chm 1045 chapter 1 prasad

8/12/2019 Chm 1045 Chapter 1 Prasad

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CHAPTER 1:CHEMISTRY AND MEASUREMENT

Vanessa Prasad-PermaulValencia College

CHM 1045


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Properties of Matter

Chemistry: The study of composition,properties, and transformations of matter

Matter: Anything that has both mass &volume

Hypothesis: Interpretation of results

Theory: Consistent explanation ofobservations

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Conservation of Mass

Law of Mass Conservation: Mass is neithercreated nor destroyed in chemical reactions.

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Example 1: Conservation of Mass

C(s)

+ O2(g)

CO2(g)

a) 12.3g C reacts with 32.8g O 2, ?g CO212.3g + 32.8g = 45.1g

a) 0.238g C reacts with ?g O 2 to make .873g CO 2 0.238g + x = 0.873g = 0.873g-0.238g = 0.635g of O 2

a) ?g C reacts with 1.63g O 2 to make 2.24g CO 2x + 1.63g = 2.24g = 2.24g - 1.63g = 0.61g C

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Example 1: Conservation of Mass

Exercise 1.11.85g of wood is placed with 9.45g of air in a sealed vessel. It isheated and the wood burns to produce ash and gases. The ashis weighed to yield 0.28g. What is the mass of the gases in thevessel?

1.85g Wood + 9.45g Air heat 0.28g Ash + ? g gases

1.85 + 9.45 - 0.28 = 11.02g of gases

What is the mass of wood that is converted to gas by the end ofthe experiment?

1.85g of Wood 0.28g of ash = 1.57g

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Matter

Matter is any substance that has mass andoccupies volume.

Matter exists in one of three physical states:solid

liquid

gas

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Solid

In a solid , the particles of matter are tightlypacked together.

Solids have a definite, fixed shape.

Solids cannot be compressed and have adefinite volume.

Solids have the least energy of the three statesof matter.

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Liquid

In a liquid , the particles of matter are looselypacked and are free to move past one another.

Liquids have an indefinite shape and assume theshape of their container.

Liquids cannot be compressed and have adefinite volume.

Liquids have less energy than gases but moreenergy than solids.

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Gases

In a gas , the particles of matter are far apartand uniformly distributed throughout thecontainer.

Gases have an indefinite shape and assume theshape of their container.

Gases can be compressed and have anindefinite volume.

Gases have the most energy of the three statesof matter.

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Phases

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A physical change is a change in the form of matterbut not in its chemical identity

A chemical change or a chemical reaction is a changein which one of more kinds of matter are transformedinto a new kind of matter or several new kinds ofmatter

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Physical Properties can be determined withoutchanging the chemical makeup of the sample.

Some typical physical properties are:

Melting Point, Boiling Point, Density, Mass, Touch,Taste, Temperature, Size, Color, Hardness,Conductivity.

Some typical physical changes are:Melting, Freezing, Boiling, Condensation, Evaporation,Dissolving, Stretching, Bending, Breaking.

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Chemical Properties are those that do changethe chemical makeup of the sample.

Some typical chemical properties are:

Burning, Cooking, Rusting, Color change, Souring ofmilk, Ripening of fruit, Browning of apples, Taking a

photograph, Digesting food.

Note: Chemical properties are actually chemicalchanges

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Exercise 1.2

Potassium (K) is a soft, silvery-colored metal that melts @ 64 oC.It reacts vigorously with water (H 2O), Oxygen (O 2) and Chlorine(Cl 2).Identify all physical properties:

Soft Silvery-colored Melting point of 64 oC

Identify all chemical properties: Metal (its chemical identity) K reacts vigorously with H 2O K reacts vigorously with O 2 K reacts vigorously with Cl 2


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Classifications of Matter

Matter can be divided into two classes:

mixtures

pure substances

Mixtures are composed of more than onesubstance and can be physically separated intoits component substances.

Pure substances are composed of only onesubstance and cannot be physically separated.

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Pure Substances

There are two types of pure substances:CompoundsElements

A compound is a substance composed of two ormore elements chemically combined

Compounds can be chemically separated intoindividual elements.Water is a compound that can be separated intohydrogen and oxygen.

An element cannot be broken down further bychemical reactions.

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Daltons Atomic Theory

Law of Definite Proportions: Differentsamples of a pure chemical substance always

contain the same proportion of elements by

mass.

Any sample of H 2O contains 2 hydrogen atoms for

every oxygen atom17


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Mixtures

There are two types of mixtures:homogeneous mixturesheterogeneous mixtures

Homogeneous mixtures have uniformproperties throughout.

Salt water is a homogeneous mixture.

Heterogeneous mixtures do not have uniformproperties throughout.

Sand and water is a heterogeneous mixture.

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Accuracy, Precision, and Significant Figures inMeasurement

Accuracy is how closeto the true value agiven measurement is.

Precision is how well anumber of independentmeasurements agreewith one another.

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l


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Example 8: Accuracy & Precision

Which of the following is precise but not

accurate?

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A P i i d Si ifi Fi i


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Rules for Significant Figures:

Zeros in the middle of a number are significant

Zeros at the beginning of a number are notsignificant

Zeros at the end of a number and following a

period are significantZeros at the end of a number and before a periodmay or may not be significant .

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E l 4 Si ifi Fi


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Example 4: Significant Figures

How many Significant Figures ?

a) 0.000459 = 3

b) 12.36 = 4

c) 36,450 = 4

d) 8.005 = 4e) 28.050 = 5

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A P i i d Si ifi t Fi i


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Rules for Calculating Numbers:

During multiplication or division, the answer canthave more significant figures than any of theoriginal numbers.

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a) 218.2 x 79 = 17237.8 = 1.7 x 104

a) 12.5 / 0.1272 = 94.33962264150943 = 94.3

b) 0.2895 x 0.29 = 0.083955 = 0.084

c) 32.567 / 22.98 = 1.417188859878155 = 1.417

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-During addition or subtraction, the answercant have more digits to the right of thedecimal point than any of the original numbers.

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Accuracy, Precision, and Significant Figures in Measurement

E ample 6: Significant Fig res


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a) 218.2 + 79 = 297.2 = 297

b) 12.5 - 0.1272 = 12.3728 = 12.4

c) 0.2895 + 0.29 = 0.5795 = 0.58

d) 32.567 - 22.98 = 55.547 = 55.55

e) 185.5+2.224 = 187.724 = 187.7

Accuracy Precision and Significant Figures in


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Rules for Rounding Numbers:

If the first digit removed is less than 5

round down (leave # same)

If the first digit removed is 5 or greaterround up

Only final answers are rounded off, do not roundintermediate calculations

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Example 7: Rounding and Significant Figures


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Example 7: Rounding and Significant Figures

Round off each of the following measurements

a) 3.774499 L to 4 sig. figs. = 3.774L

b) 255.0974 K to 3 sig. figs. = 255K

c) 55.265 kg to 4 sig. figs. = 55.27kg

d) 1.2151ml to 3 sig. figs. = 1.22ml

e) 1.2143g to 3 sig. figs. = 1.21g

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SIGNIFICANT FIGURESExercise 1.3Give answers to the following arithmetic setups. Round to thecorrect number of significant figures:

a) 5.61 x 7.891 = 4.864671 = 4.99.1

b) 8.91 - 6.435 = 2.475 = 2.48

c) 6.81 6.730 = 0.08 = 0.08

d) 38.91 x (6.81-6.730) = 38.91 x 0.08 = 3.1128 = 3

Scientific Notation


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Scientific Notation

Changing numbers into scientific notationLarge # to small #Moving decimal place to left, positive exponent

123,987 = 1.23987 x 105

Small # to large #Moving decimal place to right, negative exponent

0.000239 = 2.39 x 10 -4

How to put into calculator

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Example 3: Scientific Notation


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Example 3: Scientific Notation

Put into or take out of scientific notation

a) 1973 = 1.973 x 103

b) 5.5423 x 10-4 = 0.00055423

c) 0.775 = 7.75 x 10-1

d) 3.55 x 107 = 35,500,000

e) 8500 = 8.5 x 103

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Measurement and Units


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Physical Quantity Name of Unit AbbreviationMass kilogram kg

Length meter mTemperature kelvin K

Amount of substance mole molTime second s

Electric current ampere ALuminous intensity candela cd

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SI Units


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Factor Prefix Symbol1,000,000,000 = 10 9 giga G

1,000,000 = 10 6 mega M1,000 = 10 3 kilo k100 = 10 2 hecto h10 = 10 1 deka da

0.1 = 10 -1 deci d0.01 = 10 -2 centi c

0.001 = 10 -3 milli m0.000,001 = 10 -6 micro

0.000,000,001 = 10 -9 nano n0.000,000,000,001 = 10 -12 pico p

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*

*

* *

**

*

* Important

Some prefixes for multiples of SI units


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Exercise 1.4Express the following quantities using an SI prefix and abase unit. For instance, 1.6 x 10 -6m = 1.6mm. A quantity suchas0.000168g could be written 0.168mg or 168 mg.

a) 1.84 x 10-9 m = 1.84 nm (nanometer)b) 5.67 x 10-12 s = 5.67 ps (picosecond)c) 7.85 x 10-3 g = 7.85 mg (milligram)d) 9.7 x 103 m = 9.7 km (kilometer)e) 0.000732 s = 0.732 ms (millisecond)

= 732us (microsecond)f) 0.000000000154 m = 0.154nm (nanometer)

= 154pm (picometer)

Changes in Physical State


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Changes in Physical State

Most substances can exist as either a solid,liquid, or gas.

Water exists as a solid below 0 C; as a liquidbetween 0 C and 100 C; and as a gas above100 C.

A substance can change physical states as thetemperature changes.

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Solid Liquid


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Solid Liquid

When a solid changes to a liquid, the phasechange is called melting .

A substance melts as the temperature increases.

When a liquid changes to a solid, the phase

change is called freezing .

A substance freezes as the temperature decreases.

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Liquid Gas


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Liquid Gas

When a liquid changes to a

gas, the phase change iscalled vaporization .

A substance vaporizes asthe temperature increases.

When a gas changes to aliquid, the phase change iscalled condensation .

A substance condenses asthe temperature decreases.

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Solid Gas


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Solid Gas

When a solid changes directly to a gas,the phase change is called sublimation .

A substance sublimes as the temperatureincreases.

When a gas changesdirectly to a solid, the phasechange is called deposition .

A substance undergoesdeposition as thetemperature decreases.

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T


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Temperature

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Diagram of the various phases of temperature change

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Temperature

Temperature

Conversions:The Kelvin and Celsius

scales have equal size

units (a change of 1 oC isequivalent to a change

of 1K)

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100 K100o

C180o

F


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Temperature Conversions:

Celsius (C) Kelvin (K) temperature conversion:Kelvin (K) = t C x 1K + 273.15K

1oC

Fahrenheit (F) Celsius ( C) temperature conversions:there are exactly 9 oF for every 5oC. Knowing that 0 oC =32oF

t F = t C x 9oF + 325oC

t C = 5oC x (t oF 32)9oF

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Example 9: Temp. Conversions


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p p

Carry out the indicated temperature conversions:

a) 78C = ? K = (-78oC x 1K/1oC) +273.15K = 195.15 = 195K

b) 158C = ? F = (158oC x 9oF/5oC)+32oF = 316.4 = 316oF

c) 373.15 K = ? C = (373.15K x 1oC/1K) 273.15K = 100K

d) 98.6F = ? C = 5oC/9oF x (98.6oF 32oF) = 37oC

e) 98.6F = ? K = (37oC x 1K/1oC) +273.15K = 310.15 = 310K

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Exercise 1.5A person with a fever has temperature of 102.5 oF.What is this temperature in oC? A cooling mixtureof dry ice and isopropyl alcohol has a temperatureof -78 oC. What is the temperature in kelvins?

a) oC = 5oC x (oF 32 ) = 0.555 x (102.5 32) = 39.2oC9oF

b) K =oC + 273.15 = -78 + 273.15 = 195 K

Volume


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Volume : how much three-dimensional space asubstance (solid, liquid, gas) or shape occupies orcontains often quantified numerically using the SI

derived unit (m3

) the cubic meter.The volume of a container is generally understood to bethe capacity of the container, i. e. the amount of fluid(gas or liquid) that the container could hold, rather than

the amount of space the container itself displaces.

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Volume
http://en.wikipedia.org/wiki/SI_derived_unithttp://en.wikipedia.org/wiki/SI_derived_unithttp://en.wikipedia.org/wiki/SI_derived_unithttp://en.wikipedia.org/wiki/SI_derived_unit


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units of Volume :

m 3 or cm 3 (cc)

Traditionally chemists use liter (L)

1cm 3 = 1cc = 1mL

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Density : relates the mass of an object to itsvolume.

Density = mass / Volume D = m / V

V = m /D

m = V D

Density decreases as a substance is heatedbecause the substances volume increases.

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Density


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y

What is the density of glass (in mL) if a sampleweighing 26.43 g has a volume of 12.40 cm 3?

d = ?

m = 26.43 gV = 12.40 cm3 = 12.40 mLd = m = 26.43 g = 2.13145 = 2.131 g/mL

V 12.40 mL

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Density


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What is the volume of an unknown solution if themass is 12.567 g and the density is 14.621 g/mL ?

d = m/V V x d = m V = m/d

V = 12.567 g / 14.621 g/mL = 0.85952 mL

12.567g x 1mL = 0.85952 mL14.621g

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Density


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What is the mass of an unknown solution if the

volume is 20.2 mL and the density is 2.613 g/mL?

d = m/V m = d x V

m = 2.613g x 20.2 mL = 52.7826 = 52.8 gmL

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Exercise 1.6A piece of metal wire has a volume of 20.2 cm 3 and amass of 159 g. What is the density of the metal?

D = m = 159 g = 7.87128712 = 7.87 g /cm3V 20.2cm3

We know that the following metals have the followingdensities. Which metal is the wire made of?

Mn = 7.21 g/cm3

Fe = 7.87 g/cm3Ni = 8.90 g/cm 3


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Exercise 1.7

Ethanol (grain alcohol) has a density of 0.789 g/cm 3.What volume (mL) of ethanol must be poured intoa graduated cylinder to equal 30.3 g ?

d = m/V V x d = m V = m / d

V = 30.3 g x 1 cm3 = 38.4cm3

0.789 g

Dimensional Analysis & Units


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Dimensional-Analysis method uses a conversionfactor to express the relationship between units.

Original quantity x conversion factor = equivalentquantity

Example: express 2.50 kg lb.Conversion factor: 1.00 kg = 2.205 lb

2.50 kg x 2.205 lb = 6.00 lb1.00 kg

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Exercise 1.8The oxygen molecule (O 2) consists of two oxygenatoms a distance of 121 pm apart. How manymillimeters (mm) is this distance?

121 pm x 10-12 m x 1mm = 1.21 x 10-7 mm1 pm 10-3


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Exercise 1.9A large crystal is constructed by stacking small identicalpieces of crystal. A unit cell is the smallest piece fromwhich a crystal can be made. A unit cell of a crystal ofgold metal has a volume of 67.6 A 3. What is the volume

in dm 3?

67.6 A 3 x 10 -10 m x 10 dm = 6.76 x 10 -26 dm 3

1 A3

1 m

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Exercise 1.10

Using the following definitions, obtain theconversion factor for yards to meters. How manymeters are there in 3.54 yd?

1 in = 2.54cm (exactly) 1 yd = 36in (exactly)

1 yd x 1 in x 1 cm = 1.093613298 = 1.094 yd/m36 in 2.54 cm 10-2 m

3.54 yd x 1 m = 3.24 m1.094 yd

Conversions


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a) 1.267 km m cm1.267km x 1000m x 100cm = 126700cm = 1.267 x 1051km 1m

b) 0.784 L mL0.784L x 1000mL = 784L

1L

c) 3.67 x 105

cm in3.67 x 105cm x 1in = 144488.1889in = 1.44 x 105in2.54cm

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Conversions


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d) 79 oz g79oz x 28.35g = 2239.65g = 2.2 x 103g

1oz

e) 9.63 x 10-3 yd ft

9.63 x 10-3yd x 1m x 1km x 0.62137mile x 5280ft1.0936yd 1000m 1km 1mile= 0.0289ft

f) 23.5 cm2 m223.5cm 2 x 1m2 = 0.235m2

100cm 2

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Conversions


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g) 1.34 x 1012 pm m

1.34 x 1012pm x 1m = 1.34 x 1024m10-12pm

h) 4.67 x 10-7

nm pm4.67 x10-7nm x 1m x 10-12pm = 4.67 x 10-12pm10-9nm 1m

chm 1045 chapter 1 prasad

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