matter: properties and changes chapter 3 chapter 15.1, 15.3 chapter 12.4 1
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Matter: Properties and Changes
Chapter 3Chapter 15.1, 15.3
Chapter 12.4
1
Chapter Objectives
Distinguish between physical and chemical changes
Define and classify matter by composition
Define properties of liquids, solids, and gasses
Identify observable characteristics of a chemical reaction
Explain and apply the fundamental law of conservation of mass 2
Vocabulary – Ch. 3.1 – 3.2
Physical property Extensive property Intensive property Chemical property States of matter Solid Liquid Gas Vapor
Physical Change Chemical Change Law of
Conservation of Mass
Phase Change
3
Vocabulary – Chapter 15.1, 15.3
Energy Heat Joule Specific Heat Specific Heat Equation (q = mC ∆T)
Heat of Vaporization (∆Hvap) Heat of Fusion (∆Hfus) Heating Curve
4
Vocabulary – Ch. 12.1, 12.4
Pressure Barometer Atmosphere
Melting Point Vaporization Evaporation Vapor Pressure Boiling Point
Sublimation Freezing point Condensation Deposition Phase diagram Triple Point
5
Ch. 3.1 - What is Matter??
Matter as defined from Ch. 1 is:– Anything that takes up space and has
mass
Mass – measure of the amount of matter in an object.
6
What is a Pure Substance?
Matter that has a uniform and unchanging composition
Table salt is ALWAYS Sodium Chloride
Water is always made of 2 Hydrogen atoms and 1 Oxygen atom
Is seawater a substance?
7
Properties of Matter
Chemistry is the study of matter. Matter is classified according to its
physical and chemical properties. Physical Properties: Can be
observed without changing the identity of substance.
8
Some Physical Properties State at Room temp (liquid, solid or
gas) Melting point Boiling Point Viscosity (resistance to flow) Density Color Odor Brittle/ductile Electrical/thermal conductivity
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Physical Properties
Intensive Properties – Independent of amount of material. They property is the same no matter how much is there.– Example: Density
Extensive Properties – Dependent of the amount of material present. – Example: Length, mass, volume
10
Properties of Matter
Chemical Properties: the ability to combine or change into other substances.– Examples: flammability, oxidation,
rotting
11
States of Matter State of Matter: Its physical form. There are three physical states:
Solid: – Definite shape – Definite volume– Closely packed particles
12
States of MatterLiquid:–particles move past each other
(flow)– definite volume – takes the shape of its container
(indefinite)
13
States of MatterGas: – flows– takes the shape of
its container (indefinite shape)–Fills the container completely. (indefinite volume)
Note: A vapor refers to a gaseous state of a substance that is a solid or liquid at room temperature.
14
Physical or Chemical Property?
Bending of aluminum Salt dissolving in water Magnesium burning in air Baking soda is a white powder Fluorine is a highly reactive element
15
Ch. 3.2 - Changes in Matter
Physical changes are those which alter the substance without altering its composition. – Change of phase one physical state to
another
Melting of ice - composition unchanged, i.e. ice is water in solid form (H2O)
They generally require energy, the ability to absorb or release heat (or work).
16
Phase Changes (Ch. 12.4) What are the phase changes of water?1. Melting – changing of a solid to a liquid
(heat of fusion = ∆Hf)
2. Vaporization – changing from a liquid to a gas (heat of vaporization = ∆Hvap)
3. Sublimation – Changing from a solid to a gas (heat of sublimation = ∆Hsub)
What do these processes have in common?
17
Phase Changes Phase changes in the opposite
direction have names too.1. liquid to a solid: 2. gas to a liquid: 3. gas to a solid: What do these have in common?Answer:
18
UNITS OF ENERGYUNITS OF ENERGY
1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food
“calorie”)But we use the unit called the
JOULE1 cal = 4.184 joules
(exactly)
James Joule1818-1889
19
Heats of Fusion & Vaporization
Heat of Fusion (∆Hfus) – The amount of heat (in joules) needed to melt 1 g of substance.
For ice: 334 J/g q (heat) = ∆Hfus*m (m= mass of
ice/water) Heat of Vaporization (∆Hvap) – The
amount of heat (in joules) needed to vaporize 1 g of substance
For water: 2260 J/g q (heat) = ∆Hvap*m
(m= mass of water/steam)
20
Example Problems
How much heat does it take to melt 20.5 g of ice at 0⁰C?
q = 334 J/g * 20.5 = 6850 J (6.85 kJ)
How much heat is released when 50.0 g of steam at 100 ⁰C condenses to water at 100 ⁰C?
q = - 2260 J/g * 50.0 g = -113,000 J (-113 kJ)
21
Specific Heat Capacity
Specific Heat Capacity – amount of heat (q)
required to raise the temperature of one gram of a substance by 1 degree. C = J (energy gained or lost) mass (g) * Temp Change(⁰C)
22
Heat Capacity Values
SubstanceSpec. Heat (J/g•⁰C)Water 4.184Ethylene glycol 2.39Al uminum 0.897glass 0.84
23
Calculating Heat Gained or lost
The heat, q, gained or lost by a substance can be calculated by knowing the mass of the object, the temperature change, and the heat capacity.
q = mC∆T
24
Calculations involving Heat
Example 1: A 5.00 g piece of aluminum is heated from 25.0⁰C to 99.5⁰C. How many joules of heat did it absorb?
q = m * C * ∆T = 5.00 g * 0.897 J/g*⁰C * 74.5⁰C = 334 J
25
Calculations involving Heat
Example 2: 10.2 g of cooking oil at 25.0 ⁰C is placed in a pan and 3.34 kJ of heat is required to raise the temperature to 196.4 ⁰C. What is the specific heat of the oil?
q = m*C*∆T C = q/(m ∆T) C = 3340 J/(10.2 g * (196.4-25.0) ⁰C) C = 1.91 J/g* ⁰C
26
Calculations involving Heat
Important Points!q (heat) is a positive quantity. The sign
(+ or -) refers to whether the system you’re looking gained it (+) or lost it (-).
From the previous example, the oil would lose 3340 J of heat upon cooling back to 25.0 ⁰C. (-3340 J heat lost)
Specific heat capacity is like a bucket. It is a measure of how much energy an object absorbs before the temperature changes. 27
Heating Curve for WaterHeating Curve for WaterNote that T is constant as ice melts
Note that T is constant as ice melts
28
Heating/Cooling Curve for WaterHeating/Cooling Curve for Water
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Heat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/g
Heat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/g
What quantity of heat is required to melt 500. g of ice (at 0oC) and heat the water to steam at 100oC?
Heat & Changes of StateHeat & Changes of State
+333 J/g +2260 J/g
30
What quantity of energy as heat is required to melt 500. g of ice (at 0⁰C) and heat the water to steam at 100 oC?
1. To melt ice at 0⁰C q = (500. g)(333 J/g) = 1.67 x 105 J
2.To raise water from 0 oC to 100oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x
105 J
3.To vaporize water at 100oC q = (500. g)(2260 J/g) = 1.13 x 106 J
4. Total energy = 1.51 x 106 J = 1510 kJ
Heat & Changes of StateHeat & Changes of State
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Practice problemIf we add 6050 J of heat to 54.2g of ice at -10.0⁰C, what will it be at the end? What temperature will it be? The specific heat of ice is 2.03 J/g*⁰C.
32
• Pressure is the force acting on an object per unit area:
• Gravity exerts a force on the earth’s atmosphere• A column of air 1 m2 in cross section exerts a force of
about 105 N (101,300 N/m2).• 1 Pascal (Pa) = 1 N/m2 . So, 101,300 N/m2 = 101,300 Pa
or 101.3 kPa.• Since we are at the surface of the earth, we ‘feel’ 1
atmosphere of pressure.
FP
A
33
Pressure
Barometer A barometer
measures atmospheric pressure
The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg.
1 atm = 760 mm Hg
1 atm Pressure
760 mm Hg
Vacuum
34
Units of pressure 1 atmosphere (atm) = 760 mm Hg = 760
torr 1 atm = 101,300 Pascals = 101.3 kPa Can make conversion factors from these. What is 724 mm Hg in atm ?
What is 724 mm Hg in kPa
35
Phase ChangesVapor pressure is the pressure exerted by a vapor over a liquid.The vapor pressure increases with increasing temperature.This is why water evaporates even though it’s not 212˚F.
36
Phase Changes
However, when the vapor pressure of the water is the same as the atmospheric pressure the water is … boiling.
37
Phase Diagram
A phase diagram is a graph of pressure vs temperature that shows in which phase a substance exists under different conditions of T & P.
38
Temperature
Solid Liquid
Gas
1 Atm
Pre
ssur
e
39
Phase Diagram for water
Sublimation
Deposition
Melting
Freezing
Boiling
Condensation
Ch. 3.2 - Chemical Properties and Changes
Chemical Properties – the ability to combine or change into other substances.– Example: Water won’t react with
aluminum, but reacts with sodium and potassium (violently) and iron (slowly but surely).
A chemical property always relates to a chemical change – the changing of one or more substances into other substances.
A chemical change is also known as a chemical reaction.
40
Signs of a Chemical Change
- Color change- Odor- Gas- Formation of a precipitate (solid)-Heat
41
Chemical Properties and Changes
Chemical changes are a rearrangement of atoms in the substance.
Chemical changes follow the Law of Conservation of Mass.
This means that when there is a chemical change, matter is neither created or destroyed, just changed in form.
Massreactants = Massproducts42
Conservation of Mass Example Reactants are the substances you
start with.– Started with sugar and sulfuric acid– alcohol and air (O2)
Products are the new substances that are made.
- Carbon, water vapor, and heat
- water & carbon dioxide
43
Conservation of Mass Example Example: 10.00 g of red mercury (II) oxide
powder is placed in an open flask. It is heated until it has fully converted to
liquid mercury and oxygen gas. Written as:
– Mercury oxide mercury + oxygen
The liquid mercury has a mass of 9.26 grams
The mass of oxygen formed in this reaction is 10.0 g (total) – 9.26 g (mercury) = 0.74 g oxygen
44
Practice
57.48 g of sodium reacts with chlorine gas to form 146.10 g of sodium chloride.
Sodium + Chlorine sodium chloride
How much chlorine gas was used in the reaction?
45
Elements
An element is a pure substance that cannot be broken down into simpler substances
Each element has a symbol and they are arranged in a periodic table
46
Elements
Each element has it’s own symbol in the periodic table.
Either a single capital letter: H is for hydrogen
Or two letters, the first is capital the second is ALWAYS small: He is for helium.
Co is different than CO.
47
48
Practice
Name the element or symbol. C Calcium Cl Iodine K Mercury
49
Compounds Compounds are pure substances
that are combinations of elements in a fixed ratio.
Example: Table salt is a combination of sodium and chlorine!NaCl(1:1 ratio)
50
Mixtures
Mixture – Combination of two or more pure substances that where each component retains its own chemical properties.
Heterogeneous Mixture – individual substances remain distinct
Homogeneous Mixture – Constant composition throughout. It has a single phase.
51
Matter
Pure Substance Mixture
Can it be separated by “Physical” methods?
No Yes
Can it be separated by “Chemical” methods?
No
Element
Yes
Compound
Is it uniform?
No
Heterogen.
Homogen.
Yes
52
Separating Mixtures
Substances in mixtures are physically combined, so they can be physically separated into component substances
The separation technique you choose depends on the physical properties of the substances you want to separate
53
Size Differences
Filtration – small liquid molecules will pass through porous barrier, while solid crystals will not (Filter paper)
Screening – solids of one size are retained on screen while smaller solids pass through
54
Difference in Boiling Point Distillation:
based on difference of boiling points of liquids.– Done in refineries– refineries in US
produce 9 Mil bbl/day
gasoline 1.6 Mil bbl/day jet
fuel 4.5 Mil bbl/day fuel
oil 5 Mil bbl/day others
Also pharmaceutical, basic chemicals etc.
55
History of the Periodic Table
Dmitri Mendeleev (1834-1907) first began systematic organization of elements by their chemical behavior and physical properties in 1869.
56
History of the Periodic Table
Based on the mass of each element Mendeleev found that behavior was
periodic – that is properties within a group repeat or are “periodic” with each row or period. Hence the name “Periodic Table”
57
Law of Definite Proportions
All samples of a pure compound contain the same elements in the same proportion by mass (Joseph Proust, 1754-1826 in 1799).
Example: – 10 grams CaBr2
– 500 grams CaBr2 58
59
Example for CaBr2
Element Mass Analysis, g % by Mass
Ca 2.0051
Br 7.9949 10.000 100.000
Ca 100.252
Br 399.748
500.000 100.00
Law of Definite Proportions
18 g of water (H2O) has 2 g of hydrogen and 16 g of oxygen.
44 g of Carbon Dioxide (CO2) has 12 g of carbon and 32 g of oxygen.
28 g Carbon Monoxide (CO) has 12 g of carbon and 16 g of oxygen.
60
Law of Multiple Proportions
When different compounds are formed by a combination of the same elements, different masses of one element combine with the same relative mass of the other element in a ratio of small whole numbers (John Dalton, 1803)
Examples: CO and CO2 H2O and H2O2 CuCl and
CuCl2 61
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