Chapter 3

Chapter 3Chapter 3

The Properties of Matter and The Properties of Matter and EnergyEnergy

Malone and Dolter - Basic Concepts of Chemistry 9e

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Setting the Stage – Matter and Energy

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 AThe 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

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

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 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

Melting point (reverse: freezing point) temperature at which a substance changes

from solid to liquid (reverse: liquid to solid) Boiling point (reverse: condensation point

temperature at which a substance changes from liquid to gas (reverse: gas to liquid)


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Types of Physical Properties

Intensive properties – those properties that depend on the type or identity (but not the amount) of material present Examples: color, density, melting point

Extensive properties – those properties that depend on the amount of material present Examples: mass, volume


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Chemical Changes and Properties

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 fastreaction: Znand HCl

Slow reaction:rusting of iron


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Perform calculations involving the density of liquids and solids


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3-2 Density – A Physical Property

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

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 isthe volume in mL occupied by 485 g of table salt?

Unit map g mL Conversion factor is1 mL

2.16 g

SOLUTION 485 g x 1mL

2.16 g= 225 mL


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Determination of Density

Mass = 52.11 g Initialvolume= 12.5 mL

Finalvolume= 31.8 mL

The 52.11 g of substance occupies 31.8 - 12.5 = 19.3 mL

Therefore, its density is52.11 g

19.3 mL= 2.70 g/mL

This example shows just one way of determining the density of a solid substance


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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

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

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

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

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

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

Solder is an alloy made from 60.0 % tin and 40.0 % lead. What mass of lead is present in 72 g of solder?

72 g solder x40.0 g Pb

100 g solder= 29 g Pb


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Setting a Goal - Part BThe Properties of Energy

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

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

Energy is the capacity to do work There are many forms of energy

heat light chemical (stored energy) electrical energy mechanical nuclear


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Energy

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

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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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 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

Recall that we measure temperature in °C or K

Energy units 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)

Nutritional ‘Calorie’ is actually 1000 cal (indicated as 1 C)


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Heat Flow

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)

Brewing beer Distilling rice wine Making mist from

“dri ice” Dyeing clothes Frying chicken

C P P

C C

Label the following as physical (P) or chemical (C) changes


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Pewter Cloudy apple juice Sugar (for coffee) Clear apple juice Distilled water Tap water

Ho M He M PS Ho M PS Ho M

Classify the following as pure substances,(PS) homogeneous (Ho M) or heterogeneous (He M) mixtures


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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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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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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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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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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

Distinguish amongst different units of energy and carry out interconversions between them


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Special Topic – Units of Energy

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

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

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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KE = electronic charge x potential difference

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 witha proton and an electron separated by 1A is

x_ 2.307 x 10-18 J1 J

6.173 x 1018 eV= 14.40 eV_


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With this value, the Coulomb equation for the potential energyassociated 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 q2in meters. PE is in J

can be reduced to a much more convenient 'engineering formula',

PE = 14.40 (eVA) q1q2

r(A)

Here, q1 and q2 are merely the unit electronic charges on theparticles (e.g. -1 for an electron, +1 for a proton, +2 for a Henucleus, etc)

PE is in eV

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