Chapter 2 OutlineChapter 2 Outline

Data AnalysisAnd Measurement

I. The Metric SystemI. The Metric System

A. U.S. only industrial country in world not to use as primary system

B. Used in all sciences as system of measurement

C. Uses base units - and prefixes - that are easily manipulated

D. Base UnitsD. Base Units

1) Length Meter - the length of a meter stick2) Volume Liter - the amount of liquid that will fill a container that is one decimeter cubed. 1 dm^3 = 1 liter 1 cm^3 = 1 milliliter

3) mass - kilogram which is about equal to 2 lbs (1kg = 2.2 lbs) - grams are often used in chemistry 454 grams = 1 pound.

E. PrefixesE. Prefixes

• kids kilo (k) larger• have hecta (h)• died decka (dk)• over [ grams, liters,meters,seconds]• doing deca (d) • conversions centi (c)• metric milli (m) smaller

F. Metric to Metric F. Metric to Metric ConversionsConversions

The factor label method 1) Underline what is to be solved for 2) Write down what is given 3) Make a conversion factor (C.F.) with the units of the given on the bottom 4) The units to solve for go on top

5) Use the prefix chart to determine which unit is larger - it receives a value of one

6) Use the chart to count how many units of ten to assign to the smaller unit

7) Multiply across by the top and divide by the bottom to obtain answer.

8) Be sure to include units on your answer9) Example problems

1) How many grams are there in 400 cg?

2) How many ml are in 2 dkl?

3) How may km are in 40,000 dm?

4) How many cg are in 24 dg?

II. Metric to English II. Metric to English and Special Conversionsand Special Conversions

A. English to Metric Conversion Factors & other C.F.s. 1 yard = 36 in , 1ft = 12in 454 g = 1 pound 1 L = 1dm^3 2.54cm = 1 inch 1 ml = 1cm^3 2,000 pounds = 1 ton 1 meter = 1.09 yrd 60 sec = 1 min 1 kilometer = 0.62 miles 60 min = 1 hr 1 mile = 5280 feet

Practice ProblemsPractice Problems1) How many cm are in 2 feet ? (60.96)

2) How many meters are in a 100 yard football field? (91.4)

3) Convert 4 lbs to mg? ( 1.82x10^6)

4) How many cubic meters are in a room measuring 8ft x 10ft x 12ft? (27.2)

5) How many ml are in a box that measures 2.2 by 4 by 6 in? (865cm^3 = 865ml)

6) How many kilometers are in 143.56 yards?

7) A car is traveling 9.06 km per hour. How many meters per minute is it traveling?

(151)

8) Convert 40 miles/hr to meters/sec.(17.92)

III. Significant FiguresIII. Significant Figures

Scientists must record their data in a way that tells the reader how precise her measurements are.

Therefore, the rules of significant figures must be observed.

A. The precision of the instrument used to measure determines whether a figure is known or estimated. What is the value given to the bar below. How many significant figures are there?

1 2 3 4 5 6 7 8 9 10 11 12 13

10.5? Is the .5 in 10.5 significant?

Yes. It is an estimate and therefore significant. Estimates are often called

doubtful figures.

B. Significant figure rules.B. Significant figure rules.

1. All non-zero numbers are significant. Example 3.45 = 3 s.f. 3.556 = 4 s.f.2. Zeros at the end of a number that include a decimal

point are significant Example 3.40 = 3 s.f. 0.500 = 3 s.f. 3. Zeros between significant figures are significant.

Example 3.02 = 3 s.f. 3.04032 = 6 s.f.

4. Zeros just for spacing are not significant. Example 0.000345 = 3 s.f. 3.45 x 10^-4 If a number is listed as 7000g there is just one

s.f. Use scientific notation to list 4 s.f. 7.000 x 10^3g 5. Counting numbers and exact numbers have an

infinite number of significant numbers. Example 40 cars or 1000 mm = 1 m

Practice ProblemsPractice Problems

• Identify the number of significant figures:

1) 3.0800 2) 0.00418 3) 7.09 x 10¯5

4) 91,600 5) 0.003005

6) 3.200 x 109

7) 2508) 780,000,0009) 0.010110) 0.00800Answers1)5 2)3 3)3 4)3 5)4 6)4 7)2 8)2 9)3 10)3

D.Calculations With D.Calculations With Significant FiguresSignificant Figures

1. Multiplication and 1. Multiplication and DivisionDivision

• The following rule applies for multiplication and division:

• The LEAST number of significant figures in any number of the problem determines the number of significant figures in the answer.

• Example #1: 2.5 x 3.42.• The answer to this problem would be 8.6 (which

was rounded from the calculator reading of 8.55). Why?

• 2.5 has two significant figures while 3.42 has three. Two significant figures is less precise than three, so the answer has two significant figures.

• Example #2: How many significant figures will the answer to 3.10 x 4.520 have?

• You may have said two. This is too few. A common error is for the student to look at a number like 3.10 and think it has two significant figures. The zero in the hundedth's place is not recognized as significant when, in fact, it is. 3.10 has three significant figures.

2. Addition and 2. Addition and SubtractionSubtraction

• 1) Count the number of significant figures in the decimal portion of each number in the problem. (The digits to the left of the decimal place are not used to determine the number of decimal places in the final answer.)

• 2) Add or subtract in the normal fashion.

• 3) Round the answer to the LEAST number of places in the decimal portion of any number in the problem.

• WARNING: the rules for add/subtract are different from multiply/divide. A very common student error is to swap the two sets of rules. Another common error is to use just one rule for both types of operations.

Practice ProblemsPractice Problems

1) 3.461728 + 14.91 + 0.980001 + 5.26312) 23.1 + 4.77 + 125.39 + 3.5813) 22.101 - 0.93074) 0.04216 - 0.00041345) 564,321 - 264,321

AnswersAnswers

1. 24.61 2. 156.83. 21.1704. 0.041755. 300,000 correct 3.00000 x 105

The Structure of the AtomThe Structure of the Atom

History, Structure, Properties and Forces

Early Theories of MatterEarly Theories of Matter

I. Before the early 1800’s many Greek philosophers thought that matter was formed of air, earth, fire and water.

II. DemocratusII. Democratus

A. First to propose atomos - matters as small indivisible particles

B. Said they move through empty space

C. different properties of matter due to changes in arrangement of atoms