measurement and problem solving chemistry

Chapter 2. Measurements

2.4 problem solving and unit conversions, learning objectives.

By the end of this section, you will be able to:

• Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities
• Use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties

It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:

[latex]\large\text{speed}=\frac{\text{distance}}{\text{time}}[/latex]

An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of [latex]\large\frac{\text{100 m}}{\text{10 s}}=\text{10 m/s}[/latex].

Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:

[latex]\large\text{time}=\frac{\text{distance}}{\text{speed}}[/latex]

The time can then be computed as [latex]\large\frac{\text{25 m}}{\text{10 m/s}}=\text{2.5 s}[/latex].

Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”

These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method ). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers . This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.

Conversion Factors and Dimensional Analysis

A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor . For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,

[latex]\large\text{(2.54 cm = 1 in)}\text{ or }\frac{\text{2.54 cm}}{\text{1 in}}\text{ or }\text{ 2.54}\frac{\text{cm}}{\text{ in}} [/latex]

Several other commonly used conversion factors are given in Table 1.

When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:

[latex]\large\text{34}\cancel{\text{in}}\times \frac{\text{2.54 cm}}{1\cancel{\text{in}}}=\text{86 cm}[/latex]

Since this simple arithmetic involves quantities , the premise of dimensional analysis requires that we multiply both numbers and units . The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield [latex]\large\frac{\text{in}\times \text{cm}}{\text{in}}[/latex] . Just as for numbers, a ratio of identical units is also numerically equal to one, [latex]\large\frac{\text{in}}{\text{in}}=\text{1,}[/latex] and the unit product thus simplifies to cm . (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.

The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses.

Example 1: Using a Unit Conversion Factor

The mass of a competition frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table 1).

If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.

We write the unit conversion factor in its two forms:

[latex]\frac{\text{1 oz}}{\text{28.35 g}}\text{ and }\frac{\text{28.349 g}}{\text{1 oz}}[/latex]

The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.

[latex]\text{125}\cancel{\text{ g}}\times \frac{\text{1 oz}}{28.349\cancel{\text{ g}}}=\text{4.41 oz (three significant figures)}[/latex]

Check Your Learning

Convert a volume of 9.345 qt to liters.

Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.

Example 2: Computing Quantities from Measurement Results and Known Mathematical Relations

What is the density of common antifreeze in units of g/mL? A 4.00 qt sample of the antifreeze weighs 9.26 lb.

Since [latex]\text{density}=\frac{\text{mass}}{\text{volume}}[/latex] , we can write the density as [latex]\large\frac{\text{9.26 lb}}{\text{4.00 qt}}[/latex]

Using table 1.6 as reference, we can determine what conversion factors are required.  To convert lbs to grams we use the one conversion factor, 1 lb = 453.59 g; to convert qt to mL, we need two conversion factors, 1 L = 1.057 qt, and 1 L = 1,000 mL.

The question can be set up in a way that uses these three unit conversion factors sequentially as follows:

[latex]\frac{\text{9.26}\cancel{\text{lb}}}{\text{4.00}\cancel{\text{qt}}}\times \frac{\text{453.59 g}}{1\cancel{\text{lb}}}\times \frac{\text{1.057}\cancel{\text{qt}}}{1\cancel{\text{L}}}\times \frac{1\cancel{\text{L}}}{\text{1000 mL}}=\text{1.11 g/mL}[/latex]

What is the volume in liters of 1.000 oz, given that 1 L = 1.057 qt and 1 qt = 32 oz (exactly)?

Example 3: Computing Quantities from Measurement Results and Known Mathematical Relations

While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.

• What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
• If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?

We first convert distance from kilometers to miles:

[latex]\text{1250 }\cancel{\text{km}}\times \frac{\text{0.62137 mi}}{\text{1 }\cancel{\text{ km}}}=\text{777 mi}[/latex]

Then we convert volume from liters to gallons:

[latex]213\cancel{\text{L}}\times \frac{\text{1.0567}\cancel{\text{qt}}}{1\cancel{\text{L}}}\times \frac{\text{1 gal}}{4\cancel{\text{qt}}}=\text{56.3 gal}[/latex]

[latex]\text{(average) mileage}=\frac{\text{777 mi}}{\text{56.3 gal}}=\text{13.8 miles/gallon}=\text{13.8 mpg}[/latex]

Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:

[latex]\frac{1250\cancel{\text{km}}}{213\cancel{\text{L}}}\times \frac{\text{0.62137 mi}}{1\cancel{\text{km}}}\times \frac{1\cancel{\text{L}}}{\text{1.0567}\cancel{\text{qt}}}\times \frac{4\cancel{\text{qt}}}{\text{1 gal}}=\text{13.8 mpg}[/latex]

Using the previously calculated volume in gallons, we find: [latex]\text{56.3 }\cancel{\text{gal}}\times\frac{\$3.80}{\text{1 }\cancel{\text{gal}}}=\$214[/latex]

A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1.3 × 10 3 km.

• What (average) fuel economy, in miles per gallon, did the Prius get during this trip?
• If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?

Example 4: Conversion of a Derived Unit

Pronghorn Antelope, Cabin Lake Road, Fort Rock, Oregon. Source: Alan D. Wilson, CC BY-SA 3.0 <https://creativecommons.org/licenses/by-sa/3.0>, via Wikimedia Commons

The pronghorn, a mammal indigenous to the interior western and central North America, is the fastest land mammal in the Western Hemisphere.  The pronghorn can run at speeds of up to 88.5 km/h.

What will be this speed in km/min?

We want to change the unit in the denominator from hours to seconds. Because there are 60 minutes in 1 hour (60 min = 1 h), we construct a conversion factor so that the units we want to remove will cancel. Apply and perform the math:

[latex]\frac{\text{88.5}{\text{ km}}}{\text{1}\cancel{\text{ h}}}\times\frac{\text{1}\cancel{\text{h}}}{\text{60}{\text{ min}}}=\text{1.48 km/min}[/latex] Notice how the 88.5 automatically goes in the numerator. That’s because any number can be thought of as being in the numerator of a fraction divided by 1.

A common garden snail moves at a rate of about 0.203 m/min, what is the speed the snail in meters per second?

How Fast Is Fast?

A common garden snail moves at a rate of about 0.2 m/min, which is about 0.003 m/s, which is 3 mm/s! Source: “Grapevine snail”by Jürgen Schoneris licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.

Chemistry Is Everywhere: The Gimli Glider

On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed “the Gimli Glider.”

The Gimli Glider is the Boeing 767 that ran out of fuel and glided to safety at Gimli Airport. The aircraft ran out of fuel because of confusion over the units used to express the amount of fuel. “Aircanada.b767′′ is in the the public domain.

The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely.

What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units.

Video source: Unit conversion by keyj ( https://viuvideos.viu.ca/media/Unit+Conversion/0_o671v9j6 )

All conversion problems below can be solved using the conversion factors provided within the question or in Table 1.  Some questions can be solved in more than one way.

Volume conversions.

• How many milliliters of a soft drink are contained in a 12.0 fl oz can? (1 fl oz = 29.6 mL)
• A barrel of oil is 42.0 gal. How many liters of oil are in a barrel?
• The Reliant Robin, a small three-wheeled car produced by the Reliant Motor Company during the 1970’s in England, has a fuel tank capacity of 27 liters.  What is the tank capacity in gallons?
• A bottle of soda has a volume of 16.0 fl oz.  How many gallons does the bottle contain? (1 fl oz = 29.6 mL)
• Many medical laboratory tests are run using 5.0 μL blood serum. What is 5.0 μL in pints?
• A certain pancake recipe calls for 3.5 tsp of baking powder to be used to make the batter.  How many cups of baking powder is 3.5 tsp?

1. [latex]\text{12.0}\cancel{\text{fl oz}}\times \frac{\text{29.6}\text{ mL}}{1\cancel{\text{fl oz}}}=\text{355 mL}[/latex]

2. [latex]\text{42.0 }\cancel{\text{gal}}\times\frac{\text{3.785 }\cancel{\text{L}}}{\text{1 }\cancel{\text{gal}}}=\text{159 L}[/latex]

3. [latex]\text{27 }\cancel{\text{L}}\times\frac{\text{1 }\cancel{\text{gal}}}{\text{3.785 }\cancel{\text{L}}}=\text{7.1 L}[/latex]

4. [latex]\text{16.0 }\cancel{\text{fl oz}}\times\frac{\text{29.6 }\cancel{\text{mL}}}{\text{1 }\cancel{\text{fl oz}}}\times\frac{10^{-3}\cancel{\text{ L}}}{\text{1 }\cancel{\text{mL}}}\times\frac{\text{1}\text{ gal}}{\text{3.785 }\cancel{\text{L}}}=\text{0.125 gal}[/latex]

5. [latex]\text{5.0 }\cancel{\text{μL}}\times\frac{10^{-6}\text{ }\cancel{\text{mL}}}{10^{-3}\text{ }\cancel{\text{μL}}}\times\frac{\text{1}\text{ pt}}{\text{473.2 }\cancel{\text{mL}}}=\text{ 1.1}\times 10^{-5}\text{ pt}[/latex]

6. [latex]\text{3.5 }\cancel{\text{tsp}}\times\frac{\text{4.9 }\cancel{\text{mL}}}{\text{1 }\cancel{\text{tsp}}}\times\frac{10^{-3}\cancel{\text{ L}}}{\text{1 }\cancel{\text{mL}}}\times\frac{\text{1.057 }\cancel{\text{qt}}}{\text{1 }\cancel{\text{L}}}\times\frac{\text{4}\text{ cups}}{\text{1 }\cancel{\text{qt}}}=\text{ 7.3}\times 10^{-2}\text{ cups}[/latex]

Mass Conversions

• Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less?
• A weight lifter lifted 192 kg in a move called the clean and jerk. What was the mass of the weight lifted in pounds (lb)?
• If an aspirin tablet contains 325 mg aspirin, how many ounces (oz) of aspirin does it contain?
• A paperclip has a mass of 0.030 oz. What will be the paperclips mass in centigrams (cg)?

1. yes; [latex]\text{197 }\cancel{\text{lb}}\times \frac{\text{1}\text{ kg}}{2.205\cancel{\text{ lb}}}=\text{89.3 kg}[/latex]

2. [latex]\text{192 }\cancel{\text{kg}}\times \frac{\text{2.205}\text{ lb}}{1\cancel{\text{ kg}}}=\text{423 lb}[/latex]

3. [latex]\text{325 }\cancel{\text{mg}}\times\frac{10^{-3}\text{ }\cancel{\text{g}}}{\text{1 }\cancel{\text{mg}}}\times\frac{1\text{ oz}}{\text{28.35 }\cancel{\text{g}}}=\text{0.0115 oz}[/latex]

4. [latex]\text{0.030 }\cancel{\text{oz}}\times\frac{\text{28.35 }\cancel{\text{g}}}{\text{1 }\cancel{\text{oz}}}\times\frac{1\text{ cg}}{10^{-2}\text{ }\cancel{\text{g}}}=\text{85 cg}[/latex]

Length Conversions

• The diameter of a red blood cell is 3 × 10 −4 in. What is the diameter of a red blood cell in centimeters?
• The average distance between the Moon and the Earth measures 238,857 miles. What is this distance in kilometers?
• Lake Ontario is the third deepest Great Lake after lakes Superior and Michigan. The maximum depth of Lake Ontario is 802 ft.  What is the maximum depth in meters?
• Interstate I-90 is the longest Interstate Highway in the United States at 3021 miles.  How megameters is this distance?
• Watkins Glen International road course is 3.450 miles in length.  What is this length in yards?

1. [latex]\text{3}\times 10^{-4}\text{ }\cancel{\text{in}}\times\frac{\text{2.54}\text{ cm}}{\text{1 }\cancel{\text{in}}}=\text{ 8}\times 10^{-4}\text{ cm}[/latex]

2. [latex]\text{238,857 }\cancel{\text{mi}}\times\frac{\text{1}\text{ km}}{\text{0.6214 }\cancel{\text{mi}}}=\text{ 3.844}\times 10^{5}\text{ km}[/latex]

3. [latex]\text{802 }\cancel{\text{ft}}\times\frac{\text{12}\text{ in}}{\text{1 }\cancel{\text{ft}}}\times\frac{\text{1}\cancel{\text{m}}}{\text{39.37 }\cancel{\text{in}}}=\text{ 244}\text{ m}[/latex]

4. [latex]\text{3021 }\cancel{\text{mi}}\times\frac{\text{1}\text{ km}}{\text{0.6214 }\cancel{\text{mi}}}\times\frac{10^{3}\text{ }\cancel{\text{Mm}}}{10^{6}\text{ }\cancel{\text{km}}}=\text{ 4.862}\text{ Mm}[/latex]

5. [latex]\text{3.450 }\cancel{\text{mi}}\times\frac{\text{5280 }\cancel{\text{ft}}}{\text{1 }\cancel{\text{mi}}}\times\frac{\text{1 }\text{yd}}{\text{3 }\cancel{\text{ft}}}=\text{ 6072}\text{ yd}[/latex]

Conversions Involving Derived Units

• A person spends $50.00 on gas for their vehicle.  If the cost of the gas is $3.93 per gallon, how many liters of gas will have been purchased?
• Usain Bolt, eight-time Olympic gold medalist, ran the 100-meter event with an average speed of 10.4 m/s.  What is this speed in miles per hour?
• The distance of a transatlantic trip from Southampton, UK to New York, USA is 3185 nautical miles (nmi).  If a ship is traveling at 42.6 km/h, how many days will it take for the ship to complete the crossing?  (1 nmi = 1.151 mi)
• The track record for the fastest stock car at Talladega Speedway is 212.809 miles per hour, set by Bill Elliott in 1987. If the track length is 4.28 km, how many seconds will it take to complete two laps around the track?

1. [latex]\text{50.00 }\cancel{\text{dollars}}\times\frac{\text{1 }\cancel{\text{gal}}}{\text{3.93 }\cancel{\text{dollars}}}\times\frac{\text{3.785 }\text{L}}{\text{1 }\cancel{\text{gal}}}=\text{ 48.2 }\text{L}[/latex]

2. [latex]\frac{\text{10.4 }\cancel{\text{m}}}{\text{1 }\cancel{\text{s}}}\times\frac{\text{1 }\cancel{\text{km}}}{10^{3}\text{ }\cancel{\text{m}}}\times\frac{\text{0.6214 }\text{mi}}{\text{1 }\cancel{\text{km}}}\times\frac{\text{60 }\cancel{\text{s}}}{\text{1 }\cancel{\text{min}}}\times\frac{\text{60 }\text{min}}{\text{1 }\text{h}}=\text{ 23.3}\text{ mi/h}[/latex]

3. [latex]\text{3185 }\cancel{\text{nmi}}\times\frac{\text{1.151 }\cancel{\text{mi}}}{\text{1 }\cancel{\text{nmi}}}\times\frac{\text{1 }\cancel{\text{km}}}{\text{0.6214 }\cancel{\text{mi}}}\times\frac{\text{1 }\cancel{\text{h}}}{\text{42.6 }\cancel{\text{km}}}\times\frac{\text{1 }\text{day}}{\text{24 }\cancel{\text{h}}}=\text{ 5.77}\text{ days}[/latex]

4. [latex]\text{2.00 }\cancel{\text{laps}}\times\frac{\text{4.28 }\cancel{\text{km}}}{\text{1 }\cancel{\text{lap}}}\times\frac{\text{0.6214 }\cancel{\text{mi}}}{\text{1 }\cancel{\text{km}}}\times\frac{\text{1 }\cancel{\text{h}}}{\text{212.806 }\cancel{\text{mi}}}\times\frac{\text{60 }\cancel{\text{min}}}{\text{1 }\cancel{\text{h}}}\times\frac{\text{60 }\text{s}}{\text{1 }\cancel{\text{min}}}=\text{ 90.0}\text{ s}[/latex]

• Introductory Chemistry- 1st Canadian Edition . Authored by : Jessie A. Key and David W. Ball. Provided by : BCCampus. Located at : https://opentextbc.ca/introductorychemistry/ . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike . License Terms : Download this book for free at http://open.bccampus.ca
• Chemistry. Provided by : OpenStax College. Located at : http://openstaxcollege.org . License : CC BY: Attribution . License Terms : Download for free at https://openstaxcollege.org/textbooks/chemistry/get

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1.6: The Measurement of Matter

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Learning Objectives

• Express quantities properly, using a number and a unit.
• State the different measurement systems used in chemistry.
• Express a large number or a small number in scientific notation.
• Learn how to use SI prefixes.
• Perform unit conversions using conversion factors.

A coffee maker’s instructions tell you to fill the coffeepot with 4 cups of water and use 3 scoops of coffee. When you follow these instructions, you are measuring. When you visit a doctor’s office, a nurse checks your temperature, height, weight, and perhaps blood pressure (Figure \(\PageIndex{1}\)); the nurse is also measuring.

Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something, and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5. If you ask a friend how far he or she walks from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks 12 miles, 12 kilometers, 12 furlongs, or 12 yards...etc. Both a number and a unit must be included to express a quantity properly.

To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. The next two sections examine the rules for expressing numbers.

All measurements depend on the use of units that are well known and understood. The English system of measurement units (inches, feet, ounces, etc.) are not used in science because of the difficulty in converting from one unit to another. The metric system is used because all metric units are based on multiples of 10, making conversions very simple. The metric system was originally established in France in 1795. The International System of Units is a system of measurement based on the metric system. The acronym SI is commonly used to refer to this system and stands for the French term, Le Système International d'Unités . The SI was adopted by international agreement in 1960 and is composed of the seven base units shown in Table \(\PageIndex{1}\).

The first units are frequently encountered in chemistry. All other measurement quantities, such as volume, force, and energy, can be derived from these seven base units.

Exponential Numbers: Powers of Ten

Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. Furthermore, use of prefixes is another way to express measurements involving large and small numbers.

Scientific Notation

In scientific notation, numbers are expressed in the form

\[ N \times 10^n \nonumber \]

where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (10 0 = 1). The number 10 is called the base because it is this number that is raised to the power \(n\). Although a base number may have values other than 10, the base number in scientific notation is always 10.

A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:

• If the decimal point is moved to the left n places, n is positive.
• If the decimal point is moved to the right n places, n is negative.

Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example \(\PageIndex{1}\).

Example \(\PageIndex{1}\): Expressing Numbers in Scientific Notation

Convert each number to scientific notation.

Exercise \(\PageIndex{1}\)

Convert each ordinary number to scientific notation, or vice versa.

• 67,000,000,000

Exercise \(\PageIndex{2}\)

• 0.000006567
• 6.22 × 10 −2
• 9.9 × 10 −9

Metric Prefixes

Conversions between metric system units are straightforward because the system is based on powers of ten. For example, meters, centimeters, and millimeters are all metric units of length. There are 10 millimeters in 1 centimeter and 100 centimeters in 1 meter. Metric prefixes are used to distinguish between units of different size. These prefixes all derive from either Latin or Greek terms. For example, mega comes from the Greek word \(\mu \varepsilon \gamma \alpha \varsigma\), meaning "great". Table \(\PageIndex{2}\) lists the most common metric prefixes and their relationship to the central unit that has no prefix. Length is used as an example to demonstrate the relative size of each prefixed unit.

There are a couple of odd little practices with the use of metric abbreviations. Most abbreviations are lowercase. We use "\(\text{m}\)" for meter and not "\(\text{M}\)". However, when it comes to volume, the base unit "liter" is abbreviated as "\(\text{L}\)" and not "\(\text{l}\)". So, 3.5 milliliters is written as \(3.5 \: \text{mL}\).

As a practical matter, whenever possible, you should express the units in a small and manageable number. If you are measuring the weight of a material that weighs \(6.5 \: \text{kg}\), this is easier than saying it weighs \(6500 \: \text{g}\) or \(0.65 \: \text{dag}\). All three are correct, but the \(\text{kg}\) units in this case make for a small and easily managed number. However, if a specific problem needs grams instead of kilograms, go with the grams for consistency.

Example \(\PageIndex{2}\): Unit Abbreviations

Give the abbreviation for each unit and define the abbreviation in terms of the base unit.

• microsecond

Exercise \(\PageIndex{3}\)

Mass and weight.

Mass is a measure of the amount of matter that an object contains. The mass of an object is made in comparison to the standard mass of 1 kilogram. The kilogram was originally defined as the mass of \(1 \: \text{L}\) of liquid water at \(4^\text{o} \text{C}\) (volume of a liquid changes slightly with temperature). In the laboratory, mass is measured with a balance (see below), which must be calibrated with a standard mass so that its measurements are accurate.

Other common units of mass are the gram and the milligram. A gram is 1/1000th of a kilogram, meaning that there are \(1000 \: \text{g}\) in \(1 \: \text{kg}\). A milligram is 1/1000th of a gram, so there are \(1000 \: \text{mg}\) in \(1 \: \text{g}\).

Mass is often confused with the term weight. Weight is a measure of force that is equal to the gravitational pull on an object. The weight of an object is dependent on its location. On the moon, the force due to gravity is about one sixth that of the gravitational force on Earth. Therefore, a given object will weigh six times more on Earth than it does on the moon. Since mass is dependent only on the amount of matter present in an object, mass does not change with location. Weight measurements are often made with a spring scale, by reading the distance that a certain object pulls down and stretches a spring.

Length and Volume

Length is the measurement of the extent of something along its greatest dimension. The SI basic unit of length, or linear measure, is the meter \(\left( \text{m} \right)\). All measurements of length may be made in meters, though the prefixes listed in various tables will often be more convenient. The width of a room may be expressed as about 5 meters \(\left( \text{m} \right)\), whereas a large distance, such as the distance between New York City and Chicago, is better expressed as 1150 kilometers \(\left( \text{km} \right)\). Very small distances can be expressed in units such as the millimeter or the micrometer. The width of a typical human hair is about 10 micrometers \(\left( \mu \text{m} \right)\).

Volume is the amount of space occupied by a sample of matter. The volume of a regular object can be calculated by multiplying its length by its width and its height. Since each of those is a linear measurement, we say that units of volume are derived from units of length. The SI unit of volume is the cubic meter \(\left( \text{m}^3 \right)\), which is the volume occupied by a cube that measures \(1 \: \text{m}\) on each side. This very large volume is not very convenient for typical use in a chemistry laboratory. A liter \(\left( \text{L} \right)\) is the volume of a cube that measures \(10 \: \text{cm}\) \(\left( 1 \: \text{dm} \right)\) on each side. A liter is thus equal to both \(1000 \: \text{cm}^3\) \(\left( 10 \: \text{cm} \times 10 \: \text{cm} \times 10 \: \text{cm} \right)\) and to \(1 \: \text{dm}^3\).

1L = \(1000 \: \text{cm}^3\) \(\left( 10 \: \text{cm} \times 10 \: \text{cm} \times 10 \: \text{cm} \right)\) = \(1 \: \text{dm}^3\)

A smaller unit of volume that is commonly used is the milliliter (\(\text{mL}\)). A milliliter is the volume of a cube that measures \(1 \: \text{cm}\) on each side. Therefore, a milliliter is equal to a cubic centimeter \(\left( \text{cm}^3 \right)\).

1 mL \(\left( \text{cm}^3 \right)\)

There are \(1000 \: \text{mL}\) in \(1 \: \text{L}\), which is the same as saying that there are \(1000 \: \text{cm}^3\) in \(1 \: \text{dm}^3\).

During your studies of chemistry (and physics also), you will note that mathematical equations are used in many different applications. Many of these equations have a number of different variables with which you will need to work. You should also note that these equations will often require you to use measurements with their units. Algebra skills become very important here!

Converting Between Units with Conversion Factors

A conversion factor is a factor used to convert one unit of measurement into another. A simple conversion factor can be used to convert meters into centimeters, or a more complex one can be used to convert miles per hour into meters per second. Since most calculations require measurements to be in certain units, you will find many uses for conversion factors. Always remember that a conversion factor has to represent a fact; this fact can either be simple or much more complex. For instance, you already know that 12 eggs equal one dozen. A more complex fact is that the speed of light is \(1.86 \times 10^5\) miles/sec. Either one of these can be used as a conversion factor, depending on what type of calculation you might be working with (Table \(\PageIndex{1}\)).

*Pounds and ounces are technically units of force, not mass, but this fact is often ignored by the non-scientific community.

Of course, there are other ratios which are not listed in Table \(\PageIndex{1}\). They may include:

• Ratios embedded in the text of the problem (using words such as per or in each , or using symbols such as / or %).
• Conversions in the metric system, as covered earlier in this chapter.
• Common knowledge ratios (such as 60 seconds \(=\) 1 minute).

If you learned the SI units and prefixes described, then you know that 1 cm is 1/100th of a meter.

\[ 1\; \rm{cm} = \dfrac{1}{100} \; \rm{m} = 10^{-2}\rm{m} \nonumber \]

\[100\; \rm{cm} = 1\; \rm{m} \nonumber \]

Suppose we divide both sides of the equation by \(1 \text{m}\) (both the number and the unit):

\[\mathrm{\dfrac{100\:cm}{1\:m}=\dfrac{1\:m}{1\:m}} \nonumber \]

As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1:

\[ \dfrac{ \text{100 cm}}{\text{1 m}} = \dfrac{ \text{1000 mm}}{\text{1 m}}= \dfrac{ 1\times 10^6 \mu \text{m}}{\text{1 m}}= 1 \nonumber \]

We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units.

Performing Dimensional Analysis

Dimensional analysis is amongst the most valuable tools used by physical scientists. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit, using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others. The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis .

Here is a simple example: How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as \(\mathrm{\frac{100\:cm}{1\:m}}\) and multiply:

\[ 3.55 \; \rm{m} \times \dfrac{100 \; \rm{cm}}{1\; \rm{m}} \nonumber \]

The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out:

\[\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}} \nonumber \]

The final step is to perform the calculation that remains once the units have been canceled:

\[ \dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \nonumber \]

In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows:

quantity (in old units) × conversion factor = quantity (in new units)

You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you will encounter will not always be so simple . If you can master the technique of applying conversion factors, you will be able to solve a large variety of problems.

In the previous example, we used the fraction \(\frac{100 \; \rm{cm}}{1 \; \rm{m}}\) as a conversion factor. Does the conversion factor \(\frac{1 \; \rm m}{100 \; \rm{cm}}\) also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator (except that they are expressed in different units). Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten:

\[ 3.55 \; \rm{m} \times \dfrac{1\; \rm{m}}{100 \; \rm{cm}} = 0.0355 \dfrac{\rm{m}^2}{\rm{cm}} \nonumber \]

For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out . Figure \(\PageIndex{1}\) shows a concept map for constructing a proper conversion.

The General Steps in Performing Dimensional Analysis

• Identify the " given " information in the problem. Look for a number with units to start this problem with.
• What is the problem asking you to " find "? In other words, what unit will your answer have?
• Use ratios and conversion factors to cancel out the units that aren't part of your answer, and leave you with units that are part of your answer.
• When your units cancel out correctly, you are ready to do the math . You are multiplying fractions, so you multiply the top numbers and divide by the bottom numbers in the fractions.

Exercise \(\PageIndex{4}\)

Perform each conversion.

• 101,000 ns to seconds
• 32.08 kg to grams
• 1.53 grams to cg

Multiple Conversions

Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. We will set up a series of conversion factors so that each conversion factor produces the next unit in the sequence. W e first convert the given amount in km to the base unit, which is meters. We know that 1,000 m =1 km. Then we convert meters to mm, remembering that \(1\; \rm{mm}\) = \( 10^{-3}\; \rm{m}\).

Concept Map

Calculation

\[ \begin{align} 54.7 \; \cancel{\rm{km}} \times \dfrac{1,000 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1\; \cancel{\rm{mm}}}{\cancel{10^{-3} \rm{m}}} & = 54,700,000 \; \rm{mm} \\ &= 5.47 \times 10^7\; \rm{mm} \end{align} \nonumber \]

In each step, the previous unit is canceled and the next unit in the sequence is produced; each successive unit cancels out, until only the unit needed in the answer is left.

Example \(\PageIndex{4}\): Unit Conversion

Convert 58.2 ms to megaseconds in one multi-step calculation.

Example \(\PageIndex{5}\): Unit Conversion

How many seconds are in a day?

Exercise \(\PageIndex{5}\)

Perform each conversion in one multistep calculation.

• 43.007 ng to kg
• 1005 in to ft
• 12 mi to km
• Metric prefixes derive from Latin or Greek terms. The prefixes are used to make the units manageable.
• The SI system is based on multiples of ten. There are seven basic units in the SI system. Five of these units are commonly used in chemistry.
• Mass is a measure of the amount of matter that an object contains.
• Weight is a measure of force that is equal to the gravitational pull on an object.
• Mass is independent of location, while weight depends on location.
• Length is the measurement of the extent of something along its greatest dimension.
• Volume is the amount of space occupied by a sample of matter.
• Volume can be determined by knowing the length of each side of the item.
• Conversion factors are used to convert one unit of measurement into another.
• Dimensional analysis (unit conversions) involves the use of conversion factors that will cancel unwanted units and produce desired units.

Contributors and Attributions

Marisa Alviar-Agnew  ( Sacramento City College )

Henry Agnew (UC Davis)

Elizabeth R. Gordon (Furman University)

• Hayden Cox (Furman University)

Snapsolve any problem by taking a picture. Try it in the Numerade app?

Chemistry A Molecular Approach

Nivaldo j. tro, matter, measurement, and problem solving - all with video answers.

Chapter Questions

Explain the following statement in your own words and give an example. The properties of the substances around us depend on the atoms and molecules that compose them.

Explain the main goal of chemistry.

Describe the scientific approach to knowledge. How does it differ from other approaches?

Explain the differences between a hypothesis, a law, and a theory.

What observations did Antoine Lavoisier make? What law did he formulate?

What theory did John Dalton formulate?

What is wrong with the expression, "That is just a theory," if by theory you mean a scientific theory?

What are two different ways to classify matter?

How do solids, liquids, and gases differ?

What is the difference between a crystalline solid and an amorphous solid?

Explain the difference between a pure substance and a mixture.

Explain the difference between an element and a compound.

Explain the difference between a homogeneous and a heterogeneous mixture.

What kind of mixtures can be separated by filtration?

Explain how distillation works to separate mixtures.

What is the difference between a physical property and a chemical property?

What is the difference between a physical change and a chemical change? Give some examples of each.

Explain the significance of the law of conservation of energy.

What kind of energy is chemical energy? In what way is an elevated weight similar to a tank of gasoline?

What are the standard SI base units of length, mass, time, and temperature?

What are the three common temperature scales? Does the size of a degree differ among them?

What are prefix multipliers? Give some examples.

What is a derived unit? Give an example.

Explain the difference between density and mass.

Explain the difference between intensive and extensive properties.

What is the meaning of the number of digits reported in a measured quantity?

When multiplying or dividing measured quantities, what determines the number of significant figures in the result?

When adding or subtracting measured quantities, what determines the number of significant figures in the result?

What are the rules for rounding off the results of calculations?

Explain the difference between precision and accuracy.

Explain the difference between random error and systematic error.

What is dimensional analysis?

Classify each statement as an observation, a law, or a theory. a. All matter is made of tiny, indestructible particles called atoms. b. When iron rusts in a closed container, the mass of the container and its contents does not change. c. In chemical reactions, matter is neither created nor destroyed. d. When a match burns, heat is released.

Classify each statement as an observation, a law, or a theory. a. Chlorine is a highly reactive gas. b. If elements are listed in order of increasing mass of their atoms, their chemical reactivity follows a repeating pattern. c. Neon is an inert (or nonreactive) gas. d. The reactivity of elements depends on the arrangement of their electrons.

A chemist decomposes several samples of carbon monoxide into carbon and oxygen and weighs the resultant elements. The results are shown below: a. Do you notice a pattern in these results? Next, the chemist decomposes several samples of hydrogen peroxide into hydrogen and oxygen. The results are shown below: b. Do you notice a similarity between these results and those for carbon monoxide in part a? c. Can you formulate a law from the observations in a and b? d. Can you formulate a hypothesis that might explain your law in c?

When astronomers observe distant galaxies, they can tell that most of them are moving away from one another. In addition. the more distant the galaxies, the more rapidly they are likely to be moving away from each other. Can you devise a hypothesis to explain these observations?

Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous. a. sweat b. carbon dioxide c. aluminum d. vegetable soup

Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous. a. wine b. beef stew c. iron d. carbon monoxide

Complete the table.

Complete the table. $\begin{aligned}&\text { water } \text { pure } & \text { compound } \\&\text { coffee } & \frac{ }{\longrightarrow} \text { ice } \\&\text { carbon } & \end{aligned}$

Determine whether each molecular diagram represents a pure substance or a mixture. If it represents a pure substance, classify the substance as an element or a compound. If it represents a mixture, classify the mixture as homogeneous or heterogeneous.

Determine whether each molecular diagram represents a pure substance or a mixture. If it represents a pure substance, classify

Classify each of the listed properties of isopropyl alcohol (also known as rubbing alcohol) as physical or chemical. a. colorless b. flammable c. liquid at room temperature d. density $=0.79 \mathrm{~g} / \mathrm{mL}$ e, mixes with water

Classify each of the listed properties of ozone (a pollutant in the lower atmosphere, but part of a protective shield against UV light in the upper atmosphere) as physical or chemical. a. bluish color b. pungent odor c. very reactive d. decomposes on exposure to ultraviolet light e. gas at room temperature

Classify each property as physical or chemical. a. the tendency of ethyl alcohol to burn b. the shine of silver c. the odor of paint thinner d. the flammability of propane gas

Classify each property as physical or chemical. a. the boiling point of ethyl alcohol b. the temperature at which dry ice evaporates c. the tendency of iron to rust d. the color of gold

Classify each change as physical or chemical. a. Natural gas burns in a stove. b. The liquid propane in a gas grill evaporates because the valve was left open. c. The liquid propane in a gas grill burns in a flame. d. A bicycle frame rusts on repeated exposure to air and water.

Classify each change as physical or chemical. a. Sugar burns when heated on a skillet. b. Sugar dissolves in water. c. A platinum ring becomes dull because of continued abrasion. d. A silver surface becomes tarnished after exposure to air for a long period of time.

Based on the molecular diagram, classify each change as physical or chemical.

Convert each temperature. a. $32^{\circ} \mathrm{F}$ to ${ }^{\circ} \mathrm{C}$ (temperature at which water freezes) b. $77 \mathrm{~K}$ to ${ }^{\circ} \mathrm{F}$ (temperature of liquid nitrogen) c. $-109^{\circ} \mathrm{F}$ to ${ }^{\circ} \mathrm{C}$ (temperature of dry ice) d. $98.6^{\circ} \mathrm{F}$ to $\mathrm{K}$ (body temperature)

Convert each temperature. a. $212^{\circ} \mathrm{F}$ to ${ }^{\circ} \mathrm{C}$ (temperature of boiling water at sea level) b. $22{ }^{\circ} \mathrm{C}$ to $\mathrm{K}$ (approximate room temperature) c. $0.00 \mathrm{~K}$ to ${ }^{\circ} \mathrm{F}$ (coldest temperature possible, also known as absolute zero) d. $2.735 \mathrm{~K}$ to ${ }^{\circ} \mathrm{C}$ (average temperature of the universe as measured from background black body radiation)

The coldest temperature ever measured in the United States is $-80{ }^{\circ} \mathrm{F}$ on January 23,1971 , in Prospect Creek, Alaska. Convert that temperature to ${ }^{\circ} \mathrm{C}$ and $\mathrm{K}$. (Assume that $-80^{\circ} \mathrm{F}$ is accurate to two significant figures.)

The warmest temperature ever measured in the United States is $134{ }^{\circ} \mathrm{F}$ on July 10,1913, in Death Valley, California. Convert that temperature to ${ }^{\circ} \mathrm{C}$ and $\mathrm{K}$.

Use the prefix multipliers to express each measurement without any exponents. a. $1.2 \times 10^{-9} \mathrm{~m}$ b. $22 \times 10^{-15} \mathrm{~s}$ c. $1.5 \times 10^{9} \mathrm{~g}$ d. $3.5 \times 10^{6} \mathrm{~L}$

Use prefix multipliers to express each measurement without any exponents. a. $38.8 \times 10^{5} \mathrm{~g}$ b. $55.2 \times 10^{-10} \mathrm{~s}$ c. $23.4 \times 10^{11} \mathrm{~m}$ d. $87.9 \times 10^{-7} \mathrm{~L}$

Use scientific notation to express each quantity with only the base units (no prefix multipliers). a. $4.5 \mathrm{~ns}$ b. $18 \mathrm{fs}$ c. $128 \mathrm{pm}$ d. $35 \mu \mathrm{m}$

Use scientific notation to express each quantity with only the base units (no prefix multipliers). a. $35 \mu \mathrm{L}$ b. $225 \mathrm{Mm}$ c. $133 \mathrm{Tg}$ d. $1.5 \mathrm{cg}$

Complete the table: a. $1245 \mathrm{~kg}$ b. $515 \mathrm{~km}$ c. $122.355 \mathrm{~s}$ d. $3.345 \mathrm{~kJ}$

Complete the table: a. $355 \mathrm{~km} / \mathrm{s}$ b. $1228 \mathrm{~g} / \mathrm{L}$ c. $554 \mathrm{mK} / \mathrm{s}$ d. $2.554 \mathrm{mg} / \mathrm{mL}$

Express the quantity $254,998 \mathrm{~m}$ in each unit. a. $\mathrm{km}$ b. $\mathrm{Mm}$ c. $\mathrm{mm}$ d. $\mathrm{cm}$

Express the quantity $556.2 \times 10^{-12} \mathrm{~s}$ in each unit. a. $\mathrm{ms}$ b. $\mathrm{ns}$ c. $\mathrm{ps}$ d. $\mathrm{fs}$

How many $1-\mathrm{cm}$ squares would it take to construct a square that is $1 \mathrm{~m}$ on each side?

How many $1-\mathrm{cm}$ cubes would it take to construct a cube that is $4 \mathrm{~cm}$ on edge?

A new penny has a mass of $2.49 \mathrm{~g}$ and a volume of $0.349 \mathrm{~cm}^{3}$. Is the penny made of pure copper? Explain.

A titanium bicycle frame displaces $0.314 \mathrm{~L}$ of water and has a mass of $1.41 \mathrm{~kg} .$ What is the density of the titanium in $\mathrm{g} / \mathrm{cm}^{3}$ ?

Glycerol is a syrupy liquid often used in cosmetics and soaps. A 3.25-L sample of pure glycerol has a mass of $4.10 \times 10^{3} \mathrm{~g}$. What is the density of glycerol in $\mathrm{g} / \mathrm{cm}^{3} ?$

A supposedly gold nugget is tested to determine its density. It is found to displace $19.3 \mathrm{~mL}$ of water and has a mass of 371 grams. Could the nugget be made of gold?

Ethylene glycol (antifreeze) has a density of $1.11 \mathrm{~g} / \mathrm{cm}^{3}$. a. What is the mass in $g$ of $417 \mathrm{~mL}$ of this liquid? b. What is the volume in $L$ of $4.1 \mathrm{~kg}$ of this liquid?

Acetone (nail polish remover) has a density of $0.7857 \mathrm{~g} / \mathrm{cm}^{3} .$ a. What is the mass, in $\mathrm{g}$, of $28.56 \mathrm{~mL}$ of acetone? b. What is the volume, in $\mathrm{mL}$, of $6.54 \mathrm{~g}$ of acetone?

A small airplane takes on $245 \mathrm{~L}$ of fuel. If the density of the fuel is $0.821 \mathrm{~g} / \mathrm{mL}$, what mass of fuel has the airplane taken on?

Human fat has a density of $0.918 \mathrm{~g} / \mathrm{cm}^{3}$. How much volume (in $\mathrm{cm}^{3}$ ) is gained by a person who gains $10.0 \mathrm{lbs}$ of pure fat?

Read each measurement to the correct number of significant figures. Note: Laboratory glassware should always be read from the bottom of the meniscus.

Read each measurement to the correct number of significant figures. Note: Laboratory glassware should always be read from the bottom of the meniscus. Digital balances normally display mass to the correct number of significant figures for that particular balance.

For each number, underline the zeroes that are significant and draw an $\mathrm{x}$ through the zeroes that are not: a. $1,050,501 \mathrm{~km}$ b. $0.0020 \mathrm{~m}$ c. $0.000000000000002 \mathrm{~s}$ d. $0.001090 \mathrm{~cm}$

For each number, underline the zeroes that are significant and draw an $\mathrm{x}$ through the zeroes that are not: a. $180,701 \mathrm{mi}$ b. $0.001040 \mathrm{~m}$ c. $0.005710 \mathrm{~km}$ d. $90,201 \mathrm{~m}$

How many significant figures are in each number? a. $0.000312 \mathrm{~m}$ b. $312,000 \mathrm{~s}$ c. $3.12 \times 10^{5} \mathrm{~km}$ d. $13,127 \mathrm{~s}$

How many significant figures are in each number? a. $0.1111 \mathrm{~s}$ b. $0.007 \mathrm{~m}$ c. $108,700 \mathrm{~km}$ d. $1.563300 \times 10^{11} \mathrm{~m}$ e. 30,800

Which numbers are exact (and therefore have an unlimited number of significant figures)? a. $\pi=3.14$ b. 12 inches $=1$ foot c. EPA gas mileage rating of 26 miles per gallon d. 1 gross $=144$

Indicate the number of significant figures in each number. If the number is an exact number, indicate an unlimited number of significant figures. a. $305,435,087$ (2008 U.S. population) b. $2.54 \mathrm{~cm}=1$ in c. $11.4 \mathrm{~g} / \mathrm{cm}^{3}$ (density of lead) d. $12=1$ dozen

Round each number to three significant figures. a. $79,845.82$ b. $1.548937 \times 10^{7}$ c. $2.3499999995$ d. $0.000045389$

Calculate to the correct number of significant figures. a. $9.15 \div 4.970$ b. $\quad 1.54 \times 0.03060 \times 0.69$ c. $27.5 \times 1.82 \div 100.04$ d. $\left(2.290 \times 10^{6}\right) \div\left(6.7 \times 10^{4}\right)$

Calculate to the correct number of significant figures. a. $89.3 \times 77.0 \times 0.08$ b. $\left(5.01 \times 10^{5}\right) \div\left(7.8 \times 10^{2}\right)$ c. $4.005 \times 74 \times 0.007$ d. $453 \div 2.031$

Calculate to the correct number of significant figures. a. $43.7-2.341$ b. $17.6+2.838+2.3+110.77$ c. $19.6+58.33-4.974$ d. $5.99-5.572$

Calculate to the correct number of significant figures. a. $0.004+0.09879$ b. $1239.3+9.73+3.42$ c. $2.4-1.777$ d. $532+7.3-48.523$

Calculate to the correct number of significant figures. a. $(24.6681 \times 2.38)+332.58$ b. $(85.3-21.489) \div 0.0059$ c. $(512 \div 986.7)+5.44$ d. $\left[\left(28.7 \times 10^{5}\right) \div 48.533\right]+144.99$

Calculate to the correct number of significant figures. a. $\left[\left(1.7 \times 10^{6}\right) \div\left(2.63 \times 10^{5}\right)\right]+7.33$ b. $(568.99-232.1) \div 5.3$ c. $(9443+45-9.9) \times 8.1 \times 10^{6}$ d. $(3.14 \times 2.4367)-2.34$

Convert: a. $154 \mathrm{~cm}$ to in b. $3.14 \mathrm{~kg}$ to $\mathrm{g}$ c. $3.5 \mathrm{~L}$ to $\mathrm{qt}$ d. $109 \mathrm{~mm}$ to in

Convert: a. $1.4$ in to $\mathrm{mm}$ b. $116 \mathrm{ft}$ to $\mathrm{cm}$ c. $1845 \mathrm{~kg}$ to $\mathrm{lb}$ d. $815 \mathrm{yd}$ to $\mathrm{km}$

A runner wants to run $10.0 \mathrm{~km}$. She knows that her running pace is $7.5$ miles per hour. How many minutes must she run?

A cyclist rides at an average speed of 18 miles per hour. If she wants to bike $212 \mathrm{~km}$, how long (in hours) must she ride?

A European automobile has a gas mileage of $17 \mathrm{~km} / \mathrm{L}$. What is the gas mileage in miles per gallon?

A gas can holds $5.0$ gallons of gasoline. What is this quantity in $\mathrm{cm}^{3} ?$

A house has an area of $195 \mathrm{~m}^{2}$. What is its area in: a. $\mathrm{km}^{2}$ b. $\mathrm{dm}^{2}$ c. $\mathrm{cm}^{2}$

A bedroom has a volume of $115 \mathrm{~m}^{3}$. What is its volume in: a. $\mathrm{km}^{3}$ b. $\mathrm{dm}^{3}$ c. $\mathrm{cm}^{3}$

The average U.S. farm occupies 435 acres. How many square miles is this? ( 1 acre $=43,560 \mathrm{ft}^{2}, 1$ mile $=5280 \mathrm{ft}$ )

Total U.S. farmland occupies 954 million acres. How many square miles is this? $\left(1\right.$ acre $\left.=43,560 \mathrm{ft}^{2}, 1 \mathrm{mile}=5280 \mathrm{ft}\right)$ Total U.S. land area is $3.537$ million square miles. What percentage of U.S. land is farmland?

An acetaminophen suspension for infants contains $80 \mathrm{mg} / 0.80 \mathrm{~mL}$ suspension. The recommended dose is $15 \mathrm{mg} / \mathrm{kg}$ body weight. How many $\mathrm{mL}$ of this suspension should be given to an infant weighing $14 \mathrm{lb}$ ? (Assume two significant figures.)

Problem 100

An ibuprofen suspension for infants contains $100 \mathrm{mg} / 5.0 \mathrm{~mL}$ suspension. The recommended dose is $10 \mathrm{mg} / \mathrm{kg}$ body weight. How many mL of this suspension should be given to an infant weighing $18 \mathrm{lb}$ ? (Assume two significant figures.)

Problem 101

There are exactly 60 seconds in a minute, there are exactly 60 minutes in an hour, there are exactly 24 hours in a mean solar day, and there are $365.24$ solar days in a solar year. Find the number of seconds in a solar year. Be sure to give your answer with the correct number of significant figures.

Problem 102

Use exponential notation to indicate the number of significant figures in each statement: a. Fifty million Frenchmen can't be wrong. b. "For every ten jokes, thou hast got an hundred enemies" (Laurence Sterne, $1713-1768$ ). c. The diameter of a Ca atom is $1.8$ one hundred millionths of a centimeter. d. Sixty thousand dollars is a lot of money to pay for a car. e. The density of platinum (Table 1.4).

Problem 103

Classify each property as intensive or extensive. a. volume b. boiling point c. temperature d. electrical conductivity e. energy

Problem 104

At what temperatures will the readings on the Fahrenheit and Celsius thermometers be the same?

Problem 105

Suppose you design a new thermometer called the X thermometer. On the $X$ scale the boiling point of water is $130^{\circ} \mathrm{X}$ and the freezing point of water is $10^{\circ} \mathrm{X}$. At what temperature will the readings on the Fahrenheit and $\mathrm{X}$ thermometers be the same?

Problem 106

On a new Jekyll temperature scale, water freezes at $17{ }^{\circ} \mathrm{J}$ and boils at $97^{\circ} \mathrm{J} .$ On another new temperature scale, the Hyde scale, water freezes at $0{ }^{\circ} \mathrm{H}$ and boils at $120{ }^{\circ} \mathrm{H}$. If methyl alcohol boils at $84^{\circ} \mathrm{H}$, what is its boiling point on the Jekyll scale?

Problem 107

Force is defined as mass times acceleration. Starting with SI base units, derive a unit for force. Using SI prefixes suggest a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour and for the force resulting from the collision of a molecule of mass around $10^{-20} \mathrm{~kg}$ moving almost at the speed of light $\left(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ with the wall of its container. (Assume a 1 second deceleration time for both collisions.)

Problem 108

A temperature measurement of $25^{\circ} \mathrm{C}$ has three significant figures, while a temperature measurement of $-196^{\circ} \mathrm{C}$ has only two significant figures. Explain.

Problem 109

Do each calculation without using your calculator and give the answers to the correct number of significant figures. a. $1.76 \times 10^{-3} / 8.0 \times 10^{2}$ b. $1.87 \times 10^{-2}+2 \times 10^{-4}-3.0 \times 10^{-3}$ c. $\left[\left(1.36 \times 10^{5}\right)(0.000322) / 0.082\right](129.2)$

Problem 110

The value of the Euro is currently $\$ 1.35$ U.S. and the price of liter of gasoline in France is $0.97$ Euro. What is the price of 1 gallon of gasoline in U.S. dollars in France?

Problem 111

A thief uses a can of sand to replace a solid gold cylinder that sits on a weight-sensitive, alarmed pedestal. The can of sand and the gold cylinder have exactly the same dimensions (length $=22 \mathrm{~cm}$ and radius $=3.8 \mathrm{~cm}$ ). a. Calculate the mass of each cylinder (ignore the mass of the can itself). (density of gold $=19.3 \mathrm{~g} / \mathrm{cm}^{3}$, density of sand $=$ $\left.3.00 \mathrm{~g} / \mathrm{cm}^{3}\right)$ b. Did the thicf sct off the alarm? Explain.

Problem 112

The proton has a radius of approximately $1.0 \times 10^{-13} \mathrm{~cm}$ and a mass of $1.7 \times 10^{-24} \mathrm{~g}$. Determine the density of a proton. For a sphere $V=(4 / 3) \pi r^{3}$.

Problem 113

The density of titanium is $4.51 \mathrm{~g} / \mathrm{cm}^{3}$. What is the volume (in cubic inches) of $3.5 \mathrm{lb}$ of titanium?

Problem 114

The density of iron is $7.86 \mathrm{~g} / \mathrm{cm}^{3}$. What is its density in pounds per cubic inch $\left(\mathrm{lb} / \mathrm{in}^{3}\right)$ ?

Problem 115

A steel cylinder has a length of $2.16$ in, a radius of $0.22 \mathrm{in}$, and a mass of $41 \mathrm{~g} .$ What is the density of the steel in $\mathrm{g} / \mathrm{cm}^{3}$ ?

Problem 116

A solid aluminum sphere has a mass of $85 \mathrm{~g}$. Use the density of aluminum to find the radius of the sphere in inches.

Problem 117

A backyard swimming pool holds 185 cubic yards $\left(\mathrm{yd}^{3}\right)$ of water. What is the mass of the water in pounds?

Problem 118

An iceberg has a volume of 7655 cubic feet. What is the mass of the ice (in $\mathrm{kg}$ ) composing the iceberg?

Problem 119

The Toyota Prius, a hybrid electric vehicle, has an EPA gas mileage rating of $52 \mathrm{mi} / \mathrm{gal}$ in the city. How many kilometers can the Prius travel on 15 liters of gasoline?

Problem 120

The Honda Insight, a hybrid electric vehicle, has an EPA gas mileage rating of 57 mi/gal in the city. How many kilometers can the Insight travel on the amount of gasoline that would fit in a soda pop can? The volume of a soda pop can is $355 \mathrm{~mL}$.

Problem 121

The single proton that forms the nucleus of the hydrogen atom has a radius of approximately $1.0 \times 10^{-13} \mathrm{~cm} .$ The hydrogen atom itself has a radius of approximately $52.9 \mathrm{pm}$. What fraction of the space within the atom is occupied by the nucleus?

Problem 122

A sample of gaseous neon atoms at atmospheric pressure and $0{ }^{\circ} \mathrm{C}$ contains $2.69 \times 10^{22}$ atoms per liter. The atomic radius of neon is $69 \mathrm{pm}$. What fraction of the space is occupied by the atoms themselves? What does this reveal about the separation between atoms in the gaseous phase?

Problem 123

The diameter of a hydrogen atom is $212 \mathrm{pm}$. Find the length in kilometers of a row of $6.02 \times 10^{23}$ hydrogen atoms. The diameter of a ping pong ball is $4.0 \mathrm{~cm}$. Find the length in kilometers of a row of $6.02 \times 10^{23}$ ping pong balls.

Problem 124

The world's record in the $100 \mathrm{~m}$ dash is $9.69 \mathrm{~s}$ and in the 100 yard dash is $9.21 \mathrm{~s}$. Find the speed in miles/hr of the runners who set these records.

Problem 125

Table salt contains $39.33 \mathrm{~g}$ of sodium per $100 \mathrm{~g}$ of salt. The U.S. Food and Drug Administration (FDA) recommends that adults consume less than $2.40 \mathrm{~g}$ of sodium per day. A particular snack mix contains $1.25 \mathrm{~g}$ of salt per $100 \mathrm{~g}$ of the mix. What mass of the snack mix can you consume and still be within the FDA limit?

Problem 126

Lead metal can be extracted from a mineral called galena, which contains $86.6 \%$ lead by mass. A particular ore contains $68.5 \%$ galena by mass. If the lead can be extracted with $92.5 \%$ efficiency, what mass of ore is required to make a lead sphere with a $5.00 \mathrm{~cm}$ radius?

Problem 127

Liquid nitrogen has a density of $0.808 \mathrm{~g} / \mathrm{mL}$ and boils at $77 \mathrm{~K}$. Researchers often purchase liquid nitrogen in insulated $175-\mathrm{L}$ tanks. The liquid vaporizes quickly to gaseous nitrogen (which has a density of $1.15 \mathrm{~g} / \mathrm{L}$ at room temperature and atmospheric pressure) when the liquid is removed from the tank. Suppose that all $175 \mathrm{~L}$ of liquid nitrogen in a tank accidentally vaporized in a lab that measured $10.00 \mathrm{~m} \times 10.00 \mathrm{~m} \times 2.50 \mathrm{~m}$. What maximum fraction of the air in the room could be displaced by the gaseous nitrogen?

Problem 128

Mercury is often used as an expansion medium in a thermometer. The mercury sits in a bulb on the bottom of the thermometer and rises up a thin capillary as the temperature rises. Suppose a mercury thermometer contains $3.380 \mathrm{~g}$ of mercury and has a capillary that is $0.200 \mathrm{~mm}$ in diameter. How far does the mercury rise in the capillary when the temperature changes from $0.0{ }^{\circ} \mathrm{C}$ to $25.0{ }^{\circ} \mathrm{C} ?$ The density of mercury at these temperatures is $13.596 \mathrm{~g} / \mathrm{cm}^{3}$ and $13.534 \mathrm{~g} / \mathrm{cm}^{3}$, respectively.

Problem 129

A force of $2.31 \times 10^{4} \mathrm{~N}$ is applied to a diver's face mask that has an area of $125 \mathrm{~cm}^{2}$. Find the pressure in atm on the face mask.

Problem 130

The SI unit of force is the Newton, derived from the base units by using the definition of force, $F=m a$. The dyne is a non-SI unit of force in which mass is measured in grams and time is measured in seconds. The relationship between the two units is dyne $=10^{-5} \mathrm{~N}$. Find the unit of length used to define the dyne.

Problem 131

Kinetic energy can be defined as $1 / 2 m v^{2}$ or as $3 / 2 P V$. Show that the derived SI units of each of these terms are those of energy. (Pressure is force/area and force is mass $\times$ acceleration.)

Problem 132

In 1999, scientists discovered a new class of black holes with masses 100 to 10,000 times the mass of our sun, but occupying less space than our moon. Suppose that one of these black holes has a mass of $1 \times 10^{3}$ suns and a radius equal to one-half the radius of our moon. What is the density of the black hole in $\mathrm{g} / \mathrm{cm}^{3}$ ? The radius of our sun is $7.0 \times 10^{5} \mathrm{~km}$ and it has an average density of $1.4 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$. The diameter of the moon is $2.16 \times 10^{3}$ miles.

Problem 133

Section $1.7$ showed that in 1997 Los Angeles County air had carbon monoxide (CO) levels of $15.0$ ppm. An average human inhales about $0.50 \mathrm{~L}$ of air per breath and takes about 20 breaths per minute. How many milligrams of carbon monoxide does the average person inhale in an 8 -hour period for this level of carbon monoxide pollution? Assume that the carbon monoxide has a density of $1.2 \mathrm{~g} / \mathrm{L}$. (Hint: $15.0$ ppm CO means $15.0 \mathrm{~L} \mathrm{CO}$ per $10^{6} \mathrm{~L}$ air. $)$

Problem 134

Nanotechnology, the field of trying to build ultrasmall structures one atom at a time, has progressed in recent years. One potential application of nanotechnology is the construction of artificial cells. The simplest cells would probably mimic red blood cells, the body's oxygen transporters. For example, nanocontainers, perhaps constructed of carbon, could be pumped full of oxygen and injected into a person's bloodstream. If the person needed additional oxygen - due to a heart attack perhaps, or for the purpose of space travel - these containers could slowly release oxygen into the blood, allowing tissues that would otherwise die to remain alive. Suppose that the nanocontainers were cubic and had an edge length of 25 nanometers. a. What is the volume of one nanocontainer? (Ignore the thickness of the nanocontainer's wall.) b. Suppose that each nanocontainer could contain pure oxygen pressurized to a density of $85 \mathrm{~g} / \mathrm{L}$. How many grams of oxygen could be contained by each nanocontainer?

Problem 135

Approximate the percent increase in waist size that occurs when a 155 -lb person gains $40.0$ lbs of fat. Assume that the volume of the person can be modeled by a cylinder that is $4.0$ feet tall. The average density of a human is about $1.0 \mathrm{~g} / \mathrm{cm}^{3}$ and the density of fat is $0.918 \mathrm{~g} / \mathrm{cm}^{3}$.

Problem 136

A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the displacement of water to be $427 \mathrm{~cm}^{3}$ and the total mass is $4.36 \mathrm{~kg}$. What percentage of the spheres are copper?

Problem 137

A volatile liquid (one that easily evaporates) is put into a jar and the jar is then sealed. Does the mass of the sealed jar and its contents change upon the vaporization of the liquid?

Problem 138

The diagram represents solid carbon dioxide, also known as dry ice.

Which of the diagrams below best represents the dry ice after it has sublimed into a gas?

Problem 139

A cube has an edge length of $7 \mathrm{~cm}$. If it is divided up into $1-\mathrm{cm}$ cubes, how many $1-\mathrm{cm}$ cubes would there be?

Problem 140

Substance A has a density of $1.7 \mathrm{~g} / \mathrm{cm}^{3}$. Substance B has a density of $1.7 \mathrm{~kg} / \mathrm{m}^{3}$. Without doing any calculations, determine which substance is most dense.

Problem 141

For each box, examine the blocks attached to the balances. Based on their positions and sizes, determine which block is more dense (the dark block or the lighter-colored block), or if the relative densities cannot be determined. (Think carefully about the information being shown.)

Problem 142

142. Identify each statement as being most like an observation, a law, or a theory. a. All coastal areas experience two high tides and two low tides each day. b. The tides in Earth's oceans are caused mainly by the gravitational attraction of the moon. c. Yesterday, high tide in San Francisco Bay occurred at 2:43 A.M. and 3:07 P.M. d. Tides are higher at the full moon and new moon than at other times of the month.

• Research Matters — to the Science Teacher

Problem Solving in Chemistry

One of the major difficulties in teaching introductory chemistry courses is helping students become efficient problem solvers. Most beginning chemistry students find this one of the most difficulty aspects of the introductory chemistry course. What does research tell us about problem solving in chemistry? Just why do students have such difficulty in solving chemistry problems? Are some ways of teaching students to solve problems more effective than others? Problem solving in any area is a very complex process. It involves an understanding of the language in which the problem is stated, the interpretation of what is given in the problem and what is sought, an understanding of the science concepts involved in the solution, and the ability to perform mathematical operations if these are involved in the problem. The first requirement for successful problem solving is that the problem solver understand the meaning of the problem. In order to do so there must be an understanding of the vocabulary and its usage in the problem. There are two types of words that occur in problems, ordinary words that science teachers generally assume that students know and more technical terms that require understanding of concepts specific to the discipline. Researchers have found that many students do not know the meaning of common words such as contrast, displace, diversity, factor, fundamental, incident, negligible, relevant, relative, spontaneous and valid. Slight changes in the way a problem is worded may make a difference in whether a students is able to solve it correctly. For example, when "least" is changed to "most" in a problem, the percentage getting the question correct may increase by 25%. Similar improvements occur for changing negative to positive forms, for rewording long and complex questions, and for changing from the passive to the active voice. Although teachers would like students to solve problems in whatever way they are framed they must be cognizant of the fact that these subtle changes will make a difference in students' success in solving problems. From several research studies on problem solving in chemistry, it is clear that the major reason why students are unable to solve problems is that they do not understand the concepts on which the problems are based. Studies that compare the procedures used by students who are inexperienced in solving problems with experts show that experts were able to retrieve relevant concepts more readily from their long term memory. Studies have also shown that experts concepts are linked to one another in a network. Experts spend a considerable period of time planning the strategy that will be used to solve the problem whereas novices jump right in using a formula or trying to apply an algorithm. In the past few years, science educators have been trying to determine which science concepts students understand and which they do not. Because chemistry is concerned with the nature of matter, and matter is defined as anything that has mass and volume, students must understand these concepts to be successful problem solvers in chemistry. Research studies have shown that a surprising number of high school students do not understand the meaning of mass, volume, heat, temperature and changes of state. One reason why students do not understand these concepts is because when they have been taught in the classroom, they have not been presented in a variety of contexts. Often the instruction has been verbal and formal. This will be minimally effective if students have not had the concrete experiences. Hence, misconceptions arise. Although the very word "misconception" has a negative connotation, this information is important for chemistry teachers. They are frameworks by which the students view the world around them. If a teacher understands these frameworks, then instruction can be formulated that builds on student's existing knowledge. It appears that students build conceptual frameworks as they try to make sense out of their surroundings. In addition to the fundamental properties of matter mentioned above, there are other concepts that are critical to chemical calculations. One of these is the mole concept and another is the particulate nature of matter. There is mounting evidence that many students do not understand either of these concepts sufficiently well to use them in problem solving. It appears that if chemistry problem solving skills of students are to improve, chemistry teachers will need to spend a much greater period of time on concept acquisition. One way to do this will be to present concepts in a variety of contexts, using hands-on activities.

What does this research imply about procedures that are useful for helping students become more successful at problem solving?

Chemistry problems can be solved using a variety of techniques. Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. It has been shown that this is not the best technique for high school students of high mathematics anxiety and low proportional reasoning ability. The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. The use of analogs is not profitable for certain types of problems. When problems became complex (such as in dilution problems) students are unable to solve even the analog problems. For these types of problems, using analogs in instruction would be useless unless teachers are willing to spend additional time teaching students how to solve problems using the analog. Many students are unable to match analogs with the chemistry problems even after practice in using analogs. Students need considerable practice if analogs are used in instruction. When teaching chemistry by the lecture method, concept development needed for problem solving may be enhanced by pausing for a two minute interval at about 8 to 12 minute intervals during the lecture. This provides students time to review what has been presented, fill in the gaps, and interpret the information for others, and thus learn it themselves. The use of concept maps may also help students understand concepts and to relate them to one another. Requiring students to use a worksheet with each problem may help them solve them in a more effective way. The worksheet might include a place for them to plan a problem, that is list what is given and what is sought; to describe the problem situation by writing down other concepts they retrieve from memory (the use of a picture may integrate these); to find the mathematical solution; and to appraise their results. Although the research findings are not definitive, the above approaches offer some promise that students' problem solving skills can be improved and that they can learn to solve problems in a meaningful way.

For further information about this research area, please contact:

Dr. Dorothy Gabel Education Building 3rd and Jordan Bloomington, Indiana 47405

PERSPECTIVE article

Problem-solving is embedded in context… so how do we measure it.

• 1 University of California, Irvine, Irvine, United States
• 2 California State University, Fullerton, Fullerton, California, United States

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Problem solving encompasses the broad domain of human, goal-directed behaviors. Though we may attempt to measure problem solving using tightly controlled and decontextualized tasks, it is inextricably embedded in both reasoners’ experiences and their contexts. Without situating problem-solvers, problem contexts, and our own experiential partialities as researchers, we risk intertwining the research of information relevance with our own confirmatory biases about people, environments, and ourselves. We review each of these ecological facets of information relevance in problem-solving, and we suggest a framework to guide its measurement. We ground this framework with concrete examples of ecologically valid, culturally relevant measurement of problem solving.

Keywords: problem-solving, Measurement, Information relevance, Ecological Validity, Cultural relevance

Received: 01 Feb 2024; Accepted: 26 Apr 2024.

Copyright: © 2024 Rhodes, Richland and Alcalá. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY) . The use, distribution or reproduction in other forums is permitted, provided the original author(s) or licensor are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.

* Correspondence: Katherine Rhodes, University of California, Irvine, Irvine, United States

Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.