Transcript Chapter 1

Chapter 3
The Properties of Matter and
Energy
Setting the Stage – Matter and
Energy
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Understanding the properties of matter
allows us to draw conclusions about
compounds and elements found on other
planets
Energy, which has no mass, is a second
component of the universe, in addition to
matter
We can quantify energy and how it
interacts with matter
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Setting a Goal - Part A
The Properties of Matter

You will learn how a sample of matter
can be described by its properties
and how they can be quantitatively
expressed
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Objective for Section 3-1
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List and define several properties of
matter and distinguish them as
physical or chemical
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3-1 The Physical and Chemical
Properties of Matter
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Properties describe the particular
characteristics of a substance
Pure substances have definite
composition and definite, unchanging
properties
Physical properties - can be observed
without changing the substance
Chemical properties - require that the
substance changes into another
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The Physical States of Matter
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The three fundamental physical states are
solid, liquid and gas
 solids have a definite shape and volume
 liquids have a definite volume but not a
definite shape
 Gases have neither a definite volume
nor shape
A substance exists in a particular physical
state under defined conditions
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The Physical States of Matter
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Changes in Physical State
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Melting point (reverse: freezing point)
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temperature at which a substance changes
from solid to liquid (reverse: liquid to solid)
Boiling point (reverse: condensation point
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temperature at which a substance changes
from liquid to gas (reverse: gas to liquid)
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Types of Physical Properties
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Intensive properties – those properties
that depend on the type or identity (but
not the amount) of material present
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Examples: color, density, melting point
Extensive properties – those properties
that depend on the amount of material
present
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Examples: mass, volume
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Chemical Changes and Properties
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Chemical properties – tendency of a pure
substance to undergo chemical changes
Sometimes quite difficult to determine
Some examples of chemical changes are
burning (as opposed to boiling) and color
changes
Law of the Conservation of Mass - matter
is neither created nor destroyed in
chemical reactions
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Chemical Changes
Fast reaction:
“Thermite
Reaction”
Medium fast
reaction: Zn
and HCl
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Concepts of Chemistry 9e
Slow reaction:
rusting of iron
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Objective for Section 3-2
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Perform calculations involving the
density of liquids and solids
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3-2 Density – A Physical Property
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Density is the ratio of the mass of a
substance to the volume of that
mass – it is usually measured in
g/mL for solids and liquids; g/L for
gases
Specific gravity is the ratio of the
mass of a substance to the mass of
an equal volume of water
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Density as a Conversion Factor
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Density can also be used to convert
between mass and volume
Typically g → mL or mL → g
The density of table salt is 2.16 g/mL. What is
the volume in mL occupied by 485 g of table salt?
Unit map
SOLUTION
g
485 g x
mL
1mL
Conversion factor is
1 mL
2.16 g
=
225 mL
2.16 g
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Determination of Density
This example shows just one way of determining
the density of a solid substance
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Objective for Section 3-3
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Describe the differences in
properties between a pure substance
and a mixture
Perform calculations involving
percent as applied to mixtures
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3-3 The Properties of Mixtures
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A mixture is an aggregation of two or more pure
substances
Can be separated by physical means (filtration,
distillation, crystallization, chromatography)
Have chemical and physical properties that are
different from the substances that make them up
The percentages by mass of the components of a
mixture can be varied continuously
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Types of Mixtures
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Heterogeneous mixture - nonuniform mixture
containing two or more phases with definite
boundaries between the phases (e.g. ice and
water; sand and water)
Homogeneous mixture - same throughout and
contains only one phase (substances are mixed
at the atomic or molecular level) (e.g. air;
aqueous solution of glucose)
A phase is one physical state with distinct
boundaries and uniform properties
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Separation of Mixtures
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A heterogeneous mixture of a solid and
liquid can be separated by filtration
A heterogeneous mixture of two liquids
can be separated using a separating
funnel
Several substances in solution can be
separated by chromatography
A solution of a solid in a liquid or of two
liquids can be separated by distillation
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Basic Distillation Kit
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Distillation is Important in Industry
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Solutions
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A type of homogeneous mixture
Usually involves a liquid phase, but can be
solid-solid, gas-gas, solid-liquid, etc.
The pure substances can be in different
phases but form a homogeneous mixture
(table salt or glucose and water, for
example)
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Alloys
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Alloys are homogeneous mixtures of
metallic elements existing in one phase
Important solid solutions of two or more
metals include:
brass (copper and zinc)
dental fillings (silver and mercury)
stainless steel (iron, chromium and nickel)
solder (tin and lead)
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Analysis
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Solder is an alloy made from 60.0 % tin
and 40.0 % lead. What mass of lead is
present in 72 g of solder?
40.0 g Pb
72 g solder x
100 g solder
=
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29 g Pb
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Setting a Goal - Part B
The Properties of Energy
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You will be able to qualitatively and
quantitatively describe processes in
terms of the forms and types of
energy associated with them
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Objectives for Section 3-4
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Distinguish among the forms and
types of energy
Define the terms endothermic and
exothermic, providing several
examples of each type of process
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3-4 The Forms and Types of
Energy
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Energy is the capacity to do work
There are many forms of energy
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heat
light
chemical (stored energy)
electrical energy
mechanical
nuclear
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Energy
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Law of the Conservation of Energy - energy can
neither be created nor destroyed, but can only
transformed from one form to another
The transformation from one type to another may
not be efficient (the efficiency of transforming
chemical energy to electricity energy (Figure 3-8)
is only about 35% efficient). The other 65% is lost
as heat but the total amount of energy is constant
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Energy Flow
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Exothermic reactions - produce energy
(release energy to the surroundings)
Endothermic reactions - require energy
input (store energy)
Potential energy is that energy available
due to position or composition
Kinetic energy is that energy resulting
from motion
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Objective for Section 3-5
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Perform calculations involving the
specific heat of a substance, and use
it to identify a substance
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3-5 Energy Measurement and
Specific Heat
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Specific heat - the amount of heat required to
raise the temperature of one gram of a substance
by one degree Celsius (or Kelvin)
Reflects how some substances heat up faster
than others
Specific heat =
amount of heat energy
mass x t (oC)
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Temperature and Specific Heat
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Recall that we measure temperature in °C
or K
Energy units
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calorie (cal) - amount of heat required to raise
the temperature of one gram of water from 14.5
°C to 15.5 °C
*
 joule: 1 cal = 4.184 J (exactly)
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Nutritional ‘Calorie’ is actually 1000 cal
(indicated as 1 C)
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Heat Flow
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When two substances at different
temperatures are put in contact with each
other, or mixed, heat flows spontaneously
from the substance at higher temperature
to the substance at lower temperature
This heat flow continues until the
temperatures are the same
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Worked Example 1 (Ch. 3)
Label the following as
physical (P) or chemical (C) changes
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Brewing beer
Distilling rice wine
Making mist from
“dri ice”
Dyeing clothes
Frying chicken
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C
P
P
C
C
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Worked Example 2 (Ch. 3)
Classify the following as pure substances,
(PS) homogeneous (Ho M) or
heterogeneous (He M) mixtures
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Pewter
 Ho M
Cloudy apple juice
 He M
Sugar (for coffee)
 PS
Clear apple juice
 Ho M
Distilled water
 PS
Tap water
 Ho M
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Worked Example 3 (Ch. 3)
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An unknown metal with a mass of 23.6 g is
placed in a graduated cylinder that had an
initial water volume of 22.0 mL. After the
sample is submerged in the water, the
level of water in the cylinder was observed
to be 24.1 mL. Determine the identity of
the metal
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Worked Example 3 (Ch. 3)
Solution
23.6 g occupies 2.1 mL, hence the density of
the metal is 23.6 g/2.1 mL = 11.2 g/mL
The metal is likely to be Pb (density = 11.3
g/mL)
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Worked Example 4 (Ch. 3)
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A gas can (438 g) contains 33.5 mL of
kerosene. If the mass of the can plus the
kerosene is 465 g, what is the density of
kerosene?
Solution: The 33.5 mL of kerosene weighs
465 g – 438 g = 27.0 g, hence its density is
27.0/33.5 = 0.806 g/mL
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Worked Example 5 (Ch. 3)
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At room temperature, the specific heat of
benzene is 1.060 J/goC. If a 30.-g sample
of benzene releases 450 J of energy, what
is the change in temperature?
Solution: Energy (heat) transferred = m x
sp ht x Change in temp (T)
450 (J) = 30 (g) x 1.060 (J/goC) x T (oC)
Change in temp = 14 oC
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Worked Example 6 (Ch. 3)
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A 440 g piece of metal at 100.0 oC is placed
in 258 g of water initially at 25.0 oC. If the
final temperature is 36.5 oC, what is the
specific heat of the metal?
Solution:
Heat lost (metal) = heat gained (water)
-m(m) x sp ht (m) x t (m) = m(w) x sp ht (w) x t (w)
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Worked Example 6 (Ch. 3)…contd.
Hence,
-440 g x sp ht(m) x (36.5 - 100.0 ) oC
= 258 g x 4.184 J/g oC x (36.5 – 25.0) oC
Therefore,
sp ht (m) = 258 g x 4.184 J/g oC x 11.5 oC
440 g x 63.5 oC
= 0.444 J /g oC (iron)
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Objective for Special Topic – Units
of Energy
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Distinguish amongst different units
of energy and carry out
interconversions between them
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Special Topic – Units of Energy
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The SI (m-kg-s or MKS system) unit of
energy is the joule (J) (= kg m2 s-2)
The cgs (cm-g-s) system unit of energy is
the erg (= g cm2 s-2)
The conversion factor is 1 J = 107 ergs
The joule is a much bigger unit of energy
than the erg
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An Energy Unit for Atomic
Systems
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The potential energy associated with a proton
and an electron separated by 1 Å (= 10-10 m), a
typical atomic distance, is only –2.307 x 10-18 J
(or –2.307 x 10-11 ergs)
Hence a more convenient unit is needed to
express energies relating to atoms.
This unit is the electron-volt (eV)
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The Electron-Volt Unit
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The kinetic energy (KE) gained by an
electron falling through an electrical
potential difference (voltage) of 1 volt (V)
is defined as 1 eV
1 V is defined as the voltage that gives 1
joule of energy to 1 Coulomb (C) of
charge; that is, 1 V = 1 J C-1
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The Electron-Volt Unit
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KE = electronic charge x potential difference
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Hence, 1 eV = 1.602 x 10-19 C x 1 V
= 1.602 x 10-19 C x 1 J C-1
= 1.602 x 10-19 J (1 J = 6.173 x 1018 eV)
Therefore, the potential energy associated with
a proton and an electron separated by 1A is
18
_ 2.307 x 10-18 J x 6.173 x 10 eV
_ 14.40 eV
=
1J
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The Electron-Volt Unit
With this value, the Coulomb equation for the potential energy
associated with two particles of charge q1 and q2
PE =
kq1q2
r
q1 and q2 are in coulombs;
k is a proportionality constant;
r is the distance between q1 and q2
in meters. PE is in J
can be reduced to a much more convenient 'engineering formula',
PE =
14.40 (eVA) q1q2
PE is in eV
r(A)
Here, q1 and q2 are merely the unit electronic charges on the
particles (e.g. -1 for an electron, +1 for a proton, +2 for a He
nucleus, etc)
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