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Classsification of Matter

Phases of Matter

Heat Energy

Boyle's Law

Charles's Law

Kinetic Molecular Theory

Avogadro's Hypothesis

Separation of Mixtures

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- 1. Matter and Energy
- 2. Matter Anything that has mass and occupies space (volume) ◦ Chemistry is the study of matter. Matter is composed of atoms.
- 3. Classifying Matter
- 4. Classifying Matter Matter ◦ Substances – have constant composition and constant properties throughout a given sample. Cannot be separated by physical means. Element – substances that are composed of atoms that have the same atomic number. Cannot be broken down by chemical means. Examples; Hydrogen (H), Helium (He), Oxygen (O), Zinc(Zn), etc. Compound – a substance composed of two or more elements that are chemically combined I definite proportions by mass. Examples; H2O, CO2, CH4
- 5. Classifying Matter Matter ◦ Mixtures – a combination of two or more substances that can be separated by physical means. Homogeneous (a.k.a. Solutions)– a mixture that is uniform in its properties throughout a given sample. Examples; Coca-Cola, milk, air Heterogeneous – consists of physically distinct parts each with different properties. Examples; Soil, salad, cement
- 6. Properties of Matter Physical Change - a change in matter from one for or another (rearrangement of particles) without a change in chemical properties. ◦ Example; Salt-water can easily be physicall separated back into its components of solid NaCl crystals and H2O by a process called distillation.
- 7. Properties of Matter Chemical Change – a change in which results in the formation of different kinds of matter with changed properties. ◦ Example; Rustin of iron – Fe combines with O2 in the air to from a new material called rust. The Fe and O2 atoms cannot be separated by physical means.
- 8. Phases of Matter Solid Phase Liquid Phase Gaseous Phase Plasma Phase
- 9. Solid Phase Rigid form. Definite shape. Definite volume. Strong attractive forces hold atoms and molecules in fixed locations. True solids have a crystalline structure.
- 10. Liquid Phase Particles not held together as rigidly and are able to move past one another. Sufficient attractive forces to have definite volume. No definite shape, takes the shape of the container.
- 11. Gaseous Phase Minimal attractive forces to hold particles together, no definite volume. No definite shape. Spread out indefinitely unless confined to a container where the gas will occupy the volume of the container. Vapor is the gaseous phase of a substance that is a liquid or solid at normal conditions.
- 12. Particle Diagrams Draw the particle diagrams for water (H2O) in the three phase of matter.Solid Liquid Gaseous WaterWater Water
- 13. Diatomic molecules HOFBrINCl’s ◦ H = Hydrogen (H2) ◦ O = Oxygen (O2) ◦ F = Flourine (F2) ◦ Br = Bromine (Br2) ◦ I = Iodine (I2) ◦ N = Nitrogen (N2) ◦ Cl = Chlorine (Cl2)
- 14. Draw the particle diagrams forthe following… Solid Sodium (Na) Liquid Mercury (Hg) Helium Gas (He) Nitrogen Gas (N)
- 15. Draw the particle diagrams forthe following… Solid Sodium (Na) Liquid Mercury (Hg) Helium Gas (He) Nitrogen Gas (N)
- 16. Heating and Cooling Curves Describe the changes in energy as substances are heated and cooled. ◦ Potential Energy – energy of a material as a result of its position in an electrical, magnetic, or gravitational field. ◦ Kinetic Energy – energy available because of the motion of an object.
- 17. Heating Curve
- 18. Heating CurveBoiling Point Boiling (Vaporization) P.E. Increases K.E. ConstantMeltin g MeltingPoint (Fusion) P.E. Increases K.E. Constant
- 19. Heating Curve Summary AB: heating of a solid, one phase present, kinetic energy increases. BC: melting of a solid, two phases present, potential energy increases, kinetic energy constant. CD: heating of a liquid, one phase present, kinetic energy increases. DE: boiling of a liquid, two phases present, potential energy increases, kinetic energy remains constant. EF: heating of a gas, one phase present, kinetic energy increases.
- 20. Cooling Curve
- 21. Cooling Curve Condensation P.E. DecreasesCondensin K.E. Constant g Point Freezing P.E. Decreases Freezin K.E. Constant g Point
- 22. Cooling Curve Summary AB: cooling of a gas (vapor), one phase present, kinetic energy decreases. BC: condensation of a gas (vapor) to liquid, two phases present, potential energy decreases, kinetic energy constant. CD: cooling of a liquid, one phase present, kinetic energy decreases. DE: solidification (freezing) of a liquid, two phases present, potential energy decreases, kinetic energy remains constant. EF: cooling of a solid, one phase present, kinetic energy decreases.
- 23. Heating Curve of H2O
- 24. Cooling Curve of H2O
- 25. Temperature Temperature – a measure of the average kinetic energy of the particles within a substance. ◦ The average kinetic energy depends only upon the temperature of a substance, not on the nature or amount of the material. Ex; A 10g sample of H2O at 50oC has a greater average kinetic energy than a 500g sample of Fe(s) at 20oC.
- 26. Kelvin Scale The SI unit of temperature. A measure of absolute temperature. ◦ 0 K = Absolute Zero K = oC + 273 A change of one degree Celsius is equal to a change of one Kelvin. Celsius to Fahrenheit to Fahrenheit Celsius
- 27. Convert the following into Kelvin 0oC 25oC 37oC 100oC 200oC
- 28. Temperature vs. Heat Heat – a measure of the amount of energy transferred from one substance to another. ◦ Measures in units of calories or joules. ◦ Temperature difference between two things indicates direction of heat flow. Heat flows from an object at higher temperature to the object at a lower temperature until both object are at the same temperature.
- 29. Temperature vs. HeatTemperature measures average kinetic energy of particles within object.Heat measures the amount of energy transferred from one object toanother.
- 30. Measurement of Heat EnergyCan calculate the amount of heat given off orabsorbed during a reaction using... q= mC∆Tq = heat (in joules)m = mass of the substanceC = specific heat capacity of thesubstance∆T = (Temperaturefinal –Temperatureinitial)
- 31. Sample ProblemHow many joules are absorbed when 50.0g of water are heated from 30.2oC to 58.6oC?
- 32. Sample Problem How many joules are absorbed when 50.0g of water are heated from 30.2oC to 58.6oC? q = mC∆T m = 50.0g C = 4.18 J/g•K∆T = 58.6oC - 30.2oC = 28.4oC ~or~∆T = 331.6 K – 303.2 = 28.4
- 33. Sample ProblemHow many joules are absorbed when 50.0g of water are heated from 30.2oC to 58.6oC? q= (50.0g)(4.18J/g•oC)(28.4oC) q = 5936J = 5.94x103J
- 34. Practice When 25.0g of water are cooled from 20oC to 10oC the number of joules of heat released is?
- 35. Practice How many kilojoules of heat energy are absorbed when 100.0g of water are heated from 20oC to 30oC ?
- 36. Complex Problem A 2.70g piece of metal is heated to 98.7oC. It is then added to a beaker containing 150mL of water at 23.5oC. The final temperature of the water is 25.2oC. What is the specific heat of this metal?
- 37. Heat of Fusion ∆Hf The amount of heat needed to convert a mass of a substance from solid to liquid at its melting point. ◦ The heat of fusion of solid water (ice) at 0oC and 1atmosphere of pressure is 334 J/g q = mHf
- 38. Practice Problem How many joules are required to melt 255g of ice at 0oC?
- 39. Practice Problem How many joules are required to melt 255g of ice at 0oC? q = mHf m = 225 g Hf = 334 J/gq = (225g)(334 J/g) = 85,170 J = 85.2kJ
- 40. Heat of Vaporization ∆Hv Boiling, as substance is converted from liquid to solid at its boiling point. ◦ The heat of vaporization of water at 100oC and 1atmosphere of pressure is 2260 J/g q = mHv
- 41. Practice Problem How many joules of energy are required to vaporize 423g of water at 100oC and 1atmosphere of pressure.
- 42. Practice Problem How many joules of energy are required to vaporize 423g of water at 100oC and 1atmosphere of pressure. q = mHv m = 423 g Hv = 2260 J/gq = (423g)(2260J/g) = 955,980 J = 956
- 43. Table T – Heat Formulas
- 44. Behavior of Gases Kinetic Molecular Theory (KMT) ◦ Ideal Gases Have mass but negligible volume. In constant, random, straight-line motion. Not subject to attractive or repulsive forces. Gas Particles collide with each other and the walls of their container where a transfer of energy between particles may occur but there is no net loss of energy. Collisions are said to be perfectly elastic.
- 45. KMT and Pressure Gases exert pressure by colliding with the walls of a container. ◦ The volume occupied by an ideal gas is essentially the volume of its container. ◦ The greater the number of gas particles the higher the pressure. Example; adding to much air to a balloon ◦ Pressure and number of gas particles are directly proportional.
- 46. KMT and Pressure Relationship of Pressure and Volume ◦ Boyle’s Law Pressure and volume vary inversely. As volume decreases pressure increases. Temperature and number of particles held constant. ↑ Volume ↓Volume ↓ Pressure ↑ Pressure
- 47. Boyle’s Law 50 kPa • 6L = 300 kPa•L 100 kPa • 3L = 300 kPa•L 150 kPa • 2L = 300 kPa•L Pressure • Volume = Constant
- 48. Boyle’s Law Can re-write Boyle’s law to solve mathematical problems. P1• V1 = P2• V2 Problem ◦ According to the graph what will be the volume of the gas when the pressure is 520 kPa? P1 = 50 kPa P1• V1 = P2• V2 V1 = 6 L P2 = 520 kPa 50 kPa • 6 L = 520 kPa • V2 V2 = ? L V2 = 0.58 L
- 49. Boyle’s Law Problem The volume occupied by a gas at STP is 250 L. At what pressure (in atmospheres) will the gas occupy 1500 liters, if the temperature and number of particles remain constant? P1• V1 = P2• V2
- 50. KMT and Temperature Relationship of Temperature and Pressure of a gas. ◦ Directly related Increase Temp. = Increase Pressure Relationship of Temperature and Volume of a gas. ◦ Charles’s Law Temperature and constant directly related at constant pressure and volume.
- 51. Charles’s Law
- 52. Charles’s LawPressure and number of particles held constant.Graph is linear.Volume and Temperature (K) are directly proportional to each other.
- 53. KMT and Particle Velocity The greater the temperature the greater the velocity of its particles. Increased Kinetic Energy. Increased Temperature.
- 54. Combined Gas Law The relationship among pressure, temperature, and volume can be mathematically expressed by a single equation… ◦ The Combined Gas Law
- 55. Combined Gas LawExample Problem What volume will a gas occupy if the pressure on 244 cm3 gas at 4.0 atm is increased to 6.0 atm? Assume temperature remains constant.
- 56. Combined Gas LawExample Problem What volume will a gas occupy if the pressure on 244 cm3 gas at 4.0 atm is increased to 6.0 atm? Assume temperature remains constant. 3P1 = 244 cm P2 = ?T1 = Constant T2 = ConstantV1 = 4.0 atm V2 = 6.0 atm
- 57. Combined Gas LawExample Problem What volume will a gas occupy if the pressure on 244 cm3 gas at 4.0 atm is increased to 6.0 atm? Assume temperature remains constant. 3P1 = 244 cm P2 = ?T1 = Constant T2 = ConstantV1 = 4.0 atm V2 = 6.0 atm
- 58. Combined Gas LawExample Problem If 75 cm3 of a gas is at STP, what volume will the gas occupy if the temperature is raised to 75oC and the pressure is increased to 945 torr? 101.3 kPa = 1 atm = 760 torr
- 59. Ideal vs. Real Gases Kinetic Molecular Theory explains the behavior of gases using “ideal” gases as a model. Real gases however, do not always follow this. ◦ Ideal Gases (always behave as predicted; H, He) Negligible volume. No attractive or repulsive forces. Move randomly in straight lines. Perfectly elastic collisions. ◦ Real Gases Attractive forces cannot always be disregarded. Water vapor molecules attract one another to form rain or snow in the atmosphere. Volume of gas particles not negligible under high pressure.
- 60. Avogadro’s Hypothesis Equal volume of gases. ◦ When volume, temperature, and pressure of two or more gases are the same, they contain the SAME number of molecules. Ex; 12 L of N2(g) at STP would contain the same number of molecules as 12 L of O2(g) at STP
- 61. Avogadro’s Hypothesis 22.4 Liters of ANY gas at Standard Temperature and Pressure (STP) contains one mole (6.02 x 1023 atoms or molecules) of that gas. ◦ 22.4 liters of CO2(g) at 0oC and 1atm will have 6.02 x 1023 CO2 molecules. ◦ 22.4 liters of Ne(g) at 0oC and 1atm will have 6.02 x 1023 Ne molecules.
- 62. Separation of Mixtures Filtration Distillation Chromatography
- 63. FiltrationAllowing some particles to pass while trapping others Solids suspended in a liquid. Coffee Filter Immiscible liquids Separatory Funnel Mixtures of solids and gases A/C Filters
- 64. DistilationSeparation by boiling point. Homogeneous Mixtures ◦ Solid dissolved in liquid Purifying salt water Miscible Liquids ◦ Liquids that mix with each other. Gasoline from crude oil.
- 65. ChromatographySeparation by using the differing attractive forces of the components within a mixture to a transport medium. Paper Chromatography Gas Chromatography Column Chomatography
- 66. http://mrmartinschemistryblog.blogspot.com/

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