1. Chapter 3

2.  Matter – anything that has mass and takes up space  Everything around us  Mass: measurement that reflects the amount of matter (usually in grams)  Volume: the amount of space something takes up  Chemistry – the study of matter and the changes it undergoes

3.  Solids  particles vibrate but can’t move around  fixed shape  fixed volume  incompressible

4.  Liquids  particles can move around but are still close together  variable shape  fixed volume  Virtually incompressible

5.  Gases  particles can separate and move throughout container  variable shape  variable volume  Easily compressed  Vapor = gaseous state of a substance that is a liquid or solid at room temperature

6.  Plasma  atoms collide with enough energy to break into charged particles (+/-)  gas-like, variable shape & volume  stars, fluorescent light bulbs, TV tubes

7. II. Properties & Changes in Matter (p.73-79) Extensive vs. Intensive Physical vs. Chemical

8.  Physical Property  can be observed & measured without changing the identity of the substance

9.  Physical properties can be described as one of 2 types:  Extensive Property  depends on the amount of matter present (example: length, mass, volume)  Intensive Property  depends on the identity of substance, not the amount (example: scent, density, melting point)

10.  Chemical Property  describes the ability of a substance to be observed reacting with or changing into another substance

11.  Examples:  melting point  flammable  density  magnetic  tarnishes in air physical chemical physical chemical

12.  Physical Change  changes the form of a substance without changing its identity  properties remain the same  Examples: cutting a sheet of paper, breaking a crystal, all phase changes

13.  Evaporation =  Condensation =  Melting =  Freezing =  Sublimation =  Deposition = Liquid -> Gas Gas -> Liquid Solid -> Liquid Liquid -> Solid Solid -> Gas Gas -> Solid

14.  Process that involves one or more substances changing into a new substance  Commonly referred to as a chemical reaction  New substances have different compositions and properties from original substances  Reaction involves reactants reacting to produce products

15.  Signs of a Chemical Change  change in color or odor  formation of a gas (bubbles)  formation of a precipitate (solid)  change in light or heat

16.  Examples:  rusting iron  dissolving in water  burning a log  melting ice  grinding spices chemical physical chemical physical

18. Exothermic- heat energy EXITS the system  surroundings usually feel warmer  1 g H 2 O (g)  1 g H 2 O (l) + 2260 J  ex. Combustion, evaporation of water

19. Endothermic- heat energy ENTERS the system - heat absorbed from surroundings - surroundings usually feel cooler - 1 g H 2 O (s) + 333 J  1 g H 2 O (l) - 1 g H 2 O (l) + 2260 J  1 g H 2 O (g) - ex. Cold packs, melting ice

20.    

21.  Although chemical changes occur, mass is neither created nor destroyed in a chemical reaction  Mass of reactants equals mass of products mass reactants = mass products A + B  C

22. III. Classification of Matter (pp. 80-87) Matter Flowchart Pure Substances Mixtures

23. MATTER Can it be physically separated? Homogeneous Mixture (solution) Heterogeneous MixtureCompoundElement MIXTUREPURE SUBSTANCE yes no Can it be chemically decomposed? noyes Is the composition uniform? noyes

24.  Examples:  graphite  pepper  sugar (sucrose)  paint  soda element hetero. mixture compound hetero. mixture solution

25.  Element  composed of one type of identical atoms  Atom : Composed of protons, electrons, and neutrons. Smallest particle of matter that can be identified as one element  EX: copper wire, aluminum foil

26.  Compound  composed of 2 or more elements in a fixed ratio (bonded together)  properties differ from those of individual elements  EX: table salt (NaCl)

27.  Variable combination of 2 or more pure substances, each retains its chemical identity & properties. HeterogeneousHomogeneous

28.  Homogeneous: are uniform throughout  Solutions  very small particles  particles don’t settle  EX: rubbing alcohol, gasoline, soda

29.  Heterogeneous  medium-sized to large-sized particles  particles may or may not settle  EX: milk, fresh- squeezed lemonade

30.  Examples:  tea  muddy water  fog  saltwater  Italian salad dressing  Answers:  Solution  Heterogeneous  Solution  Heterogeneous

31. + Separation Methods Ways to separate mixtures – Chapter 3: Matter & Its Properties

32. + Separating Mixtures Substances in a mixture are physically combined, so processes bases on differences in physical properties are used to separate component Numerous techniques have been developed to separate mixtures to study components Visually Magnetism Filtration Distillation Crystallization Chromatography

33. + Filtration Used to separate heterogeneous mixtures composed of solids and liquids Uses a porous barrier to separate the solid from the liquid Liquid passes through leaving the solid in the filter paper

34. + Distillation Used to separate homogeneous mixtures Based on differences in boiling points of substances involved

35. + Crystallization Separation technique that results in the formation of pure solid particles from a solution containing the dissolved substance As one substance evaporates, the dissolved substance comes out of solution and collects as crystals Produces highly pure solids Rocky candy is an example of this

36. + Chromatography Separates components of a mixture based on ability of each component to be drawn across the surface of another material Mixture is usually liquid and is usually drawn across chromatography paper Separation occurs because various components travel at different rates Components with strongest attraction for paper travel the slowest

37. Thermochemistry Chapter 17:1 Pages 505 – 510

38. A. Vocabulary The study of energy changes that occur during chemical reactions and changes in state Thermochemistry: The study of energy changes that occur during chemical reactions and changes in state Energy: The capacity to do work or produce heat

39. A. Vocabulary  TEMPERATURE is a measure of the amount of kinetic energy an object/substance has  HEAT is energy that transfers from one object/substance to another (thermal energy) Represented by the symbol q Transfers because of difference in temperature Always flows from a warmer to a cooler object

40. A. Vocabulary  Which has more thermal energy? 200 mL 80ºC A 400 mL 80ºC B

41. B. Heat Transfer  Why does A feel hot and B feel cold? 80ºC A 10ºC B Heat flows from A to your hand = hot Heat flows from your hand to B = cold

42. C. Types of Energy  Potential: due to position or composition – can be converted to work  Kinetic: due to motion of the object KE = ½ mv 2 (m = mass, v = velocity)  Law of Conservation of Energy – energy can be neither created nor destroyed

43.  Energy that is stored in the chemical bonds of a substance is called CHEMICAL POTENTIAL ENERGY  Types of atoms and their arrangement determine amount of energy stored in substance C. Types of Energy

44. D. Exothermic and Endothermic  System = the reaction (our focus)  Surroundings = everything around the reaction (rxn container, room, etc) Surroundings Universe = System + Surroundings System

45. D. Exothermic and Endothermic  Exothermic process – heat is released into the surroundings Exo = Exit  Exothermic processes are represented by a negative “ q ” HEAT

46. B. Exothermic and Endothermic  Combustion of Methane

47. D. Exothermic and Endothermic  Endothermic Process – heat is absorbed from the surroundings Endo = Into  Endothermic processes are represented by a positive “ q ” HEAT

48. D. Exothermic and Endothermic  Formation of Nitric Oxide

49. D. Exothermic and Endothermic  Sign of ( q ) is a ‘signal’ to indicate the direction of the heat transfer Exothermic - q heat transferred from a substance Endothermic +q heat transferred into a substance

50. E. Measuring Heat Flow  Two Common Units Joule calorie  4.184 J = 1 cal  1J = 0.2390 cal  1Calorie = 1 kilocal = 1000 cal

51. F. Heat Capacity  “the amount of heat needed to increase the temperature of an object by 1 o C”  Heat Capacity depends on: The mass of the object The chemical composition of the object

52.  Specific heat capacity – amount of heat needed to raise the temperature of 1g of a substance by 1 o C  Specific Heat of H 2 O (l) = 4.184 J/g o C  Specific Heat of H 2 O (s) = 2.02 J/g o C  Molar heat capacity – amount of heat needed to raise the temperature of 1 mole of a substance by 1 o C G. Specific Heat Capacity

53. q = m  C   T q:heat (J) m:mass (g) C:specific heat (J/g·K) or (J/g o C)  T:change in temperature (K or °C)  T = T f - T i – q = heat loss + q = heat gain

54. G. Specific Heat Capacity  q = m x C X ΔT  q = heat (joules or calories)  m = mass (grams)  C = Specific Heat  ΔT = change in temperature The change in temperature can be measure in Kelvin or degrees Celsius

55. G. Heat Transfer Problem  A 32.0-g silver spoon cools from 60.0°C to 20.0°C. How much heat is lost by the spoon? GIVEN: m = 32.0 g T i = 60.0°C T f = 20.0°C q = ? C =.235 J/g·C WORK: q = m·C·  T  T = 20°C - 60°C = – 40°C q =(32.0g)(-40°C)(.235J/g·C) q = – 301 J Exo

56.  The temperature of a 95.4g piece of copper increases from 25.0 o C to 48.0 o C when it absorbs 849 J of heat. What is the specific heat of copper? GIVEN: m = 95.4 g T i = 25.0°C T f = 48.0°C q = 849 J C = ? WORK: q = m·C ·  T C = q/m ·  T  T = 48°C – 25°C = 23°C C = 849J/(95.4g)(23°C) C = 0.387 J/g°C Endo G. Heat Transfer Problem

57.  What is the molar heat capacity of copper? GIVEN: C = 0.387 J/g°C MM = 63.55 g/mol WORK: C = 0.387 J G. Heat Transfer Problem g o C 63.55 g Cu 1 mol Cu = 24.6 J/mol o C

58. Thermochemistry Chapter 17:2 Pages 511 – 517

59. A. Calorimetry  Measures the heat flow into or out of a system  Heat released by the system is equal to heat absorbed by the surroundings  Calorimeter = Insulated device used to measure absorption or release of heat Coffee cup Calorimeter

60. A. Calorimetry  Enthalpy: ( H ) the heat absorbed or released of a system at constant pressure  Δ H is the heat of a reaction  Heat or enthalpy change are used interchangeably here q = Δ H

61. B. Thermochemical Equations  In a thermochemical equation, the enthalpy of change for the reaction can be written as either a reactant or a product  Endothermic (positive ΔH) 2NaHCO 3 + 129kJ Na 2 CO 3 + H 2 O + CO 2 ΔH = 129 kJ  Exothermic (negative ΔH) CaO + H 2 O Ca(OH) 2 + 65.2kJ ΔH = - 65.2 kJ

62. A. Calorimetry ∆H = H final - H initial OR ∆H reaction (rxn) = H products – H reactants For endothermic reactions H final >H initial & ∆H is positive (+∆H) For exothermic reaction H final <H initial and ∆H is negative (-∆H)

63. A. Calorimetry When solving for the heat transfer of a system (between 2 objects), assume that: q initial soln = - q final rxn heat goes in = heat goes out +∆H = -∆H or C H20 x m H20 x ∆T H20 = -1(C metal x m metal x ∆T metal )

64. A. Calorimetry  Example:  2 H 2 + O 2 → 2 H 2 O + 483.6 kJ  ∆H = - 483.6 kJ  Exothermic

65. A. Calorimetry  How to measure Δ H for a reaction in aqueous solution Dissolve chemicals in water Place in calorimeter Measure temperature change

66. A. Calorimetry  q surr = m x C x ΔT  q sys = Δ H = - q surr = -m x C x ΔT Negative enthalpy = exothermic Positive enthalpy = endothermic

67. When 25.0mL of water containing HCl at 25.0 o C is added to 25.0mL of water containing NaOH at 25.0 o C in a calorimeter a rxn occurs. Calculate the enthalpy change (in kJ) during the rxn if the highest temperature observed was 32.0 o C. Assume all densities =1.00g/mL KNOWN: C water = 4.184 J/g o C V = 25.0mL + 25.0mL ΔT = 32.0 – 25.0 = 7.0 o C Density= 1.00g/mL M = (50mL) x (1.00g/mL) = 50g ΔH = ?  ΔH= -mCΔT=  -(50.0g)(4.184J/g o C) (7.0 o C)  ΔH= -1463 J = -1460J Exo

68. C. Heat of Combustion  The heat of reaction for the complete burning of one mole of a substance  Written the same way as change in enthalpy

69. Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ Fe(s) + O 2 (g)→ Fe 2 O 3 (s) + 1652 kJ How much heat is evolved when 10.00g of Iron is reacted with excess oxygen? 4 3 2 10.00g Fe 55.85g Fe 1 mol 4 mol Fe 1652 kJ =73.97 kJ of heat Exo

70. Thermochemistry Chapter 17:3 Pages 520 – 526

71. A. Heat of Fusion  Heat of Fusion (  H fus ) Heat absorbed by one mole of a solid substance when it melts to a liquid at a constant temperature  H fus of ice = 6.009 kJ/mol  Heat of Solidification (  H solid ) Heat lost by one mole of a liquid substance when it solidifies at a constant temperature  H fus = -  H solid  H solid of water = -6.009 kJ/mol

72. B. Heating Curves Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

73. B. Heating Curves  Temperature Change change in KE (molecular motion) depends on heat capacity  Phase Change change in PE (molecular arrangement) temp remains constant

74. C. Heat of Vaporization  Heat of Vaporization (  H vap ) energy required to boil 1 gram of a substance at its b.p.  H vap for water = 40.79 kJ/mol usually larger than  H fus …why?  EX: sweating, steam burns

75. How much heat energy is required to melt 25 grams of ice at 0 o C to liquid water at a temperature of 0 o C? 25 g H 2 O1 mol H 2 O 18.02 g H 2 O = 8.3 kJ 6.009 kJ 1 mol H 2 O ice D. Practice Problems

76. How much heat energy is required to change 500.0 grams of liquid water at 100 o C to steam at 100 o C? = 1132 kJ 500.0 g H 2 O1 mol H 2 O 18.02 g H 2 O 40.79 kJ 1 mol H 2 O steam

77. D. Practice Problems How many kJ are absorbed when 0.46g of C 2 H 5 Cl vaporizes at its normal boiling point? The molar  H vap is 26.4 kJ/mol. = 0.19 kJ 0.46 g C 2 H 5 Cl 1 mol C 2 H 5 Cl 64.52 g C 2 H 5 Cl 26.4 kJ 1 mol C 2 H 5 Cl

78. E. Heat of Solution  During the formation of a solution, heat is either released or absorbed  Enthalpy change caused by dissolution of 1 mol of a substance is the molar heat of solution  H soln  Examples: hot packs, cold packs

79. E. Heat of Solution  NaOH(s) → Na + (aq) + OH - (aq)  H soln = -445.1 kJ/mol  NH 4 NO 3 → NH 4 + (aq) + NO 3 - (aq)  H soln = 25.7 kJ/mol