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1.4Important LinksDigital Support Online quizzes, study notes, and morewww.edvantagescience.comOrder your own copy of AP Chemistry 1…

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1.4Important LinksDigital Support Online quizzes, study notes, and morewww.edvantagescience.comOrder your own copy of AP Chemistry 1 www.edvantageinteractive.com1.4 Analysis of Units and Conversions in Chemistry Warm Up Place a check by the larger quantity in each row of the table. Metric QuantityImperial QuantityA kilogram of butterA pound of butterA five-kilometer hiking trailA five-mile mountain bike trailOne liter of milkOne quart of milkA twelve-centimeter rulerA twelve-inch rulerA fifteen-gram piece of chocolateA fifteen-ounce chocolate barA temperature of 22°CA temperature of 22°FMeasurement Through History Units of measurement were originally based on nature and everyday activities. The grain was derived from the mass of one grain of wheat or barley a farmer produced. The fathom was the distance between the tips of a sailor’s fingers when his arms were extended on either side. The origin of units of length like the foot and the hand leave little to the imagination. The inconvenient aspect of units such as these was, of course, their glaring lack of consistency. One “Viking’s embrace” might produce a fathom much shorter than another. These inconsistencies were so important to traders and travellers that over the years most of the commonly used units were standardized. Eventually, they were incorporated into what was known as the English units of measurement. Following the Battle of Hastings in 1066, Roman measures were added to the primarily Anglo-Saxon ones that made up this system. These units were standardized by the Magna Carta of 1215 and were studied and updated periodically until the UK Weights and Measures Act of 1824 resulted in a major review and a renaming to the Imperial system of measurement. It is interesting to note that the United States had become independent prior to this and did not adopt the Imperial system, but continued to use the older English units of measure. Despite the standardization, its development from ancient agriculture and trade has led to a rather vast set of units that are quite complicated. For example, there are eight different definitions for the amount of matter in a ton. The need for a simpler system, a system based on decimals, or multiples of 10, was recognized as early 1585 when Dutch mathematician, Simon Stevin published a book called The Tenth. However, most authorities credit Gabriel Mouton of Lyon, France, as the originator of the metric system nearly 100 years later. Another 100 years would pass before the final development and adoption of the metric system in France in 1795. Recall the International Bureau of Weights and Measures (BIPM) was established in Sévres, France, in 1825. The BIPM has governed the metric system ever since. Since 1960, the metric system has become the SI system. Its use and acceptance has grown to the point that only three countries in the entire world have not adopted it: Burma, Liberia, and the United States (Figure 1.4.1).44 Chapter 1 Skills and Processes of Chemistry© Edvantage Interactive 2017U.S.A.Burma LiberiaCountries using the metric system Countries not using the metric systemFigure 1.4.1 Map of the world showing countries that have adopted the SI/metric systemDimensional Analysis Dimensional analysis is a method that allows you to easily solve problems by converting from one unit to another through the use of conversion factors. The dimensional analysis method is sometimes called the factor label method. It may occasionally be referred to as the use of unitary rates. A conversion factor is a fraction or factor written so that the denominator and numerator are equivalent values with different units.One of the most useful conversion factors allows the user to convert from the metric to the imperial system and vice versa. Since 1 inch is exactly the same length as 2.54 cm, the factor may be expressed as: 2.54 cm 1 inch 2.54 cm or 1 inch0 cm 10 in23These two lengths are identical so multiplication of a given length by the conversion factor will not change the length. It will simply express it in a different unit. Now if you wish to determine how many centimeters are in a yard, you have two things to consider. First, which of the two forms of the conversion factor will allow you to cancel the imperial unit, converting it to a metric unit? Second, what other conversion factors will you need to complete the task? Assuming you know, or can access, these equivalencies: 1 yard = 3 feet and 1 foot = 12 inches, your approach would be as follows:1 1.00 yards ×Figure 1.4.2 A ruler with bothimperial and metric scales shows that 1 inch = 2.54 cm.© Edvantage Interactive 20172.54 cm 3 feet 12 inches × × = 91.4 cm 1 inch 1 yard 1 footNotice that as with the multiplication of any fractions, it is possible to cancel anything that appears in both the numerator and the denominator. We’ve simply followed a numerator-todenominator pattern to convert yards to feet to inches to cm. Chapter 1 Skills and Processes of Chemistry 45The number of feet in a yard and inches in a foot are defined values. They are not things we measured. Thus they do not affect the number of significant figures in our answer. This will be the case for any conversion factor in which the numerator and denominator are in the same system (both metric or both imperial). Interestingly enough, the BIPM has indicated that 2.54 cm will be exactly 1 inch. So it is the only multiple-system conversion factor that will not influence the number of significant figures in the answer to a calculation. As all three of the conversion factors we used are defined, only the original value of 1.00 yards influences the significant figures in our answer. Hence we round the answer to three sig figs.Converting Within the Metric System The metric system is based on powers of 10. The power of 10 is indicated by a simple prefix. Table 1.4.1 is a list of SI prefixes. Your teacher will indicate those that you need to commit to memory. You may wish to highlight them. Metric conversions require either one or two steps. You will recognize a one-step metric conversion by the presence of a base unit in the question. The common base units in the metric system include:Table 1.4.1 SI PrefixesMeasuresSymbol10nyottaY1024zettaZ1021exaEpetaPrefixUnit NameSymbollengthmetermmassgramg1018volumeliterLP1015timesecondsteraT1012gigaG109megaM106kilok103hectoh102decada101decid10-1centic10–2millim10–3microμ10–6nanon10–9picop10–12femtof10–15attoa10–18zeptoz10–21yoctoy10–24One-step metric conversions involve a base unit (meters, liters, grams, or seconds) being converted to a prefixed unit or a prefixed unit being converted to a base unit. Two-step metric conversions require the use of two conversion factors. Two factors will be required any time there are two prefixed units in the question. In a two-step metric conversion, you must always convert to the base unit first.46 Chapter 1 Skills and Processes of Chemistry© Edvantage Interactive 2017Sample Problems — One- and Two-Step Metric Conversions 1. Convert 9.4 nm into m. 2. Convert 6.32 μm into km.What to Think aboutHow to Do ItQuestion 1 1. In any metric conversion, you must decide whether you need one step or two. There is a base unit in the question and only one prefix. This problem requires only one step. Set the units up to convert nm into m. Let the units lead you through the problem. You are given 9.4 nm, so the conversion factor must have nm in the denominator so it will cancel. 2. Now determine the value of nano and fill it in appropriately. 1 nm = 10–9 m Give the answer with the appropriate number of significant figures and the correct unit. Because the conversion factor is a defined equality, only the given value affects the number of sig figs in the answer.9.4 nm ×m = nmm–9 9.4 nm × 10 m = 9.4 × 10–9 m 1 nmQuestion 2 1. This problem presents with two prefixes so there must be two steps. The first step in such a problem is always to convert to the base unit. Set up the units to convert from μm to m and then to km. 2. Insert the values for 1 μm and 1 km. 1 μm = 10–6 m 1 km = 103 m 3. Give the answer with the correct number of significant figures and the correct unit. m × μm6.32 μm ×km = mkm–6 6.32 μm × 10 m × 1 km = 6.32 × 10–9 km 103 m 1 μmPractice Problems — One- and Two-Step Metric Conversions 1. Convert 16 s into ks. 2. Convert 75 000 mL into L.3. Convert 457 ks into ms.4. Convert 5.6 × 10–4 Mm into dm.© Edvantage Interactive 2017Chapter 1 Skills and Processes of Chemistry 47Derived Unit ConversionsA derived unit is composed of more than one unit. Units like those used to express rate (km/h) or density (g/mL) are good examples of derived units. Derived unit conversions require cancellations in two directions (from numerator to denominator as usual AND from denominator to numerator).Sample Problem — Derived Unit Conversions Convert 55.0 km/h into m/sWhat to Think aboutHow to Do It1. The numerator requires conversion of a prefixed metric unit to a base metric unit. This portion involves one step only and is similar to sample problem one above.55.0 km × h2. The denominator involves a time conversion from hours to minutes to seconds. The denominator conversion usually follows the numerator. Always begin by putting all conversion factors in place using units only. Now that this has been done, insert the appropriate numerical values for each conversion factor.=m × kmh × minmin sm s55.0 km × 103 m × 1 h × 1 min 60 s h 1 km 60 min3. As always, state the answer with units and round to the correct number of significant figures (in this case, three). = 15.3 m sPractice Problems — Derived Unit Conversions 1. Convert 2.67 g/mL into kg/L. Why has the numerical value remained unchanged?2. Convert the density of neon gas from 8.9994 × 10–4 mg/mL into kg/L.3. Convert 35 mi/h (just over the speed limit in a U.S. city) into m/s. (Given: 5280 feet = 1 mile)48 Chapter 1 Skills and Processes of Chemistry© Edvantage Interactive 2017Use of a Value with a Derived Unit as a Conversion Factor A quantity expressed with a derived unit may be used to convert a unit that measures one thing into a unit that measures something else completely. The most common examples are the use of rate to convert between distance and time and the use of density to convert between mass and volume. These are challenging! The keys to this type of problem are determining which form of the conversion factor to use and where to start. Suppose we wish to use the speed of sound (330 m/s) to determine the time (in hours) required for an explosion to be heard 5.0 km away. It is always a good idea to begin any conversion problem by considering what we are trying to find. Begin with the end in mind. This allows us to decide where to begin. Do we start with 5.0 km or 330 m/s? First, consider: are you attempting to convert a unit ➝ unit, or aunit unit ➝ ? unit unitAs the answer is unit ➝ unit, begin with the single unit: km. The derived unit will serve as the conversion factor. Second, which of the two possible forms of the conversion factor will allow 330 m 1s conversion of a distance in km into a time in h? Do we require or ? As 1s 330 m the distance unit must cancel, the second form is the one we require. Hence, the correct approach is 5.0 km ×1 min 1h 103 m 1s × × × = 4.2 × 10–3 h 60 s 60 min 1 km 330 mSample Problem — Use of Density as a Conversion Factor What is the volume in L of a 15.0 kg piece of zinc metal? (Density of Zn = 7.13 g/mL)What to Think about 1. Decide what form of the conversion factor to use: g/mL or the reciprocal, mL/g. Always begin by arranging the factors using units only. As the answer will contain one unit, begin with one unit, in this case, kg. 2. Insert the appropriate numerical values for each conversion factor. In order to cancel a mass and convert to a volume, use the reciprocal of the density: 1mL 7.13 gHow to Do It15.0 kg ×15.0 kg ×g× kgmL × gL =L mL–3 1 mL 103 g × × 10 L = 2.10 L 7.13 g 1 kg 1 mL3. Calculate the answer with correct unit and number of significant digits.© Edvantage Interactive 2017Chapter 1 Skills and Processes of Chemistry 49Practice Problems — Use of Rate and Density as Conversion Factors 1. The density of mercury metal is 13.6 g/mL. What is the mass of 2.5 L?2. The density of lead is 11.2 g/cm3. The volumes 1 cm3 and 1 mL are exactly equivalent. What is the volume in L of a 16.5 kg piece of lead?3. The speed of light is 3.0 × 1010 cm/s. Sunlight takes 8.29 min to travel from the photosphere (light-producing region) of the Sun to Earth. How many kilometers is Earth from the Sun? Conversions Involving Units with Exponents (Another Kind of Derived Unit)If a unit is squared or cubed, it may be cancelled in one of two ways. It may be written more than once to convey that it is being multiplied by itself or it may be placed in brackets with the exponent applied to the number inside the brackets as well as to the unit. Hence, the use of the equivalency 1 L = 1 dm3 to convert 1 m3 to L might appear in either of these formats:1 m3 ×1 dm 1 dm 1 dm 1L = 1 m3 × × × × 10–1 m 10–1 m 10–1 m 1 dm33( 101 dmm ) × –11L = 1000 L 1 dm3Sample Problem – Use of Conversion Factors Containing Exponents Convert 0.35 m3 (cubic meters) into mL. (1 mL = 1 cm3)What to Think about 1. The unit cm must be cancelled three times. Do this by multiplying the conversion factor by itself three times or through the use of brackets. 2. Once the units have been aligned correctly, insert the appropriate numerical values. 3. Calculate the answer with the correct unit and number of significant figures.How to Do ItmL = mL cm × cm × cm × 3 cm m m m mL = mL cm 3 × or 0.35 m3 × m cm3 1 cm 1 cm 1 cm 0.35 m3 × –2 × –2 × –2 × 1 mL 10 m 10 m 10 m 1 cm3 = 3.5 × 105 mL 3 or 0.35 m3 × 1 cm × 1 mL3 = 3.5 × 105 mL –2 m 10 1 cm 0.35 m3 ×()()Practice Problems — Use of Conversion Factors Containing Exponents 1. Convert 4.3 dm3 into cm3. 2. Atmospheric pressure is 14.7 lb/in2. Convert this to the metric unit, g/cm2. (Given 454 g = 1.00 lb) 3. Convert a density of 8.2 kg/m3 to lb/ft3 using factors provided in this section.50 Chapter 1 Skills and Processes of Chemistry© Edvantage Interactive 2017Measurement of TemperatureKelvin Celsius Fahrenheit373°100°212°310°37°98.6°273°0°32° 0°0°Temperature is a measure of the intensity of heat. It is the average kinetic energy of the particles is sample of matter. Heat is the energy transferred between two objects in contact with one another at different temperatures. There are several different scales for measuring temperature. Three of these scales are commonly used, two in physical sciences and one in engineering. The earliest instrument for measuring the intensity of heat involved measuring the expansion and contraction of water and was called a thermoscope. This device was invented by an Italian, named Santorio in the late 1500’s. The thermoscope, later called the thermometer, was refined around the same time by Galileo Galilei. It wasn’t until the early 1700s that the Swede Andres Celsius and the German Daniel Gabriel Fahrenheit applied measured scales of temperature to tubes containing alcohol and later, for more precision, mercury. Celsius devised a scale of temperature giving the value of 0°C to the temperature at which water freezes and 100°C to the temperature at which water boils. Fahrenheit, apparently in an attempt to avoid the use of negative numbers on really cold German days, used the coldest temperature he could produce with rock salt and ice as his zero point. His original scale was later adjusted so that the freezing point of water was 32°F and the boiling point of water was 212°F. In the mid 1800s, Lord William Thomson Kelvin of Scotland, a student at Cambridge University who eventually became a professor at the University of Glasgow, developed the idea of an absolute temperature scale. His theory was that there was an absolute coldest temperature that could ever be attained. This temperature was called absolute zero and would be 273.15°C colder than the freezing point of water. The boiling point of water on the absolute scale would be 100°C and 100 Kelvin units hotter than the freezing point. Comparison of these scales helps you see how to convert from one scale to another (Figure 1.4.3). Not only are the conversion factors of 100°C/180°F and 100°C/100 K degrees, which result in 1°C /1.8°F and 1 oC/1 K important, but the difference at the freezing point of water is also important. This requires the addition or subtraction of 32°F and 273.15 K (or slightly less precisely, 273 K) as well. Consequently, you could simply memorize the following equations for conversion:Figure 1.4.3 The threecommonly used temperature scalesTF =1.8°F 1°C (TC) + 32 OF and TC = (TF – 32°F) × 1°C 1.8°FOr, even easier (your teacher may elect to use 273 exactly as the addition/subtraction factor):1°C TK = TC (1 K) + 273.15 K and Tc = (Tk – 273.15 K) × 1K (1°C)The number of significant figures in a temperature is generally determined by using the rule of precision. That is, if the given temperature is good to the tenths, the temperature in the new units is good to the tenths. Similarly, if the given temperature was a whole number (good to the ones place), the new temperature should also be stated to the ones place. It is interesting to note by convention there is no degree symbol (°) placed before K. Some people prefer to memorize formulas and apply them. However, it is nearly always easier to apply logic to derive a method for converting from one unit to another. Notice this method of derivation in the next sample problem.© Edvantage Interactive 2017Chapter 1 Skills and Processes of Chemistry 51Sample Problem — Conversion of the Celsius to Fahrenheit Temperature Scale Convert a warm day temperature from 25.2°C to °F.What to Think aboutHow to Do It1. Is the Fahrenheit temperature bigger or smaller than 25.2°C? Based on our common sense understanding, °F should be larger. This means the difference of 32°F should be added to the 25.2°C. 2. Should the 32°F be added to 25.2°C now, or following a conversion to °F? Units of different types cannot be added to one another, so the conversion must be completed first.25.2°C ×3. The conversion to °F must occur first and then 32°F should be added on. There are 1.8 F degrees for each 1 C degree. Needless to say, the answer is calculated to the tenths place as that is the precision of the original temperature given in the problem! The final temperature is rounded to the TENTHS of a degree.25.2°C × 1.8 °F (+ 32°F) = 77.4 °C 1°C°F °CPractice Problems — Conversion Between Units of Temperature 1. Cesium metal is very soft. It has a melting point below body temperature. Convert cesium’s melting point of 28.4°C to °F.2. Air can be liquefied at a temperature of –319°F. Convert this to °C.3. Convert absolute zero (the coldest possible temperature) from 0 K to °F. Hint: convert to °C first.52 Chapter 1 Skills and Processes of Chemistry© Edvantage Interactive 2017

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