1. 1 CHAPTER ONE Matter and Measurement

2. 2 Matter and Energy - Vocabulary ChemistryChemistry MatterMatter EnergyEnergy Natural Law-(scientific law)Natural Law-(scientific law) Scientific MethodScientific Method –Observation, Hypothesis, Experiment, and Theory

3. 3 States of Matter SolidsSolids

4. 4 States of Matter SolidsSolids LiquidsLiquids

5. 5 States of Matter SolidsSolids LiquidsLiquids GasesGases

6. 6 States of Matter Change StatesChange States –heating –cooling

7. 7 States of Matter Illustration of changes in stateIllustration of changes in state –requires energy

8. 8 Substances, Compounds, Elements and Mixtures SubstanceSubstance –matter that all samples have identical composition and properties ElementsElements –Pure substances that cannot be decomposed into simpler substances via chemical reactions –Special elemental forms of atoms (diatomic) Elemental symbols –found on periodic chart

9. 9 Substances, Compounds, Elements and Mixtures

10. 10 CompoundsCompounds –Pure substances composed of two or more elements in a definite ratio by mass –can be decomposed into the constituent elements REVIEW –Element cannot be broken down –Compound can be broken down into its elements!

11. 11 Substances, Compounds, Elements and Mixtures MixturesMixtures –composed of two or more substances –homogeneous mixtures Uniform throughoutUniform throughout Example: solutionsExample: solutions –heterogeneous mixtures NonuniformNonuniform Example: rocksExample: rocks

12. 12 Classify the following substances as an element, compound or a mixture (homogeneous or heterogeneous). Which are pure substances? Lightly scrambled eggLightly scrambled egg WaterWater Lava lampLava lamp SeawaterSeawater Freshly opened root beerFreshly opened root beer Flat root beerFlat root beer Sucrose (C 12 H 22 O 11 )Sucrose (C 12 H 22 O 11 )

13. 13 Separating Mixtures DistillationDistillation

14. 14 Separating Mixtures ChromatographyChromatographypaper

15. 15 Chemical and Physical Properties Extensive Properties - depend on quantity of materialExtensive Properties - depend on quantity of material Ex. mass Ex. mass Intensive Properties - do not depend on quantity of materialIntensive Properties - do not depend on quantity of material Ex. boiling point

16. 16 Chemical and Physical Properties Chemical Properties - chemical changesChemical Properties - chemical changes –Observed during change of material to new material Iron rustingIron rusting Physical Properties - physical changesPhysical Properties - physical changes –No change to the identity of the substance changes of statechanges of state densitydensity colorcolor solubilitysolubility

17. 17 Physical Properties DensityDensity –mass / volumeintensive property –Mass and volume extensive properties SolubilitySolubility –Amount of substance dissolved in the solvent at a given temperature Saturated solutionSaturated solution Unsaturated solutionUnsaturated solution Supersaturated solutionSupersaturated solution

18. 18 Identify the following as either a chemical or physical change. Combination of sodium and chlorine to give sodium chloride.Combination of sodium and chlorine to give sodium chloride. Liquefaction of gaseous nitrogen.Liquefaction of gaseous nitrogen. Separation of carbon monoxide into carbon and oxygen.Separation of carbon monoxide into carbon and oxygen. Freezing of water.Freezing of water.

19. 19 Measurements in Chemistry length meter mlength meter m volume liter lvolume liter l mass gram gmass gram g time second stime second s current ampere Acurrent ampere A temperature Kelvin Ktemperature Kelvin K amt. substance mole molamt. substance mole mol

20. 20 Measurements in Chemistry mega M 10 6mega M 10 6 kilo k 10 3kilo k 10 3 deka da 10deka da 10 deci d 10 -1deci d 10 -1 centi c 10 -2centi c 10 -2 milli m 10 -3milli m 10 -3 micro  10 -6micro  10 -6 nano n 10 -9nano n 10 -9 pico p 10 -12pico p 10 -12 femto f 10 -15femto f 10 -15

21. 21 Units of Measurement MassMass –measure of the quantity of matter in a body WeightWeight –measure of the gravitational attraction for a body LengthLength 1 m = 39.37 inches 2.54 cm = 1 inch VolumeVolume 1 liter = 1.06 qt 1 qt = 0.946 liter

22. 22 The Use of Numbers Exact numbers 1 dozen = 12 thingsExact numbers 1 dozen = 12 things AccuracyAccuracy –how closely measured values agree with the correct value PrecisionPrecision –how closely individual measurements agree with each other

23. 23 The Use of Numbers

24. 24 Exact numbers 1 dozen = 12 thingsExact numbers 1 dozen = 12 things –Counted numbers ex. 3 beakers Significant figuresSignificant figures –digits believed to be correct by the person making the measurement –measure a mile with a 6 inch ruler vs. surveying equipment Scientific notationScientific notation –Way of signifying the significant digits in a number

25. 25 Significant Figures - rules leading zeroes - never significantleading zeroes - never significant 0.000357 has three sig fig trailing zeroes - may be significanttrailing zeroes - may be significant must specify (after decimal – significant before decimal - ambiguous) 1300 nails - counted or weighed? Express 26800 in scientific notation with 4 sig figs3 sig figs2 sig figs

26. 26 Significant Figures - rules imbedded zeroes are always significantimbedded zeroes are always significant 3.0604 has five sig fig How many significant figures are in the following numbers? 0.01240.1241.2401240

27. 27 Significant Figures - rules multiply & divide rule - easy product has the smallest number of sig. fig. of multipliers

28. 28 Significant Figures - rules multiply & divide rule - easymultiply & divide rule - easy product has the smallest number of sig. fig. of multipliers

29. 29 Significant Figures - rules multiply & divide rule - easymultiply & divide rule - easy product has the smallest number of sig. fig. of multipliers

30. 30 Practice 142 x 2 =142 x 2 = 4.180 x 2.0 =4.180 x 2.0 = 0.00482 / 0.080 =0.00482 / 0.080 = 3.15x10 -2 / 2.00x10 5 =3.15x10 -2 / 2.00x10 5 = 24.8x10 6 / 6.200x10 -2 =24.8x10 6 / 6.200x10 -2 =

31. 31 Practice 142 x 2 = 300142 x 2 = 300 4.180 x 2.0 =4.180 x 2.0 = 0.00482 / 0.080 =0.00482 / 0.080 = 3.15x10 -2 / 2.00x10 5 =3.15x10 -2 / 2.00x10 5 = 24.8x10 6 / 6.200x10 -2 =24.8x10 6 / 6.200x10 -2 =

32. 32 Practice 142 x 2 = 300142 x 2 = 300 4.180 x 2.0 = 8.44.180 x 2.0 = 8.4 0.00482 / 0.080 =0.00482 / 0.080 = 3.15x10 -2 / 2.00x10 5 =3.15x10 -2 / 2.00x10 5 = 24.8x10 6 / 6.200x10 -2 =24.8x10 6 / 6.200x10 -2 =

33. 33 Practice 142 x 2 = 300142 x 2 = 300 4.180 x 2.0 = 8.44.180 x 2.0 = 8.4 0.00482 / 0.080 = 0.0600.00482 / 0.080 = 0.060 3.15x10 -2 / 2.00x10 5 =3.15x10 -2 / 2.00x10 5 = 24.8x10 6 / 6.200x10 -2 =24.8x10 6 / 6.200x10 -2 =

34. 34 Practice 142 x 2 = 300142 x 2 = 300 4.180 x 2.0 = 8.44.180 x 2.0 = 8.4 0.00482 / 0.080 = 0.0600.00482 / 0.080 = 0.060 3.15x10 -2 / 2.00x10 5 = 1.58x10 -73.15x10 -2 / 2.00x10 5 = 1.58x10 -7 24.8x10 6 / 6.200x10 -2 =24.8x10 6 / 6.200x10 -2 =

35. 35 Practice 142 x 2 = 300142 x 2 = 300 4.180 x 2.0 = 8.44.180 x 2.0 = 8.4 0.00482 / 0.080 = 0.0600.00482 / 0.080 = 0.060 3.15x10 -2 / 2.00x10 5 = 1.58x10 -73.15x10 -2 / 2.00x10 5 = 1.58x10 -7 24.8x10 6 / 6.200x10 -2 = 4.00x10 824.8x10 6 / 6.200x10 -2 = 4.00x10 8

36. 36 Significant Figures - rules add & subtract rule - subtleadd & subtract rule - subtle answer contains smallest decimal place of the addends

37. 37 Significant Figures - rules add & subtract rule - subtleadd & subtract rule - subtle answer contains smallest decimal place of the addends

38. 38 Significant Figures - rules add & subtract rule - subtleadd & subtract rule - subtle answer contains smallest decimal place of the addends

39. 39 Practice 416.2 – 10.18 =416.2 – 10.18 = 16.78 + 10. =16.78 + 10. = 422.501 – 420.4 =422.501 – 420.4 = 25.5 + 21.1 + 3.201 =25.5 + 21.1 + 3.201 = 42.00x10 -4 + 1.8x10 -6 =42.00x10 -4 + 1.8x10 -6 =

40. 40 Practice 416.2 – 10.18 = 406.0416.2 – 10.18 = 406.0 16.78 + 10. =16.78 + 10. = 422.501 – 420.4 =422.501 – 420.4 = 25.5 + 21.1 + 3.201 =25.5 + 21.1 + 3.201 = 42.00x10 -4 + 1.8x10 -6 =42.00x10 -4 + 1.8x10 -6 =

41. 41 Practice 416.2 – 10.18 = 406.0416.2 – 10.18 = 406.0 16.78 + 10. = 2716.78 + 10. = 27 422.501 – 420.4 =422.501 – 420.4 = 25.5 + 21.1 + 3.201 =25.5 + 21.1 + 3.201 = 42.00x10 -4 + 1.8x10 -6 =42.00x10 -4 + 1.8x10 -6 =

42. 42 Practice 416.2 – 10.18 = 406.0416.2 – 10.18 = 406.0 16.78 + 10. = 2716.78 + 10. = 27 422.501 – 420.4 = 2.1422.501 – 420.4 = 2.1 25.5 + 21.1 + 3.201 =25.5 + 21.1 + 3.201 = 42.00x10 -4 + 1.8x10 -6 =42.00x10 -4 + 1.8x10 -6 =

43. 43 Practice 416.2 – 10.18 = 406.0416.2 – 10.18 = 406.0 16.78 + 10. = 2716.78 + 10. = 27 422.501 – 420.4 = 2.1422.501 – 420.4 = 2.1 25.5 + 21.1 + 3.201 = 49.825.5 + 21.1 + 3.201 = 49.8 42.00x10 -4 + 1.8x10 -6 =42.00x10 -4 + 1.8x10 -6 =

44. 44 Practice 416.2 – 10.18 = 406.0416.2 – 10.18 = 406.0 16.78 + 10. = 2716.78 + 10. = 27 422.501 – 420.4 = 2.1422.501 – 420.4 = 2.1 25.5 + 21.1 + 3.201 = 49.825.5 + 21.1 + 3.201 = 49.8 42.00x10 -4 + 1.8x10 -6 = 4.2 x 10 -342.00x10 -4 + 1.8x10 -6 = 4.2 x 10 -3

45. 45 More Practice 4.18 – 58.16 x (3.38 – 3.01) =

46. 46 More Practice 4.18 – 58.16 x (3.38 – 3.01) = 4.18 – 58.16 x (0.37) =

47. 47 More Practice 4.18 – 58.16 x (3.38 – 3.01) = 4.18 – 58.16 x (0.37) = 4.18 – 21.5192 =

48. 48 More Practice 4.18 – 58.16 x (3.38 – 3.01) = 4.18 – 58.16 x (0.37) = 4.18 – 21.5192 = -17.3392 Round off correctly

49. 49 More Practice 4.18 – 58.16 x (3.38 – 3.01) = 4.18 – 58.16 x (0.37) = 4.18 – 21.5192 = -17.3392 Round off correctly to 2 sig. figs -17

50. 50 Unit Factor Method Dimensional Analysis simple but important way to always get right answersimple but important way to always get right answer way to change from one unit to anotherway to change from one unit to another make unit factors from statementsmake unit factors from statements 1 ft = 12 in becomes 1 ft/12 in or 12in/1 ft 1 ft = 12 in becomes 1 ft/12 in or 12in/1 ft 3 ft = 1 yd becomes 3ft/1yd or 1yd/3ft

51. 51 Unit Factor Method Dimensional Analysis simple but important way to always get right answersimple but important way to always get right answer way to change from one unit to anotherway to change from one unit to another make unit factors from statementsmake unit factors from statements 1 ft = 12 in becomes 1 ft/12 in or 12in/1 ft 1 ft = 12 in becomes 1 ft/12 in or 12in/1 ft Example: Express 12.32 yards in millimeters.Example: Express 12.32 yards in millimeters.

52. 52 Unit Factor Method

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54. 54 Example: Express 323. milliliters in gallonsExample: Express 323. milliliters in gallons

55. 55 Unit Factor Method Express 323. milliliters in gallons.Express 323. milliliters in gallons.

56. 56 Unit Factor Method Example: Express 5.50 metric tons in pounds. 1 metric ton = 1 MegagramExample: Express 5.50 metric tons in pounds. 1 metric ton = 1 Megagram

57. 57 Unit Factor Method Example: Express 5.50 metric tons in pounds.Example: Express 5.50 metric tons in pounds.

58. 58 Unit Factor Method area is two dimensionalarea is two dimensional Example: Express 4.21 x 10 6 square centimeters in square feetExample: Express 4.21 x 10 6 square centimeters in square feet

59. 59 Unit Factor Method area is two dimensionalarea is two dimensional express 4.21 x 10 6 square centimeters in square feet

60. 60 Unit Factor Method area is two dimensionalarea is two dimensional express 4.21 x 10 6 square centimeters in square feet

61. 61 Unit Factor Method area is two dimensionalarea is two dimensional express 4.21 x 10 6 square centimeters in square feet

62. 62 Unit Factor Method volume is three dimensionalvolume is three dimensional Example: Express 3.61 cubic feet in cubic centimeters.Example: Express 3.61 cubic feet in cubic centimeters.

63. 63 Unit Factor Method volume is three dimensionalvolume is three dimensional Example: Express 3.61 cubic feet in cubic centimeters.Example: Express 3.61 cubic feet in cubic centimeters.

64. 64 Percentage Percentage is the parts per hundred of a sample.Percentage is the parts per hundred of a sample. Example: A 500. g sample of ore yields 27.9 g of sulfur. What is the percent of sulfur in the ore?Example: A 500. g sample of ore yields 27.9 g of sulfur. What is the percent of sulfur in the ore?

65. 65 Percentage Percentage is the parts per hundred of a sample.Percentage is the parts per hundred of a sample. Example: A 500. g sample of ore yields 27.9 g of sulfur. What is the percent of sulfur in the ore?Example: A 500. g sample of ore yields 27.9 g of sulfur. What is the percent of sulfur in the ore?

66. 66 Derived Units - Density density = mass/volumedensity = mass/volume What is density?What is density? Example: Calculate the density of a substance if 123. grams of it occupies 57.6 cubic centimeters.Example: Calculate the density of a substance if 123. grams of it occupies 57.6 cubic centimeters.

67. 67 Derived Units - Density density = mass/volumedensity = mass/volume What is density?What is density? Example: Calculate the density of a substance if 123. grams of it occupies 57.6 cubic centimeters.Example: Calculate the density of a substance if 123. grams of it occupies 57.6 cubic centimeters.

68. 68 Derived Units - Density density = mass/volumedensity = mass/volume What is density?What is density? Example: Calculate the density of a substance if 123. grams of it occupies 57.6 cubic centimeters.Example: Calculate the density of a substance if 123. grams of it occupies 57.6 cubic centimeters.

69. 69 Derived Units - Density Example: Suppose you need 175. g of a corrosive liquid for a reaction. What volume do you need?Example: Suppose you need 175. g of a corrosive liquid for a reaction. What volume do you need? –liquid’s density = 1.02 g/mL

70. 70 Derived Units - Density Example: Suppose you need 175. g of a corrosive liquid for a reaction. What volume do you need?Example: Suppose you need 175. g of a corrosive liquid for a reaction. What volume do you need? –liquid’s density = 1.02 g/mL

71. 71 Derived Units - Density Example: Suppose you need 175. g of a corrosive liquid for a reaction. What volume do you need?Example: Suppose you need 175. g of a corrosive liquid for a reaction. What volume do you need? –liquid’s density = 1.02 g/mL

72. 72 Heat & Temperature heat and T are not the same thingheat and T are not the same thing T is a measure of the intensity of heat in a body 3 common T scales - all use water as a reference3 common T scales - all use water as a reference

73. 73 Heat & Temperature MPBP Fahrenheit 32 o F 212 o FFahrenheit 32 o F 212 o F Celsius 0 o C 100 c CCelsius 0 o C 100 c C Kelvin 273 K 373 KKelvin 273 K 373 K

74. 74 Relationships of the 3 T Scales

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77. 77 Heat and Temperature Example: Convert 111. o F to degrees Celsius.Example: Convert 111. o F to degrees Celsius.

78. 78 Heat and Temperature Example: Convert 111. o F to degrees Celsius.Example: Convert 111. o F to degrees Celsius.

79. 79 Heat and Temperature Example: Express 757. K in Celsius degrees.Example: Express 757. K in Celsius degrees.

80. 80 Heat and Temperature Example: Express 757. K in Celsius degrees.Example: Express 757. K in Celsius degrees.

81. 81 The Measurement of Heat SI unit J (Joule)SI unit J (Joule) caloriecalorie 1 calorie = 4.184 J English unit = BTUEnglish unit = BTU

82. 82 Synthesis Question It has been estimated that 1.0 g of seawater contains 4.0 pg of Au. The total mass of seawater in the oceans is 1.6x10 12 Tg, If all of the gold in the oceans were extracted and spread evenly across the state of Georgia, which has a land area of 58,910 mile 2, how tall, in feet, would the pile of Au be?It has been estimated that 1.0 g of seawater contains 4.0 pg of Au. The total mass of seawater in the oceans is 1.6x10 12 Tg, If all of the gold in the oceans were extracted and spread evenly across the state of Georgia, which has a land area of 58,910 mile 2, how tall, in feet, would the pile of Au be? Density of Au is 19.3 g/cm 3. 1.0 Tg = 10 12 g.

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86. 86 Group Activity On a typical day, a hurricane expends the energy equivalent to the explosion of two thermonuclear weapons. A thermonuclear weapon has the explosive power of 1.0 Mton of nitroglycerin. Nitroglycerin generates 7.3 kJ of explosive power per gram of nitroglycerin. The hurricane’s energy comes from the evaporation of water that requires 2.3 kJ per gram of water evaporated. How many gallons of water does a hurricane evaporate per day?On a typical day, a hurricane expends the energy equivalent to the explosion of two thermonuclear weapons. A thermonuclear weapon has the explosive power of 1.0 Mton of nitroglycerin. Nitroglycerin generates 7.3 kJ of explosive power per gram of nitroglycerin. The hurricane’s energy comes from the evaporation of water that requires 2.3 kJ per gram of water evaporated. How many gallons of water does a hurricane evaporate per day?