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Properties of Matter Chapter 4 Properties of Matter Chapter 4
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Chapter 4 - Properties of Matter
4.1 Properties of Substances
4.2 Physical Changes
4.3 Chemical Changes
4.4 Conservation of Mass
4.5 Energy
4.6 Heat: Quantitative Measurement
4.7 Energy in Chemical Changes
4.8 Conservation of Energy
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A property is a characteristic of a substance. Each substance has a set of properties that
are characteristic of that substance and give it a unique identity.
Can be classified as either physical or chemical
Properties of a Substance
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The inherent characteristics of a substance that are determined without changing its composition.
Examples: taste color & odor physical state melting point boiling point density
Properties of a SubstancePhysical Properties
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Physical Properties
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Chemical PropertiesChemical Properties
Describe the ability of a substance to form new substances, either by reaction with other substances or by decomposition.
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It will not burn in oxygen.
It will support the combustion of certain other substances.
It can be used as a bleaching agent.
It can be used as a water disinfectant.
It can combine with sodium to form sodium chloride.
Chemical Properties of Chlorine
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Physical ChangesPhysical ChangesPhysical ChangesPhysical Changes
tearing of paper change of ice into water change of water into steam heating platinum wire
Changes in physical properties (such as size shape and density) or changes in the state of matter without an accompanying change in composition.
Examples:
• No new substances are formed.
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Chemical ChangesChemical ChangesChemical ChangesChemical Changes
In a chemical change new substances are formed that have different properties and composition from the original material.
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Heating a copper wire in a Bunsen burner causes the copper to lose its original appearance and become a black material.
Formation of Copper(II) Oxide
Heating a copper wire in a Bunsen burner causes the copper to lose its original appearance and become a black material.
The black material is a new substance called copper(II) oxide.
Copper is 100% copper by mass.
Copper (II) oxide is: 79.94% copper by mass 20.1% oxygen by mass.
The formation of copper(II) oxide from copper and oxygen is a chemical change. The copper (II) oxide is a new substance with properties that are different from copper.
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Formation of Copper(II) Oxide
Copper(II) oxide is made up of Cu2+ and O2-
4.2
Neither Cu nor O2 contains Cu2+ or O2-A chemical change has occurred.
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Water is decomposed into hydrogen and oxygen by passing electricity through it.
Decomposition of Water
The composition and physical appearance of hydrogen and oxygen are different from water.The hydrogen explodes with a pop upon the addition of a burning splint.The oxygen causes the flame of a burning splint to intensify.
They are both colorless gases.But the burning splint is extinguished when placed into the water sample.
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Water decomposes into hydrogen and oxygen when electrolyzed.
reactant productsyields
Chemical EquationsChemical Equations
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Water decomposes into hydrogen and oxygen when electrolyzed.
reactant yields
2H2O 2H2 O2
products
Chemical symbols can be used to express chemical reactions
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Copper plus oxygen yields copper(II) oxide.
yield productreactants
heat
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Copper plus oxygen yields copper(II) oxide.
yield productreactants
heat
2Cu O2 2Cu2O
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No change is observed in the total mass of the substances involved in a chemical change.
Conservation of MassConservation of MassConservation of MassConservation of Mass
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sodium + sulfur sodium sulfide
46.0 g 32.1 g 78.1 g
78.1 g product
mass productsmass products
78.1 g reactant →
mass reactantsmass reactants ==
Conservation of MassConservation of MassConservation of MassConservation of Mass
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Energy is the capacity to do work
EnergyEnergyEnergyEnergy
Potential EnergyEnergy that an object possesses due to its relative position.
Kinetic Energy
Energy matter possesses due to its motion.
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Types of Energy
mechanicalchemicalelectricalheatnuclearradiant
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increasing potential energy
50 ft
20 ft
The potential energy of the ball increases with increasing height.
increasing potential energy
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The heat released when gasoline burns is associated with a decrease in its chemical potential energy.
The new substances formed by burning have less chemical potential energy than the gasoline and oxygen.
• Gasoline is a source of chemical potential energy.
Potential EnergyStored energy
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Moving bodies possess kinetic energy. The flag waving in the
wind.
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Moving bodies possess kinetic energy. A bouncing ball. The running man.
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The SI unit for heat energy is the joule (pronounced “jool”).
Another unit is the calorie.
4.184 J = 1 cal
(exactly) 4.184 Joules = 1 calorie
This amount of heat energy will raise the temperature of 1 gram of water 1oC.
Units of Heat Units of Heat EnergyEnergyUnits of Heat Units of Heat EnergyEnergy
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A form of energy associated with small particles of matter.
A measure of the intensity of heat, or of how hot or cold a system is.
An Example of the Difference Between Heat and Temperature
Heat vs. TemperatureHeat vs. Temperature
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Twice as much heat energy is required to raise the temperature of 200 g of water 10oC as compared to 100 g of water.
200 g water
20oC
A
100 g water
20oC
B
100 g water
30oC
200 g water
30oC
heat beakers 4184 J 8368 Jtemperaturerises 10oC
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The specific heat of a substance is the quantity of heat required to change the temperature of 1 g of that substance by 1oC.
Specific HeatSpecific Heat
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The units of specific heat in joules are:
o
Joulesgram Celcius
o
Jg C
The units of specific heat in calories are:
o
caloriesgram Celcius
o
calg C
Units of Specific HeatUnits of Specific Heat
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The relation of mass, specific heat, temperature change (Δt), and quantity of heat lost or gained is expressed by the general equation:
Δt = heatmass of substance)(specific heat
of substance)(
General Equation - Specific HeatGeneral Equation - Specific Heat
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o
1638 J125 g x 27.6 C
Calculate the specific heat of a solid in J/goC and in cal/ goC if 1638 J raise the temperature of 125 g of the solid from 25.0oC to 52.6oC.
(mass of substance)(specific heat of substance)Δt = heat
(g)(specific heat of substance)Δt = heatheat
specific heat = g x Δt
heat = 1638 J
mass = 125 g
Δt = 52.6oC – 25.0oC = 27.6oC
specific heat = o
0.475 J=
g C
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Calculate the specific heat of a solid in J/goC and in cal/ goC if 1638 J raise the temperature of 125 g of the solid from 25.0oC to 52.6oC.
o
0.114 cal=
g Co
0.475 Jg C
1.000 cal4.184 J
specific heat =
Convert joules to calories using 1.000 cal/4.184 J
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A sample of a metal with a mass of 212 g is heated to 125.0oC and then dropped into 375 g of water at 240.0oC. If the final temperature of the water is 34.2oC, what is the specific heat of the metal?
When the metal enters the water, it begins to cool, losing heat to the water. At the same time, the temperature of the water rises. This process continues until the temperature of the metal and the temperature of the water are equal, at which point (34.2oC) no net flow of heat occurs.
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A sample of a metal with a mass of 212 g is heated to 125.0oC and then dropped into 375 g of water at 240.0oC. If the final temperature of the water is 34.2oC, what is the specific heat of the metal?
Calculate the heat gained by the water.
Calculate the final temperature of the metal.
Calculate the specific heat of the metal.
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A sample of a metal with a mass of 212 g is heated to 125.0oC and then dropped into 375 g of water at 240.0oC. If the final temperature of the water is 34.2oC, what is the specific heat of the metal?
Δt = 34.2oC – 24.0oC = 10.2oCtemperature rise of the water
Heat Gained by the Water
o(10.2 C) = (375 )g o
4.184 Jg C
heat gained by the water
= 41.60 x 10 J
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A sample of a metal with a mass of 212 g is heated to 125.0oC and then dropped into 375 g of water at 240.0oC. If the final temperature of the water is 34.2oC, what is the specific heat of the metal?
Δt = 125.0oC – 34.2oC = 90.8oCtemperature drop of the metal
Once the metal is dropped into the water, its temperature will drop until it reaches the same temperature as the water (34.2oC).
Heat Lost by the Metal
heat lost by the metal
heat gained by the water
= = 41.60 x 10 J
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A sample of a metal with a mass of 212 g is heated to 125.0oC and then dropped into 375 g of water at 240.0oC. If the final temperature of the water is 34.2oC, what is the specific heat of the metal?
heatspecific heat =
mass x Δ t
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o
1.60 x 10 J(212g)(90.8 C)
o
0.831 Jg C)
specific heatof the metal
=
The heat lost or gained by the system is given by:
(mass) (specific heat) (Δt) = energy change
rearrange
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In all chemical changes, matter either absorbs or releases energy.
Energy in Chemical ChangesEnergy in Chemical ChangesEnergy in Chemical ChangesEnergy in Chemical Changes
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Energy Release From Chemical Sources
Type of Energy Energy Source
Electrical Storage batteries
Light A lightstick. Fuel combustion.
Heat and Light Combustion of fuels.
BodyChemical changes occurring within body cells.
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Chemical Changes Caused byAbsorption of Energy
Type of Energy Chemical Change
ElectricalElectroplating of metals. Decomposition of water into hydrogen and oxygen
Light Photosynthesis in green plants.
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An energy transformation occurswhenever a chemical change occurs.
If energy is absorbed during a chemical change, the products will have more chemical potential energy than the reactants.
• If energy is given off in a chemical change, the products will have less chemical potential energy than the reactants.
Conservation of EnergyConservation of EnergyConservation of EnergyConservation of Energy
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H2 + O2 have higher potential energy than H2O
energy is given offenergy is absorbed
Electrolysis of Water Burning of Hydrogen in Air
higher potential energy lower potential energy
Conservation of EnergyConservation of EnergyConservation of EnergyConservation of Energy
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Law of Conservation of Energy
Energy can be neither created nor destroyed, though it can be transformed from one form of energy to another form of energy.
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Concepts
1. Physical Properties
2. Distinguish Chemical from Physical Properties
3. Classify changes – Chemical or Physical
4. Kinetic vs. Potential Energy
5. Law of Conservation of Mass
6. Law of Conservation of Energy
7. Heat vs. Temperature
8. Use equation:
(mass) (specific heat) (Δt) = heat