Properties of Matter

• 1. Matter and Energy

2. Matter Anything that has mass and occupiesspace (volume) Chemistry is the study of matter. Matter is composed of atoms. 3. Classifying Matter 4. Classifying Matter Matter Substances have constant composition andconstant properties throughout a givensample. Cannot be separated by physicalmeans. 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 moresubstances that can be separated byphysical 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 matterfrom one for or another(rearrangement of particles) without achange in chemical properties. Example; Salt-water can easily bephysicall separated back into itscomponents of solid NaCl crystals andH2O by a process called distillation. 7. Properties of Matter Chemical Change a change in whichresults in the formation of differentkinds of matter with changedproperties. Example; Rustin of iron Fe combineswith O2 in the air to from a new materialcalled rust. The Fe and O2 atoms cannotbe 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 atomsand molecules in fixed locations. True solids have a crystallinestructure. 10. Liquid Phase Particles not held together as rigidlyand are able to move past oneanother. Sufficient attractive forces to havedefinite volume. No definite shape, takes the shape ofthe container. 11. Gaseous Phase Minimal attractive forces to hold particlestogether, no definite volume. No definite shape. Spread out indefinitely unless confined toa container where the gas will occupythe volume of the container. Vapor is the gaseousphase of a substancethat is a liquid or solidat normal conditions. 12. Particle Diagrams Draw the particle diagrams for water (H2O) in the three phase of matter.SolidLiquidGaseous WaterWaterWater 13. Diatomic molecules HOFBrINCls 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 assubstances are heated and cooled. Potential Energy energy of a material asa result of its position in anelectrical, magnetic, or gravitational field. Kinetic Energy energy availablebecause of the motion of an object. 17. Heating Curve 18. Heating CurveBoiling Point Boiling (Vaporization) P.E. Increases K.E. ConstantMeltin gMeltingPoint (Fusion)P.E. IncreasesK.E. Constant 19. Heating Curve Summary AB: heating of a solid, one phasepresent, kinetic energy increases. BC: melting of a solid, two phasespresent, potential energyincreases, kinetic energy constant. CD: heating of a liquid, one phasepresent, kinetic energy increases. DE: boiling of a liquid, two phasespresent, potential energyincreases, kinetic energy remainsconstant. EF: heating of a gas, one phasepresent, kinetic energy increases. 20. Cooling Curve 21. Cooling CurveCondensationP.E. DecreasesCondensin K.E. ConstantgPoint Freezing P.E. Decreases Freezin K.E. ConstantgPoint 22. Cooling Curve Summary AB: cooling of a gas (vapor), one phasepresent, kinetic energy decreases. BC: condensation of a gas (vapor) toliquid, two phases present, potentialenergy decreases, kinetic energyconstant. CD: cooling of a liquid, one phasepresent, kinetic energy decreases. DE: solidification (freezing) of aliquid, two phases present, potentialenergy decreases, kinetic energyremains constant. EF: cooling of a solid, one phasepresent, kinetic energy decreases. 23. Heating Curve of H2O 24. Cooling Curve of H2O 25. Temperature Temperature a measure of theaverage kinetic energy of the particleswithin a substance. The average kinetic energy depends onlyupon the temperature of a substance, noton 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 isequal to a change of one Kelvin.Celsius toFahrenheit toFahrenheitCelsius 27. Convert the following into Kelvin 0oC 25oC 37oC 100oC 200oC 28. Temperature vs. Heat Heat a measure of the amount ofenergy transferred from onesubstance to another. Measures in units of calories or joules. Temperature difference between twothings 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= mCTq = heat (in joules)m = mass of the substanceC = specific heat capacity of thesubstanceT = (Temperaturefinal Temperatureinitial) 31. Sample ProblemHow many joules are absorbed when 50.0g of water are heated from 30.2oC to 58.6oC? 32. Sample ProblemHow many joules are absorbed when 50.0g of water are heated from 30.2oC to 58.6oC? q = mCT m = 50.0g C = 4.18 J/gKT = 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/goC)(28.4oC) q = 5936J = 5.94x103J 34. Practice When 25.0g of water are cooled from20oC to 10oC the number of joules ofheat released is? 35. Practice How many kilojoules of heat energyare absorbed when 100.0g of waterare heated from 20oC to 30oC ? 36. Complex Problem A 2.70g piece of metal is heated to 98.7oC. It isthen added to a beaker containing 150mL of waterat 23.5oC. The final temperature of the water is25.2oC. What is the specific heat of this metal? 37. Heat of Fusion Hf The amount of heat needed to converta mass of a substance from solid toliquid at its melting point. The heat of fusion of solid water (ice) at0oC and 1atmosphere of pressure is 334J/g q = mHf 38. Practice Problem How many joules are required to melt255g of ice at 0oC? 39. Practice Problem How many joules are required to melt 255g of ice at 0oC? q = mHfm = 225 gHf = 334 J/gq = (225g)(334 J/g) = 85,170 J = 85.2kJ 40. Heat of Vaporization Hv Boiling, as substance is convertedfrom liquid to solid at its boiling point. The heat of vaporization of water at 100oCand 1atmosphere of pressure is 2260 J/gq = mHv 41. Practice Problem How many joules of energy arerequired to vaporize 423g of water at100oC and 1atmosphere of pressure. 42. Practice Problem How many joules of energy arerequired to vaporize 423g of water at100oC and 1atmosphere of pressure.q = mHvm = 423 gHv = 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 withthe walls of a container. The volume occupied by an ideal gas isessentially the volume of its container. The greater the number of gas particlesthe higher the pressure. Example; adding to much air to a balloon Pressure and number of gas particles aredirectly proportional. 46. KMT and Pressure Relationship of Pressure and Volume Boyles Law Pressure and volume vary inversely. As volume decreases pressure increases. Temperature and number of particles held constant. Volume Volume Pressure Pressure 47. Boyles Law50 kPa 6L = 300kPaL 100 kPa 3L = 300 kPaL150 kPa 2L = 300 kPaLPressure Volume = Constant 48. Boyles Law Can re-write Boyles law to solvemathematical problems. P1 V1 = P2 V2 Problem According to the graph what will be thevolume of the gas when the pressure is520 kPa? P1 = 50 kPaP1 V1 = P2 V2 V1 = 6 L P2 = 520 kPa 50 kPa 6 L = 520 kPa V2 V2 = ? LV2 = 0.58 L 49. Boyles Law Problem The volume occupied by a gas at STP is 250 L. Atwhat pressure (in atmospheres) will the gas occupy1500 liters, if the temperature and number ofparticles remain constant?P1 V1 = P2 V2 50. KMT and Temperature Relationship of Temperature andPressure of a gas. Directly related Increase Temp. = Increase Pressure Relationship of Temperature andVolume of a gas. Charless Law Temperature and constant directly related at constant pressure and volume. 51. Charless Law 52. Charless 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 thegreater the velocity of its particles. Increased Kinetic Energy. Increased Temperature. 54. Combined Gas Law The relationship amongpressure, temperature, and volume can bemathematically expressed by a singleequation The Combined Gas Law 55. Combined Gas LawExample Problem What volume will a gas occupy if thepressure on 244 cm3 gas at 4.0 atm isincreased to 6.0 atm? Assume temperatureremains constant. 56. Combined Gas LawExample Problem What volume will a gas occupy if thepressure on 244 cm3 gas at 4.0 atm isincreased to 6.0 atm? Assume temperatureremains constant.3P1 = 244 cm P2 = ?T1 = Constant T2 = ConstantV1 = 4.0 atmV2 = 6.0 atm 57. Combined Gas LawExample Problem What volume will a gas occupy if thepressure on 244 cm3 gas at 4.0 atm isincreased to 6.0 atm? Assume temperatureremains constant.3P1 = 244 cm P2 = ?T1 = Constant T2 = ConstantV1 = 4.0 atmV2 = 6.0 atm 58. Combined Gas LawExample Problem If 75 cm3 of a gas is at STP, what volumewill the gas occupy if the temperature israised to 75oC and the pressure isincreased to 945 torr?101.3 kPa = 1 atm = 760torr 59. Ideal vs. Real Gases Kinetic Molecular Theory explains thebehavior of gases using ideal gases as amodel. Real gases however, do not alwaysfollow 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. Avogadros Hypothesis Equal volume of gases. When volume, temperature, and pressureof two or more gases are the same, theycontain 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. Avogadros Hypothesis 22.4 Liters of ANY gas at StandardTemperature and Pressure (STP) containsone mole (6.02 x 1023 atoms or molecules)of that gas. 22.4 liters of CO2(g) at 0oC and 1atm willhave 6.02 x 1023 CO2 molecules. 22.4 liters of Ne(g) at 0oC and 1atm willhave 6.02 x 1023 Ne molecules. 62. Separation of Mixtures Filtration Distillation Chromatography 63. FiltrationAllowing some particles topass while trappingothers Solids suspended in aliquid. Coffee Filter Immiscible liquids Separatory Funnel Mixtures of solids andgases A/C Filters 64. DistilationSeparation by boiling point. HomogeneousMixtures Solid dissolved inliquid Purifying salt water Miscible Liquids Liquids that mix witheach other. Gasoline from crude oil. 65. ChromatographySeparation by using thediffering attractive forces ofthe components within amixture to a transportmedium. Paper Chromatography Gas Chromatography ColumnChomatography 66. http://mrmartinschemistryblog.blogspot.com/