1-1 Keys to the Study of Chemistry Chapter 1. 1-2 Chapter 1 : Keys to the Study of Chemistry 1.1 Some Fundamental Definitions 1.2 Chemical Arts and the.

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Keys to the Study of ChemistryKeys to the Study of Chemistry

Chapter 1

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Chapter 1 : Keys to the Study of Chemistry

1.1 Some Fundamental Definitions

1.2 Chemical Arts and the Origins of Modern Chemistry

1.3 The Scientific Approach: Developing a Model

1.4 Chemical Problem Solving

1.5 Measurement in Scientific Study

1.6 Uncertainty in Measurement: Significant Figures

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CHEMISTRY

is the study of matter, its properties,

the changes that matter undergoes, and

the energy associated with these changes.

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Definitions

Chemical Properties Those which the substance shows as it interacts with, or transforms into, other substances (such as flammability, corrosiveness)

Matter anything that has mass and volume -the “stuff” of the universe: books, planets, trees, professors, students

Composition the types and amounts of simpler substances that make up a sample of matter

Properties the characteristics that give each substance a unique identity

Physical Properties Those which the substance shows by itself without interacting with another substance (such as color, melting point, boiling point, density)

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Sample Problem 1.1 Distinguishing Between Physical and Chemical Change

PROBLEM: Decide whether each of the following processes is primarily a physical or a chemical change, and explain briefly.

PLAN: Does the substance change composition or just change form?

SOLUTION:

(a) Frost forms as the temperature drops on a humid winter night.

(b) A cornstalk grows from a seed that is watered and fertilized.

(c) Dynamite explodes to form a mixture of gases.

(d) Perspiration evaporates when you relax after jogging.

(e) A silver fork tarnishes in air.

(a) physical change (b) chemical change (c) chemical change

(d) physical change (e) chemical change

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Figure 1.2

The Physical States of Matter

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Figure 1.3

Energy is the capacity to do work.

energy due to the position of the object or energy from a chemical reaction

Potential Energy

Kinetic Energy

energy due to the motion of the object

Potential and kinetic energy can be interconverted.

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Figure 1.3(continued) Energy is the capacity to do work.

Potential Energy

energy due to the position of the object or energy from a chemical reaction

Kinetic Energy

energy due to the motion of the object

Potential and kinetic energy can be interconverted.

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Scientific Approach: Developing a Model

Observations :

Hypothesis:

Experiment:

Model (Theory):

Further Experiment:

Natural phenomena and measured events; universally consistent ones can be stated as a natural law

Tentative proposal that explains observations

Procedure to test hypothesis; measures one variable at a time

Set of conceptual assumptions that explains data from accumulated experiments; predicts related phenomena

Tests predictions based on model

revised if experiments do not support it

altered if predictions do not support it

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Sample Problem 1.2 Converting Units of Length

PROBLEM: What is the price of a piece of copper wire 325 centimeters (cm) long that sells for $0.15/ft?

PLAN: Know length (in cm) of wire and cost per length (in ft)

Need to convert cm to inches and inches to ft followed by finding the cost for the length in ft.

SOLUTION:length (cm) of wire

length (ft) of wire

length (in) of wire

Price ($) of wire

2.54 cm = 1 in

12 in = 1 ft

1 ft = $0.15

Length (in) = length (cm) x conversion factor

= 325 cm x in2.54 cm

= 128 in

Length (ft) = length (in) x conversion factor

= 128 in x ft12 in

= 10.7 ft

Price ($) = length (ft) x conversion factor

= 10.7 ft x $0.15ft

= $1.60

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Table 1. 2 SI - Base Units

Physical Quantity Unit Name Abbreviation

mass

meter

kg

length

kilogram

m

time second s

temperature kelvin K

electric current ampere A

amount of substance mole mol

luminous intensity candela cd

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Common Decimal Prefixes Used with SI Units

Prefix PrefixSymbol

Number Word ExponentialNotation

tera T 1,000,000,000,000 trillion 1012

giga G 1,000,000,000 billion 109

mega M 1,000,000 million 106

kilo k 1,000 thousand 103

hecto h 100 hundred 102

deka da 10 ten 101

----- ---- 1 one 100

deci d 0.1 tenth 10-1

centi c 0.01 hundredth 10-2

milli m 0.001 thousandth 10-3

micro μ 0.000001 millionth 10-6

nano n 0.000000 001 billionth 10-9

pico p 0.000000 000001 trillionth 10-12

femto f 0.000000 0000000 01 quadrillionth 10-15

Table 1.3

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Common SI-English Equivalent Quantities

English Equivalent

Mass

Length 1 kilometer (km) 1000 (103) m 0.6214 mi

1 meter (m) 100 (102) cm 1.094 yd

1000 (103) mm 39.37 in

1 centimeter (cm) 0.01 (10-2 ) m 0.3937 in

Volume 1,000,000 (106) cm3 35.31 ft31 cubic meter (m3)

1 cubic decimeter(dm3)

1000 cm3 0.2642 gal 1.057 qt

1 cubic centimeter (cm3)

0.001 dm3 0.03381 fluid ounce

1 kilogram (kg) 1000 g 2,205 lb

1 gram (g) 1000 mg 0.03527 oz

Table 1.4

Quantity SI Unit SI EquivalentEnglish to

SI Equivalent

1 mi = 1.609 km

1 yd = 0.9144 m

1 ft = 0.3048 m

1 in = 2.54 cm (exactly)

1 ft3 = 0.02832 m3

1 gal = 3.785 dm3

1 qt = 0.9464 dm3

1 qt = 946.4 cm3

1 fluid ounce = 29.57 cm3

1 lb = 0.4536 kg1 lb = 453.6 g1 oz = 28.35 g

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Sample Problem 1.3 Determining the Volume of a Solid by Displacement of Water

PROBLEM: The volume of an irregularly shaped solid can be determinedfrom the volume of water it displaces. A graduated cylinder contains 19.9 mL water. When a small piece of galena, an oreof lead, is submerged in the water, the volume increases to24.5 mL. What is the volume of the piece of galena in cm3 and in L?

PLAN: The volume of galena is equal to the change in the water volume before and after submerging the solid.

volume (mL) before and after addition

volume (mL) of galena

volume (cm3) of galena

subtract

volume (L) of galena

1 mL = 1 cm3 1 mL = 10-3 L

SOLUTION:(24.5 - 19.9)mL = volume of galena

4.6 mL xmL

1 cm3

= 4.6 cm3

4.6 mL xmL

10-3 L = 4.6x10-3 L

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Sample Problem 1.4 Converting Units of Mass

PROBLEM: International computer communications will soon be carried by optical fibers in cables laid along the ocean floor. If one strand of optical fiber weighs 1.19 x 10 -3 lbs/m, what is the total mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris (8.84 x 103km)?

PLAN: The sequence of steps may vary but essentially you have to find the length of the entire cable and convert it to mass.

length (km) of fiber

mass (lb) of fiber

length (m) of fiber

1 km = 103 m

mass (kg) of cablemass (lb) of cable

6 fibers = 1 cable

1 m = 1.19x10-3 lb

2.205 lb = 1 kg

SOLUTION:

8.84 x 103 km xkm

103 m

8.84 x 106 m x m

1.19 x 10 -3lbs

1.05 x 104 lb xcable

6 fibers

= 1.05 x 104 lb

= 8.84 x 106 m

2.205 lb

1 kgx

6.30x 104 lb

cable

6.30x 104 lb =cable

2.86x104 kgcable

=

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Some Interesting Quantities

Length Volume Mass

Figure 1. 10

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Substance Physical State Density (g/cm3)

Densities of Some Common SubstancesTable 1.5

Hydrogen Gas 0.000089

Oxygen Gas 0.0014

Grain alcohol Liquid 0. 789

Water Liquid 1.000

Table salt Solid 2.16

Aluminum Solid 2.70

Lead Solid 11.3

Gold Solid 19.3

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Sample Problem 1.5 Calculating Density from Mass and Length

PROBLEM: Lithium (Li) is a soft, gray solid that has the lowest densityof any metal. If a slab of Li weighs 1.49 x 103 mg and has sides that measure 20.9 mm by 11.1 mm by 12.0 mm, whatis the density of Li in g/cm3 ?

PLAN: Density is expressed in g/cm3 so we need the mass in grams and the volume in cm3.

mass (mg) of Li

lengths (mm) of sides

mass (g) of Li

density (g/cm3) of Li

103mg = 1g

10 mm = 1 cm

lengths (cm) of sides

volume (cm3)

multiply lengths

SOLUTION:

20.9 mm x

mg10-3g

= 1.49 g1.49x103 mg x

10 mmcm

= 2.09 cm

Similarly the other sides will be 1.11 cm and 1.20 cm, respectively.

2.09 x 1.11 x 1.20 = 2.76 cm3

density of Li =1.49 g

2.76cm3= 0.540 g/cm3

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Figure 1.11

Some Interesting Temperatures

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Figure 1.12

The Freezing and Boiling Points of Water

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Temperature Scales and Interconversions

Kelvin ( K ) - The “absolute temperature scale” begins at absolute zero and only has positive values.

Celsius ( oC ) - The temperature scale used by science, formally called centigrade and most commonly used scale around the world. Water freezes at 0oC and boils at 100oC.

Fahrenheit ( oF ) - Commonly used scale in the US for weather reports. Water freezes at 32oF and boils at 212oF.

T (in K) = T (in oC) + 273.15 T (in oC) = T (in K) - 273.15

T (in oF) = 9/5 T (in oC) + 32T (in oC) = [ T (in oF) - 32 ] 5/9

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Sample Problem 1.6 Converting Units of Temperature

PROBLEM: A child has a body temperature of 38.70C.

PLAN: We have to convert 0C to 0F to find out if the child has a fever and we use the 0C to kelvin relationship to find the temperature in kelvins.

(a) If normal body temperature is 98.60F, does the child have a fever?

(b) What is the child’s temperature in kelvins?

SOLUTION:

(a) Converting from 0C to 0F9

5(38.70C) + 32 = 101.70F

(b) Converting from 0C to K 38.70C + 273.15 = 311.8K

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The Number of Significant Figures in a Measurement Depends Upon the Measuring Device

Figure 1.14

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Rules for Determining Which Digits are

Significant All digits are significant

•Make sure that the measured quantity has a decimal point.•Start at the left of the number and move right until you reach

the first nonzero digit.•Count that digit and every digit to its right as significant.

Zeros that end a number and lie either after or before the decimal point are significant; thus 1.030 mL has four significant figures, and 5300. L has four significant figures also. Numbers such as 5300 L are assumed to only have 2 significant figures. A terminal decimal point is often used toclarify the situation, but scientific notation is the best!

except zeros that are used only toposition the decimal point.

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Sample Problem 1.7 Determining the Number of Significant Figures

PROBLEM: For each of the following quantities, underline the zeros that are significant figures(sf), and determine the number of significant figures in each quantity. For (d) to (f) express each in exponential notation first.

PLAN: Determine the number of sf by counting digits and paying attention to the placement of zeros.

SOLUTION:

(b) 0.1044 g(a) 0.0030 L (c) 53.069 mL

(e) 57,600. s(d) 0.00004715 m (f) 0.0000007160 cm3

(b) 0.1044 g(a) 0.0030 L (c) 53.069 mL

(e) 57,600. s(d) 0.00004715 m (f) 0.0000007160 cm3

2sf 4sf 5sf

(d) 4.715x10-5 m 4sf (e) 5.7600x104 s 5sf (f) 7.160x10-7 cm3 4sf

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Rules for Significant Figures in Answers

1. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places.

106.78 mL = 106.8 mL

Example: subtracting two volumes

863.0879 mL = 863.1 mL

865.9 mL - 2.8121 mL

Example: adding two volumes 83.5 mL

+ 23.28 mL

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2. For multiplication and division. The number with the least

certainty limits the certainty of the result. Therefore, the answer

contains the same number of significant figures as there are in the

measurement with the fewest significant figures.

Rules for Significant Figures in Answers

Multiply the following numbers:

9.2 cm x 6.8 cm x 0.3744 cm = 23.4225 cm3 = 23 cm3

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Issues Concerning Significant Figures

graduated cylinder < buret ≤ pipet

numbers with no uncertainty

1000 mg = 1 g

60 min = 1 hr

These have as many significant digits as the calculation requires.

be sure to correlate with the problem

FIX function on some calculators

Electronic Calculators

Choice of Measuring Device

Exact Numbers

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Rules for Rounding Off Numbers1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if three significant figures are retained and to 5.4 if two significant figures are retained.

2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if three significant figures are retained and to 0.24 if two significant figures are retained.

3.If the digit removed is 5, the preceding number increases by 1 if it is odd and remains unchanged if it is even.17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by nonzeros, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7

4. Be sure to carry two or more additional significant figures through a multistep calculation and round off only the final answer.

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Sample Problem 1.8 Significant Figures and Rounding

PROBLEM: Perform the following calculations and round the answer to the correct number of significant figures.

PLAN: In (a) we subtract before we divide; for (b) we are using an exact number.

SOLUTION:

7.085 cm2

16.3521 cm2 - 1.448 cm2(a)

11.55 cm3

4.80x104 mg(b)

1 g

1000 mg

7.085 cm2

16.3521 cm2 - 1.448 cm2(a) =

7.085 cm2

14.904 cm2

= 2.104 cm2

11.55 cm3

4.80x104 mg(b)

1 g

1000 mg=

48.0 g

11.55 cm3

= 4.16 g/ cm3

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Precision and Accuracy Errors in Scientific Measurements

Random Error In the absence of systematic error, some values that are higher and some that are lower than the actual value

Precision Refers to reproducibility or how close the measurements are to each other

Accuracy Refers to how close a measurement is to the actual value

Systematic errorValues that are either all higher or all lower than the actual value

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Figure 1.16

precise and accurate

precise but not accurate

Precision and Accuracy in the Laboratory

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systematic error

random error

Precision and Accuracy in the LaboratoryFigure 1.16 continued