1-Ch1(pengantar).ppt

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INTRODUCTION

andCHAPTER ONE

Read the Introduction and Chapter 1.

Chemistry is NOT a spectator sport.

Work out complete solutions for all

the bold numbered problems.


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Welcome to the

World ofChemistry


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Beginning Chemistry 2008

Read the book before class!

Study the examples and do the

practice exercises.

Be prepared for each class

session, lecture and laboratory. Dont get behind.


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Outline for Chapter 1

DefinitionsHomogenous and Heterogeneous

Matter ?

3 states of matter

Chemicalvs.- Physical Change

Calculations

Density

Temperature

Kelvin Celsius FahrenheitFahrenheit Celsius Kelvin

Significant Figures (Sig Figs)


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Matter


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Mixtures

(a)Heterogeneous

(b)Suspension Blood

Salt water(c)Homogenous

Cookie


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The Language of Chemistry

CHEMICAL ELEMENTS - puresubstances that cannot be decomposed byordinary means to other substances.

SodiumBromineAluminum


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The Language of Chemistry

The elements, theirnames, and theirsymbols are givenon the

PERIODICTABLE

How manyelements arethere?


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The Periodic Table

Dmitri Mendeleev (1834 - 1907)


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Sodium

Find sodium, Na,

on the chart.


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Copperatoms onsilicasurface.

An atomis the smallest particle ofan element that has the chemical

properties of the element.

Find copper on the chart.


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An atom consists of a nucleus (of protonsand neutrons) and electrons in spaceabout the nucleus.

The Atom

Electron cloud

Nucleus


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Thered compound is

composed of

Ni- Nickel

C- Carbon

O- Oxygen N- Nitrogen

CHEMICAL COMPOUNDS

are composed of atoms and so can

be decomposed to those atoms.

Fixedcomposition


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MOLECULEThe smallest unit of a compound thatretains the chemical characteristics of

the compound.

MOLECULAR FORMULA

Composition of molecules

H2O C8H10N4O2- Caffeine


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The Nature of Matter

Chemists are interested in the nature ofmatter and how this is related to its atoms

and molecules.

Gold

Au

Mercury

Hg


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Graphitelayerstructure of

carbonatomsreflectsphysical

properties.


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Chemistry & Matter

We can explore the MACROSCOPIC world what we can see

Understand the PARTICULATE world

we cannot see

We can write SYMBOLS to describe these worlds.


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A Chemists View

2 H2(g) + O2 (g) --> 2 H2O(g)

Macroscopic

Symbolic-

Particulate


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STATES OF MATTER

SOLIDS have rigid shape, fixedvolume. External shape can reflect theatomic and molecular arrangement.

Reasonably well understood. LIQUIDS have no fixed shape and

may not fill a container completely.

Not well understood.

GASES expand to fill theircontainer.

Good theoretical understanding.

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THE THREE STATES OFMATTER

Bromine (gas) Aluminum (solid) Water orH2O (liquid)

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KINETIC NATURE OF MATTER

Matter consists of atoms and moleculesin motion.

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Physical

Properties

What are some physical

properties?Color

Melting and boiling

pointOdor

Conductivity

Density

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Physical Changes

Some physical changeswould be

boiling of a liquid

melting of a solid dissolving a solid in a

liquid to give a

homogeneous mixture

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Density= mass(g)

volume (cm3)

DENSITY- an important and useful

physical property

13.6 g/cm3

Gold

19.3 g/cm3

Mercury

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Which is more dense?

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Relative Densities of the Elements

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Sig Figs

Sig Fig PPT

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Problem: A piece of

copper has a mass of

57.54 g. It is 9.36 cmlong, 7.23 cm wide,

and 0.95 mm thick.

Calculate density(g/cm3).

Density= mass(g)

volume (cm3)

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SOLUTION

1. Get dimensions in common units.

2. Calculate volume in cubic centimeters.

3. Calculate the density.

. =095mm

1cm

10 mm 0.095 cm3

(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm3

6.4 cm3

57.54g=9.0g/cm3

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PROBLEM: Mercury (Hg) has a densityof 13.6 g/cm3. What is the mass of 95 mL

of Hg? In grams? In pounds?

Solve the problem using

DIMENSIONAL

ANALYSIS.

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Then, use dimensional analysis tocalculate mass.

PROBLEM: Mercury (Hg) has a densityof 13.6 g/cm3. What is the mass of 95 mLof Hg?

First, note that1 cm3 = 1 mL

See next s l ide95cm

3

13.6 g

cm3 =1.3 x 103

g

What is the mass in pounds?

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The milliliter

and the cubic

centimeter are

equivalent.

Notice the units

of 10s.

back

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What is the mass in pounds? (1 lb = 454 g)

PROBLEM: Mercury (Hg) has a densityof 13.6 g/cm3. What is the mass of 95 mLof Hg?

1.3 x 103g 1 lb454 g

=2.8 lb

35Density


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35DensityPROBLEM: An object weighing 15.67 g isplaced in water starting at 6.8 mL, then

displaces water to 20.2 mL. What is thedensity of the object?

6.8

20.2

20.2mL - 6.8mL = 13.4mL

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15.67g1

(20.2 - 6.8)ml = 1.17g/cm3

PROBLEM: An object weighing 15.67 g isplaced in water starting at 6.8 mL, then

displaces water to 20.2 mL. What is thedensity of the object?

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Ch i l P ti d


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Chemical Properties andChemical Change

Burning hydrogen (H2) in

oxygen (O2) gives H2O.

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Chemical Properties

Similar to Physical Properties onlywith reference to a Chemical

reactionsHeat and or light produced

Color

Oder

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Ch i l P ti d


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Chemical Properties andChemical Change

Chemical change orchemical reaction

involves the

transformation of one or

more atoms or molecules

into one or more different

molecules.

Burning hydrogen (H2) in

oxygen (O2) gives H2O.

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2 Al + 3 Br2 Al2Br6

Chemical Change

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Electrolyzing water

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UNITS OF MEASUREMENT

We make QUALITATIVEobservations of reactions

changes in color and physical

state.

We also make QUANTITATIVE

MEASUREMENTS, which involve

numbers. Use SI units based on the

metric system

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UNITS OF MEASUREMENT

Use SI units based on themetric system

length (meter, m)

mass (kilogram, kg,

and gram, g)

time (second)

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U it f L th


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Units of Length 1 kilometer (km) = ? meters (m)

1 meter (m) = ? centimeters (cm)

1 centimeter (cm) = ? millimeter (mm)

1 nanometer (nm) = 1.0 x 10-9 meter

OH distance =

9.4 x 10-11 m

9.4 x 10-9 cm0.094 nm

94 pm

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M t


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Measurement Learn the prefixes in Table 1.4

Other Relationships1 cm3 = 1 mL = 0.001 L

1.00 lb = 454 g

1.00 in = 2.54 cm

1.06 qt = 1.00 L

Significant figures Page 47Precision and accuracy Page 43

Examples

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Temperature Scales

Fahrenheit Celsius

Kelvin

Anders Celsius

1701-1744

Lord Kelvin

(William Thomson)

1824-1907

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T t S l


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Temperature Scales

Notice that1 Kelvin degree = 1 degree Celsius

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Calculations Using

Temperature

Generally require temps in Kelvin

T (K) = t (C) + 273

Body temp = 37 oC + 273 = 310. K

Liquid nitrogen = -196 oC + 273 = 77 K

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In Class Problems

A rectangular box has dimensions of20.0 cm 15.0 cm 8.00 mm. Calculatethe volume of the box in liters.

A standard sheet of paper has dimensionsof 8.5 inch by 11 inch. A sheet of paperweighs on the average 0.150 g and has adensity of 0.710 g/cm3. Calculate thethickness of the paper in cm.

A gallon (3.78 L) of latex paint can cover385 ft2 of the surface of a wall. What isthe average thickness of one coat of paint(in micrometers)?

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S l bl


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Sample problems

Calculate the volume of 525 g of

mercury, d=13.534g/cm3. The melting point of tin is 505.5 K.

Calculate the Celsius temperature.

Find the symbol for gold and theelement name for K.

An iron sheet is 3.50 cm square and

has a mass of 15.396 g. The densityof iron is 7.87 g/cm3. Calculate thethickness of the iron sheet in mm.

The End!

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Dimensional Analysis

English-English (one conversion)

1) 6 in = ? ft

2) 3.5 gal = ? qt

6 in 1 ft

12 in= 0.5 ft

3.5 gal 4 qt

1 gal= 14 qt

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Metric-Metric (one conversion)

1) 5.0 cm = ? mm

4.0 dg 1 kg

104 dg = 4.0 x 10-4 kg

5.0 cm 10 mm1 cm

= 50. mm

2) 4.0 dg = ? kg

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Metric-English (one conversion)

1) 200.0 cm = ? in

2) 34 qt = ? L

34 qt 1.00 L

1.06 qt= 32 L

200.0 cm 1.00 in

2.54 cm= 78.74 in

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English-English (two or more conversions)

1) 6 in = ? mile

2) 3.5 gal = ? oz

= 450 oz

6 in 1 ft

12 in= 9 x 10-5 mile1 mile

5280 ft

3.5 gal 4 qt

1 gal

32 oz

1 qt

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Metric-Metric (two or more conversions)

1) 5.0 cm = ? km

2) 4 kg = ? pg

= 4 x 1015 pg

5.0 cm 1 m

100cm= 5.0 x 10-5 km1 km

1000 m

4 kg 103 g

1 kg

1012 pg

1 g

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Metric-English (two or more conversions)

1) 200 m = ? in

2) 34 qt = ? mL

= 3.2 x 104 mL

200 m 100 cm

1 m= 8000 in1.00 in

2.54 cm

34 qt 1.00 L

1.06 qt

103 mL

1 L

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Derived Unit Conversions Area

1) 8.0 ft2 = ? cm2

2) 2.3 cm2

= ? nm2

= 2.3 x 1014 nm2

8.0 ft2 144 in2

1 ft2= 7400 cm2

(2.54)2 cm2

1.00 in2

2.3 cm2 1014 nm2

1 cm2

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Derived Unit Conversions Volume

1) 445 dm3 = ? mL

2) 5 cm3

= ? mm3

= 5 x 103 mm3

445 dm3 103 cm3

1 dm3= 4.45 x 105mL1 mL

1 cm3

5 cm3 103 mm3

1 cm3

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Other Conversions Problems

1) 100. km/hr = ? mile/hr

2) 25 m/gal = ? nm/qt

= 6.2 x 109 nm/qt

100. kmhr

105 cm1 km

= 62.1 mile/hr

1.00 in2.54 cm

25 m

gal

109 nm

1 m

1 ft12 in

1 mile5280 ft

1 gal

4 qt

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Di i l


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Dimensional AnalysisOther Conversions Problems (continue)

3) Calculate the density of a material if 45mL of it has a mass of 128 g.

4) Calculate the volume in mL of 2.5 g of a

material that has a density of 3.65 g/mL.

= 0.68 mL2.5 g mL

3.65 g

= 2.8 g/mL128 g45 mL

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Other Conversions Problems (continue)

5) Calculate the mass of 20 L of a materialthat has a density of 8.54 g/mL.

6) How many grams of gold are in 48 g of

an alloy that is 22.1% gold?

= 2 x 105 g20 L 103 ml1 L

8.54 gml

= 11 g gold48 g alloy 22.1 g gold

100.0 g alloy

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Practice Problems

1) 6.45 m = ? cm

2) 12.4 kg = ? mg

3) 184 oz = ? g

4) 24 oz/hr = ? L/day5) Determine the volume (in L) of 2 kg of sodium

chloride. (density = 2.17 g/mL)

6) How many grams of brass contain 50.0 g ofzinc? (This brass contains 15% Zn)

1) 645 cm 2) 1.24 x 107 mg 3) 5220 g 4) 17 L/day

5) 0.9 L 6) 330 g

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