chapter 1 summary chemistry

7/29/2019 Chapter 1 Summary Chemistry

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Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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1.1 Some Fundamental Definitions

1.2 The Scientific Approach: Developing a Model

1.3 Chemical Problem Solving

1.4 Measurement in Scientific Study

1.5 Uncertainty in Measurement: Significant Figures

Chapter 1 : Keys to the Study of Chemistry


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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 Propertiesthose which the substance shows

as it interacts with, or transformsinto, other substances such asflammability, corrosiveness

Matter anything that has mass and volume -the stuff of theuniverse: 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 uniqueidentity

Physical Propertiesthose which the substance

shows by itself withoutinteracting with anothersubstance such as color, meltingpoint, boiling point, density
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Figure 1.1

A Physical change B Chemical change

The distinction between physical and chemical change.


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Figure 1.2 The physical states of matter.


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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

Energy is the capacity to do work.

Potential and kinetic energy can be interconverted.
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Energy is the capacity to do work. Figure1.3A

less stable

more stable

change in potential energy

EQUALS

kinetic energy

A gravitational system. The potential energy gained when alifted weight is converted to kinetic energy as the weight falls.
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Energy is the capacity to do work. Figure1.3B

less stable

more stable


EQUALS

kinetic energy

A system of two balls attached by a spring. The potential energygained by a stretched spring is converted to kinetic energy when themoving balls are released.
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Energy is the capacity to do work. Figure1.3C

less stable

more stable


EQUALS

kinetic energy

A system of oppositely charged particles. The potential energy gainedwhen the charges are separated is converted to kinetic energy as theattraction pulls these charges together.

C i ht Th M G Hill C i I P i i i d f d ti di l


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Energy is the capacity to do work. Figure1.3D

less stable

more stable


EQUALS

kinetic energy

A system of fuel and exhaust. A fuel is higher in chemical potentialenergy than the exhaust. As the fuel burns, some of its potential energy isconverted to the kinetic energy of the moving car.



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Scientific Approach: Developing a Model

Observations :Natural phenomena and measured events;

universally consistent ones can be stated as anatural law.

Hypothesis: Tentative proposal that explains observations.

Experiment: Procedure to test hypothesis; measures one

variable at a time.

Model (Theory):

Set of conceptual assumptions that explains

data from accumulated experiments; predictsrelated phenomena.

Further Experiment: Tests predictions based on model.

revised if

experiments do

not support it

altered if

predictions do

not support it



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A Systematic Approach to Solving Chemistry Problems

Problem statement

Plan

Clarify the known and unknown.

Suggest steps from known to unknown.

Prepare a visual summary of steps.

Solution

Check

Comment and Follow-up Problem


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Table 1. 1 SI Base Units

Physical Quantity

(Dimension) Unit Name

Unit

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 UnitsTable 1.2

Prefix PrefixSymbol

Word ConventionalNotation

ExponentialNotation

tera T trillion 1,000,000,000,000 1x1012

giga G billion 1,000,000,000 1x10

9

mega M million 1,000,000 1x106

kilo k thousand 1,000 1x103

hecto h hundred 100 1x102

deka da ten 10 1x101

----- ---- one 1 1x100

deci d tenth 0.1 1x10-1

centi c hundredth 0.01 1x10-2

milli m thousandth 0.001 1x10-3

micro millionth 0.000001 1x10-6

nano n billionth 0.000000001 1x10-9

pico p trillionth 0.000000000001 1x10-12

femto f quadrillionth 0.000000000000001 1x10-15

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English EquivalentLength

1 kilometer(km) 1000(103)m0.62mi

Common SI-English Equivalent QuantitiesTable 1.3

Quantity SI Unit

SI EquivalentEnglish to

SI Equivalent1 kilometer(km)

1000(103)m0.62miles(mi)

1 mi = 1.61km

1 meter(m)100(102)cm

1.094yards(yd)1000(103)mm

39.37inches(in) 1 yd = 0.9144m

1 foot (ft) = 0.3048m

1 centimeter(cm) 0.01(10-2)m0.3937in

1 in = 2.54cm(exactly!)

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Volume

1,000,000(106

)cubic centimeters35.2cubic feet (ft3)

1 cubic meter(m3)

1 ft3 = 0.0283m3

1 cubic decimeter(dm3)1000cm3

0.2642 gallon (gal)1.057 quarts (qt) 1 gal = 3.785 dm3

1 qt = 0.9464 dm3

1 cubic centimeter (cm3)

0.001 dm3

0.0338 fluid ounce 1 qt = 946.4 cm3

1 fluid ounce = 29.6 cm3

English Equivalent

Quantity SI Unit

SI Equivalent

English to

SI Equivalent




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English Equivalent

Quantity SI Unit

SI Equivalent

English to

SI Equivalent

Mass

1 kilogram (kg)

1000 grams

2,205 pounds (lb)1 (lb) = 0.4536 kg

1 gram (g)

1000 milligrams

0.03527 ounce(oz)

1 lb = 453.6 g

1 ounce = 28.35 g




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Sample Problem 1.3 Converting Units of Volume

PROBLEM: The volume of an irregularly shaped solid can be determinedfrom the volume of water it displaces. A graduated cylindercontains 19.9mL of water. When a small piece of galena, anore of lead, is submerged in the water, the volume increases to

24.5mL. 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 watervolume before and after submerging the solid.

volume (mL) before and after addition

volume (mL) of galena

volume (cm3) ofgalena

subtract

volume (L) ofgalena

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

SOLUTION:

(24.5 - 19.9)mL = volume of galena

4.6 mL x mL

1 cm3

= 4.6 cm3

4.6 mL xmL

10-3 L = 4.6x10-3 L



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py g p , q p p y

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Sample Problem 1.4 Converting Units of Mass

PROBLEM: International computer communications are often carried by opticalfibers in cables laid along the ocean floor. If one strand of opticalfiber weighs 1.19 x 10-3lbs/m, what is the total mass (in kg) of acable made of six strands of optical fiber, each long enough to linkNew York and Paris (8.84 x 103km)?

PLAN: The sequence of steps may vary but essentially you have to findthe 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 103km x km103

m

8.84 x 106m xm

1.19 x 10 -3lbs

1.05 x 104lb xcable

6 fibers

= 1.05 x 104lb

= 8.84 x 106m

2.205 lb

1kgx

6.30x 104lb

cable

6.30x 104lb=cable

2.86x104 kg

cable

=



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py g p , q p p y

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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 hassides that measure 20.9 mm by 11.1 mm by 11.9 mm, whatis the density of Li in g/cm3 ?

PLAN: Density is expressed in g/cm3

so we need the mass in gramsand the volume in cm3.

mass (mg) of Li

lengths (mm) of sides

mass (g) of Li

density (g/cm3) of Li

103 mg = 1 g

10 mm = 1 cm

lengths (cm) of sides

volume (cm3)

multiply lengths

SOLUTION:

20.9mm x

1mg

10-3g = 1.49g1.49x103mg x

10mm

1cm= 2.09cm

Similarly the other sides will be 1.11cm and 1.19 cm, respectively.

2.09 x 1.11 x 1.20 = 2.76cm3

density of Li =

1.49g

2.76 cm3= 0.540 g/cm3



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Figure 1.8 The freezing and boiling points of water.

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Temperature Scales and Interconversions

Kelvin ( K ) - The Absolute temperature scale begins atabsolute zero and only has positive values.

Celsius (o

C ) - The temperature scale used by science,formally called centigrade, most commonly used scale around theworld; water freezes at 0oC, and boils at 100oC.

Fahrenheit (o

F ) - Commonly used scale in the U.S. for ourweather reports; water freezes at 32oF and boils at 212oF.

T (in K) = T (in oC) + 273.15T (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 feverand we use the 0C to kelvin relationship to find the temperaturein kelvins.

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

(b) What is the childs 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 dependsupon the measuring device.

Figure 1.9A

32.30C32.330C



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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 firstnonzero digit.Count that digit and every digit to its right as significant.

Numbers such as 5300 L are assumed to only have 2 significantfigures. A terminal decimal point is often used to clarify thesituation, but scientific notation is the best!

except zeros that are used only to position the

decimal point.

Zeros that end a number and lie either after or before the decimalpoint are significant; thus 1.030 ml has four significant figures,and 5300. L has four significant figures also.



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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 aresignificant figures(sf), and determine the number of significantfigures in each quantity. For(d) to (f), express each inexponential notation first.

PLAN: Determine the number of sf by counting digits and paying attentionto 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 cm32sf 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 thesame 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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= 23.4225 cm3 = 23 cm39.2 cm x 6.8 cm x 0.3744 cm

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:



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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 retainedand to 5.4 if two significant figures are retained.

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

3.If the digit removed is 5, the preceding number increases by1 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 finalanswer.


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Sample Problem 1.8 Significant Figures and Rounding

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

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

number.

SOLUTION:

7.085 cm

16.3521 cm2 - 1.448 cm2

(a)11.55 cm3

4.80x104 mg

(b)

1 g

1000 mg

7.085 cm

16.3521 cm2 - 1.448 cm2(a) =

7.085 cm

14.904 cm2

= 2.104 cm

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 AccuracyErrors in Scientific Measurements

Random Error-In the absence of systematic error, some values that are higher andsome 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 real value.

Systematic Error-

Values that are either all higher or all lower than the actual value.



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Figure

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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 laboratory.Figure 1.10

continued

chapter 1 summary chemistry

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