1. 1 Basic Concepts

2. 2 States of Matter Chemical and Physical Properties Chemical and Physical Changes Mixtures, Substances, Compounds, and Elements Measurements in Chemistry Units of Measurement Chapter Outline

3. 3 Use of Numbers The Unit Factor Method (Dimensional Analysis) Density and Specific Gravity Heat and Temperature Heat Transfer and the Measurement of Heat Chapter Outline

4. 4 Mixtures, Substances, Compounds, and Elements Matter-Anything that occupies space and has mass. Pure Substances or Substances-Cannot be separated by physical processes. -Elements-A substance which cannot be broken down into simpler substances. e.g. Na, He, C, (atoms) or N 2, Cl 2 (molecules) -Compounds-A pure substance made up of two or more elements. e.g. NaCl, H 2 O. Mixtures-Can be separated by physical processes. –composed of two or more substances –homogeneous mixtures-A mixture that is uniform throughout-e.g. white wine, grape juice. Clear. Solutions. –heterogeneous mixtures-A mixture that is not uniform throughout-e.g. oil and water, orange juice. Cloudy.

5. 5 Mixtures, Substances, Compounds, and Elements

6. 6 Substances and Mixtures Homogeneous Mixture Matter Pure Substances Mixtures Physical process Elements Chemical Reaction Compounds Heterogeneous Mixture

7. 7 Fig. 1-7, p. 10

8. 8 States of Matter Change States –heating –cooling Vaporization: Evaporation Boiling Freezing: Solidification Crystallization Melting: Fusion

9. 9 States of Matter SolidLiquid Gas heat cool heat cool Less attractive force and more disordered.

10. 10 States of Matter Illustration of changes in state –requires energy

11. 11 Mixtures, Substances, Compounds, and Elements

12. 12 Elements you Need to Know:

13. 13 Types of Solutions: Liquid solutions are the most common, but there are also gas and solid solutions. Solutions have two components: a)Solute - Solution component(s) present in lesser amounts. b)Solvent - Solution component present in the greatest amount.

14. 14 Types of Solutions:

15. 15 Characteristics of Solutions Uniform distribution Components do not separate upon standing. Components cannot be separated by filtration. Within certain limits its composition can vary. Almost always transparent. (i.e. one can see through it). An alloy is a homogeneous mixture of metals. i.e. brass, bronze, sterling silver.

16. 16 Chemical and Physical Properties Physical Properties – A property that can be observed in the absence of any change in composition. e.g. color, odor, taste, melting point, boiling point, freezing point, density, length, specific heat, density, solubility. Physical Changes-Changes observed without a change in composition. i.e. cutting wood, melting of solids and boiling of liquids. water, ice water, liquid water, steam –changes of state –dissolving –polishing

17. 17 Chemical and Physical Properties Chemical Properties-A property that matter exhibits as it undergoes changes in composition. e.g. coal and gasoline can burn in air to form carbon dioxide and water; iron can react with oxygen in the air to form rust; bleach can turn blond hair blonde. Chemical Changes-Changes observed only when a change in composition is occurring. e.g. reaction of sodium with chlorine, rusting of iron, dying of hair, burning of wood, cooking an egg, rotting food. Extensive Properties - depends on the amount of material present. e.g. volume and mass. Intensive Properties – does not depend on the amount of material present. e.g. melting point, boiling point, freezing point, color, density.

18. 18 Natural Laws Law of Conservation of Mass-Mass is neither created nor destroyed. Law of Conservation of Energy-Energy is neither created nor destroyed, only converted from one form to another. Law of Definite Proportions-Different samples of any pure compound contain the same element in the same proportion by mass. e.g. water (H 2 O) contains 11.1 % H and 88.9% O by mass. Thus, a 25.0 sample of water would contain 2.78 g of H and 22.2 g of O.

19. 19 Law of Definite Proportions 11.1 % H and 88.9% O by mass, 25.0 g sample of water:

20. 20 Use of Numbers Exact numbers – 1 dozen = 12 things for example

21. 21 Rounding off Numbers Previous digit Next digit 1.If the next digit is less than 5 the previous.9946.99 digit remains the same. 1.294 1.29 2. If the next digit is greater than 5 or 5 followed by non zeros then the previous digit.999 1.00 is increased by one. 1.2951 1.30 3. If the next digit is 5 or 5 followed by all zeros 1.285 1.28 then the previous digit remains the same if it 1.295 1.30 is even or increased by one if it is odd. 1.22500 1.22 1.294

22. 22 Scientific Notation Used to handle very large and very small numbers. Any number that is from + or – 1 to 9 N. X 10 x For example: 3.21 x 10 3 -9.9 x 10 -4 1.0 x 10 0 (Note that 10 0 is 1) Exponent-Power of 10

23. 23 Scientific Notation To convert numbers to scientific notation use the following guidelines: 1750.0 = 1750.0 x 10 0 = 1.7500 x 10 3 Number decreases by 3 powers of 10 Exponent increases by 3 powers of 10 A you move the decimal place to the left (i.e. make the number smaller), the power of ten (i.e., exponent) must increase by the same amount.

24. 24 Scientific Notation 0.050 = 0.050 x 10 0 = 5.0 x 10 -2 The number gets larger by 2 powers of 10 The exponent gets smaller by 2 powers of 10. As you move the decimal place to the right (i.e. make the number larger), the power of ten (i.e. exponent) must decrease by the same amount.

25. 25 Use of Numbers Significant figures –digits believed to be correct by the person making the measurement Measure a mile with a 6 inch ruler vs. surveying equipment Exact numbers have an infinite number of significant figures 12.000000000000000 = 1 dozen because it is an exact number

26. 26 Use of Numbers Significant Figures - Rules Leading zeroes are never significant 0.000357 has three significant figures Trailing zeroes may be significant must specify significance by how the number is written Use scientific notation to remove doubt 2.40 x 10 3 has ? significant figures

27. 27 Use of Numbers 3,380 ? significant figures 3.38 x 10 3 3,380. has ? significant figures 3.380 x 10 3 Imbedded zeroes are always significant 3.0604 has ? significant figures

28. 28 Use of Numbers Piece of Paper Side B – enlarged –How long is the paper to the best of your ability to measure it? 13.36 in. The second decimal place is estimated

29. 29 Use of Numbers Piece of Paper Side A – enlarged –How wide is the paper to the best of your ability to measure it? 8.3 in The first decimal place is estimated

30. 30 Manipulating Powers of 10 a) When multiplying powers of ten, the exponents are added. For example: 10 5 x 10 -4 = 10 5+(-4) =10 1 b) When dividing powers of ten, the exponents are subtracted. For example: 10 4 = 10 4-(-4) = 10 8 10 -4 c) When raising powers of ten to an exponent, the exponents are multiplied. For example: (10 4 ) 3 = 10 (4 x 3) = 10 12

31. 31 Use of Numbers Multiplication & Division rule Easier of the two rules Product has the smallest number of significant figures of multipliers

32. 32 Use of Numbers Multiplication & Division rule Easier of the two rules Product has the smallest number of significant figures of multipliers

33. 33 Use of Numbers Multiplication & Division rule Easier of the two rules Product has the smallest number of significant figures of multipliers

34. 34 Multiplying and Dividing Numbers with Powers of Ten 1.When using scientific notation: a.) Place the powers of ten together. (1.76 x 10 200 ) x (2.650 x 10 200 )= (1.76 x 2.650) x (10 200 + 200 )= b.) The final answer has the same number of significant figures as the number with the least number of significant figures. 4.66 x 10 400 c.) You must round off correctly. d.) Preferably report the answer in scientific notation.

35. 35 Multiplying and Dividing Numbers with Powers of Ten (1.760 x 10 2 ) /(2.65 x 10 -2 )= (1.760 / 2.65) x (10 2 – (-2) )= 0.664 x 10 4 = 6.64 x 10 3

36. 36 Use of Numbers Addition & Subtraction rule More subtle than the multiplication rule Answer contains smallest decimal place of the addends

37. 37 Use of Numbers Addition & Subtraction rule More subtle than the multiplication rule Answer contains smallest decimal place of the addends

38. 38 Addition and Subtraction with Powers of Ten a.) All numbers must have the same power or ten before addition or subtraction is performed. b.) Once the powers of ten are the same, the coefficients can then be added or subtracted while the power of ten remains the same. c.) After adding or subtracting the coefficients, the answer must have the same number of decimal places as the coefficient with the fewest decimal places at the time of the operation. d.) You must round off correctly. e.) Preferably report the answer in scientific notation.

39. 39 Addition and Subtraction with Powers of Ten 4.76 x 10 200 + 9.6 x 10 201 = ? 0.4 76 x 10 201 + 9.6 x 10 201 10.0 76 x 10 201 1.01 x 10 202 (written in scientific notation and rounded off to the correct number of significant figures)

40. 40 Addition and Subtraction with Powers of Ten 2.95 x 10 -15 – 1.00 x 10 -14 = ? -1.00 x 10 -14 0.29 5 x 10 -14 -0.70 5 x 10 -14 -7.0 x 10 -15 (written in scientific notation and rounded off to the correct number of significant figures)

41. 41 Mixing Addition/Subtraction with Multiplication/Division 7.54 x 10 -5 (99. x 10 200 + 1.25 x 10 201 ) (1.75 x 10 -3 ) 3 7.54 x 10 -5 (9.9 x 10 201 + 1.25 x 10 201 ) = 1.75 x 10 -3 x 1.75 x 10 -3 x 1.75 x 10 -3 7.54 x 10 -5 [(9.9 + 1.25) x 10 201 ) = 1.75 x 1.75 x 1.75 x 10 -3 x 10 -3 x 10 -3 7.54 x 10 -5 (11.2 x 10 201 ) = 5.36 x 10 -9 7.54 x 11.2 x 10 -5 x 10 201 = 1.58 x 10 206 5.36 10 -9

42. 42 Measurements in Chemistry QuantityUnitSymbol lengthmeter m masskilogram kg timesecond s currentampere A temperatureKelvin K amt. substancemole mol

43. 43 Measurements in Chemistry Metric Prefixes NameSymbolMultiplier mega M 10 6 kilo k 10 3 deka da 10 deci d 10 -1 centi c 10 -2

44. 44 Measurements in Chemistry Metric Prefixes NameSymbolMultiplier milli m 10 -3 micro  10 -6 nano n 10 -9 pico p 10 -12 femto f 10 -15

45. 45 Metric Conversions 1 km = 10 3 m 1 dL = 10 -1 L 1 msec = 10 -3 sec 1  m = 10 -6 m

46. 46 Fig. 1-20, p. 24

47. 47 Metric English Conversions Common Conversion Factors Length –2.54 cm = 1 inch (exact conversion) Volume –1 qt = 0.946 liter (Rounded off) Mass _ 1 lb = 454 g (Rounded off)

48. 48 Use of Conversion Factors in Calculations Commonly known relationship (i.e. equality): 1 ft = 12 in Respective conversion factors to above equality: 1 ft or 12 in 12 in 1 ft Use the conversion factor that allows for the cancellation of units. Convert 24 in to ft: ? ft = 24 in x

49. 49 Conversion Factors Example 1-1: Express 9.32 yards in millimeters.

50. 50 Conversion Factors

51. 51 Conversion Factors

52. 52 Conversion Factors

53. 53 Conversion Factors

54. 54 The Unit Factor Method Area is two dimensional thus units must be in squared terms. Example 1-3: Express 2.61 x 10 4 cm 2 in ft 2.

55. 55 The Unit Factor Method  Area is two dimensional thus units must be in squared terms. Example 1-3: Express 2.61 x 10 4 cm 2 in ft 2. common mistake

56. 56 The Unit Factor Method  Area is two dimensional thus units must be in squared terms. Example 1-3: Express 2.61 x 10 4 cm 2 in ft 2.

57. 57 The Unit Factor Method  Area is two dimensional thus units must be in squared terms. Example 1-3: Express 2.61 x 10 4 cm 2 in ft 2.

58. 58 The Unit Factor Method  Area is two dimensional thus units must be in squared terms. Example 1-3: Express 2.61 x 10 4 cm 2 in ft 2.

59. 59 The Unit Factor Method Volume is three dimensional thus units must be in cubic terms. Example 1-4: Express 2.61 ft 3 in cm 3.

60. 60 The Unit Factor Method Example 1-2. Express 627 milliliters in gallons.

61. 61 Conversions of Double Units

62. 62 Density and Specific Gravity density = mass/volume What is density? Why does ice float in liquid water?

63. 63 Density and Specific Gravity Example 1-6: Calculate the density of a substance if 742 grams of it occupies 97.3 cm 3.

64. 64 Density and Specific Gravity Example 1-6: Calculate the density of a substance if 742 grams of it occupies 97.3 cm 3.

65. 65 Density and Specific Gravity Example 1-7 Suppose you need 125 g of a corrosive liquid for a reaction. What volume do you need? –liquid’s density = 1.32 g/mL

66. 66 Density and Specific Gravity Water’s density is essentially 1.00 g/mL at room T. Thus the specific gravity of a substance is very nearly equal to its density. Specific gravity has no units.

67. 67 Density and Specific Gravity The density of lead is 11.4 g/cm 3. What volume, in ft 3, would be occupied by 10.0 g of lead?

68. 68 Density and Specific Gravity What is the density (in g/mL) of a rectangular bar of lead that weighs 173 g and has the following dimensions: length = 2.00 cm, w = 3.00 cm, h = 1.00 in?

69. 69 Density and Specific Gravity An irregularly shaped piece of metal with a mass of 0.251 lb was placed into a graduated cylinder containing 50.00 mL of water; this raised the water level to 67.50 mL. What is the density of the metal? What is the density (in g/cm 3 ) of the metal? Will the metal float or sink in water? V (disp)l = 67.50 mL – 50.00 mL = 17.50 mL The metal will sink in water because its density is greater than that of water. (1.00 g/mL)

70. 70 Density and Specific Gravity Example1-8: A 31.0 gram piece of chromium is dipped into a graduated cylinder that contains 5.00 mL of water. The water level rises to 9.32 mL. What is the specific gravity of chromium?

71. 71 Density and Specific Gravity Example1-8: A 31.0 gram piece of chromium is dipped into a graduated cylinder that contains 5.00 mL of water. The water level rises to 9.32 mL. What is the specific gravity of chromium?

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

73. 73 Heat and Temperature Heat and Temperature are not the same thing T is a measure of the intensity of heat in a body 3 common temperature scales - all use water as a reference

74. 74 Heat and Temperature MP waterBP water Fahrenheit 32 o F 212 o F Celsius 0.0 o C 100 o C Kelvin 273 K 373 K

75. 75 Relationships of the Three Temperature Scales

76. 76 Relationships of the Three Temperature Scales

77. 77 Relationships of the Three Temperature Scales

78. 78 Heat and Temperature Example 1-10: Convert 211 o F to degrees Celsius. = 99.4

79. 79 Heat and Temperature Example 1-11: Express 548 K in Celsius degrees.

80. 80 Heat Transfer and the Measurement of Heat Chemical reactions and physical changes occur with either the simultaneous evolution of heat (exothermic process), or the absorption of heat (endothermic process). The amount of heat transferred is usually expressed in calories (cal) or in the SI unit of joules (J). 1 cal = 4.184 J Specific heat is defined as the amount of heat necessary to raise the temperature of 1 g of substance by 1 o C. Each substance has a specific heat, which is a physical intensive property, like density and melting point.

81. 81 Heat Transfer and the Measurement of Heat From a knowledge of a substance’s specific heat, the heat (q) that is absorbed or released in a given process can be calculated by use of the following equation: q = s x m x  T q (heat energy) cal, kcal, J or kJ m (mass) g s (specific heat) cal g o C (kcal, J, or kJ can be used in lieue of cal).  T = T 2 – T 1 (change in temp-make  T a positive #) o C

82. 82 Heat Transfer and the Measurement of Heat Substances with large specific heats require more heat to raise their temperature. Water has one of the highest specific heats, 1.00 cal/g o C. The high specific heat of water (which constitutes ~60% of our body weight) makes our body’s task of maintaining a constant body temperature of ~37 o C much easier. Thus, our body has the ability to absorb or release considerable amounts of energy with little change in temperature.

83. 83 Heat Transfer and the Measurement of Heat Substancespecific heat (cal/g o C) water1.00 wood0.421 gold0.0306 graphite0.172

84. 84 Heat Transfer and the Measurement of Heat Calculate the amount of heat to raise T of 200.0 g of water from 10.0 o C to 55.0 o C. You need to know that the specific heat for water (s water ) is 1.00 cal/g o C

85. 85 Heat Transfer and the Measurement of Heat Example 1-13: Calculate the amount of heat to raise T of 200.0 g of Hg from 10.0 o C to 55.0 o C. Specific heat for Hg is 0.138 J/g o C. Requires 30.3 times more heat for water 4.184 is 30.3 times greater than 0.138

86. 86 Heat Transfer and the Measurement of Heat If we add 450 cal of heat to 37 g of ethyl alcohol (s=0.59 cal/g o C) at 20 o C, what would its final temperature be? q = m x s x   cal = 37 g x 0.59 x   o  C Since heat was added, the final temperature must be greater than the initial temperature.  T=T 2 - T 1 21 o C = T 2 – 20 o C T 2 = 21 o C + 20 o C = 41 o C