cm First digit dropped is 4.0.112 g Significant zero is added. 8.20 L CH104 57

59. An answer can’t have greater significance than the quantities used to produce it. speed = 1.00 km 3.0 min 0.3333333333 ? = ? Example How fast did you run if you went 1.00 km in 3.0 minutes? CH104 58

60. Multiplication & Division Problems: Do calculations. Look at sig figs for each value in calculation. (Constants don’t count.) Report answer with same sig figs as least significant value. Round off as needed. speed = 1.00 km 3.0 min km = 0.333333333 km min min 0.33 km = 0.33 km min min 2 sig figs 3 sig figs CH104 59

61.  The measurement containing the fewest significant figures determines the number of significant figures in the answer.  Example 1: 2.8 m x 0.2 m = 0.56 = 0.6 m 2  Example 2: 252 mi / 3.2 hr = 78.75 = 79 mi/hr  The measurement containing the fewest significant figures determines the number of significant figures in the answer.  Example 1: 2.8 m x 0.2 m = 0.56 = 0.6 m 2  Example 2: 252 mi / 3.2 hr = 78.75 = 79 mi/hr CH104 60

62. Simplified rules for significant figures Addition & Subtraction Problems: Do calculations. place Look at least significant place for each value in calculation. Report answer to least significant place. Round off as needed. 1.9 + 18.65 20.55 = 20.6 Significant to.1 Significant to.01 Significant to.1 CH104 61

63. (1.9 + 18.65 ) = 2.153 Add & Sub mixed w/ Mult & Div Problems: ( 20.55 ) = 2.153 Do Addition & Subtraction calculations 1st. Make note of the least significant place. 3 sig figs (after addition) 4 sig figs CH104 62

64. Do Multiplication & Division calculations. 9.54482118 Round to least # sig fig. 9.54 (1.9 + 18.65 ) = 2.153 ( 20.55 ) = 2.153 3 sig figs (after addition) Add & Sub mixed w/ Mult & Div Problems: 4 sig figs CH104 63

65. Give an answer for the following with the correct number of significant figures: A. 2.19 x 4.2 = 1) 9 2) 9.2 3) 9.198 B. 4.311 ÷ 0.07 = 1) 61.59 2) 62 3) 60 C. 2.54 x 0.0028 = 0.0105 x 0.060 1) 11.32) 11 3) 0.041 CH104 64

66. 2.54 x 0.0028  0.0105  0.060 = = 11.28888889 = 11 (rounded) CH104 65 Give an answer for the following with the correct number of significant figures: A. 2.19 x 4.2 = 1) 9 2) 9.2 3) 9.198 B. 4.311 ÷ 0.07 = 1) 61.59 2) 62 3) 60 C. 2.54 x 0.0028 = 0.0105 x 0.060 1) 11.32) 11 3) 0.041

67. For each calculation, round the answer to give the correct number of decimal places. A. 235.05 + 19.6 + 2 = 1) 257 2) 256.7 3) 256.65 B. 58.925 – 18.2 = 1) 40.725 2) 40.73 3) 40.7 CH104 66

68. For each calculation, round the answer to give the correct number of decimal places. A. 235.05 + 19.6 + 2 = 1) 257 2) 256.7 3) 256.65 B. 58.925 – 18.2 = 1) 40.725 2) 40.73 3) 40.7 235.05 +19.6 + 2 256.65 rounds to 257 58.925 –18.2 40.725 rounds to 40.7 CH104 67

69. 2.4 Prefixes and Equalities CH10468

70. A prefix in front of a unit increases or decreases the size of that unit by one or more factors of 10 indicates a numerical value PrefixValue 1 kilometer=1000 meters 1 kilogram=1000 grams 69 CH104

71. Prefix (Symbol) Factor (multiple) Common Conversion mega (M) kilo (k) deci (d) centi (c) milli (m) micro (  ) nano (n) 1,000,000 = (10 6 ) 1Mm = 1,000,000 m 1,000 = (10 3 ) 1km = 1,000 m 0.1 = (10 -1 ) 1m = 10 dm 0.01 = (10 -2 ) 1m = 100 cm 0.001 = (10 -3 ) 1m = 1,000 mm 0.000001 = (10 -6 ) 1m = 1,000,000  m 0.000,000,001 = (10 -9 ) 1m = 1,000,000,000 nm

72. 71 CH104

73. 72 CH104

74. Indicate the unit that matches the description: 1. a mass that is 1000 times greater than 1 gram 1) kilogram2) milligram3) megagram 2. a length that is 1/100 of 1 meter 1) decimeter2) centimeter3) millimeter 3. a unit of time that is 1/1000 of a second 1) nanosecond 2) microsecond 3) millisecond CH104 73

75. = 0.01 of 1 meter = 0.001 of a sec CH104 74 Indicate the unit that matches the description: 1. a mass that is 1000 times greater than 1 gram 1) kilogram2) milligram3) megagram 2. a length that is 1/100 of 1 meter 1) decimeter2) centimeter3) millimeter 3. a unit of time that is 1/1000 of a second 1) nanosecond 2) microsecond 3) millisecond

76. Select the unit you would use to measure A. your height 1) millimeters2) meters 3) kilometers B. your mass 1) milligrams2) grams 3) kilograms C. the distance between two cities 1) millimeters2) meters 3) kilometers D. the width of an artery 1) millimeters2) meters 3) kilometers CH104 75

77. CH104 76 Select the unit you would use to measure A. your height 1) millimeters2) meters 3) kilometers B. your mass 1) milligrams2) grams 3) kilograms C. the distance between two cities 1) millimeters2) meters 3) kilometers D. the width of an artery 1) millimeters2) meters 3) kilometers

78. States the same measurement in two different units  Length: 1 meter is the same as 100 cm or 1000 mm. 1 m = 100 cm 1 m = 1000 mm Volume: 1 L is the same as 1000 cm 3. 1 L = 10 cm X 10cm X 10 cm 1 L = 1000 mL Mass: 1 kg = 1000 g 1 g = 1000 mg 1 mg = 0.001 g 1 mg = 1000 µg CH104 77

79. Indicate the unit that completes each of the following equalities: A. 1000 m = 1) 1 mm 2) 1 km3) 1dm B. 0.001 g = 1) 1 mg2) 1 kg3) 1dg C. 0.1 s = 1) 1 ms2) 1 cs3) 1ds D. 0.01 m = 1) 1 mm 2) 1 cm3) 1dm CH104 78

80. Indicate the unit that completes each of the following equalities: A. 1000 m = 1) 1 mm 2) 1 km3) 1dm B. 0.001 g = 1) 1 mg2) 1 kg3) 1dg C. 0.1 s = 1) 1 ms2) 1 cs3) 1ds D. 0.01 m = 1) 1 mm 2) 1 cm3) 1dm CH104 79

81. Complete each of the following equalities: A. 1 kg = 1) 10 g2) 100 g 3) 1000 g B. 1 mm =1) 0.001 m2) 0.01 m 3) 0.1 m CH104 80

82. Complete each of the following equalities: A. 1 kg = 1) 10 g2) 100 g 3) 1000 g B. 1 mm =1) 0.001 m2) 0.01 m 3) 0.1 m CH104 81

83. 2.5 Writing Conversion Factors 82 CH104

84. See Handout Sheet of Common conversion factors & Handout of Conversion Problems CH104 83

85. 84 CH104

86. Write equalities and conversion factors for each pair of units: A. liters and mL B. hours and minutes C. meters and kilometers 85 CH104

87. Write equalities and conversion factors for each pair of units: A. liters and mL B. hours and minutes C. meters and kilometers 86 Equality: 1 L = 1000 mL 1 L and 1000 mL 1000 mL 1 L Equality: 1 hr = 60 min 1 hr and 60 min 60 min 1 hr Equality: 1 km = 1000 m 1 km and 1000 m 1000 m 1 km CH104

88. Write the equality and conversion factors for each of the following: A. meters and centimeters B. jewelry that contains 18% gold C. one liter of gas is $ 0.95 87 CH104

89. Write the equality and conversion factors for each of the following: A. meters and centimeters B. jewelry that contains 18% gold C. one liter of gas is $ 0.95 88 CH104 1 m and 100 cm 100 cm 1 m 18 g gold and 100 g jewelry 100 g jewelry 18 g gold 1 L and $0.95 $0.95 1 L

90. 2.6 Problem Solving 89 CH104

91. Example: Metric Conversion How many milligrams (mg) are in 5 kilograms (kg)? Factor label method Identify your conversions factors. 1 kg = 1 1000 g 1000 g = 1 1 kg 1 g = 1 1000 mg 1000 mg = 1 1 g CH104 90

92. Identify what is to the problem. want Identify how you want the answer to look. 5 kg = mg Example:Metric Conversion Example: Metric Conversion How many milligrams are in 5 kilograms? CH104 91

93. Multiply by conversion factors until units cancel. If the words work, the numbers will work. 5 kg 1 = mg 1000 g 1 kg 1000 mg 1 g 5,000,000 Example:Metric Conversion Example: Metric Conversion How many milligrams are in 5 kilograms? CH104 92

94. How many decimeters are there in 5.5 meters? How many meters are there in 25 centimeters? CH104 93

95. How many meters are in 3 kilometers? 3000 m = 3 km How many milliliters are in 0.5 liters? 500 mL = 0.5 L How many grams are in 2.5 kg? 2500 g = 2.5 kg How many millimeters are in 1 meter? 1000 mm = 1 m CH104 94

96. How many teaspoons in a barrel of oil? 42. 1 barrel of oil= 42. gallons 1 gallon= 4quarts 1 quart= 4cups 1 cup= 16tablespoons 1 tablespoon= 3 teaspoons 42 gal 1 bal 4 qt 1 gal 4 cup 1 qt 16 Tbl 1 cup 3 tsp 1 Tbl = tsp 1 bal 32,256 32,000 tsp CH104 95

97. How many grams are there in 125 pounds? 454 g = 1 lb1 L = 1.06 qt2.54 cm = 1 in How many inches are there in 8.7 meters? CH104 96

98. Example: English-Metric Conversion You have a pen of rats each with an average 0.75 lb weight of 0.75 lb. How many mg rubbing alcohol will it take to kill ½ of the population if the LD 50 is 5000. mg/kg ? Identify your conversions factors.Identify your conversions factors. 1 kg Bw = 1 5000 mg Alc 5000 mg Alc = 1 1 kg Bw 1.0 kg Bw = 1 2.2 lb Bw 2.2 lb Bw = 1 1.0 kg Bw CH10497

99. 0.75 lbBW Example: English-Metric Conversion You have a pen of rats each with an average 0.75 lb weight of 0.75 lb. How many mg rubbing alcohol will it take to kill ½ of the population if the LD 50 is 5000. mg/kg ? 1.0 kgBW 2.2 lbBW 5000. mgAlc 1 kg BW = mgAlc 1704.545 1704.545 1700 mg = 1.7 x 10 3 unique Identify what is unique to the problem. want Identify how you want the answer to look. CH10498

100. Given (unique) = 2.5 h Needed unit = ? min Plan =h  min How many minutes are 2.5 h? Set up problem to cancel hours (h). Given Conversion Needed unit factor unit 2.5 h x 60 min = 150 min (2 SigFigs) 1 h CH10499

101. A rattlesnake is 2.44 m long. How many centimeters long is the snake? 1) 2440 cm 2) 244 cm 3) 24.4 cm 100 CH104

102. 2.44 m x 100 cm = 244 cm 1 m 101 CH104 A rattlesnake is 2.44 m long. How many centimeters long is the snake? 1) 2440 cm 2) 244 cm 3) 24.4 cm

103. Given (unique) = 1.60 days Needed unit = ? min Plan =days  hours  min How many minutes are 1.60 days? Set up problem to cancel hours (h). 1.60 days x 24 hrs x 60 min = 1 day 1 hr 3 SigFigs Exact Exact = 3 SigFigs 2300 min = 2.3 x 10 3 2304 min CH104102

104.  Be sure to check your unit cancellation in the setup.  The units in the conversion factors must cancel to give the correct unit for the answer.  Example: What is wrong with the following setup?  1.4 day x 1 day x 1 h  24 h 60 min  Units = day 2 /min, which is not the unit needed  Units don’t cancel properly. Therefore, setup is wrong. CH104 103

105. A bucket contains 4.65 L of water. How many gallons of water is that? 104 CH104

106. Given : 4.65 L Need: gal Plan: L  qt  gallon Equalities: 1.06 qt = 1 L; 1 gal = 4 qt Set up problem: 4.65 L x 1.06 qt x 1 gal = 1.23 gal 1 L 4 qt 3 SF 3 SF exact 3 SF 105 CH104 A bucket contains 4.65 L of water. How many gallons of water is that?

107. If a ski pole is 3.0 feet in length, how long is the ski pole in mm? 106 CH104

108. Given: 3.0 ft Need: mm Plan: ft  in.  cm  mm Equalities: 1 ft = 12 in. 2.54 cm = 1 in. 1 cm = 10 mm Set up problem: 3.0 ft x 12 in. x 2.54 cm x 10 mm = 910 mm 1 ft 1 in. 1 cm (2SigFigs, rounded) CH104 107 If a ski pole is 3.0 feet in length, how long is the ski pole in mm?

109. If your pace on a treadmill is 65 meters per minute, how many minutes will it take for you to walk a distance of 7500 feet? 108 CH104

110. Given: 7500 ft, 65 m/minNeed: min Plan: ft  in.  cm  m  min Equalities: 1 ft = 12 in. 1 in. = 2.54 cm 1 m = 100 cm 1 min = 65 m (walking pace) Set up problem: 7500 ft x 12 in. x 2.54 cm x 1m x 1 min 1 ft 1 in.100 cm 65 m = 35 min (2SF) 109 CH104 If your pace on a treadmill is 65 meters per minute, how many minutes will it take for you to walk a distance of 7500 feet?

111. 13 males x 100 = 37.1429 % 35 Students Part x 100 = Whole % ___100 Secret code for 37% male CH104 110

112. Example: 37%75 Example: The population of the automotive repair course is 37% male. Of the 75 students in the class how many are men? Secret code for 37% male = 37 male 100 students 37 male CH104111 Identify your conversions factors.Identify your conversions factors.

113. = males Percentages as Conversion Factors Example: 37%75 Example: The population of the automotive repair course is 37% male. Of the 75 students in the class how many are men? Identify what is to the problem. want Identify how you want the answer to look. 75 students 1 37 males 100 students 27.75 28 males CH104112

114. How many pounds (lb) of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? 113 CH104

115. How many pounds (lb) of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? percent factor 120 g candy x 1 lb candy x 25 lb sugar 454 g candy 100 lb candy = 0.066 lb of sugar 114 CH104

116. = 100 mL 100 mL 10 % Alcohol Part x 100 = Whole % ___100 Secret code for 10 mL Alcohol 10 mL Alcohol Solution Solution CH104115

117. Example:15 % 74 Example: An athlete normally has 15 % body fat. How many lbs of fat does a 74 kg athlete have? Secret code for 15% Body Fat = 15 lb Fat 100 lb BW 15 lb Fat CH104116 Identify your conversions factors.Identify your conversions factors.

118. 2.2 lbBw 1.0 KbBw = lb fat Percentages as Conversion Factors Identify what is to the problem. want Identify how you want the answer to look. 74 KgBw 24.42 24 lb fat Example:15 % 74 Example: An athlete normally has 15 % body fat. How many lbs of fat does a 74 kg athlete have? 15 lb Fat 100 lb BW CH104117

119. If the thickness of the skin fold at the waist indicates an 11% body fat, how much fat is in a person with a mass of 86 kg? CH104118

120. 11% body fat means 11 kg fat 100 kg 86 kg x 11 kg fat = 9.5 kg of fat 100 kg If the thickness of the skin fold at the waist indicates an 11% body fat, how much fat is in a person with a mass of 86 kg? CH104119

121. 2.7 Density 120 CH104

122. Water 1.0Urine1.01 - 1.03 Air 0.0013Bone1.7 - 2.0 Gold19.3Gasoline0.66 - 0.69 Density = Mass Volume cccm 3 mlg 1cc = 1 cm 3 = 1 ml = 1 g water g cm 3 gml or At 4 o C CH104 121

123. 122 CH104

124. 5.00 ml 5.230 grams What is the density of 5.00 ml of serum if it has a mass of 5.230 grams? d = m V d = 5.230 g 5.00 ml 5.00 ml = 1.05 g ml ml CH104 123

125. Osmium is a very dense metal. What is its density in g/cm 3 if 50.0 g of osmium has a volume of 2.22 cm 3 ? 1) 2.25 g/cm 3 2) 22.5 g/cm 3 3) 111 g/cm 3 124 CH104

126. Given: mass = 50.0 g,volume = 22.2 cm 3 Need: Density D = mass = 50.0 g volume2.22 cm 3 Calculator = 22.522522 g/cm 3 Final answer (2 SF) = 22.5 g/cm 3 CH104 125 Osmium is a very dense metal. What is its density in g/cm 3 if 50.0 g of osmium has a volume of 2.22 cm 3 ? 1) 2.25 g/cm 3 2) 22.5 g/cm 3 3) 111 g/cm 3

127. The mass of a 6.85 mL sample of a liquid is found to weigh 7.453 grams. What is the density of the liquid? The specific gravity equals 1.09 and as a ratio, has no units. CH104 126

128. The mass of a 4.96 cm 3 sample of gold is found to weigh 95.78 grams. What is the density of the liquid? (1.00mL = 1.00cm 3 ) CH104 127

129.  A solid completely submerged in water displaces its own volume of water.  The volume of the solid is calculated from the volume difference. Volume of solid = 45.0 mL – 35.5 mL = 9.5 mL = 9.5 cm 3 128 CH104 The density of the zinc object is calculated from its mass and volume. mass = 68.60 g = 7.2 g/cm 3 volume 9.5 cm 3

130. What is the density (g/cm 3 ) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added? 1) 0.17 g/cm 3 2) 6.0 g/cm 3 3) 380 g/cm 3 CH104 129 33.0 mL 25.0 mL object 25.0 mL

131. Given: 48.0 g Volume of water = 25.0 mL Volume of water + metal = 33.0 mL Need: Density (g/cm 3 ) Volume of metal = 33.0 mL – 25.0 mL= 8.0 mL 8.0 mL x 1 cm 3 = 8.0 cm 3 1 mL Set up problem: Density = 48.0 g = 6.0 g = 6.0 g/cm 3 8.0 cm 3 1 cm 3 (2 SF) CH104 130 What is the density (g/cm 3 ) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added? 1) 0.17 g/cm 3 2) 6.0 g/cm 3 3) 380 g/cm 3

132.  Ice floats in water because the density of ice is less than the density of water.  Aluminum sinks in water because its density is greater than the density of water. 131 CH104

133. Which diagram correctly represents the liquid layers in the cylinder? Karo (K) syrup (1.4 g/mL), vegetable (V) oil (0.91 g/mL), water (W) (1.0 g/mL) 1 2 3 K K W W W V V V K CH104 132

134. 1) vegetable oil (0.91 g/mL) water (1.0 g/mL) Karo syrup (1.4 g/mL) 133 K W V CH104

135. Density as a Conversion A liquid sample with a density of 1.09 g/mL is found to weigh 7.453 grams. What is the volume of the liquid in mLs? 1.09 g 1 ml 1.09 g Identify any conversion factors. How should the answer look? 7.453 g = ml What is unique to the problem? 1 ml 1.09 g 6.837614 = 6.84 ml CH104134

136. The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? 1) 0.614 kg2) 614 kg3) 1.25 kg 135 CH104

137. Given: D = 0.702 g/mL V = 875 mL Need: kg Plan: mL  g  kg Equalities: density 0.702 g = 1 mL and 1 kg = 1000 g Set up problem: 875 mL x 0.702 g x 1 kg = 0.614 kg 1 mL 1000 g The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? 1) 0.614 kg2) 614 kg3) 1.25 kg CH104 136

138. If olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil? 1) 0.26 L2) 0.31 L3) 310 L CH104 137

139. Given: D = 0.92 g/mL mass = 285 g Need: volume in L Plan: g  mL  L Equalities: 1 mL = 0.92 g 1 L = 1000 mL Set up: 285 g x 1 mL x 1 L = 0.31 L 0.92 g 1000 mL density metric factorfactor If olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil? 1) 0.26 L2) 0.31 L3) 310 L CH104 138

140. A group of students collected 125 empty aluminum cans to take to the recycling center. If 21 cans contain 1.0 lb of aluminum, how many liters of aluminum (D = 2.70 g/cm 3 ) are obtained from the cans? 1) 1.0 L2) 2.0 L3) 4.0 L CH104 139

141. 125 cans x 1.0 lb x 454 g x 1 cm 3 x 1 mL x 1 L 21 cans 1 lb 2.70 g 1 cm 3 1000 mL = 1.0 L CH104 140 A group of students collected 125 empty aluminum cans to take to the recycling center. If 21 cans contain 1.0 lb of aluminum, how many liters of aluminum (D = 2.70 g/cm 3 ) are obtained from the cans? 1) 1.0 L2) 2.0 L3) 4.0 L

142. Which of the following samples of metals will displace the greatest volume of water? 1 2 3 25 g of aluminum 2.70 g/mL 45 g of gold 19.3 g/mL 75 g of lead 11.3 g/mL CH104 141

143. 1) Plan: Calculate the volume for each metal, and select the metal sample with the greatest volume. 1) 25g x 1 mL = 9.3 mL of aluminum 2.70 g 2) 45 g x 1 mL = 2.3 mL of gold 19.3 g 3) 75 g x 1 mL = 6.6 mL of lead 11.3 g 25 g of aluminum 2.70 g/mL CH104 142

144. Specific Gravity = density of substance g ml density of reference g ml Reference commonly water at 4 o C Specific Gravity is unitless. Specific Gravity is unitless. density = specific gravity (if 4 o C) density = specific gravity (if at 4 o C) CH104 143

145. Specific gravity Commonly used to test sugar in urine. Hydrometer Float height will be based on Specific Gravity.