Significant Figues - Units - Density

Significant Figues - Units - Density - Rules for SF...

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Unformatted text preview: Rules for SF Significant Figures, Units, Percentages, and Density J.P. Harris CH104/Fall 2003 Lecture 3 • Exact numbers have an infinite number of SF (ex. 100 cm = 1 m EXACTLY) • SF must be counted for measurements – Nonzero digits are always SF – Zeros • Leading zeros are never SF • In the middle zeros are always SF • Trailing zeros – with a decimal point are SF – without a decimal point are not SF How many significant figures are in each of the following numbers? 1. 2. 3. 4. 5. 6. 7. 8. 0.02050 0.205 2.05 20.5 205 2050 2.05 × 104 2.050 × 106 1. 2. 3. 4. 5. 6. 7. 8. 4 SF; 0.02050 3 SF; 0.205 3 SF; 2.05 3 SF; 20.5 3 SF; 205 3 SF; 2050 3 SF; 2.05 × 104 4 SF; 2.050 × 106 SF, ppd, and Calculations • × and ÷ – The product or quotient of this operation will have the same number of SF as the number ×’d or ÷’d with the fewest SF. • + and – The sum or difference of this operation will have the same number of places past the decimal (ppd) as the number +’d or -’d with the fewest ppd. Calculate the answers for the following problems and express the answer using an appropriate number of significant figures. Assume all of the numbers are the results of measurements. SF, ppd, and Calculations (part 2) • Calculations with more than one step – Don’t round until the end of the problem! – Trace SF and ppd through the problem from the beginning of the calculation • If the result of a multiplication step is used in an addition step, use SF for the multiplication step, then “translate” the SF into ppd. • If the result of an addition step is used in a multiplication step, use ppd for the addition step, then “translate” the ppd into SF. 1. (1.02 × 10-21)(1.1 × 109)2 = 1.2 × 10-3 2. (251)(3.1×10-1) = 1.1 (24)(3.0) 3. (2.93 × 10-1) + (6.2 × 10-2) = 3.55× 10-1 4. 471.19 - 365.09 = 106.10 5. 17.76 - 0.0479 = 17.71 1 Calculate the answers for the following problems and express the answer using an appropriate number of significant figures. Assume all of the numbers are the results of measurements. When will I use this? • When performing calculations for this class. 1. Set-up of the problem (SHOW YOUR WORK!) 2. Answer to the problem (number and units) 3. Significant Figures 1. 27.65 - 21.71 = 1.29 4.97 - 0.36 2. 12.47 – 203.4 = 0.78 6.97 201.8 3. 19.37 – 18.49 = 1.1 0.822 • In lab, the last digit of measurements should contain some uncertainty. – balances have uncertainty in the last digit already – Always guesstimate one digit past the smallest markings on glassware. • For example, if the smallest markings are 1 mL markings, then the reading should be made to the nearest 0.1 mL. Calculations and Conversion Factors • Last time we met, we made a list of equalities to memorize. – Example: 1000 mL = 1 L Express 275 mL in liters. • Identify what we know. • Identify what we want to know. • Do we know of a relationship between milliliters and liters? • Write a conversion factor with the desired units in the numerator (top) and the original units in the denominator (bottom). • Multiply the original number by the conversion factor. The original units should “cancel” and leave the new units. • volume in milliliters • volume in liters • Yes, 1000 mL = 1 L. • 1L 1000 mL • We can rewrite this equality as two different fractions that equal 1. 1000 mL = 1 1L 1L =1 1000 mL • We can multiply by these conversion factors to change units because they are “fancy versions of 1.” 1L 275 mL = 0.275 L 1000 mL Express 4.87 inches in meters. • Identify what we know. • Identify what we want to know. • Do we know of a relationship between inches and meters? • Make a plan. • Write conversion factors with the desired units in the numerator (top) and the original units in the denominator (bottom). • length in inches • length in meters • No, but we do know that 1 inch = 2.54 cm and 100 cm = 1 m. • Convert from in. to cm and from cm to m. 2.54 cm 1m and 1 inch 100 cm • Multiply the original number by the conversion factors. The original units should “cancel” and leave the new units. 2.54 cm 1 m 4.87 inches = 0.124 m 1 inch 100 cm • Remember, conversion factors are exact numbers and will not limit the significant figures of the answer! 2 Calculations and Conversion Factors • Last time we met, we made a list of equalities to memorize. – Example: 1000 mL = 1 L Express 275 mL in liters. • Identify what we know. • Identify what we want to know. • Do we know of a relationship between milliliters and liters? • Write a conversion factor with the desired units in the numerator (top) and the original units in the denominator (bottom). • Multiply the original number by the conversion factor. The original units should “cancel” and leave the new units. • volume in milliliters • volume in liters • Yes, 1000 mL = 1 L. • 1L 1000 mL • We can rewrite this equality as two different fractions that equal 1. 1000 mL = 1 1L 1L =1 1000 mL • We can multiply by these conversion factors to change units because they are “fancy versions of 1.” 1L 275 mL = 0.275 L 1000 mL Express 4.87 inches in meters. • Identify what we know. • Identify what we want to know. • Do we know of a relationship between inches and meters? • Make a plan. • Write conversion factors with the desired units in the numerator (top) and the original units in the denominator (bottom). • length in inches • length in meters • No, but we do know that 1 inch = 2.54 cm and 100 cm = 1 m. • Convert from in. to cm and from cm to m. 2.54 cm 1m and 1 inch 100 cm • Multiply the original number by the conversion factors. The original units should “cancel” and leave the new units. 2.54 cm 1 m 4.87 inches = 0.124 m 1 inch 100 cm • Remember, conversion factors are exact numbers and will not limit the significant figures of the answer! Express 35 kJ in calories. Use scientific notation. • • • • know energy in kJ want energy in calories 1 kJ = 1000 J and 4.184 J = 1 calorie Plan = kJ → J → calories 1000 J 1 calorie and 1 kJ 4.184 J 1000 J 1 calorie 3 35 kJ = 8.4 × 10 cal 1 kJ 4.184 J How do calories relate to Calories? • calorie = scientific calorie = amount of energy required to raise the temperature of 1 g of water by 1°C • kcal = 1000 scientific calories • Calorie = Nutritional Calorie • 1 Calorie = 1 kcal • 1 Nutritional Calorie is the amount of energy required to raise the temperature of 1000 g of water by 1°C • 1 Calorie = 1000 calories 1 Express 35 kJ in Nutritional Calories. • know energy in kJ • want energy in Nutritional Calories • 1 kJ = 1000 J, 4.184 J = 1 sci. calorie, 1000 cal = 1 kcal, & 1 kcal = 1 Nutritional Calorie • Plan N- - VFL FDORULHV NFDO 1XW &DORULH 1000 J 1 calorie 1 kcal 1 Calorie and and and 1 kJ 4.184 J 1000 calorie 1 kcal § 1000 J · § 1 calorie · § · § 1 Calorie · 1 kcal 35 kJ ¨ ¸¨ ¸¨ ¸¨ ¸ © 1 kJ ¹ © 4.184 J ¹ © 1000 calories ¹ © 1 kcal ¹ = 8.4 Calories Percentages %= part × 100 whole • In a class of 64 students, 28 people are wearing white socks. What is the percentage of people NOT wearing white socks? 64 people − 28 people × 100 = 56.25% 64 people • What about the SF? • These are exact numbers, so we can report all the numbers from our calculators! If a sample of 6.022 × 1023 iron atoms is 91.75% iron-56, how many iron-56 atoms are present in the sample? 91.75% = part × 100 6.022 × 1023 atoms atoms 91.75% = part × 100 23 Density density = mass volume ( 6.022 × 10 23 ) • Ratio between mass and volume • Can be used as a conversion factor between mass and volume – Example: The density of iron is 7.874 g/cm3. 7.874 g 1 cm3 and 1 cm3 7.874 g ( 6.022 × 10 atoms 91.75% ) 100 = part – What is the mass of an 84 cm3 iron statue? 23 part = 5.525 × 10 iron − 56 atoms 7.874 g 84 cm3 = 661.416 g ≈ 660 g = 6.6 x 102 g 3 1 cm What is the volume of a 825 g iron paper weight? 1 cm3 825 g = 7.874 g 1 cm3 3 825 g = 104.77520955 cm 7.874 g 1 cm3 3 3 825 g = 104.77520955 cm ≈ 105 cm 7.874 g A sample of course rock salt is found to have a mass of 11.7 g. The rock salt is placed into a graduated cylinder that contains kerosene. The rock salt does not float or dissolve. The kerosene level in the cylinder was at a level of 20.7 mL before adding the salt and 26.1 mL after adding the salt. What is the volume of the salt? V = 26.1 mL − 20.7 mL = 5.4 mL What is the density of the salt? d= 11.7 g 11.7 g = 26.1 mL − 20.7 mL 5.4 mL g g = 2.16 = 2.2 mL mL 2 ...
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This note was uploaded on 12/05/2010 for the course CHM 238 taught by Professor Arney during the Spring '07 term at Sam Houston State University.

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