DIMENSIONAL ANALYSIS (FACTOR LABEL M ETHOD)

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DIMENSIONAL ANALYSIS (FACTOR LABEL M ETHOD) Name Using this method, it is possible to solve many problems by using the relationship of one unit to another. For example, 12 inches =one foot. Since these two numbers represent the same value, the fractions 12 in/1 ft and 1 ft/12 in are both equal to one. When you multiply another number by the number one, you do not change its value. However, you may change its unit. Example 1: Convert 2 miles to inches. 2 miles x 5.280 ft x 12 inches = 126,720 in 1 mile 1 ft (Using significant figures, 2 mi = 100,000 in.) Example 2: How many seconds are in 4 days? 4 days x 24 hrs x 60 min x 60 sec = 345,600 sec 1 day 1 hr 1 min (Using significant figures, 4 days =300,000 sec.) Solve the following problems. Write the answers in significant figures. 1. 3 hrs = sec 2. 0.035 mg = cg 3. 5.5 kg = Ibs 4. 2.5 yds = in 5. 1.3 yrs = hr (1 yr =365 days) 6. 3 moles = molecules (1 mole =6.02 x 10 23 molecules) 7. 2.5 x 10 24 molecules = moles 8. 5 moles = liters (1 mole =22.4 liters) 9. 100. liters = moles 10. 50. liters = molecules 11. 5.0 x 10 24 molecules = liters 12. 7.5x103mL = liters Chemistry ff8766 6 Instructional Fair, Inc.

METRICS AND MEASUREMENT Name _---:::. In the chemistry classroom and lab, the metric system of measurement is used, so it is important to be able to convert from one unit to another. mega kilo hecto deca (M) (k) (da) 1,000,000 1 000 100 10 10 6 10 3 10 2 _ 10'.._-- Basic Unit gram (g) liter (L) meter (m) deci centi milli micro (d) (c) (m) (p.).1.01.001.000001 10" 10.2 10.3 10.6 - - Factor Label Method 1. Write the given number and unilt. 2. Set up a conversion factor (fraction used to convert one unit to another). a. Place the given unit as denominator of conversion factor. b. Place desired unit as numerator. c. Place a \\ 1" in front of the larger unit. d. Determine the number of smaller units needed to make \\ 1" of the larger unit. 3. Cancel units. Solve the problem. Example 1: 55 mm = m Example 2: 88 km = m 55 rnm 11m =0.055 m I 1000 m = 88,000 m 1000Jru'll 1jmi Example 3: 7000 em = hm Example 4: 8 dol = dl 7000)::R'r1 1J>'I' I 1 hm = 0.7 hm 8.de1t I lox 10 dl =800 dl 100cm loop>( 1.dert 1): The factor label method can be used to solve virtually any problem including changes in units. It is especially useful in making complex conversions dealing with concentrations and d erived units. Convert the following. 1. 35 ml = dl 6. 4,500 mg = g 2. 950 g = kg 7. 25cm = mm 3. 275 mm = cm 8. 0.005 kg = dag 4. 1,000 L = kl 9. 0.075 m = cm 5. 1,000 ml = L 10. 15 g = mg Chemistry IF8766 7 Instructlonal Fair, Inc.

SCI NTIFIC NOTATION Name Scientists very often deal with very small and very large numbers, which can lead to a lot of confusion when counting zeros! We have learned to express these numbers as powers of 10. Scientific notation takes the form of M x 10 n where 1 M < 10 and "nil represents the number of decimal places to be moved. Positive n indicates the standard form is a 'large number. Negative n indicates a number between zero and one. Example 1: Convert 1,500,000 to scientific notation. We move the decimal point so that there is only one digit to its left, a total of 6 places. 1,500,000 = 1.5 x 10 6 Example 2: Convert 0.000025 to scientific notation. For this, we move the decimal point 5 places to the right. 0.000025 = 2.5 x 1Q-5 (Note that when a number starts out less than one, the exponent is always negative.) Convert the fo'llowing to scientific notation. 1. 0.005 = 6. 0.25 = 2. 5,050 = 7. 0.025 = 3. 0.0008 = 8. 0.0025 = 4. 1,000 = 9. 500 = 5. 1,000,000 = 10. 5,000 = Convert the following to standard notation. 1. 1.5 X 10 3 = 6. 3.35 X 10-1 = 2. 1.5 X 10-3 = 7. 1.2 x 10-4 = 3. 3.75 X 10-2 = 8. lxl()4= 4. 3.75 x 10 2 = 9. 1 X 10-1 = 5. 2.2x los = 10. 4xl00 = Chemistry IF8766 8 Instructional Fair, Inc.

SIGNIFICANT -FIGURES Name _----'-_"----- ---'" A measurement can only be as accurate and precise as the instrument that produced it. A scientist must be able to express the accuracy of a number, not just its numerical value. We can determine the accuracy of a number by the number of significant figures it contains. 1) All digits 1-9 inclusive are significant. Example: 129 has 3 significant figures. 2) Zeros between significant digits are always significant. Example: 5.007 has 4 significant figures. 3) Trailing zeros in a number are significant QDjy if the number contains a decimal point. Example: 100.0 has 4 significant figures. 100 has 1 significant figure. 4) Zeros in the beginning of a number whose only function is to place the decimal point are not significant. Exam ple: 0.0025 has 2 significant figures. 5) Zeros following a decimal significant figure are significant. Example: 0.000470 has 3 significant figures. 0.47000 has 5 significant figures. Determine the number of significant figures in the following numbers. 1. 0.02 6. 5,000. 2. 0.020 7. 6,051.00 3. 501 B. 0.0005 4. 501.0 9. 0.1020 5. 5,000 10. 10,001 Determine the location of the last significant place value by placing a bar over the digit. (Example: 1.700) 1. B040 6. 90,100 2. 0.0300 7. 4.7 x 10-8 3. 699.5 B. 10,BOO,000. 4. 2.000 x 10 2 9. 3.01 X 10 21 5. 0.90100 10. 0.000410 Chemistry IF8766 9 Instructional Fair, Inc.

CALCULATIONS USING SIGNIFICANT FIGURES When multiplying and dividing, limit and round to any of the factors. Name least number of significant figures in Example 1: 23.0 cm x 432 cm x 19 cm = 188,784 cm 3 The answer is expressed as 190,000 cm 3 since 19 cm has only two significant figures. When adding and subtracting, limit and round your answer to the least number of decimal places in any of the numbers that make up your answer. Example 2: 123.25 ml + 46.0 ml + 86.257 ml = 255.507 ml II The answer is expressed as 255.5 ml since 46.0 ml has only lone decimal place. Perform the following operations expressing the answer in the correct number of significant figures. 1. 1.35 m x 2.467 m = 2. 1,035 m 2 + 42 m = 3. 12.01 ml + 35.2 ml + 6 ml = 4. 55.46 g - 28.9 g = 5..021 cm x 3,2 cm x 100.1 cm = 6. 0.15cm + 1.15 cm + 2,051 cm = 7, 150L3 + 4L = 8. 505 kg - 450.25 kg = 9. 1.252 mm x 0.115 mm x 0.012 mm = 10, 1.278 x 10 3 m 2 + 1,4267 x 10 2 m = Chemistry IF8766 10 Instructional Fair, Inc.