# Chapter 2 Standards of Measurement Objectives:  Understand Mass and Weight (2.1)  Identify the metric units of measurement (2.6)  Explain what causes.

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Chapter 2 Standards of Measurement Objectives:  Understand Mass and Weight (2.1)  Identify the metric units of measurement (2.6)  Explain what causes uncertainty in measurements (2.7, 2.8 – 2.12)  Learn how to use significant digits and scientific notation (2.2 – 2.5)  Dimensional Analysis (2.8)  Density (2.12)

The Metric System (2.6) The International System of Units Standards of measurement Base units (7) – see Table 2.2 (pg 20) 1.MASS: kilogram (kg) 2.LENGTH: meter (m) 3.TIME: second (s) 4.COUNT, QUANTITY: mole (mol) 5.TEMPERATURE: Kelvin (K) 6.ELECTRIC CURRENT: ampere (A) 7.LUMINOUS INSTENSITY: candela (cd)

The Metric System Derived Units:  AREA: square meter, m 2  VOLUME: cubic meter, m 3  ENERGY: joule, J  FORCE: newton, N  PRESSURE: pascal, Pa  POWER: watt, W  VOLTAGE: volt, V  FREQUENCY: hertz, Hz  ELECTRIC CHARGE: coulomb, C

The Metric System Metric Prefixes – make base unit larger or smaller Table 2.1 – must know bolded prefixes Based on 10 Math method vs. “Stairs”

Convert a volume of 12 microliters into centiliters Express a distance of 15 meters in kilometers Convert 83 cm into meters Which is the longer amount of time, 1351 ps or 1.2 ns? Convert 16 dL into L Conversion Practice Answer: 0.0012 cL Answer: 0.015 km Answer: 0.83 m Answer: 1351 ps (1.2 ns = 1200 ns) Answer: 1.6 L

Uncertainty in Measurement Why are digits in measurements uncertain? 1.Instruments never completely free of flaws 2.Always involves estimation  Choose the right instrument for the job  May be estimated for you (electronic scales)  Scale is marked but you estimate the in- between

Uncertainty in Measurement Precision: getting the same result again and again under same conditions Accuracy: close to accepted value

Significant Digits All digits known with certainty plus one final digit which is uncertain (or estimated) All non-zeros are significant (143.34) A zero is significant when : –It is between nonzero digits (2004) –It is at the end of a number that includes a decimal point (23.560) A zero is not significant when: –It is before the first nonzero digit (0.25) –It is at the end of a number without a decimal point (2500)

Significant Digits - PRACTICE How many significant digits? 1. 54.23 2. 23.00005 3. 0.0004 4. 35000 5. 0.000504 6. 45.623200 7. 5,000,000 8. 4,000,000.1 ANSWERS: 1.4 sig figs 2.7 3.1 4.2 5.3 6.8 7.1 8.8

Significant Digits - Calculations Addition and Subtraction –Round answer to have final digit in the SAME PLACE as the last digit in the LEAST ACCURATE MEASUREMENT 1.21 + 5.002 + 10. = 16.212 becomes 16 34.5 + 12.45 + 23.0505 = 186.31 + 11.1 = 12.0231 + 3.86 = 0.100012 + 120. = 1200 + 12 + 15 + 0.5 = 1200 + 12 + 15 + 0.5 = 70.0005 becomes 70.0 197.41 becomes 197.4 15.8831 becomes 15.88 120.100012 becomes 120. 1227.5 becomes 1200

Significant Digits - Calculations Multiplication and Division –The answer has as many sig figs as the number with the fewest sig figs 14.8 x 3.1 = 45.88 becomes 46 18.2 x 3.0 = 52/1.5 = 321.868783 x 1 = 2400 x 2.123 = 15000/12.354 = 54.6 becomes 55 34.66666 becomes 35 321.868783 becomes 300 5095.2 becomes 5100 1214.181641 becomes 1200

Scientific Notation Convenient way of writing very large or very small numbers and showing only significant figures Number between 1 & 10 with a power of ten 5120 becomes 5.12 x 10 3 Move decimal point in original number to make number 1-10 Move left = +; move right = -

Scientific Notation Practice 123,000 = 0.000045 = 23.45 = 0.0000000003 = 1,000,000 = 1.23 x 10 5 4.5 x 10 -5 2.345 x 10 1 1 x 10 6 3 x 10 -10

Scientific Notation Adding and subtracting –Numbers must be the SAME POWER –1.4 x 10 4 + 2.1 x 10 5 (must change to 21.0 x 10 4 ) and then = 2.24 x 10 4 –3.2 x 10 3 + 1.8 x 10 2 = 3.38 x 10 3

Scientific Notation Multiplying –Add exponents –(2.0 x 10 3 ) x (3.0 x 10 4 ) = 6.0 x 10 7 Dividing –Subtract exponents –(8.2 x 10 8 ) x (4.1 x 10 4 ) = 2.0 x 10 4

Types of Measurements Mass – amount of matter in a body –Expressed in grams, kilograms, etc. –Does not change Weight – measure of earth’s gravitational attraction for that object –Expressed in same units –Changes with location (position on earth or distance from earth)

Types of Measurements Volume – the amount of space occupied by matter –Cubic meter or liter –Many instruments to measure Temperature – measure of the intensity of heat (figure 2.6) –Kelvin –Degrees Celsius –Degress Farenheit

Conversion Factors Enable movement between metric system and “English” system See back cover of book and Appendix III Common conversions you should memorize –1 inch = 2.54 cm –1 mile = 1.609 km –1 kg = 2.20 pounds –1 mL = 1 cm 3 –0 K = -273.15 0 C – 0 F = 1.8( 0 C) + 32

Dimensional Analysis (Problem Solving) Remember: ALWAYS use UNITS OF MEASUREMENT in your work!!! A technique of converting between units –Same system (metrics) –Different systems (inches to meters) –Chemical equations….later chapters…

Dimensional Analysis (Problem Solving) Conversion Factors: ratio derived from the equality between 2 different units 3 feet = 1 1 dollar = 1 1 yard4 quarters 1 yard4 quarters CF can be written either way 1 minute = 1 60 seconds = 1 1 minute = 1 60 seconds = 1 60 seconds 1 minute

Dimensional Analysis (Problem Solving) The “t” method unit given unit wanted = unit wanted unit given Example: How many liters are in 125.6 gallons? 125.6 gallons3.785 Liters = 1 gallon Conversion Factor 475.4 L

4.15 hours 60 minutes60 seconds = 1 hour 1 minute 1.5 mL 1 L 1 gal 4 qts 4 cups = 1000 mL 3.785 L 1 gal 1 qt How many seconds are in 4.15 hours? Dimensional Analysis (Problem Solving) If a student needs 1.5 mL of water, how many cups does he need? 14900 s 0.0063 cups

Density Common ratio used in chemistry Physical property of a substance Mass/volume D = m v SI units: kg/m 3 Solid –g/cm 3Liquid –g/mLGas –g/L Can change due to temperature and/or pressure changes

m = d x v = 0.24 g/cm 3 x 2 cm 3 = Density 1.Find the density of a piece of metal with a volume of 2.7 cm 3 and a mass of 10.8 g. D = m v = 10.8 g 2.7 cm 3 = 4.0 g/cm 3 2. Determine the mass of an object with a density of 0.24 g/cm 3 and a volume of 2 cm 3. SIG FIGS!!! 0.5 g 0.48

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