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Week 4

by: Mikaela Notetaker

Week 4 CHEM 110

Mikaela Notetaker

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About this Document

Covers conversion and Temperature Scale
Introduction to Chemistry
Melissa G. Ely
Class Notes
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This 2 page Class Notes was uploaded by Mikaela Notetaker on Sunday February 21, 2016. The Class Notes belongs to CHEM 110 at West Virginia University taught by Melissa G. Ely in Spring 2016. Since its upload, it has received 21 views. For similar materials see Introduction to Chemistry in Chemistry at West Virginia University.


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Date Created: 02/21/16
Conversion Between Different Measurement Systems If multiplying a number by one of these conversion factors you are not changing the value of the number in anyway just changing the unit. Use dimensional analysis to convert (let the units do the work) 63.5 in to ft (63.5 in / 1) x (1 ft / 12 in) (On paper the in would be diagonal from one another therefor cancelling each other out leaving only ft behind. Note that 1 ft is an exact number just like 12 in therefor they do not have sig fig like 63.5 so only 3 sig fig.) = 5.29 ft 10.070 qt -> L (10.070 qt / 1) x (.9463 L / 1 qt) = (10.070 x .9463 L) = 9.529 L Metric conversions used to express very large or small numbers. Length/Mass Conversion Kilometer 1 km = 1,000 m Meter base unit Centimeter 100 cm = 1 m Millimeter 1,000 mm = 1 m Micrometer 1,000,000 um = 1 m Decimeter 10 dc = 1 m Kilogram 1,000 g = 1 kg Decigram 10 dc = 1 g Centigram 100 cg = 1 g Milligram 1,000 mg = 1 g Complex Conversion 2.78 mi -> m (2.78 mi / 1) x (5280 ft / 1 mi) x (12 in / 1 ft) x (2.54 cm /31 in) x (1 m / 100 cm) = ((2.78 x 5280 x 12 x 2.54) / 100) = 4473.97 m ->(3 sf) 4.47 x 10 m 45 mi/g -> km/L (45 mi / 1 g) x (5280 ft / 1 mi) x (12 in / 1 ft) x (2.54 cm / 1 in) x (1 m / 100 cm) x (1 km / 1000 m) x (1 gal / 4 qt) x (1 qt / .9463 L) = 19.1625 -> (2 sf) 19 km/L Temperature Scales (sf is based off decimal place!) O 1. Fahrenheit (F ) 2. Celsius ( C) 3. Kelvin (K) 1. Degrees on Fahrenheit scale are approximately 2 times as large as those on celsius scale 2. Fahrenheit scale offset from Celsius by 32 Fahrenheit O TF= ((9/5) x C) + 32 Tc= (5/9) x (F - 32) = (((F-32) x 5) / 9) TK= C + 273.15 Tc= K – 273.15 211 C - > K 211 C + 273.15 = 484 K 83.12 F -> C (5/9) x (83.12 - 32) = (5 x 51.12) / 9 = 28.40 C 118 F -> K (5/9) x (118 -32) = (5 x 86) / 9 = 47.(77777778) + 273.15 = 321 K (No decimals since the 7’s and 8 were cut off since the last equation didn’t have a decimal place but we needed them for rounding purpose)


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