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 Pure Substance Element Compound Matter Mixture Homogeneous Heterogeneous solution suspension SG pg.5 #10 – 19 MC pg.119-120 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

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

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 Freezing Evaporations Condensation Sublimation Deposition Solid to liquid Liquid to solid Liquid to gas Gas to liquid Solid to gas Gas to solid H 2 O (s) ---> H 2 O (l) H 2 O (l) ---> H 2 O (s) C 2 H 5 OH (l) ---> C 2 H 5 OH (g) C 2 H 5 OH (g) ---> C 2 H 5 OH (l) CO 2(s) ---> CO 2(g) CO 2(g) ---> CO 2(s) 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

(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  SG pg.10 #32 – 40 MC pg.130-131 Sets 14 & 15 Kelvin = ° C + 273 °

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 – Increases Kinetic energy – Increases Potential energy Remains the same

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

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 – Decreases Potential energy Decreases Remains the same

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

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 = 146.3 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.134-135 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.136-137 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

  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 ?

   P total = P gas X + VP H2O (at temp) VP H2O is the vapor pressure of water at the given water temperature. (Table H) 26 kPa = P gas 26 kPa = P gas 16kPa = P gas O O + VPH 2 O at 45 ° C O + 10

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 2 6.0 = 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) / 600 45 mL = V 2 SG pg.25 #87 – 91 MC pg.142-143 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 = 101.3 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 = 101.3 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) 101.3

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 V T P 1 P 2 1 2 1 T 2 = 100 mL = ?

= 273 K = = = 546 K 1 atm 0.5 atm P 1 V 1 T 1 (1)(100) 273 = = P 2 V 2 T 2 400 mL = V 2 (0.5)(V 2 ) 546 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