Matter and Energy

A. Introduction: 1.Chemistry The study of matter, its compositions, structures, properties, changes it undergoes, and energy accompanying these changes 2.Matter - Anything that has mass and takes up space –Can be pure substances or mixtures

B. Types of Matter: 1.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

2.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

3.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

4.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

(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

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

B. Separation of Mixtures: 1.Boiling, distillation, evaporation  Every separates homogeneous mixtures (solutions) in which the components have different boiling points  Example: salt and water 2.Filtration  Used when the mixture has different particle sizes

3.Centrifugation  Spinning objects to separate heavier substances from a mixture  Example: blood 4.Dialysis  (Diffusion) – when a mixture moves from an area where it has a high concentration to an area where its concentration is lower

Element / compound / mixtureChemical symbolHow many sample of each Element (diatomic) Element Compound Mixture X2X2 5 X 2 Y XY 8 Y 5 XY XY 2 6 XY 2 X 2 and Y 5 X 2 and 4 Y XY 2 and Y 4 XY 2 and 4 Y Atom X Atom Y SG pg.6 #20 – 22 MC pg.121 Set 3 W/S 1: Types of Matter

A. Phases of Matter: 1.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)

2.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)

3.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

B. Phases Changes: 1.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

2.Change in Phase Melting H 2 O (s) ---> H 2 O (l) Freezing H 2 O (l) ---> H 2 O (s) Evaporations C 2 H 5 OH (l) ---> C 2 H 5 OH (g) CondensationC 2 H 5 OH (g) ---> C 2 H 5 OH (l) SublimationCO 2(s) ---> CO 2(g) DepositionCO 2(g) ---> CO 2(s) Solid to liquid Liquid to solid Liquid to gas Gas to liquid Solid to gas Gas to solid MC pg.123 Set 6

3.Phase changes and energy  A phase change occurs when it has absorbed or released enough 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

(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

SG pg.9 #27 – 31 MC pg.124 Sets 7 & 8

4.Phase changes and temperature (a)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

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

(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 Sets 14 & 15

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

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

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

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

A. Introduction: 1.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

2.Exothermic  A process that releases (emits or loses) heat 3.Endothermic  A process that absorbs (or gains) heat 4.Joules & Calories  unites for measuring heat 5.Calorimeter  Device used for measuring heat during physical and chemical changes SG pg.14 #51 – 53 MC pg.131 Set 16

B. Heat constants and heat equations: 1.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

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 = J SG pg.17 #54 – 56 MC pg.133 Set 18

2.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

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

3.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

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 Sets 21-22

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

A. Introduction: 1.Behavior of gases  Influenced by three key factors  Volume (space)  Pressure  Temperature

2.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)

3.An ideal gas  A theoretical (or assumed) gas that has all properties previously summarized 4.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 Sets 23-25

B. Gas Laws: 1.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 molecules of oxygen in B. SG pg.21 #77 – 80 MC pg.138 Set 26

2.Dalton’s Law of Partial Pressure  The total pressure (P total ) 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  P gas Δ =.2 atm  P gas O =.4 atm  P gas =.5 atm Ptotal = PgasA + PgasB + Pgas C.2 atm +.4 atm +.5 atm = 1.1 atm

Total pressure when gas X is collected over water  P total = P gas X + VP H2O (at temp)  VP H2O 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 = P gas O + VPH 2 O at 45°C  26 kPa = P gas O + 10  16kPa = P gas O

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 N 2 = 1.2 (0.5) 2.0 = 0.3 atm

3.Graham’s Law of Diffusion  a lighter gas will diffuse faster than a heavier gas 4.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

Boyle’s Law (continued)  equation to calculate the new volume of a gas when pressure on the gas is changed atconstant temperature  P 1 V 1 = P 2 V 2 [ 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? P 1 V 1 = P 2 V 2 (0.5 atm)(3 L) = (0.25 atm)(V 2 ) 1.5 = 0.25 V = V 2 SG pg.22 #81 – 82 MC pg.139 Sets 28-29

5.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  V 1 /T 1 = V 2 / T 2

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 ? V 1 /T 1 = V 2 /T 2 (25mL) / (280K) = (75mL) / (T 2 ) 840 K = T 2 SG pg.23 #83 – 84 MC pg.140 Sets 30-31

6.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  P 1 /T 1 = P 2 /T 2

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? P 1 /T 1 = P 2 /T 2 (45 kPa) / (T 1 ) = (50 kPa) / (340 °K) T 1 = 306 ° K SG pg.24 #85 – 86 MC pg.141 Sets 32-33

7.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  P 1 V 1 /T 1 = P 2 V 2 /T 2

Combined Gas Law (continued) Example: A 30 mL sample of H 2 is gas at 1 atm and 200 K. What will be its new volume at 2.0 atm and 600 K P 1 V 1 /T 1 = P 2 V 2 /T 2 (1) (30) / 200 = (V 2 ) (2) / mL = V 2 SG pg.25 #87 – 91 MC pg Sets 34-36

C. Pressure, Volume, and Temperature: 1.Pressure  Pressure of gas is a measure of how much force is put on a confined gas Units: Atmosphere (atm) or Kilopascal (kPa) 1 atm = kPa

2.Volume  Volume of a gas is the space of the container the gas is placed. Units: Milliliters (mL) or centimeters cube (cm 3 ) 1 atm = kPa 1 mL = 1 cm 3

3.Temperature (gas)  a measure of the average kinetic energy of the gas particles  As temperature increases, gas particles move faster, and their average kinetic energy increases. Units: degrees Celsius (°C) or Kelvin (K) °K = °C + 273

D. Standard Temperature and Pressure: STP Reference Table A Standard Temperature 273 °K (or) 0 °C Standard Pressure 1 atm (or) kPa

NOTE: Always use Kelvin temperature in all gas law calculations. Example: Hydrogen gas has a volume of 100 mL at STP. If temperature and pressure are changed to 546 K and 0.5 atm respectively, what will be the new volume of the gas? V 1 = 100 mL V 2 = ? T 1 = 273 K T 2 = 546 K P 1 = 1 atm P 2 = 0.5 atm P1V1T1P1V1T1 = P2V2T2P2V2T2 (1)(100) 273 = (0.5)(V 2 ) mL = V 2 W/S 4: Gasses

A. Introduction: 1.Properties  set of characteristics that can be used to identify and classify matter  Two types of properties of matter are physical and chemical properties.

2.Physical Property  can be observed or measured without changing chemical composition (a) extensive properties  depend on sample size or amount  Example: mass, weight and volume (b) intensive properties  does not depend on sample size or amount  Example: Melting, freezing and boiling points, density, solubility, color, odor, conductivity, luster, and hardness

3.Physical change  a change of a substance from one form to another without changing its chemical composition Examples: Phase change Size change Dissolving NaCl(s) Na + (aq) + Cl – (aq)

4.Chemical Property  a characteristic of a substance that is observed or measured through interaction with other substances. (a)Examples:  it burns, it combusts, it decomposes, it reacts with, it combines with, or, it rusts

4.Chemical Change  a change in composition and properties of one substance to those of other substances.  Chemical reactions are ways by which chemical changes of substances occur SG pg.26 #92 – 96 MC pg.144 Set 37