ConversionFactors

ConversionFactors - Module 4 – Conversion Factors Module...

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Unformatted text preview: Module 4 – Conversion Factors Module 4 – Conversion Factors Prerequisites: Module 4 requires knowledge of exponential math and metric fundamentals in Lessons 1A, 1B, 2A, and 2B. The other lessons in Modules 1-3 will be helpful, but not essential, for most of Module 4. ***** Introduction The material in Module 4 may be a review of what you have learned previously. Each lesson will include suggestions for how you can complete this review quickly. However, some rules in Module 4 that you may not know, such as “writing the wanted unit first,” and using the “starting template,” will be important in solving problems in later lessons. Therefore, be sure to read each section and do at least the last two problems in each practice set. ***** Lesson 4A: Conversion Factor Basics Conversion factors can be used to change from one unit of measure to another, or to find measures of substances or processes that are equivalent. Conversion factor is a term for a ratio (a fraction) made from two measured quantities that are equal or equivalent in a problem. A conversion factor is a fraction that equals one. Conversion factors equal unity (1) because they are made from equalities. For any fraction in which the top and bottom are equal, its value is one. For example: 7 =1 7 Or, since 1 milliliter = 10─3 liters ; 10─3 L = 1 1 mL and 1 mL =1 10─3 L These last two fractions are typical conversion factors. Any fraction that equals one rightside up will also equal one up-side down. Any conversion factor can be inverted (flipped over) for use if necessary, and it will still be equal to one. When converting between liters and milliliters, all of these are legal conversion factors: 1 mL 10─3 L 1000 mL 1L 103 mL 1L 3,000 mL 3L All are equal to one. All of those fractions are mathematically equivalent, because they all represent the same ratio. Upside down, each fraction is also legitimate conversion factor, because the top and bottom are equal, and its value is one. In these lessons, our preferred conversions will be those based on fundamental definitions, such as “milli- means 10─3.” However, the other forms above are legal and may be used in places in these lessons and in other textbooks. ©2009 www.ChemReview.Net v. s7 Page 53 Module 4 – Conversion Factors Let’s try an example of conversion-factor math. Try the following problem. Show your work on the page or in your problem notebook, then check your answer below. 7.5 kilometers • 103 meters = 1 kilometer Multiply ***** Answer ( * * * mean: cover below, write your answer, then check below.) 7.5 kilometers • 103 meters = (7.5 • 103) meters = 7.5 x 103 meters 1 kilometer 1 When these terms are multiplied, the “like units” on the top and bottom cancel, leaving meters as the unit on top. Since the conversion factor multiplies the given quantity by one, the answer equals the given amount that we started with. This answer means that 7,500 meters is the same as 7.5 km. Multiplying a quantity by a conversion factor changes the units that measure a quantity but does not change the original amount of the given quantity. The result is what we started with, measured in different units. This process answers a question posed in many science problems: From the units we are given, how can we obtain the units we want? Our method of solving problems will focus on finding equal or equivalent quantities. Using those equalities, we will construct conversion factors to solve problems. ***** Summary • Conversion factors are made from two measured quantities that are either defined as equal or are equivalent or equal in the problem. • Conversion factors have a value of one, because the top and bottom terms are equal or equivalent. • Any equality can be made into a conversion (a fraction or ratio) equal to one. • When the units are set up to cancel correctly, given numbers and units multiplied by conversions will result the WANTED numbers and units. Units tell you where to write the numbers to solve a calculation correctly. To check the metric conversion factors encountered most often, use these rules. Since milli- = m- = 10─3, and conversions must be equal on the top and bottom, in metric conversions, • milli- or m- must be above or below 10─3 ; • centi- or c- must be above or below 10─2 ; • kilo- or k- must be above or below 103 . ©2009 www.ChemReview.Net v. s7 Page 54 Module 4 – Conversion Factors Practice: Try every other lettered problem. Check your answers frequently. If you miss one on a section, try a few more. Answers are on the next page. 1. Multiply the conversion factors. Cancel units that cancel, then group the numbers and do the math. Write the answer number and unit in scientific notation. a. b. 10─2 gram 1 centigram 225 centigrams • • 1 kilogram = 103 grams • 60 seconds = 1 minute 1.5 hours • 60 minutes 1 hour 2. To be legal, the top and bottom of conversion factors must be equal. Label these conversion factors as legal or illegal. b. 1000 L 1 mL a. 1000 mL 1 liter e. f. 103 cm3 1L 1 mL 1 cc c. 1.00 g H2O 1 mL H2O g. 103 kilowatts 1 watt d. h. 10─2 volt 1 centivolt 1 kilocalorie 103 calories 3. Add numbers to make legal conversion factors, with at least one of the numbers in each conversion factor being a 1. a. d. grams kilograms b. centijoules joules e. mole nanomole liters cubic cm c. f. cm3 mL curie picocurie 4. Finish these. a. 27A • 2T • 4W 8A 3T b. 2.5 meters • 1 cm 10─2 meter = c. 33 grams • 1 kilogram 103 grams = d. 95 km • hour e. 60 s • 60 min. • 1 kilometer = 27 meters • seconds 1 min. 1 hour 103 meters = 0.625 miles 1 km ©2009 www.ChemReview.Net v. s7 = Page 55 Module 4 – Conversion Factors ANSWERS 1 10─2 gram 1 centigram a. 225 centigrams • • 1 kilogram = 225 x 10─2 x 1 kg = 2.25 x 10─3 kg 103 grams 1 x 103 The answer means that 2.25 x 10─3 kg is equal to 225 cg. b. 1.5 hours • 60 minutes • 60 seconds = 1.5 x 60 x 60 s = 5,400 s or 5.4 x 103 s 1 hour 1 minute 1 Recall that s is the abbreviation for seconds. This answer means that 1.5 hours is equal to 5,400 s. b. 2. a. 1000 mL 1 liter Legal e. 3. a. c. 4. a. Illegal f. 103 cm3 1L Legal 1 mL_ 1 cc Legal 103 grams 1 kilogram b. 1 cm3 1 mL e. 1 liter 1000 cc. c. 1.00 g H2O 1 mL H2O 1000 L 1 mL d. or 27A • 2T • 4W 8A 3T 10─2 volt___ 1 centivolt Legal IF liquid water g. 103 kilowatts 1 watt Illegal 10─9 mole 1 nanomole 1 centijoule 10─2 joules 10─3 liters 1 cubic cm d. h. 1 kilocalorie_ 103 calories Legal 1 mole or 109 nanomole 100 centijoules 1 joule or f. Legal 10─12 curie 1 picocurie = 27A • 2T • 4W 8A 3T = or 1 curie 1012 picocurie 27 • 2 • 4 • W = 9W 8• 3 = 2.5 x 102 = 250 cm b. 2.5 meters • c. 33 grams • 1 kilogram 103 grams d. 95 km • 0.625 miles = 95 • 0.625 = 59 miles 1 hour hour 1 km e. 27 meters • 60 s • 60 min • 1 kilometer = 27 • 60 • 60 = 97 km 1 min 1 hour 103 meters 103 hour seconds 1 cm ─2 meter 10 = 33 = 0.033 kg 103 ***** ©2009 www.ChemReview.Net v. s7 Page 56 Module 4 – Conversion Factors Lesson 4B: Single Step Conversions In the previous lesson, conversion factors were supplied. In this lesson, you will learn to make your own conversion factors to solve problems. Let’s use this simple word problem as an example. Q. How many years is 925 days? In your notebook, write an answer to each step below. Steps for Solving with Conversion Factors 1. Begin by writing a question mark (?) and then the unit you are looking for in the problem, the answer unit. 2. Next write an equal sign. It means, “OK, that part of the problem is done. From here on, leave the answer unit alone.” You don’t cancel the answer unit, and you don’t multiply by it. 3. After the = sign, write the number and unit you are given (the known quantity). At this point, in your notebook should be ? years = 925 days 4. Next, write a • and a line _______ for a conversion factor to multiply by. 5. A key step: write the unit of the given quantity in the denominator (on the bottom) of the conversion factor. Leave room for a number in front. Do not put the given number in the conversion factor -- just the given unit. ? years = 925 days • ______________ days 6. Next, write the answer unit on the top of the conversion factor. ? years = 925 days • year days 7. Add numbers that make the numerator and denominator of the conversion factor equal. In a legal conversion factor, the top and bottom quantities must be equal or equivalent. 8. Cancel the units that you set up to cancel. 9. If the unit on the right side after cancellation is the answer unit, stop adding conversions. Write an = sign. Multiply the given quantity by the conversion factor. Write the number and the un-canceled unit. Done! Finish the above steps, then check your answer below. ***** ? years = 925 days • 1 year = 925 years = 2.53 years 365 365 days (SF: 1 is exact, 925 has 3 sf, 365 has 3 sf (1 yr. = 365.24 days is more precise), round to 3 sf.) You may need to look back at the above steps, but you should not need to memorize them. By doing the following problems, you will quickly learn what you need to know. ©2009 www.ChemReview.Net v. s7 Page 57 Module 4 – Conversion Factors Practice: After each numbered problem, check your answers at the end of this lesson. Look back at the steps if needed. In the problems in this practice section, write conversions in which one of the numbers (in the numerator or the denominator) is a 1. If these are easy, do every second or third letter. If you miss a few, do a few more. 1. Add numbers to make these conversion factors legal, cancel the units that cancel, multiply the given by the conversion, and write your answer. a. ? cm = 0.35 meters • 1 cm = meter b. ? days = 96 hours • day = 24 hours c. ? mL = 3.50 liters • 1 mL liter = minutes = seconds d. ? minutes = 330 s • 2. To start these, put the unit of the given quantity where it will cancel. Then finish the conversion factor, do the math, and write your answer with its unit. a. ? seconds = 0.25 minutes • sec. = 1 kilogram = b. ? kilograms = 250 grams • 10 3 c. ? meters = 14 cm • = cm d. ? days = 2.73 years • e. ? years = 200. days • 1 = 365 = 3. You should not need to memorize the written rules for arranging conversion factors, however, it is helpful to memorize this “single unit starting template.” When solving for single units, begin with ? unit WANTED = number and UNIT given • _________________ UNIT given ©2009 www.ChemReview.Net v. s7 Page 58 Module 4 – Conversion Factors The template emphasizes that your first conversion factor puts the given unit (but not the given number) where it will cancel. = a. ? months = 5.0 years • b. ? liters = 350 mL • = c. ? minutes = 5.5 hours = 4. Use the starting template to find how many hours equal 390 minutes. ? 5. ? milligrams = 0.85 kg • gram • _____________ = kg gram ANSWERS Some but not all unit cancellations are shown. For your answer to be correct, it must include its unit. Your conversions may be in different formats, such as 1 meter = 100 cm or 1 cm = 10─2 meters , as long as the top and bottom are equal and the result is the same answers as below. 1. a. ? cm = 0.35 meters • = 0.35 • 102 cm = 35 cm 1 cm 10─2 meter (c- = centi- = 10─2. SF: 0.35 has 2 sf, prefix definitions are exact with infinite sf, round to 2 sf) b. ? days = 96 hours • 1 day 24 hours = 96 days = 4.0 days 24 1 mL = 3.50 • 103 mL = 3.50 x 103 mL 10─3 liter (SF: 3.50 has 3 sf, prefix definitions are exact with infinite sf, answer is rounded to 3 sf) c. ? mL = 3.50 liters • d ? minutes = 330 sec. • 1 minute = 330 minutes = 5.5 minutes 60 60 seconds (SF: 330 has 2 sf, this time definition is exact with infinite sf, answer is rounded to 2 sf) ©2009 www.ChemReview.Net v. s7 Page 59 Module 4 – Conversion Factors 2. Your conversions may be different (for example, you may use 1,000 mL = 1 L or 1 mL = 10─3 L), but you must get the same answer. 60 sec. = 0.25 • 60 sec. = 15 s 1 minute (SF: 0.25 has 2 sf, 1 min = 60 sec. is a definition with infinite sf, answer is rounded to 2 sf) a. ? seconds = 0.25 minutes • b. ? kilograms = 250 grams • 1 kilogram 103 grams c. ? meters = 14 cm • 10─2 meter 1 cm = 250 kg = 0.25 kg 103 = 0.14 meters d. ? days = 2.73 years • 365 days 1 year = 2.73 • 365 days = 996 days e. ? years = 200. days • 1 year 365 days = 200 years = 0.548 years 365 = 60. months 12 months 1 year (SF: 5.0 has 2 sf, 12 mo. = 1 yr. is a definition with infinite sf, round to 2 sf , the 60. decimal means 2 sf) 3. a. ? months = 5.0 years • b. ? liters = 350 mL • 10─3 liter 1 mL = 350 x 10─3 liters = 0.35 L (m- = milli- = 10─3. SF: 350 has 2 sf, prefix definitions are exact with infinite sf, round to 2 sf) 60 minutes = 330 minutes 1 hour = 6.5 hours 4. ? hours = 390 minutes • 1 hour 60 minutes c. ? minutes = 5.5 hours • 1 mg = 0.85 x 106 mg = 8.5 x 105 mg 5. ? milligrams = 0.85 kg • 103 gram • 1 kg 10─3 gram ***** Lesson 4C: Multi-Step Conversions In Problem 5 at the end of the previous lesson, we did not know a direct conversion from kilograms to milligrams. However, we knew a conversion from kilograms to grams, and another from grams to milligrams. In most problems, you will not know a single conversion from the given to wanted unit, but there will be known conversions that you can chain together to solve. Try this two-step conversion, based on Problem 5 above. Answer in scientific notation. Q. ? milliseconds = 0.25 minutes ***** ©2009 www.ChemReview.Net v. s7 Page 60 Module 4 – Conversion Factors A. ? milliseconds = 0.25 minutes • 60 s • 1 ms = 15 x 103 ms = 1.5 x 104 ms 1 min. 10─3 s The 0.25 has two sf, both conversions are exact definitions that do not affect the significant figures in the answer, so the answer is written with two sf. The rules are, when Solving With Multiple Conversions • If the unit on the right after you cancel units is not the answer unit, get rid of it. Write it in the next conversion factor where it will cancel. • Finish the next conversion with a known conversion, one that either includes the answer unit, or gets you closer to the answer unit. • In making conversions, set up units to cancel, but add numbers that make legal conversions. • The numbers in any definition or equality that is exactly true have infinite significant figures and do not restrict the sf in an answer. Practice Write the seven metric fundamentals from memory. Use those fundamentals for the problems below. Convert your final answers to scientific notation. These are in pairs. If Part A is easy, go to Part A of the next question. If you need help with Part A, do Part B for more practice. 1. a. ? gigagrams = 760 milligrams • b. ? cg = 4.2 kg • gram • ______________ = • _______________ = g 2. a. ? years = 2.63 x 104 hours • b. ? seconds = 1.00 days • 3. a. ? µg H2O (l) = 1.5 cc H2O(l) • b. ? kg H2O(liquid) = 5.5 liter H2O(l)• ©2009 www.ChemReview.Net v. s7 • = hr • • = g H2O(l) • _____________ = • • = Page 61 Module 4 – Conversion Factors Always check metric prefix-conversions to be sure: • m- is above or below 10─3 , c- is above or below 10─2 , k- is above or below 103. ANSWERS For visibility, not all cancellations are shown, but cancellations should be marked on your paper. Your conversions may be different (for example, you may use 1,000 mL = 1 L or 1 mL = 10─3 L ), but you must get the same answer. 1a. ? gigagrams = 760 milligrams • 10─3 g • 1 mg b. ? cg = 4.2 kg • 103 g 1 kg • 2a. ? years = 2.63 x 104 hours • 1 Gg_ = 760 x 10─12 Gg = 7.6 x 10─10 Gg 109 g 1 cg _ = 4.2 x 105 cg 10─2 g 1 day 24 hr. • 1 yr = 2.63 x 104 = 3.00 x 100 years 365 days 24 • 365 b. ? seconds = 1.00 days • 24 hr • 60 min • 60 s 1 day 1 hr 1 min = 8.64 x 104 s 3a. ? µg H2O(l) = 1.5 cc H2O(l) • 1.00 g H2O(l) • 1 µg = 1.5 x 106 µg H2O(l) 1 cc H2O(l) b. ? kg H2O(l) = 5.5 liter H2O(l) • 10─6 g 0 1 mL _• 1.00 g H2O(l) • 1 kg = 5.5 x 10 kg H2O(l) 1 mL H2O(l) 10─3 L 103 g ***** Lesson 4D: English/Metric Conversions Using Familiar Conversions All of the unit conversions between units that we have used so far have had the number 1 on either the top or the bottom, but a one is not required in a legal conversion. Both “1 kilometer = 1,000 meters” and “3 kilometers = 3,000 meters” are true equalities, and both equalities could be used to make legal conversion factors. In most cases, however, conversions with a 1 are preferred. Why? We want conversions to be familiar, so that we can write them automatically, and quickly check that they are correct. Definitions are usually based on one of one component, such as “1 km = 103 meters.” Definitions are the most frequently encountered conversions and are therefore familiar and preferred. ©2009 www.ChemReview.Net v. s7 Page 62 Module 4 – Conversion Factors However, some conversions may be familiar even if they do not include a 1. For example, many cans of soft drinks are labeled “12.0 fluid ounces (355 mL).” This supplies an equality for English-to-metric volume units: 12.0 fluid ounces = 355 mL. That is a legal conversion and, because its numbers and units are seen often, it is a good conversion to use because it is easy to remember and check. Bridge Conversions Science problems often involve a key bridge conversion between one unit system, quantity, or substance, and another. For example, a bridge conversion between metric and English-system distance units is 2.54 centimeters ≡ 1 inch In countries that use English units, this is now the exact definition of an inch. Using this equality, we can convert between metric and English measurements of distance. Any metric-English distance equality can be used to convert between distance measurements in the two systems. Another metric-English conversion for distance that is frequently used (but not exact) is 1 mile = 1.61 km . (When determining the significant figures in an answer, for conversions based on equalities that are not exact, assume that an integer 1 is exact, but other numbers are precise only to the number of sf shown.) In problems that require bridge conversions, our strategy to begin will be to “head for the bridge,” to begin by converting to one of the two units in the bridge conversion. When a problem needs a bridge conversion, use these steps. 1) First convert the given unit to the unit in the bridge conversion that is in the same system as the given unit. 2) Next, multiply by the bridge conversion. The bridge conversion crosses over from the given system to the WANTED system. 3) Multiply by other conversions in the WANTED system to get the answer unit WANTED. Conversions between the metric and English systems provide a way to practice the bridgeconversion methods that we will use in chemical reaction calculations. Add these English distance-unit definitions to your list of memorized conversions. 12 inches ≡ 1 foot 3 feet ≡ 1 yard 5,280 feet ≡ 1 mile Memorize this metric to English bridge conversion for distance. 2.54 centimeters ≡ 1 inch Then cover the answer below and apply the steps and conversions above to this problem. Q. * * ? feet = 1.00 meter * ** ©2009 www.ChemReview.Net v. s7 Page 63 Module 4 – Conversion Factors Answer Since the wanted unit is English, and the given unit is metric, an English/metric bridge is needed. Step 1: Head for the bridge. Since the given unit (meters) is metric system, convert to the metric unit used in the bridge conversion (2.54 cm = 1 inch) -- centimeters. ? feet = 1.00 meter • 1 cm 10─2 m • _______ cm Note the start of the next conversion. Since cm is not the wanted answer unit, cm must be put in the next conversion where it will cancel. If you start the “next unit to cancel” conversion automatically after finishing the prior conversion, it helps to arrange and choose the next conversion. Adjust and complete your work if needed. ***** Step 2: Complete the bridge that converts to the system of the answer: English units. ? feet = 1.00 meter • * 1 cm • 10─2 m **** 1 inch __________ 2.54 cm inch Step 3: Get rid of the unit you’ve got. Get the unit you want. ? feet = 1.00 meter • 1 cm • 1 inch • 1 foot 10─2 m 2.54 cm 12 inches = 3.28 feet The answer tells us that 1.00 meter (the given quantity) is equal to 3.28 feet. Some science problems take 10 or more conversions to solve. However, if you know that a bridge conversion is needed, “heading for the bridge” breaks the problem into pieces, which will simplify your navigation to the answer. Practice: Use the inch-to-centimeter bridge conversion above. Start by doing every other problem. Do more if you need more practice. 1. ? cm = 12.0 inches • __________ = 2. ? inches = 1.00 meters • __________ • __________ 3. For ? inches = 760. mm a. To what unit to you aim to convert the given in the initial conversions? Why? b. Solve: ? inches = 760. mm ©2009 www.ChemReview.Net v. s7 Page 64 Module 4 – Conversion Factors 4. ? mm = 0.500 yards 5. For ? km = 1.00 mile , to convert using 1 inch = 2.54 cm , a. To what unit to you aim to convert the given in the initial conversions? Why? b. Solve: ? km = 1.00 mile 6. Use as a bridge for metric mass and English weight units, 1 kilogram = 2.2 lbs. ? grams = 7.7 lbs 7. Use the “soda can” volume conversion (12.0 fluid ounces = 355 mL). ? fluid ounces = 2.00 liters ANSWERS In these answers, some but not all of the unit cancellations are shown. The definition 1 cm = 10 mm may be used for mm to cm conversions. Doing so will change the number of conversions but not the answer. 1. ? cm = 12.0 inches • 2.54 cm = 12.0 • 2.54 = 30.5 cm (check how many cm are on a 12 inch ruler) 1 inch 2. ? inches = 1.00 meters • 1 cm • 1 inch = 1 x 102 = 0.394 x 102 in. = 39.4 inches 2.54 cm 2.54 10─2 m 3a. Aim to convert the given unit (mm) to the one unit in the bridge conversion that is in the same system (English or metric) as the given. Cm is the bridge unit that is in the same measurement system as mm. 3b. ? inches = 760. mm • 10─3 meter • 1 cm • 1 inch = 760 x 10─1 in. = 29.9 inches 2.54 1 mm 10─2 m 2.54 cm SF: 760., with the decimal after the 0, means 3 sf. Metric definitions and 1 have infinite sf. The answer must be rounded to 3 sf (see Module 2). 4. ? mm = 0.500 yd. • 3 ft. • 12 in. • 2.54 cm • 10─2 meter • 1 mm = 457 mm 1 yd. 1 ft. 1 inch 1 cm 10─3 m 5a. Aim to convert the given unit (miles) to the bridge unit in the same system (English or metric) as the given. Inches is in the same system as miles. 5b. ? km = 1.00 mile • 5,280 ft. • 12 in. • 2.54 cm • 10─2 m • 1 km 1 inch 1 cm 1 mile 1 ft. 103 m = 1.61 km SF: Assume an integer 1 that is part of any equality or conversion is exact, with infinite sf. ©2009 www.ChemReview.Net v. s7 Page 65 Module 4 – Conversion Factors 6. ? grams = 7.7 lbs • 1 kg • 103 grams 2.2 lb 1 kg = 3.5 x 103 grams SF: 7.7 and 2.2 have 2 sf. A 1 has infinite sf. Definitions, including metric-prefix definitions, have infinite sf. Round the answer to 2 sf. 7. ? fluid ounces = 2.00 liters • 1 mL • 12.0 fl. oz. = 67.6 fl. oz. 355 mL 10─3 L (Check this answer on any 2-liter soda bottle.) ***** Lesson 4E: Ratio Unit Conversions Long Distance Cancellation The order in which numbers are multiplied does not affect the result. For example, 1 x 2 x 3 has the same answer as 3 x 2 x 1. The same is true when multiplying symbols or units. While some sequences may be easier to set up or understand, from a mathematical perspective the order of multiplication does not affect the answer. The following problem is an example of how units can cancel in separated as well as adjacent conversions. Try Problem 1: Multiply these conversions. Cancel units that cancel. Write the answer number and its unit. 12 meters • 60 sec. • 60 min. • 1 kilometer • 0.62 miles = sec. 1 min 1 hour 1000 meters 1 kilometer ***** 12 meters • 60 sec. • 60 min. • 1 kilometer • 0.62 miles = 27 miles sec. 1 min. 1 hour 1000 meters 1 kilometer hr. This answer means that a speed of 12 meters/sec is the same as 27 miles/hour. ***** Ratio Units in the Answer In these lessons, we will use the term single unit to describe a unit that has a numerator but no denominator (which means the denominator is 1). Single units measure amounts. Meters, grams, minutes, milliliters are all single units. We will use the term ratio unit to describe a fraction that has one unit in the numerator and one unit in the denominator. If a problem asks you to find meters per second or meters/second or meters second or m • s─1 all of those terms are identical, and the problem is asking for a ratio unit. In conversion calculations, all ratio units must be written in the fraction form with a clear top and bottom. ©2009 www.ChemReview.Net v. s7 Page 66 Module 4 – Conversion Factors In Module 11, after we have worked with a wider variety of units, we will address in detail the different characteristics of single units and ratio units. For now, the distinctions above between single and ratio units will allow us to solve problems. Converting the Denominator In solving for single units, we have used a starting template that includes canceling a given single unit. When solving for single units, begin with ? unit WANTED = # and UNIT given • __________________ UNIT given When solving for ratio units, we may need to cancel a denominator (bottom) unit to start a problem. To do so, we will loosen our starting rule to say this. When Solving With Conversion Factors If a unit to the right of the equal sign, in or after the given, on the top or the bottom, • matches a unit in the answer unit, in both what it is and where it is, circle that unit on the right side and do not convert it further; • is not what you WANT, put it where it will cancel, and convert until it matches what you WANT. After canceling units, if the unit or units to the right of the equal sign match the answer unit, stop adding conversions, do the math, and write the answer. ***** Problem 2: Use the rule above for these. Do Part A first, and check your answer on the next page. Then do Part C. Do Part B if you need more practice. a. ? cm = 0.50 cm • __________ = min. s b. ? g = 355 g dL L • __________ = c. ? meters = 4.2 x 105 meters • __________ • __________ = second hour ***** ©2009 www.ChemReview.Net v. s7 Page 67 Module 4 – Conversion Factors Answers to Problem 2: a. ? cm = 0.50 cm min s • 60 s 1 min = 30. cm min Start by comparing the wanted units to the given units. Since you WANT cm on top, and are given cm on top, circle cm to say, “The top is done. Leave the top alone.” On the bottom, you have seconds, but you WANT minutes. Put seconds where it will cancel. Convert to minutes on the bottom. When the units on the right match the units you WANT on the left for the answer, stop conversions and do the math. b. ? g = 355 g dL L • 10─1 L 1 dL = 35.5 g dL c. ? meters = 4.2 x 105 meters • 1 hour • 1 min s hour 60 min 60 s = 1.2 x 102 meters s ***** Converting Both Top and Bottom Units Many problems require converting both numerator and denominator units. Which you convert first — the top or bottom unit — makes no difference. The order in which you multiply factors does not change the answer. ***** Problem 3: In these, an order of conversion is specified. Write what must be placed in the blanks to make legal conversions, cancel units, do the math, and write then check your answers below. a. ? meters = 740 cm • s min. meters • b. ? miles hour minutes_ = = 80.7 feet • sec. mile • __________ • __________= min. c. ? meters = s 250. feet • min. min. • ________ • _________ • __________= 1 inch Check your metric prefix- conversions: • m- is opposite 10─3 , c- is opposite 10─2 , k- is opposite 103. ***** ©2009 www.ChemReview.Net v. s7 Page 68 Module 4 – Conversion Factors ANSWERS to Problem 3: 3a. ? meters = 740 cm • 10─2 meters • s min 1 cm 1 min 60 s = 0.12 meters s In the given on the right, cm is not the unit WANTED on top, so put it where it will cancel, and convert to the unit you want on top. Next, since minutes are on the bottom on the right, but seconds are WANTED, put minutes where it will cancel. Convert to the seconds you WANT. 3b. ? miles hour = 80.7 feet • sec. 1 mile • 60 sec. • 60 min. = 55.0 miles 5,280 feet 1 min. 1 hour hour 3c. ? meters = 250. feet • 1 min. • 12 inches • 2.54 cm • 10─2 meter = 1.27 meters sec. min. 60 sec. 1 foot 1 inch 1 cm s (SF: 250. due to the decimal has 3 sf, all other conversions are definitions, answer is rounded to 3 sf) ***** Problem 4: In these, no order for the conversions is specified. Add legal conversions in any order, solve, and check your answers on the next page. Before doing the math, double check each conversion, one at a time, to make sure it is legal. a. ? centigrams liter b. ? km = hour c. ? ng = 0.550 x 10─2 g mL 1.17 x 104 mm sec mL d. ? 47 x 102 mg dm3 feet = sec. 95 meters minute = Do the prefix check. ***** ANSWERS to Problem 4: Your conversions may be in a different order. a. ? centigrams = 0.550 x 10─2 g • liter ©2009 www.ChemReview.Net mL v. s7 1 cg • 10─2 g 1 mL 10─3 L = 5.50 x 102 cg L Page 69 Module 4 – Conversion Factors b. ? km = 1.17 x 104 mm • 10─3 m • hour sec 1 mm c. ? ng = 47 x 102 mg • 1 dm3 • mL dm3 1L 1 km • 60 sec • 60 min = 42.1 km 1 min 1 hour hr 103 m 10─3 L 1 mL • 10─3 g • 1 ng 1 mg 10─9 g = 4.7 x 106 ng mL d. Hint: an English/metric bridge conversion for distance units is needed. Head for the bridge: convert the given metric distance unit to the metric distance unit used in your bridge conversion. ***** ? feet = sec. 95 meters • 1 min • 1 cm • 1 inch • 1 foot min 60 s 10─2 m 2.54 cm 12 in. = 5.2 feet s ***** Lesson 4F: Review Quiz For Modules 1-4 Use a calculator and scratch paper, but no notes or tables. State your answers to calculations in proper significant figures. Except as noted, convert your answers to scientific notation. To answer multiple choice questions, it is suggested that you • Solve as if the question is not multiple choice, • Then circle your answer among the choices provided. Set a 20-minute limit, then check your answers after the Summary that follows. ***** 1. 1023 = (1.25 x 1010)(4.0 x 10―6) a. 2.0 x 1018 b. 5.0 x 1018 c. 0.20 x 1019 d. 2.0 x 1020 e. 5.0 x 10―19 2. (─ 60.0 x 10―16) ─ (─ 4.29 x 10―14) = a. 4.8 x 10―16 b. 3.69 x 10―14 c. 3.7 x 10―14 d. 4.8 x 10―16 e. 4.89 x 10―14 3. 15.0 mL of liquid water has what mass in kg? a. 1.5 x 10―3 kg b. 15 x 10―4 kg c. 1.5 x 10―4 kg 2.0 s3 4. 5.00 x 10─2 L3 • m • 2.00 m • s 8.00 x 10─5 L2 a. 1.00 x 10―4 m2 • s2 • L d. 1.0 x 10―3 m • s2 • L ©2009 www.ChemReview.Net v. s7 d. 1.5 x 10―4 kg e. 1.5 x 10―2 kg • (an exact 2) = b. 5.00 x 103 m2 • s2 • L c. 5.0 x 103 m2 • s2 • L e. 5.0 x 10―3 m2 • s2 • L Page 70 Module 4 – Conversion Factors 5. State your answer in proper significant figures but do not convert to scientific notation. a. 255.00 b. 255.0 6. If 1 kg = 2.20 lb., a. 8.8 x 10―7 mg 7. ? kg = c. 255.008 d. 255.1 + e. 255.01 1 .008 238.00 16.00 ? mg = 4.0 x 10─2 lb. b. 8.8 x 104 mg c. 1.8 x 10―7 mg d. 1.8 x 104 mg e. 8.8 x 101 mg 2.4 x 105 μg dm3 mL a. 2.4 x 10―7 kg b. 2.4 x 105 kg mL mL ***** c. 2.4 x 10―10 kg d. 2.4 x 10―5 kg mL mL e. 2.4 x 10―4 kg mL SUMMARY: Conversion Factors 1. Conversion factors are fractions or ratios made from two entities that are equal or equivalent. Conversion factors have a value of one. 2. An equality can be written as a conversion or fraction or ratio equal to one. 3. In solving a problem, first write the unit WANTED, then an = sign. 4. Solving for single units, start conversion factors with ? unit WANTED = # and UNIT given • ________________ UNIT given 5. Finish each conversion with the answer unit or with a unit that takes you closer to the answer unit. 6. In making conversions, set up units to cancel, but add numbers that make legal conversions. 7. Chain your conversions so that the units cancel to get rid of the unit you’ve got and get to the unit you WANT. 8. When the unit on the right is the unit of the answer on the left, stop conversion factors. Evaluate the numbers. Write the answer and its unit. 9. Units determine the placement of the numbers to get the right answer. 10. If you plan on a career in a science-related field, add these to your flashcard collection. Front-side (with notch at top right): 1 inch = ? cm 1 kg = ? pounds 12 fluid oz. = ? mL Back Side -- Answers 2.54 cm 2.2 lb. 355 mL ***** ©2009 www.ChemReview.Net v. s7 Page 71 Module 4 – Conversion Factors ANSWERS – Module 1-4 Review Quiz Only partial solutions are provided below. 1. a. 2.0 x 1018 1/5 x 1023―10+6 = 0.20 x 1019 = 2.0 x 1018 (See Lesson 1C) 2. b. 3.69 x 10―14 (+ 4.29 x 10―14) ─ (0.600 x 10―14) = net doubt in hundredths place 3. e. 1.5 x 10―2 kg 1.00 g H2O = 1 mL H2O ; 1.00 kg H2O = 1 L H2O (Lessons 2A and 4D) 4. c. 5.0 x 103 m2 • s2 • L 5. e. 255.01 (2 sf and Lesson 2D) (Adding and subtracting, round to highest place with doubt. See Lesson 3A, 3B) 6. d. 1.8 x 104 mg 7. a. 2.4 x 10―7 kg/mL (Lessons 2B and 4D) L = dm3 (Lessons 2B and 4E) ##### ©2009 www.ChemReview.Net v. s7 Page 72 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.

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