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Chemistry

Chemistry. Matter and Energy. I. Types of Matter. Introduction: Chemistry The study of matter, its compositions, structures, properties , changes it undergoes, and energy accompanying these changes Matter - Anything that has mass and takes up space Can be pure substances or mixtures.

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Chemistry

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  1. Chemistry Matter and Energy

  2. I. Types of Matter • Introduction: • ChemistryThe study of matter, its compositions, structures, properties, changes it undergoes, and energy accompanying these changes • Matter -Anything that has mass and takes up space • Can be pure substances or mixtures

  3. Types of Matter: • Pure Substance • Every sample has a definite and fixed composition • Has a unique set of properties • Each sample is the same (homogeneous) • Elements or compounds SG pg.1 #1 & 2

  4. Elements • Pure substance composed of identical atoms with the same atomic number • Cannot be decomposed into simpler substances by physical or chemical methods SG pg.1 #3 & 4

  5. Compounds • Pure substances that are composed of two or more different elements chemically combined • Can be decomposed into simpler substances by chemical methods • Properties of a compound are different from those of the elements that make up the compound • Law of definite composition Elements in a compound are combined in a definite ratio by mass SG pg.2 #5 & 7

  6. Mixtures • Composed of two or more substances that are physically combined • Composition may vary from one sample to another • Can be separated by physical methods • Retains properties of the individual components SG pg.2 #8 & 9

  7. (a) Homogeneous Mixtures • Uniformly and evenly mixed throughout • Samples have definite and fixed composition • Aqueous solutions are homogeneous mixtures that are made with water • Example: salt water NaCl(aq) • Solution - its components are all in the same phase (dissolve) • Suspension - its components are in different phases

  8. (b) Heterogeneous Mixtures • Not uniformly nor evenly mixed throughout • Samples have different and varying composition Matter Mixture Pure Substance Homogeneous Element Heterogeneous Compound solution suspension SG pg.5 #10 – 19 MC pg.119-120 Sets 1 & 2

  9. Separation of Mixtures: • Boiling, distillation, evaporation • Every separates homogeneous mixtures (solutions) in which the components have different boiling points • Example: salt and water • Filtration • Used when the mixture has different particle sizes

  10. Centrifugation • Spinning objects to separate heavier substances from a mixture • Example: blood • Dialysis • (Diffusion) – when a mixture moves from an area where it has a high concentration to an area where its concentration is lower

  11. Atom X Atom Y X2 5 X2 Element (diatomic) Element Y 8 Y Compound 5 XY XY XY2 Compound 6 XY2 Mixture 5 X2 and 4 Y X2 and Y 4 XY2 and 4 Y XY2 and Y Mixture SG pg.6 #20 – 22 MC pg.121 Set 3 W/S 1: Types of Matter

  12. II. Phases of Matter, Energy and Temperature • Phases of Matter: • Solid • Definite volume and definite shape • Particles arranged orderly in a “regular geometric pattern” • Particles with strong attractive force to one another • Particles vibrating around fixed points • Particles that cannot be easily compressed (incompressible)

  13. Liquid • definite volume, but no definite shape (it takes the shape of its container) • particles flow over each other • particles that cannot be easily compressed (incompressible)

  14. Gas • No definite shape and no definite volume (it takes volume and shape of its container) • Particles that are most random • Particles move fast and freely • Particles have very weak attractive force to each other • Particles that can be easily compressed (compressible) SG pg.7 #23 – 26 MC pg.122 Sets 4 & 5

  15. Phases Changes: • Phase change • A physical change • A substances changes form (state) without changing its chemical composition • Depends of temperature and pressure • Water can be found in nature in all three phases

  16. Change in Phase Solid to liquid Liquid to solid Liquid to gas Gas to liquid Solid to gas Gas to solid MC pg.123 Set 6

  17. Phase changes and energy • A phase change occurs when it has absorbed or releasedenough heat energy to rearrange its particles (atoms or molecules) from one form to another • Latent Heat – amount of heat needed to be absorbed or released during a change in phase • Units for heat energy: calories or joules

  18. (a) Endothermic • describes a process that absorbs heat energy • includes fusion, evaporation, sublimation (b) Exothermic • describes a process that releases heat energy • includes freezing, condensation and deposition

  19. SG pg.9 #27 – 31 MC pg.124 Sets 7 & 8

  20. Phase changes and temperature • Temperature • A measure of the average kinetic energy of particles in matter (b) Kinetic energy • Due to movements of particles in matter • The higher the temperature the greater its kinetic energy

  21. As temperature increases, the average kinetic energy increases. Average kinetic energy Temperature

  22. (c) Thermometer • Instrument used to measure temperature • Degree Celsius (°C) and Kelvin (°K) • Phase change of water is used as reference points • 0°C and 273°K = freezing point • 100°C and 373°K = boiling point • Kelvin = °C + 273° SG pg.10 #32 – 40 MC pg.130-131 Sets 14 & 15

  23. Heating curve - Shows changes of a substance starting from the solid phase liquid / gas solid / liquid Heat is being absorbed / endothermic Time (minutes) While the substance is a solid, liquid or gas, the Temperature – Kinetic energy – Potential energy - Increases Increases Remains the same

  24. liquid / gas solid / liquid Time (minutes) When two phases are present (solid/liquid or liquid/gas), the Temperature – Kinetic energy – Potential energy - Remains the same Remains the same Increases

  25. Cooling curve - Shows changes of a substance starting from the gas phase liquid / gas solid / liquid Heat is being released / exothermic Time (minutes) While the substance is a gas, liquid or solid, the Temperature – Kinetic energy – Potential energy - Decreases Decreases Remains the same

  26. solid / liquid liquid / gas Time (minutes) When two phases are present (solid/liquid or liquid/gas), the Temperature – Kinetic energy – Potential energy - Remains the same Remains the same Decreases SG pg.13 #41 – 50 MC pg.125-129 Sets 9 – 13 W/S 2: Phases of Matter

  27. III. Heat and heat calculations • Introduction: • Heat • A form of energy that can flow (or transfer) from one object to another • Heat flows from an area of higher temperature to an area of lower temperature until equilibrium is reached • Energy is either absorbed or released during a chemical or physical changes

  28. Exothermic • A process that releases (emits or loses) heat • Endothermic • A process that absorbs (or gains) heat • Joules & Calories • unites for measuring heat • Calorimeter • Device used for measuring heat during physical and chemical changes SG pg.14 #51 – 53 MC pg.131 Set 16

  29. Heat constants and heat equations: • Specific Heat capacity (C) • The amount of heat needed to change the temperature of one gram sample of the substance by adding one degree Celsius • Depends on the substances • Specific heat for water is 4.18 J/g ∙ °C MC pg.132 Set 17

  30. Calculating Heat gained or released (Table T) • Determining the amount of heat absorbed or released by a substance • Heat (q) = m × C × ΔT Example: How much heat is released by a 7 gram sample of water to change its temperature from 15°C to 10°C? q = m × C × ΔT = 7g × 4.18 J/g ∙ °C × 5 °C = 146.3 J SG pg.17 #54 – 56 MC pg.133 Set 18

  31. Heat of fusion (Table B) • Amount of heat needed to melt or freeze one gram of the substance at a constant temperature • Heat of fusion for water is 334 J/g • 334 J/g – absorbed to melt one gram of ice • 334 J/g – released to freeze one gram of water

  32. Calculating Heat of fusion (Table T) • Determining the amount of heat absorbed or released by freezing or melting • Heat (q) = m × Hf Example: What is the number of joules needed to melt a 16g sample of ice to water at °C? q = m × Hf q = 16g 334 J/g = 5344 J SG pg.17 #57 – 59 MC pg.133 Set 19

  33. Heat of vaporization (Table B) • Amount of heat needed to vaporize (evaporate) or condense one gram of the substance at a constant temperature • Heat of fusion for water is 2260 J/g • 2260 J/g – absorbed needed to vaporize one gram of ice • 2260 J/g – released to condense one gram of water

  34. Calculating Heat of vaporization (Table T) • Determining the amount of heat absorbed or released by vaporizing or condensing • Heat (q) = m × Hv Example:Liquid ammonia has a heat of vaporization of 1.35 KJ/g. How many kilojoules of heat are needed to evaporate a 5 gram sample of ammonia at its boiling point? q = m × Hv q = 5g x 1.35 KJ/g = 6.75 KJ SG pg.18 #60 – 62 MC pg.134-135 Sets 21-22

  35. Review of Equations (Table B and Table T) If two temperatures are given, change temperature from q = mCΔT To melt/freeze, changes from liquid to solid, at 0°C q = mHf To boil/condense/evaporate, at 100°C q = mHv SG pg.18 #63 – 64 W/S 3: Heat & Heat Calculations

  36. IV. Gas characteristics and Gas laws • Introduction: • Behavior of gases • Influenced by three key factors • Volume (space) • Pressure • Temperature

  37. Kinetic molecular theory • Gas is composed of individual particles • Distances between particles are far apart • Gas particles are in continuous, random, straight-line motion • When two particles collide energy is transferred from one particle to another • Particles of gasses have no attraction to each other • Individual gas particle has no volume (negligible or insignificant)

  38. An ideal gas • A theoretical (or assumed) gas that has all properties previously summarized • An real gas • A gas that actually exists • Oxygen, carbon dioxide, hydrogen, helium, etc. • Has particles that attract each other • Does have volume • Real gases with small molecular mass behave most like an ideal gas (H and He) SG pg.20 #65 – 72 MC pg.136-137 Sets 23-25

  39. Gas Laws: • Avogadro’s Law • Under the same conditions of temperature and pressure equal volume of different gasses contains equal number of molecules (particles) If the number of helium gas molecules are counted in Container A and the number of oxygen gas molecules are counted in Container B, you will find that the number of molecules of helium in A is the same as the number of moleculesof oxygen in B. SG pg.21 #77 – 80 MC pg.138 Set 26

  40. Dalton’s Law of Partial Pressure • The total pressure (Ptotal) of a gas mixture is the sum of all the partial pressures.) • Partial Pressure (P) is a pressure exerted by individual gas in a gas mixture • Total Pressure from Partial Pressures: • A three gas mixture • PgasΔ= .2 atm • PgasO = .4 atm • Pgas = .5 atm Ptotal = PgasA + PgasB + Pgas C .2 atm + .4 atm + .5 atm = 1.1 atm

  41. Total pressure when gas X is collected over water • Ptotal= Pgas X + VPH2O (at temp) • VPH2O is the vapor pressure of water at the given water temperature. (Table H) Example: Oxygen gas is collected over water at 45oC in a test tube. If the total pressure of the gas mixture in the test tube is 26 kPa, what is the partial pressure of the oxygen gas ? • 26 kPa = Pgas O + VPH2O at 45°C • 26 kPa = Pgas O + 10 • 16kPa = PgasO

  42. Partial Pressure of gas X from mole fraction: • P gas X = Moles of gas X (P total) Total moles Example: A gas mixture contains 0.8 moles of O2 and 1.2 moles of N2. If the total pressure of the mixture is 0 5 atm, what is the partial pressure of N2 in this mixture? P gas N2 =1.2 (0.5) 2.0 = 0.3 atm

  43. Graham’s Law of Diffusion • a lighter gas will diffuse faster than a heavier gas • Boyle’s Law • At constant temperature, volume of a gas is inversely proportional to the pressure on the gas. • As pressure increases, volume (space) of the gas decreases by the same factor MC pg.138 Set 27

  44. Boyle’s Law (continued) • equation to calculate the new volume of a gas when pressure on the gas is changed atconstanttemperature • P1V1 = P2V2 [ P= pressure, V = volume, T = Kelvin temperature, 1 = initial condition, 2 = new condition] Example: At constant temperature, what is the new of volume of a 3 L sample of O gas if its pressure is changed from 0.5 atm to 0.25 atm? P1V1 = P2V2 (0.5 atm)(3 L) = (0.25 atm)(V2) 1.5 = 0.25 V2 6.0 = V2 SG pg.22 #81 – 82 MC pg.139 Sets 28-29

  45. Charle's Law • At constant pressure, the volume of a gas is directly proportional to the Kelvin temperature of the gas. • as temperature increases, volume (space) increases by the same factor • equation to calculate the new volume of a gas when temperature of the gas is changed at constant pressure • V1/T1 = V2 / T 2

  46. Charle's Law (continued) Example: The volume of a confined gas is 25 ml at 280 K. At what temperature would the gas volume be 75 ml if the pressure is held constant? V1/T1= V2/T2 (25mL) / (280K) = (75mL) / (T2) 840 K = T2 SG pg.23 #83 – 84 MC pg.140 Sets 30-31

  47. Gay-Lussac’s Law • At constant volume, pressure of a gas is directly proportional to the Kelvin temperature of the gas. • As temperature increases, pressure increases by the same factor • Equation to calculate the new pressure of a gas when temperature of the gas is changed at constant volume • P1/T1= P2/T2

  48. Gay-Lussac’s Law (continued) Example: At constant volume, pressure on a gas changes from 45 kPa to 50 kPa when the temperature of the gas is changed to 340°K.What was the initial temperature of the gas? P1/T1 = P2/T2 (45 kPa) / (T1) = (50 kPa) / (340 °K) T1 = 306 °K SG pg.24 #85 – 86 MC pg.141 Sets 32-33

  49. Combined Gas Law • describes a gas behavior when all three factors (volume, pressure, and temperature) of the gas are changing: • the only constant is the mass of the • equation to solve any problem related to the above three gas • P1 V1/T1 = P2V2/T2

  50. Combined Gas Law (continued) Example: A 30 mL sample of H2 is gas at 1 atm and 200 K. What will be its new volume at 2.0 atm and 600 K P1 V1/T1 = P2V2/T2 (1) (30) / 200 = (V2) (2) / 600 45 mL = V2 SG pg.25 #87 – 91 MC pg.142-143 Sets 34-36

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