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Copyright 2011 Pearson Education, Inc.

Composition of MatterAtoms and Molecules

Scientific Method

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Copyright 2011 Pearson Education, Inc.3

Structure Determines Properties


1. composed of one carbon

atom and one oxygen atom

2. colorless, odorless gas3. burns with a blue flame

4. binds to hemoglobin

carbon monoxide

1. composed of one carbon

atom and two oxygen atoms

2. colorless, odorless gas3. incombustible

4. does not bind to hemoglobin

carbon dioxide

The properties of matter are determined by the

atoms and molecules that compose it


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Atoms and Molecules

Atomsare submicroscopic particles

are the fundamental building blocks of ordinary matter

Molecules

are two or more atoms attached together in a specificgeometrical arrangementattachments are called bonds

attachments come in different strengths

come in different shapes and patterns

Chemistryis the science that seeks to understandthe behavior of matter by studying the behavior ofatoms and molecules

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The Scientific Approach to Knowledge

Philosophers try to understand the universe byreasoning and thinking about idealbehavior

Scientists try to understand the universethrough empirical knowledge gained throughobservation and experiment



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Gathering Empirical Knowledge

Observation

Tro: Chemistry: A Molecular Approach

Some observations are descriptions of thecharacteristics or behavior of nature

qualitative

The soda pop is a liquid with a brown color and a

sweet taste. Bubbles are seen floating up through it.

Some observations compare a characteristic to

a standard numerical scalequantitativeA 240 mL serving of soda pop contains 27 g of sugar.



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From Observation to Understanding


Hypothesisa tentative interpretation orexplanation for an observation

The sweet taste of soda pop is due to the

presence of sugar.

A good hypothesis is one that can be testedto be proved wrong!

falsifiable

one test may invalidate your hypothesis




Testing Ideas

Ideas in science are tested with experiments

An experimentis a set of highly controlled

procedures designed to test whether an ideaabout nature is valid

The experiment generates observations thatwill either validate or invalidate the idea




From Specific to General Observations


A scientific lawis a statement thatsummarizes all past observations and predicts

future observations

Law of Conservation of Mass

In a chemical

reaction matter is neither created nor destroyed.

A scientific law allows you to predict futureobservations

so you can test the law with experiments

Unlike state laws, you cannot choose to violatea scientific law!



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11/122Copyright 2011 Pearson Education, Inc.Tro: Chemistry: A Molecular Approach 11

Scientific Method

Careful noting and

recording of natural

phenomena

Procedure

designed totest an idea Tentative explanation of asingle or small number of

observations

General explanation of

natural phenomena

Generally observed

occurence in nature




Relationships Between Pieces of the

Scientific Method



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Classification of MatterStates of Matter

Physical and Chemical PropertiesPhysical and Chemical Changes




Classification of Matter

Matteris anything that occupies space andhas mass

We can classify matter based on its stateand its composition

whether its solid, liquid, orgas

its basic components




Classifying Matter

by Physical State

Matter can be classified as solid, liquid, or gasbased on the characteristics it exhibits



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Solids

The particles in a solid are packedclose together and are fixed in position

though they may vibrate

The close packing of the particlesresults in solids being incompressible

The inability of the particles to movearound results in solids retaining their

shape and volume when placed in anew container, and prevents the solid

from flowing

16Tro: Chemistry: A Molecular Approach, 2/e



Crystalline Solids

Some solids have theirparticles arranged in

patterns with long-range

repeating order we callthese crystalline solids

salt

diamonds

sugar




Amorphous Solids

Some solids have theirparticles randomly

distributed without any

long-range pattern wecall these amorphous

solids

plastic

glass

charcoal




Liquids

The particles in a liquid areclosely packed, but they havesome ability to move around

The close packing results inliquids being incompressible

The ability of the particles tomove allows liquids to take the

shape of their container and toflow however, they dont have

enough freedom to escape or

expand to fill the container




Gases

In the gas state, the particleshave freedom of motion and arenot held together

The particles are constantly flyingaround, bumping into each otherand the container

In the gas state, there is a lot of

empty space between theparticles

on average




Gases

Because there is a lot ofempty space, the particlescan be squeezed closertogether therefore gases

are compressible Because the particles are

not held in close contactand are moving freely,

gases expand to fill andtake the shape of theircontainer, and will flow



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Copyright 2011 Pearson Education, Inc.Tro: Chemistry: A Molecular Approach 22

Classifying Matter by Composition

Another way to classify matter is to examine itscomposition

Composition includestypes of particlesarrangement of the particles

attractions and attachments between the particles



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by Composition Matter whose composition does not change from

one sample to another is called a pure substancemade of a single type of atom or molecule

because the composition of a pure substance is always

the same, all samples have the same characteristics

Matter whose composition may vary from onesample to another is called a mixture two or more types of atoms or molecules combined in

variable proportionsbecause composition varies, different samples have

different characteristics



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by Composition

1. made of one type of

particle2. all samples show

the same intensive

properties

1. made of multiple

types of particles2. samples may show

different intensive

properties



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Classification of Pure Substances

Elements Pure substances that cannot be

decomposed into simpler substances by

chemical reactions are called elementsdecomposed = broken down

basic building blocks of matter

composed of single type of atomthough those atoms may or may not be combined

into molecules



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Compounds

Pure substances that can be decomposed arecalled compounds

chemical combinations of elements

composed of molecules that contain two or more

different kinds of atoms

all molecules of a compound are identical, so all

samples of a compound behave the same way

Most natural pure substances are compounds



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1. made of onetype of atom

(some

elements

found as multi-atom

molecules in

nature)

2. combine

together to

make

compounds

1. made of one

type of

molecule, or

an array of

ions2. units contain

two or more

different kinds

of atoms



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Classification of Mixtures

Homogeneous mixturesare mixtures that haveuniform composition throughout

every piece of a sample has identical characteristics,

though another sample with the same components

may have different characteristics

atoms or molecules mixed uniformly

Heterogeneous mixturesare mixtures that do

not have uniform composition throughout regions within the sample can have different

characteristics

atoms or molecules not mixed uniformly



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Classification of Mixtures

1. made ofmultiple

substances,

but appears to

be onesubstance

2. all portions of

an individual

sample have

the samecomposition

and

properties

1. made ofmultiple

substances,

whose

presence canbe seen

2. portions of a

sample have

different

compositionand

properties



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Changes in Matter

Changes that alter the state or appearance ofthe matter without altering the composition are

called physical changes

Changes that alter the composition of thematter are called chemical changes

during the chemical change, the atoms that are

present rearrange into new molecules, but all of theoriginal atoms are still present



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Physical Changes in Matter

The boiling ofwater is a

physical change.

The water

molecules are

separated from

each other, but

their structureand composition

do not change.



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Chemical Changes in Matter

The rusting of iron isa chemical change.

The iron atoms in

the nail combine

with oxygen atoms

from O2in the air to

make a new

substance, rust, witha different

composition.



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Properties of Matter

Physical propertiesare the characteristics ofmatter that can be changed without changingits composition

characteristics that are directly observable

Chemical propertiesare the characteristicsthat determine how the composition of matter

changes as a result of contact with other matteror the influence of energy

characteristics that describe the behavior of matter



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CO2(s)

CO2(g)

Dry Ice

Subliming of dry ice

Tro: Chemistry: A Molecular Approach 35

Common Physical Changes

Processes that causechanges in the matterthat do not change its

composition

State changesboiling / condensing

melting / freezing

subliming

Dissolving

Dissolving of sugarC12H22O11(s)

C12H22O11(aq)



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Common Chemical Changes

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)

Processes that causechanges in the matterthat change its

composition Rusting Burning Dyes fading or changing

color



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Energy



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Energy Changes in Matter

Changes in matter, both physical and chemical, resultin the matter either gaining or releasing energy

Energyis the capacity to do work

Work is the action of a force applied across a distancea force is a push or a pull on an object

electrostatic force is the push or pull on objects that have

an electrical charge



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Energy of Matter

All matter possesses energy

Energy is classified as either kineticorpotential

Energy can be converted from one formto another

When matter undergoes a chemical orphysical change, the amount of energy in

the matter changes as well



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Energy of Matter Potential

Potential energyis energy that is storedin the matter

due to the composition of the matter and its

position relative to other things

chemical potential energy arises from

electrostatic attractive forces between atoms,

molecules, and subatomic particles



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Conversion of Energy

You can interconvert kinetic energy andpotential energy

Whatever process you do that convertsenergy from one type or form to another,

the total amount of energy remains the

same

Law of Conservation of Energy



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Spontaneous Processes

Materials that possess high

potential energy are lessstable

Processes in nature tend tooccur on their own when the

result is material with lowertotal potential energyprocesses that result in

materials with higher totalpotential energy can occur, butgenerally will not happenwithout input of energy from anoutside source



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Changes in Energy

If a process results in the system having lesspotential energy at the end than it had at the

beginning, the lostpotential energy was

converted into kinetic energy, which is releasedto the environment

During the conversion of form, energy that is

released can be harnessed to do work



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Potential to Kinetic Energy



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Standard Units of

Measure



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The Standard Units

Scientists have agreed on a set of internationalstandard units for comparing all ourmeasurements called the SI unitsSystme International = International System



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Length Measure of the two-dimensional distance an object covers

often need to measure lengths that are very long (distancesbetween stars) or very short (distances between atoms)

SI unit = meterabout 3.37 inches longer than a yard

1 meter = distance traveled by light in a specific period of time Commonly use centimeters (cm)1 m = 100 cm

1 cm = 0.01 m = 10 mm

1 inch = 2.54 cm (exactly)



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Mass Measure of the amount of matter present in

an objectweight measures the gravitational pull on anobject, which depends on its mass

SI unit = kilogram (kg)

about 2 lbs. 3 oz. Commonly measure mass in grams (g) ormilligrams (mg)1 kg = 2.2046 pounds, 1 lb. = 453.59 g

1 kg = 1000 g = 10

3

g1 g = 1000 mg = 103mg

1 g = 0.001 kg = 103kg

1 mg = 0.001 g = 103 g



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Time

Measure of the duration of an event

SI units = second (s)

1 s is defined as the period of time ittakes for a specific number of radiation

events of a specific transition from

cesium133


T t


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Copyright 2011 Pearson Education, Inc.Tro: Chemistry: A Molecular Approach

Temperature

Measure of the average amount of kinetic energy

caused by motion of the particleshigher temperature = larger average kinetic energy

Heat flows from the matter that has high thermalenergy into matter that has low thermal energy until

they reach the same temperature

heat flows from hot object to cold

heat is exchanged through molecular collisions between

the two materials

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( F 32)C

1.8

K C 273.15


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Temperature Scales

Fahrenheit scale, F used in the U.S.

Celsius scale, C used in all other countries

Kelvin scale, K absolute scale

no negative numbers

directly proportional toaverage amount of kinetic

energy 0 K = absolute zero



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Fahrenheit vs. Celsius

A Celsius degree is 1.8 times larger than aFahrenheit degree

The standard used for 0 on the Fahrenheitscale is a lower temperature than the

standard used for 0 on the Celsius scale



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Kelvin vs. Celsius

The size of a degreeon the Kelvin scale isthe same as on the Celsius scale

though technically, we dont call the divisions on

the Kelvin scale degrees; we call them kelvins!so 1 kelvin is 1.8 times larger than 1F

The 0 standard on the Kelvin scale is a muchlower temperature than on the Celsius scale


Example 1 2: Convert 40 00 C into K and F


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Example 1.2: Convert 40.00 C into K and F

Substitute and compute

Solve the equation for thequantity you want to find

40.00 C

F

Given:

Find:

Equation:

Find the equation thatrelates the given quantity to

the quantity you want to find

K = C + 273.15

K = 40.00 + 273.15

K = 313.15 K

Because the equation issolved for the quantity you

want to find, substitute and

compute

40.00 C

K

K = C + 273.15

Given:

Find:

Equation:

Find the equation thatrelates the given quantity to


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PracticeConvert 0.0F into Kelvin


Practice Convert 0 0F into Kelvin


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PracticeConvert 0.0 F into Kelvin

Because kelvin temperatures are

always positive and generally between

250 and 300, the answer makes sense

Check: Check

255.37 K = 255 KRound: Sig. figs. andround

Solution: Follow theconcept plan

to solvethe

problem

Concept

Plan:

Equations:

Strategize

0.0 F

Kelvin

Given:

Find:

Sortinformation

F C K



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Related Units in the SI System

All units in the SI system are related to thestandard unit by a power of 10

The power of 10 is indicated by a prefixmultiplier

The prefix multipliers are always the same,regardless of the standard unit

Report measurements with a unit that is closeto the size of the quantity being measured


Common Prefix Multipliers in the


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Common Prefix Multipliers in the

SI System


V l


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Volume Derived unit

any length unit cubed

Measure of the amount of spaceoccupied

SI unit = cubic meter (m3) Commonly measure solid volume in

cubic centimeters (cm3

)1 m3= 106cm3

1 cm3= 106m3 = 0.000 001 m3

Commonly measure liquid or gasvolume in milliliters (mL)1 L is slightly larger than 1 quart

1 L = 1 dm3= 1000 mL = 103mL

1 mL = 0.001 L = 103L

1 mL = 1 cm3



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Common Units and Their Equivalents

Length

1 kilometer (km) = 0.6214 mile (mi)

1 meter (m) = 39.37 inches (in.)

1 meter (m) = 1.094 yards (yd)

1 foot (ft) = 30.48 centimeters (cm)

1 inch (in.) = 2.54 centimeters (cm)

exactly



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Common Units and Their Equivalents

Volume

1 liter (L) = 1000 milliliters (mL)

1 liter (L) = 1000 cubic centimeters (cm3

)1 liter (L) = 1.057 quarts (qt)

1 U.S. gallon (gal) = 3.785 liters (L)

Mass1 kilogram (km) = 2.205 pounds (lb)

1 pound (lb) = 453.59 grams (g)

1 ounce (oz) = 28.35 grams (g)


Practice which of the following units would


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Practice which of the following units wouldbe best used for measuring the diameter of a

quarter?

a) kilometer

b) meter

c) centimeter

d) micrometer

e) megameters

a) kilometer

b) meter

c) centimeter

d) micrometer

e) megameters



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Density



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Intensive and Extensive Properties

Extensive propertiesare properties whose

value depends on the quantity of matterextensive properties cannot be used to identify

what typeof matter something is

if you are given a large glass containing 100 g of a

clear, colorless liquid and a small glass containing25 g of a clear, colorless liquid, are both liquids thesame stuff?

Intensive propertiesare properties whose

value is independent of the quantity of matter intensive properties are often used to identify the

type of matter

samples with identical intensive properties are

usually the same materialTro: Chemistry: A Molecular Approach 65Tro: Chemistry: A Molecular Approach, 2/e

M & V l


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Mass & Volume Two main physical properties of matter

Mass and volume are extensive properties Even though mass and volume are individual

properties, for a given type of matter they arerelated to each other!

Volume vs. Mass of Brass

y= 8.38x

020406080

100120140160

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0Volume, cm3

M

ass,g



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Density

Densityis the ratio of mass to volume is an intensive property

Solids = g/cm3

1 cm3= 1 mL

Liquids = g/mL

Gases = g/L Volume of a solid can be determined by water

displacement Archimedes principle

Density : solids > liquids >>> gasesexcept ice is less dense than liquid water!



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Density

For equal volumes, denserobject has larger mass

For equal masses, denserobject has smaller volume

Heating an object generally

causes it to expand,therefore the density

changes with temperature


Example 1.3: Decide if a ring with a mass of 3.15 g


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that displaces 0.233 cm3of water is platinum

Density of platinum =

21.4 g/cm3

therefore not platinum

Compare to accepted valueof the intensive property

Solve the equation for thequantity you want to find,

check the units are correct,

then substitute and compute

mass = 3.15 g

volume = 0.233 cm3

density, g/cm3

Given:

Find:

Equation: Find the equation that

relates the given quantity to


69Tro: Chemistry: A Molecular Approach

Write down the givenquantities and the quantity

you want to find


Calculating Density


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Calculating Density

What is the density of a brass sample if 100.0 g

added to a cylinder of water causes the waterlevel to rise from 25.0 mL to 36.9 mL?


Practice What is the density of the brasssample?


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sample?

units and number make senseCheck:Check

8.4033 g/cm3 = 8.40 g/cm3Round:Sig. figs. and

round

Solution:

V = 36.925.0

= 11.9 mL

= 11.9 cm3

Solvethe

equation for

the unknown

variable

Concept

Plan:

Equation:

Strategize

mass = 100 g

vol displ: 25.0 36.9 mL

d, g/cm3

Given:

Find:

Sort

information

m, V d



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Measurement

and Significant Figures



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What Is a Measurement?

Quantitative observation

Comparison to an

agreed standard Every measurement has

a number and a unit



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A Measurement

The unit tells you what standard you arecomparing your object to

The number tells you1. what multiple of the standard the object

measures

2. the uncertainty in the measurement

Scientific measurements are reported so thatevery digit written is certain, except the last

one, which is estimated



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Estimating the Last Digit

For instruments marked with a scale, you getthe last digit by estimating between the

marks

if possible

Mentally divide the space into ten equalspaces, then estimate how many spaces

over the indicator the mark is



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Significant Figures

The non-place-holding digits ina reported measurement arecalled significant figures

some zeros in a written number

are only there to help you locatethe decimal point

Significant figures tell us therange of values to expect for

repeated measurements the more significant figures there

are in a measurement, the smaller

the range of values is

12.3 cm

has 3sig. figs.

and its range is

12.2 to 12.4 cm

12.30 cm

has 4sig. figs.

and its range is12.29 to 12.31 cm



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Counting Significant Figures

1. All non-zero digits are significant 1.5 has 2 sig. figs.

2. Interior zeros are significant

1.05 has 3 sig. figs.

3. Leading zeros are NOTsignificant 0.001050 has 4 sig. figs.

1.050 x 103



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Counting Significant Figures

4. Trailing zeros may or may not be significanta) Trailing zeros after a decimal point are significant 1.050 has 4 sig. figs.

b) Trailing zeros before a decimal point are significant if

the decimal point is written 150.0 has 4 sig. figs.

c) Zeros at the end of a number without a written

decimal point are ambiguous and should be avoided

by using scientific notation if 150 has 2 sig. figs. then 1.5 x 102

but if 150 has 3 sig. figs. then 1.50 x 102



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Significant Figures and Exact Numbers

A number whose value is known with completecertainty is exactfrom counting individual objects

from definitions

1 cm is exactly equal to 0.01 mfrom integer values in equations

in the equation for the radius of a circle, the 2 is exact

Exact numbers have an unlimited number ofsignificant figures


Example 1 5: Determining the Number of


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Example 1.5: Determining the Number of

Significant Figures in a Number

How many significant figures are in each of the following?

0.04450 m

5.0003 km

10 dm = 1 m

1.000 105s

0.00002 mm

10,000 m

4 sig. figs.; the digits 4 and 5, and the trailing 0

5 sig. figs.; the digits 5 and 3, and the interior 0s

infinite number of sig. figs., exact numbers

4 sig. figs.; the digit 1, and the trailing 0s

1 sig. figs.; the digit 2, not the leading 0s

Ambiguous, generally assume 1 sig. fig.


Practice Determine the number of significantfigures the expected range of precision and


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figures, the expected range of precision, and

indicate the last significant figure

0.00120

120.

12.00

1.20 x 103


Practice determine the number of significantfigures the expected range of precision and


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figures, the expected range of precision, and

indicate the last significant figure

0.00120 3 sig. figs. 0.00119 to 0.00121

120. 3 sig. figs. 119 to 121

12.00 4 sig. figs. 11.99 to 12.01

1.20x 103 3 sig. figs. 1190 to 1210


Multiplication and Division with


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Multiplication and Division with

Significant Figures

When multiplying or dividing measurementswith significant figures, the result has the same

number of significant figures as the

measurement with the lowest number ofsignificant figures

5.02 89.665 0.10 = 45.0118 = 45

3 sig. figs. 5 sig. figs. 2 sig. figs. 2 sig. figs.5.892 6.10 = 0.96590 = 0.966

4 sig. figs. 3 sig. figs. 3 sig. figs.



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Rounding


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Rounding When rounding to the correct number of significant

figures, if the number after the place of the last significant

figure isa) 0 to 4, round down

drop all digits after the last sig. fig. and leave the last

sig. fig. alone

add insignificant zeros to keep the value if necessary

b) 5 to 9, round up

drop all digits after the last sig. fig. and increase the

last sig. fig. by one

add insignificant zeros to keep the value if necessary

To avoid accumulating extra error from rounding, roundonly at the end, keeping track of the last sig. fig. for

intermediate calculationsTro: Chemistry: A Molecular Approach, 2/e

R di


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Rounding

Rounding to 2 significant figures 2.34 rounds to 2.3because the 3 is where the last sig. fig. will be

and the number after it is 4 or less

2.37 rounds to 2.4because the 3 is where the last sig. fig. will be

and the number after it is 5 or greater

2.349865 rounds to 2.3because the 3 is where the last sig. fig. will be



R di


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Rounding

Rounding to 2 significant figures 0.0234 rounds to 0.023 or 2.3 102because the 3 is where the last sig. fig. will be and

the number after it is 4 or less

0.0237 rounds to 0.024 or 2.4 102because the 3 is where the last sig. fig. will be and

the number after it is 5 or greater

0.02349865 rounds to 0.023 or 2.3 102because the 3 is where the last sig. fig. will be and

the number after it is 4 or less


R di


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Rounding

Rounding to 2 significant figures 234 rounds to 230 or 2.3 102because the 3 is where the last sig. fig. will be


237 rounds to 240 or 2.4 102because the 3 is where the last sig. fig. will be

and the number after it is 5 or greater

234.9865 rounds to 230 or 2.3 102because the 3 is where the last sig. fig. will be



Both Multiplication/Division and


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Both Multiplication/Division and

Addition/Subtraction

with Significant Figures When doing different kinds of operations with

measurements with significant figures, do

whatever is in parentheses first, evaluate thesignificant figures in the intermediate answer,

then do the remaining steps

3.489 (5.67 2.3) =

2 dp 1 dp3.489 3.37 = 12

4 sf 1 dp & 2 sf 2 sf


Example 1.6: Perform the Following Calculations


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Example 1.6: Perform the Following Calculations

to the Correct Number of Significant Figures


Example 1.6 Perform the Following Calculations


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Example 1.6 Perform the Following Calculations

to the Correct Number of Significant Figures



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Precision

and Accuracy


Uncertainty in Measured Numbers


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Uncertainty in Measured Numbers Uncertainty comes from limitations of the instruments used for

comparison, the experimental design, the experimenter, andnatures random behavior

To understand how reliable a measurement is, we need tounderstand the limitations of the measurement

Accuracyis an indication of how close a measurement comesto the actualvalue of the quantity

Precisionis an indication of how close repeatedmeasurements are to each other

how reproducible a measurement is


Precision


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Precision

Imprecision in measurements is caused by

random errorserrors that result from random fluctuations

no specific cause, therefore cannot be corrected

We determine the precision of a set ofmeasurements by evaluating how far they are

from the actual value and each other

Even though every measurement has somerandom error, with enough measurementsthese errors should average out


Accuracy


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y Inaccuracy in measurement caused by

systematic errors

errors caused by limitations in the instruments ortechniques or experimental design

can be reduced by using more accurateinstruments, or better technique or experimental

design We determine the accuracy of a measurement

by evaluating how far it is from the actual value

Systematic errors do not average out with

repeated measurements because theyconsistently cause the measurement to beeither too high or too low


Accuracy vs. Precision


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Looking at the graph of the results shows that Student A is

neither accurate nor precise, Student B is inaccurate, but is

precise, and Student C is both accurate and precise.

y

Suppose three students are asked to

determine the mass of an object whoseknown mass is 10.00 g

The results they report are as follows



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Solving

ChemicalProblems

Equations &Dimensional Analysis



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Units

Always write every number with itsassociated unit

Always include units in your calculations

you can do the same kind of operations onunits as you can on numbers

cm cm = cm2

cm + cm = cm

cm cm = 1

using units as a guide to problem solving is

called dimens ional analys is



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Problem Solving and

Di i l A l i


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Dimensional Analysis

Arrange conversion factors so the starting unit cancels

arrange conversion factors so the starting unit is on

the bottom of the first conversion factor

May string conversion factors

so you do not need to know every relationship, aslong as you can find something else the starting and

desired units are related to



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Systematic Approach to Problem Solving


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y pp g

Sortthe information from the problem identify the given quantity and unit, the quantity and unit you

want to find, any relationships implied in the problem

Design a strategyto solve the problem devise a conceptual plan

sometimes may want to work backward

each step involves a conversion factor or equation

Apply the steps in the conceptual planto solvetheproblem check that units cancel properly

multiply terms across the top and divide by each bottom term

Checkthe answer double-check the set-up to ensure the unit at the end is the oneyou wished to find

check to see that the size of the number is reasonablebecause centimeters are smaller than inches, converting inches to

centimeters should result in a larger numberTro: Chemistry: A Molecular Approach, 2/e

Example 1.7: Convert 1.76 yd to centimeters


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units are correct; number makes

sense: cm


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(1 mL = 0.001 L; 1 L = 1.057 qt)



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Conceptual Plans for


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p

Units Raised to Powers

Convert cubic inches into cubic centimeters1. Find relationship equivalence: 1 in. = 2.54 cm

2. Write concept planin.3 cm3

3. Change equivalence into conversion factors

with given unit on the bottom


Example 1.9: Convert 5.70 L to cubic inches


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sense: in.3


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y

centimeters are there in 2.11 yd3?


Practice 1.9Convert 2.11 yd3to cubic centimeters


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Sortinformation

Given:

Find:

2.11 yd3

volume, cm3

Strategize ConceptualPlan:

Relationships:

1 yd = 36 in.

1 in. = 2.54 cm

Follow theconceptual

plan to solve

the problem

Solution:

Sig. figs. andround

Round: 1613210.75 cm3

= 1.61 x 106cm3

Check Check: units and number make sense

yd3 in3 cm3


Density as a Conversion Factor


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y

Can use density as a conversion factorbetween mass and volume!! density of H2O = 1.0 g/mL 1.0 g H2O = 1 mL H2O

density of Pb = 11.3 g/cm3 11.3 g Pb = 1 cm3Pb

How much does 4.0 cm3of lead weigh?


Example 1.10: What is the mass in kg of 173,231 Lof jet fuel whose density is 0.768 g/mL?


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units and number make senseCheck: Check

1.33041 x 105 = 1.33 x 105kgRound: Sig. figs. andround

Solution: Follow theconceptual

plan to solve

the problem

1 mL = 0.768 g, 1 mL = 103L

1 kg = 1000 g

Conceptual

Plan:

Relationships:

Strategize

173,231 L

density = 0.768 g/mL

mass, kg

Given:

Find:

Sortinformation

L mL g kg


Practice Calculate the Following


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How much does 3.0 x 102mL of etherweigh? (d= 0.71 g/mL)

What volume does 100.0 g of marbleoccupy? (d= 4.0 g/cm3)


Practice - How much does 3.0 x 102mL of ether weigh?

3 0 x 102 mLGiven:Sort


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sense: if density < 1, mass < volumeCheck:Check

2.1 x 102 gRound:Sig. figs. and

round

Solution:Follow the

conceptual

plan to solve

the problem

1 mL = 0.71 g

Conceptual

Plan:

Relationships:

Strategize

3.0 x 102mL

density = 0.71 g/mL

mass, g

Given:

Find:

Sort

information

mL g


PracticeWhat volume does 100.0 g of marble occupy?

100 0GiS t


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Check:Check

25 cm3Round:Sig. figs. and

round

Solution:Follow the

conceptual

plan to solve

the problem

1 cm3

= 4.0 g

Conceptual

Plan:

Relationships:

Strategize

m = 100.0 g

density = 4.0 g/cm3

volume, cm3

Given:

Find:

Sort

information

g cm3


sense: if density > 1, mass > volume


Order of Magnitude Estimations


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Copyright 2011 Pearson Education, Inc.Tro, Chemistry: A Molecular Approach 117

Order of Magnitude Estimations

Using scientific notation

Focus on the exponent on 10

If the decimal part of the number is less than 5,just drop it

If the decimal part of the number is greaterthan 5, increase the exponent on 10 by 1

Multiply by adding exponents, divide bysubtracting exponents


Estimate the Answer


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Copyright 2011 Pearson Education, Inc.Tro, Chemistry: A Molecular Approach 118

Estimate the Answer

Suppose you count 1.2 x 105atoms per secondfor a year. How many would you count?

1 s = 1.2 x 105105atoms

1 minute = 6 x 101102s

1 hour = 6 x 101 102min

1 day = 24 101hr

1 yr = 365 102days


Problem Solving with Equations


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Problem Solving with Equations

When solving a problem involves using anequation, the concept plan involves being given

all the variables except the one you want to

find Solve the equation for the variable you wish to

find, then substitute and compute


Example 1.12: Find the density of a metal cylinderwith mass 8.3 g, length 1.94 cm, and radius 0.55 cm


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units and number make senseCheck:Check

V= (0.55 cm)2 (1.94 cm)

V= 1.8436 cm3

Solution:Follow the

conceptual

plan to solve

the problemSig. figs. and

round

V= r2 l

d = m/V

Conceptual

Plan:

Relationships:

Strategize

m = 8.3 g

l= 1.94 cm, r= 0.55 cm

density, g/cm3

Given:

Find:

Sort

information

l, r V m, V d


PracticeWhat is the mass in kilograms of a cube oflead that measures 0.12 m on each side?


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(dPb = 11.3 g/cm3)


PracticeWhat is the mass in kilograms of acube of lead that measures 0.12 m on each side?

3S


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V= (0.12 m)3

V= 1.728 x 103m3

Solution:Follow theconceptual

plan to solve

the problemSig. figs. and

round

V= l3

, 11.3 g = 1 cm3

,1 cm = 10-2 m, 1 kg = 103g

Conceptual

Plan:

Relationships:

Strategize

l= 0.12 m, d= 11.3 g/cm3

mass, kg

Given:

Find:

Sort

information

l V

m3 cm3 g kg

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