METRIC AND MEASUREMENTS Scientific Notation Significant Digits Metric System Dimensional Analysis

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METRIC AND MEASUREMENTS

Scientific Notation Significant Digits Metric SystemDimensional Analysis

SCIENTIFIC NOTATION

Makes very large or small numbers easy to useTwo parts:

1 x < 10 (including 1 but NOT 10)

x 10 exponent

WRITING SCIENTIFIC NOTATION

EXAMPLES:

1) 2,000,000,000

2) 5430

3) 0.000000123

6) 0.0000600

4) 0.007872

5) 966,666,000

= 2 X 10 9

= 5.43 X 10 3

= 1.23 X 10 -7

= 7.872 X 10 -3

= 6.00 X 10 -5

= 9.66666 X 10 8

LARGE NUMBERS (>1)

POSITIVE EXPONENTSEQUAL TO 1 or itself ZERO EXPONENTS

SMALL NUMBERS (<1)NEGATIVE EXPONENTS

WRITING STANDARD FORMEXAMPLES:

1) 4.32 X 10 7

2) 3.45278 X 10 3

3) 8.45 X 10 -5

6) 1.123 X 10 5

4) 5.0010 X 10 -9

5) 7.00 X 10 -1

= 43,200,000

= 3452.78

= 0.0000845

= 0.0000000050010

= 112,300

= 0.700

POSITIVE EXPONENTS MOVE TO RIGHT

NEGATIVE EXPONENTS MOVE TO LEFT

SIGNIFICANT DIGITS

Exact numbers are without uncertainty and error Measured numbers are measured using instruments and have some degree of uncertainty and errorDegree of accuracy of measured quantity depends on the measuring instrument

RULES1) All NONZERO digits are significant

Examples:

a) 543,454,545

b) 34,000,000

Examples:

c) 65,945

2) Trailing zeros are NOT significant

= 9

= 2= 5

b) 234,500

= 1a) 1,000= 4

c) 34,288,900,000= 6

RULES CON’T3) Zero’s surrounded by significant digits are significant

Examples:a) 1,000,330,134

b) 534,001,000

Examples:

c) 7,001,000,100

4) For scientific notations, all the digits in the first part are significant

= 10

= 6

= 8

b) 2.34 x 10 -

16

= 4a) 1.000 x 10 9

= 3c) 3.4900 x 10 23 = 5

RULES CON’T5) Zero’s are significant if a) there is a decimal present (anywhere) b) AND a significant digit in front of the zeroZero’s at beginning of a number are not significant (placement holder)

Examples:

a) 0.00100

b) 0.1001232

c) 1.00100

= 3

= 7

= 6

e) 0.0000007

= 9d) 8900.00000

= 1

f) 0.003400 = 4

g) 0.0700 = 3

= 5h) 0.040100

Rules for Rounding in Calculations

Rounding with 5’s: UP ____ 5 greater than zero

10.257 = 10.3

34.3591 = 34.4

ODD 5 zero

99.750 = 99.8

101.15 = 101.2

Rounding with 5’s:

DOWN

EVEN 5 zero

6.850 = 6.8

101.25 = 101.2

CALCULATIONS1) Multiply and Divide: Least number of significant digits

Examples:

a) 0.102 x 0.0821 x 273b) 0.1001232 x 0.14 x 6.022 x 10 12c) 0.500 / 44.02

= 2.2861566

= 8.4412 x1010

= 0.011358473

e) 150 / 4

= 2958.770205d) 8900.00000 x 4.031 x 0.08206 0.995 = 37.5

f) 4.0 x 104 x 5.021 x 10–3 x 7.34993 x 102= 147615.9941

g) 3.00 x 10 6 / 4.00 x 10 -7 = 7.5 x 1012

CALCULATIONS2) Add and Subtract: Least precise decimal position

Examples:

a) 212.2 + 26.7 + 402.09

212.2 26.7402.09640.99

212.2 26.7402.09640.99

212.2 26.7402.09640.99

212.2 26.7402.09640.99

= 641.0

ADD AND SUBTRACT CON’T

Examples:

b) 1.0028 + 0.221 + 0.10337

1.00280.2210.103371.32717

1.00280.2210.103371.32717

1.00280.2210.103371.32717

1.00280.2210.103371.32717

= 1.327

ADD AND SUBTRACT CON’T

Examples:

c) 102.01 + 0.0023 + 0.15

102.01 0.0023 0.15102.1623

102.01 0.0023 0.15102.1623

102.01 0.0023 0.15102.1623

102.01 0.0023 0.15102.1623

= 102.16

ADD AND SUBTRACT CON’T

Examples:

d) 1.000 x 104 - 1

10000- 1 9999

10000- 1 9999

10000- 1 9999

= 1.000 x 104

ADD AND SUBTRACT CON’T

Examples:

e) 55.0001 + 0.0002 + 0.104

55.0001 0.0002 0.10455.1043

55.0001 0.0002 0.10455.1043

55.0001 0.0002 0.10455.1043

= 55.104

ADD AND SUBTRACT CON’T

Examples:

f) 1.02 x 103 + 1.02 x 102 + 1.02 x 101

1020 102 10.21132.2

1020 102 10.21132.2

1020 102 10.21132.2

1020 102 10.21132.2

= 1130

MIX PRACTICEExamples:

a) 52.331 + 26.01 - 0.9981 = 77.34= 77.3429b) 2.0944 + 0.0003233 + 12.22

7.001= 2.04466

= 2.04

c) 1.42 x 102 + 1.021 x 103

3.1 x 10 -1

= 3751.613

= 3.8 x 102

d) (6.1982 x 10-4) 2 = 3.841768 x 10-7= 3.8418 x 10-7

e) (2.3232 + 0.2034 - 0.16) x 4.0 x 103

= 9480

= 9500

Why the Metric System?

International unit of measurement: SI units Base units Derived units

Based on units of 10’s

LENGTHMeasure distances or dimensions in spaceMeter (m)Length traveled by light in a vacuum in 1/299792458 seconds.

MASSMeasure of quantity of matterKilogram (kg)Mass of a prototype platinum-iridium cylinder

TIMEForward flow of eventsSecond (s)Time is the radiation frequency of the cesium-133 atom.

VOLUMEAmount of space an object occupiesCubic meter (m3) Derived unit1 mL = 1 cm3

METRIC PREFIXESPREFIX SYMBO

LDEFINITION

MEGA- M 106 = 1,000,000

KILO- k 103 = 1000

HECTO-

h 102 = 100

DECA- da 101 = 10

BASE 100 = 1

DECI- d 10-1 = 0.1 = 1/10

CENTI- c 10-2 = 0.01 = 1/100

MILLI- m 10-3 = 0.001 = 1/1000

MICRO- μ 10-6 = 0.000001 = 1/1,000,000

NANO- n 10-9 = 0.000000001 = 1/1,000,000,000

DIMENSIONAL ANALYSIS

Process to solve problemsFactor-Label MethodDimensions of equation may be checked

DIMENSIONAL ANALYSIS

Examples:a) 3 years = _______seconds 1 year = 365 days 1 day = 24 hours 1 hour = 60 minutes 1 min = 60 seconds3 years

1 year

365 days

1 day

24 hours

1 hour

60 minutes 1

minute

60 seconds

= 94608000 seconds

= 9 x 10 7 seconds

DIMENSIONAL ANALYSIS

Examples:b) 300.100 mL = ________kL 1 L = 1000 mL 1 kL = 1000 L 300.100 mL

1000 mL

1 L

1000 L

1 kL

= 3.001 x 10-4 kL = 3.00100 x 10 –4

kL

DIMENSIONAL ANALYSIS

Examples:c) 9.450 x 109 Mg = _________dg 1 Mg = 10 6 g 1 g = 10 dg 9.450 x 109 Mg

1 Mg

10 6 g

1 g

10 dg

= 9.450 x 1016 dg

DIMENSIONAL ANALYSIS

Examples:d) 2.356 g OH- = __________ molecules OH-

1 mole = 17 g OH-

1 mole = 6.022 x 10 23 molecules 2.356 g OH -

17 g OH -

1 mole OH -

1 mole OH -

6.022 x 1023 molecules

= 8.34578 x 1022 molecules = 8.346 x 10 22

molecules

DIMENSIONAL ANALYSIS

Examples:e) 45.00 km = __________cm 1 km = 1000 m 1 m = 100 cm 45.00 km

1 km

1000 m

1 m

100 cm

= 4500000 cm= 4.500 x 10 6 cm

DIMENSIONAL ANALYSIS

Examples:f) 6.7 x 1099 seconds = _______years 1 year = 365 days 1 day = 24 hours 1 hour = 60 minutes 1 min = 60 seconds6.7 x 1099 seconds

60 seconds

1 minute

60 minutes

1 hours

24 hours

1 day

365 days

1 year

= 2.124556 x 1092 years

= 2.1 x 10 92 years

DIMENSIONAL ANALYSIS

Examples:g) 1.2400 g He = __________ Liters He 1 mole = 4 g He 1 mole = 22.4 L 1.2400 g He

4 g He

1 mole He

1 mole He

22.4 Liters He

= 6.944 Liters He= 6.9440 Liters He

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