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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
solution suspension
SG pg.5 #10 – 19MC 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
Element / compound /
mixtureChemical symbol
How many sample of each
Element (diatomic)
Element
Compound
Compound
Mixture
Mixture
X2 5 X2
Y
XY
8 Y
5 XY
XY2 6 XY2
X2 and Y 5 X2 and 4 Y
XY2 and Y 4 XY2 and 4 Y
Atom X
Atom Y
SG pg.6 #20 – 22MC 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 – 26MC 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 H2O(s) ---> H2O(l)
Freezing H2O(l) ---> H2O(s)
Evaporations
C2H5OH(l) ---> C2H5OH(g)
Condensation
C2H5OH(g) ---> C2H5OH(l)
Sublimation CO2(s) ---> CO2(g)
Deposition CO2(g) ---> CO2(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 – 31MC 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
Kelvin = °C + 273°
SG pg.10 #32 – 40MC pg.130-131 Sets 14 & 15
While the substance is a solid, liquid or gas, the Temperature –Kinetic energy – Potential energy -
Increases
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
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
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
Remains the same
Decreases
solid / liquid
liquid / gas
Time (minutes)
SG pg.13 #41 – 50MC pg.125-129 Sets 9 – 13W/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
2. Calorimeter Device used for measuring heat during physical
and chemical changes
SG pg.14 #51 – 53MC 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 – 56MC 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 – 59MC 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 – 62MC 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 – 64W/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 – 72MC 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 moleculesof oxygen in B.
SG pg.21 #77 – 80MC pg.138 Set 26
2. 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
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 = Pgas 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 N2 = 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
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 = V2SG pg.22 #81 – 82MC 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
V1/T1 = V2 / 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?
V1/T1 = V2/T2
(25mL) / (280K) = (75mL) / (T2)
840 K = T2
SG pg.23 #83 – 84MC 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
P1/T1 = P2/T2
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 – 86MC 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
P1 V1/T1 = P2 V2/T2
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 = P2 V2/T2
(1) (30) / 200 = (V2) (2) / 600
45 mL = V2
SG pg.25 #87 – 91MC 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 (cm3)
1 atm = 101.3 kPa
1 mL = 1 cm3
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?
V1= 100 mL
V2 = ?
T1 = 273 K
T2 = 546 K
P1 = 1 atm
P2 = 0.5 atm
P1V1
T1
=P2V2
T2
(1)(100)273
=(0.5)(V2)
546
400 mL = V2
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
2. does not depend on sample size or amount
3. 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 – 96MC pg.144 Set 37