Section 8.2 Unit Conversion
Unit conversion is a systematic method for converting from one kind of unit of measurement to another. It is used extensively in chemistry and other health- or science-related fields. It is a valuable skill to learn, and necessary for success in many applications.
Subsection 8.2.1 Unit Ratios
Example 8.2.2.
When building a staircase, a step typically has a rise of \(7\) inches (7 in). An inch is a unit of length in the imperial unit system, used in the United States, Canada, the United Kingdom, and a few other places. Many parts of the world do not use this unit of measurement, and the people there do not have a sense of how long \(7\) inches is. Instead, much of the world would measure a length like this using centimeters (cm). How many centimeters is \(7\) inches?
To convert from one unit of measurement to another (like inches to centimeters), we use what are called unit ratios. A unit ratio is a ratio (or fraction) where the numerator and denominator are quantities with units that equal each other. They equal each other as measurements, but they are measured with different units. For example, Appendix A tells us that \(1\) inch is equal to \(2.54\) centimeters. Knowing that, we can build the unit ratios \(\frac{1\,\text{in}}{2.54\,\text{cm}}\) and \(\frac{2.54\,\text{cm}}{1\,\text{in}}\text{.}\) Each of these unit ratios are equivalent to \(1\text{,}\) because their numerator equals their denominator.
With a unit ratio, we can work out a conversion by taking what we would like to convert (7 in) and multiplying by a unit ratio in such a way that the “old” units cancel and the “new” units remain.
\begin{align*}
7\,\text{in}\amp=\frac{7\,\text{in}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp=\frac{7\,\text{in}}{1}\cdot\frac{2.54\,\text{cm}}{1\,\text{in}}\amp\amp1\,\text{in equals }2.54\,\text{cm.}\\
\amp=\frac{7\,\cancelhighlight{\text{in}}}{1}\cdot\frac{2.54\,\text{cm}}{1\,\cancelhighlight{\text{in}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{7}{1}\cdot\frac{2.54\,\text{cm}}{1}\\
\amp=7\cdot2.54\,\text{cm}\\
\amp=17.78\,\text{cm}
\end{align*}
So \(7\) inches is equal to \(17.78\) centimeters. In practice, anyone talking about the rise of a stair might simply round to 18 cm.
Note there was another unit ratio, \(\frac{1\,\text{in}}{2.54\,\text{cm}}\text{,}\) but using that would not have been helpful, since it would not have arranged units such that the inches canceled.
Remark 8.2.3.
When you are comfortable, you might do the steps from Example 2 on one line, like:
\begin{equation*}
7\,\text{in}=\frac{7\,\cancelhighlight{\text{in}}}{1}\cdot\frac{2.54\,\text{cm}}{1\,\cancelhighlight{\text{in}}}=\frac{7}{1}\cdot\frac{2.54}{1}\,\text{cm}=17.78\,\text{cm}
\end{equation*}
The examples in this section will continue to show the steps completely drawn out, to give a better sense of what you would write first, second, and so on.
Example 8.2.4.
A canned beverage typically contains \(12\) fluid ounces (12 fl oz). A fluid ounce is a unit of volume used in the United States. (The United Kingdom also has a fluid ounce, but it is a slightly different amount.) In the rest of the world, people do not have a sense of how much \(12\) fluid ounces is. Most of the world would measure a canned beverage’s volume using milliliters (mL). How many milliliters is \(12\) fluid ounces?
Appendix A tells us that 1 fl oz is (almost) equal to 29.57 mL. Knowing that, we can build the unit ratios \(\frac{1\,\text{fl}\,\text{oz}}{29.57\,\text{mL}}\) and \(\frac{29.57\,\text{mL}}{1\,\text{fl}\,\text{oz}}\text{.}\) Each of these unit ratios are (almost) equivalent to \(1\text{,}\) because their numerator (almost) equals their denominator.
Using the appropriate unit ratio to enable cancellation of fluid ounces:
\begin{align*}
12\,\text{fl}\,\text{oz}\amp=\frac{12\,\text{fl}\,\text{oz}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp\approx\frac{12\,\text{fl}\,\text{oz}}{1}\cdot\frac{29.57\,\text{mL}}{1\,\text{fl}\,\text{oz}}\amp\amp1\,\text{fl}\,\text{oz approximately equals }29.57\,\text{mL.}\\
\amp=\frac{12\,\cancelhighlight{\text{fl}\,\text{oz}}}{1}\cdot\frac{29.57\,\text{mL}}{1\,\cancelhighlight{\text{fl}\,\text{oz}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{12}{1}\cdot\frac{29.57\,\text{mL}}{1}\\
\amp=12\cdot29.57\,\text{mL}\\
\amp\approx354.8\,\text{mL}
\end{align*}
So \(12\) fluid ounces is approximately equal to \(354.8\) milliliters. In practice, you might round to 355 mL.
Notice that each conversion fact from Appendix A gives two possible unit ratios. Deciding which one to use will depend on where units need to be placed in order to cancel the appropriate units. In unit conversion, we multiply ratios together and cancel common units the same way we can cancel common factors when multiplying fractions.
Example 8.2.5.
It’s \(1760\) feet (1760 ft) to walk from Jonah’s house to where he works. How many miles is that?
Explanation.
Since we are converting feet to miles, we use the conversion fact that there are \(5280\) feet in \(1\) mile. In this conversion, we need to use a unit ratio that will allow the feet units to cancel. So we need to use \(\frac{1\,\text{mi}}{5280\,\text{ft}}\text{.}\) This is different from previous examples in that the \(1\) is in the numerator this time. But the process is not all that different.
\begin{align*}
1760\,\text{ft}\amp=\frac{1760\,\text{ft}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp=\frac{1760\,\text{ft}}{1}\cdot\frac{1\,\text{mi}}{5280\,\text{ft}}\amp\amp1\,\text{mi equals }5280\,\text{ft.}\\
\amp=\frac{1760\,\cancelhighlight{\text{ft}}}{1}\cdot\frac{1\,\text{mi}}{5280\,\cancelhighlight{\text{ft}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{1760}{1}\cdot\frac{1\,\text{mi}}{5280}\\
\amp=\frac{1760}{5280}\,\text{mi}\\
\amp=\frac{1}{3}\,\text{mi}\approx0.3333\,\text{mi}
\end{align*}
So Jonah walks \(\frac{1}{3}\) of a mile, or about 0.3333 mi, to get from his house to where he works.
Checkpoint 8.2.6.
Convert \(60\) inches to feet.
Explanation.
We start by writing what it is that we are converting as a ratio, by placing it over a \(1\text{.}\) This is similar to writing a whole number as a fraction when we want to multiply it by a fraction. Next we multiply that ratio by a unit ratio, one that will have inches in the denominator so that inches will cancel. Multiply what’s left just as we multiply fractions (multiply the numerators together and multiply the denominators together), including the units, and simplify by dividing.
\begin{equation*}
\begin{aligned}
60\,\text{in}\amp=\frac{60\,\text{in}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp=\frac{60\,\text{in}}{1}\cdot\frac{1\,\text{ft}}{12\,\text{in}}\amp\amp1\,\text{ft equals }12\,\text{in.}\\
\amp=\frac{60\,\cancel{\text{in}}}{1}\cdot\frac{1\,\text{ft}}{12\,\cancel{\text{in}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{60}{1}\cdot\frac{1\,\text{ft}}{12}\\
\amp=\frac{60}{12}\,\text{ft}\\
\amp=5\,\text{ft}
\end{aligned}
\end{equation*}
We find that \(60\) inches is equivalent to \(5\) feet.
Example 8.2.7. Why Do We Convert Units?
Converting from one unit to another can be necessary when you are given information where the units don’t quite match. Cassidy was driving at a speed of 32 mph for seven minutes. How far did they travel in that time span?
Normally, to find a distance traveled, you would multiply speed by how much time passed. For example if Cassidy had been driving 50 mph for two hours, we would find \(50\cdot2=100\text{,}\) and conclude they had driven \(100\) miles.
But in this example, Cassidy’s speed is \(32\) miles per hour, but the time elapsed is seven minutes. The time units do not match. It will help to convert the 7 min into hours. So let’s do that.
\begin{align*}
7\,\text{min}\amp=\frac{7\,\text{min}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp=\frac{7\,\text{min}}{1}\cdot\frac{1\,\text{h}}{60\,\text{min}}\amp\amp1\,\text{h equals }60\,\text{min.}\\
\amp=\frac{7\,\cancelhighlight{\text{min}}}{1}\cdot\frac{1\,\text{h}}{60\,\cancelhighlight{\text{min}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{7}{1}\cdot\frac{1\,\text{h}}{60}\\
\amp=\frac{7}{60}\,\text{h}\\
\amp\approx0.1167\,\text{h}
\end{align*}
Now we can multiply Cassidy’s speed (32 mph) by their elapsed time (\(\frac{7}{60}\,\text{h}\)). We find \(32\cdot\frac{7}{60}\approx3.733\text{,}\) so Cassidy has traveled about \(3.733\) miles.
Actually we can do this multiplication with units and the units will cancel appropriately:
\begin{align*}
32\,\frac{\text{mi}}{\text{h}}\cdot\frac{7}{60}\,\text{h}\amp=\frac{32\,\text{mi}}{1\,\text{h}}\cdot\frac{7\,\text{h}}{60}\\
\amp=\frac{32\,\text{mi}}{1\,\cancelhighlight{\text{h}}}\cdot\frac{7\,\cancelhighlight{\text{h}}}{60}\\
\amp=\frac{32\cdot7}{60}\,\text{mi}\\
\amp\approx3.733\,\text{mi}
\end{align*}
Checkpoint 8.2.8.
The density of oil is \(6.9\) pounds per gallon. You have a 2.5-liter bottle of oil. How much does this much oil weigh? (To find weight, multiply density with volume when the units match.)
Explanation.
The density is in pounds per gallon, but the volume is in liters. So first let’s convert the 2.5 L to gallons.
\begin{equation*}
\begin{aligned}
2.5\,\text{L}\amp=\frac{2.5\,\text{L}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp\approx\frac{2.5\,\text{L}}{1}\cdot\frac{1\,\text{gal}}{3.785\,\text{L}}\amp\amp1\,\text{gal approximately equals }3.785\,\text{L.}\\
\amp=\frac{2.5\,\cancel{\text{L}}}{1}\cdot\frac{1\,\text{gal}}{3.785\,\cancel{\text{L}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{2.5}{1}\cdot\frac{1\,\text{gal}}{3.785}\\
\amp=\frac{2.5}{3.785}\,\text{gal}\\
\amp\approx0.6605\,\text{gal}
\end{aligned}
\end{equation*}
Now we can multiply the density (6.9 lb/gal) by the volume (\(\approx0.6605\,\text{gal}\)). We find \(6.9\cdot0.6605\approx4.557\text{,}\) so the oil weighs about \(4.557\) pounds.
With units:
\begin{equation*}
\begin{aligned}
6.9\,\frac{\text{lb}}{\text{gal}}\cdot0.6605\,\text{gal}\amp=\frac{6.9\,\text{lb}}{1\,\text{gal}}\cdot\frac{0.6605\,\text{gal}}{1}\\
\amp=\frac{6.9\,\text{lb}}{1\,\cancel{\text{gal}}}\cdot\frac{0.6605\,\cancel{\text{gal}}}{1}\\
\amp=6.9\cdot0.6605\,\text{lb}\\
\amp\approx4.557\,\text{lb}
\end{aligned}
\end{equation*}
Subsection 8.2.2 Using Multiple Unit Ratios
In previous examples, we used only one unit ratio to make a conversion. However, sometimes there is a need to use more than one unit ratio in a conversion. This may happen when your reference guide for conversions does not directly tell you how to convert from one unit to another. In those situations, we’ll have to consider the conversion facts that are available and then make a plan.
Example 8.2.9.
Convert \(350\) yards to miles.
Explanation.
In Appendix A, there is not a conversion that relates yards to miles. But notice that we can convert yards to feet (using the fact that one yard is three feet) and then we can convert feet to miles (using the fact that one mile is \(5280\) feet). So we will use two unit ratios. The unit ratio \(\frac{3\,\text{ft}}{1\,\text{yd}}\) can be used to cancel the yards in 350 yd. Then the unit ratio \(\frac{1\,\text{mi}}{5280\,\text{ft}}\) can be used to cancel the feet that are left over from the first conversion.
\begin{align*}
350\,\text{yd}\amp=\frac{350\,\text{yd}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp=\frac{350\,\text{yd}}{1}\cdot\frac{3\,\text{ft}}{1\,\text{yd}}\cdot\frac{1\,\text{mi}}{5280\,\text{ft}}\amp\amp\text{Both unit ratios are needed.}\\
\amp=\frac{350\,\cancelhighlight{\text{yd}}}{1}\cdot\frac{3\,\secondcancelhighlight{\text{ft}}}{1\,\cancelhighlight{\text{yd}}}\cdot\frac{1\,\text{mi}}{5280\,\secondcancelhighlight{\text{ft}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{350}{1}\cdot\frac{3}{1}\cdot\frac{1\,\text{mi}}{5280}\\
\amp=\frac{350\cdot3}{5280}\,\text{mi}\\
\amp\approx0.1989\,\text{mi}
\end{align*}
So \(350\) yards is about \(0.1989\) miles.
Checkpoint 8.2.10.
Convert \(4.5\) months into hours.
Explanation.
Notice that we can convert months to days (using the fact that one month is approximately \(30\) days) and then we can convert days to hours (using the fact that one day is \(24\) hours).
\begin{equation*}
\begin{aligned}
4.5\,\text{mo}\amp=\frac{4.5\,\text{mo}}{1}\amp\amp\text{We are about to do fraction-like multiplication.}\\
\amp\approx\frac{4.5\,\text{mo}}{1}\cdot\frac{30\,\text{d}}{1\,\text{mo}}\cdot\frac{24\,\text{h}}{1\,\text{d}}\amp\amp\text{Two unit ratios are needed.}\\
\amp=\frac{4.5\,\cancel{\text{mo}}}{1}\cdot\frac{30\,\cancel{\text{d}}}{1\,\cancel{\text{mo}}}\cdot\frac{24\,\text{h}}{1\,\cancel{\text{d}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{4.5}{1}\cdot\frac{30}{1}\cdot\frac{24\,\text{h}}{1}\\
\amp=4.5\cdot30\cdot24\,\text{h}\\
\amp=3240\,\text{h}
\end{aligned}
\end{equation*}
So \(4.5\) months is about \(3240\) hours.
Subsection 8.2.3 Converting Squared or Cubed Units
When calculating the area or volume of a geometric figure, units of measurement are multiplied together, resulting in squared units (when calculating area) or cubed units (when calculating volume). Thus, there may be circumstances where you may need to convert either squared or cubed units. For example, suppose you are carpeting a room in your home and you know the square footage of the room, but the carpet is sold in square yards. In that case, you would need to convert the square feet of the room into square yards.
Example 8.2.11.
Jin’s bedroom is \(153\) square feet (153 ft2). How many square yards is that?
We start the process the same as in the previous examples. That is, we write what we are converting in ratio form with a denominator of \(1\text{.}\)
\begin{align*}
153\,\text{ft}^2\amp=\frac{153\,\text{ft}^2}{1}\amp\amp\text{We are about to do fraction-like multiplication.}
\end{align*}
Now, we do want feet to be replaced with yards, so the unit ratio \(\frac{1\,\text{yd}}{3\,\text{ft}}\) will be useful. But using it once is not enough:
\begin{align*}
153\,\text{ft}^2\amp=\frac{153\,\text{ft}^2}{1}\\
\amp=\frac{153\,\text{ft}^2}{1}\cdot\frac{1\,\text{yd}}{3\,\text{ft}}\amp\amp1\text{ yd equals }3\text{ feet.}
\end{align*}
The ft2 in the first numerator do not fully cancel with the ft in the second denominator. We need to use this unit ratio twice.
\begin{align*}
153\,\text{ft}^2\amp=\frac{153\,\text{ft}^2}{1}\\
\amp=\frac{153\,\text{ft}^2}{1}\cdot\frac{1\,\text{yd}}{3\,\text{ft}}\cdot\frac{1\,\text{yd}}{3\,\text{ft}}\amp\amp1\text{ yd equals }3\text{ feet.}
\end{align*}
Now there is \(\text{ft}^2\) in the overall numerator, and \(\text{ft}\cdot\text{ft}\) in the overall denominator. They will fully cancel.
Here is the complete process from the beginning.
\begin{align*}
153\,\text{ft}^2\amp=\frac{153\,\text{ft}^2}{1}\\
\amp=\frac{153\,\text{ft}^2}{1}\cdot\frac{1\,\text{yd}}{3\,\text{ft}}\cdot\frac{1\,\text{yd}}{3\,\text{ft}}\\
\amp=\frac{153\,\cancelhighlight{\text{ft}^2}}{1}\cdot\frac{1\,\text{yd}}{3\,\cancelhighlight{\text{ft}}}\cdot\frac{1\,\text{yd}}{3\,\cancelhighlight{\text{ft}}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{153}{1}\cdot\frac{1\,\text{yd}}{3}\cdot\frac{1\,\text{yd}}{3}\\
\amp=\frac{153}{9}\,\text{yd}\cdot\text{yd}\\
\amp=17\,\text{yd}^2
\end{align*}
So Jin’s bedroom has \(17\) square yards of area.
Alternatively, we can set up conversions with squared or cubed units this way:
\begin{align*}
153\,\text{ft}^2\amp=\frac{153\,\text{ft}^2}{1}\\
\amp=\frac{153\,\text{ft}^2}{1}\cdot\left(\frac{1\,\text{yd}}{3\,\text{ft}}\right)^2\amp\amp\text{The ft in the denominator will be squared.}\\
\amp=\frac{153\,\text{ft}^2}{1}\cdot\frac{1\,\text{yd}^2}{9\,\text{ft}^2}\amp\amp\text{Using }[cross-reference to target(s) "fact-quotient-to-a-power-property" missing or not unique]\text{.}\\
\amp=\frac{153\,\cancelhighlight{\text{ft}^2}}{1}\cdot\frac{1\,\text{yd}^2}{9\,\cancelhighlight{\text{ft}^2}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{153}{1}\cdot\frac{1\,\text{yd}^2}{9}\\
\amp=\frac{153}{9}\,\text{yd}^2\\
\amp=17\,\text{yd}^2
\end{align*}
When using this setup where the unit ratio is raised to a power, you must be careful to remember that everything inside the parentheses is raised to that power: the units and the numbers alike.
Checkpoint 8.2.12.
Convert \(85\) cubic inches into cubic centimeters.
Explanation.
\begin{equation*}
\begin{aligned}
85\,\text{in}^3\amp=\frac{85\,\text{in}^3}{1}\\
\amp=\frac{85\,\text{in}^3}{1}\cdot\left(\frac{2.54\,\text{cm}}{1\,\text{in}}\right)^3\amp\amp\text{The inches in the denominator will be cubed.}\\
\amp=\frac{85\,\text{in}^3}{1}\cdot\frac{2.54^3\,\text{cm}^3}{1\,\text{in}^3}\amp\amp\text{Using the quotient to a power property.}\\
\amp=\frac{85\,\cancel{\text{in}^3}}{1}\cdot\frac{2.54^3\,\text{cm}^3}{1\,\cancel{\text{in}^3}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{85}{1}\cdot\frac{2.54^3\,\text{cm}^3}{1}\\
\amp=85\cdot2.54^3\,\text{cm}^3\\
\amp\approx1393\,\text{cm}^3
\end{aligned}
\end{equation*}
So \(85\) cubic inches is about \(1393\) cubic centimeters.
Subsection 8.2.4 Converting Rates
A rate unit has a numerator and a denominator. For example, speed is a rate, and speed can be measured in mi⁄h. The numerator unit is a mile and the denominator unit is an hour.
Suppose we wanted to convert a speed rate, such as 65 mi⁄h, into m⁄s. Or a concentration rate, such as 180 mg⁄L, into g⁄dL. We can use the same process that we’ve used before to do these conversions. That is, we start by writing what we want to convert as a ratio, which will have units in both the numerator and denominator, and then we multiply by unit ratios until both units have been converted into the units we want. It helps to focus on converting one unit at a time and to make sure that the units in our unit ratios are placed so that the proper units will cancel.
Example 8.2.13.
Convert 65 mi⁄h into m⁄min.
Explanation.
We start by writing what we are converting, which is 65 mi⁄h, as a ratio. Then, our job is to convert the miles to meters and the hours to minutes, one at a time. It doesn’t matter which unit ratio we use first, as long as the units line up to cancel appropriately.
\begin{align*}
65\,\frac{\text{mi}}{\text{h}}\amp=\frac{65\,\text{mi}}{1\,\text{h}}\amp\amp\text{Write the rate as a ratio.}\\
\amp\approx\frac{65\,\text{mi}}{1\,\text{h}}\cdot\frac{1.609\,\text{km}}{1\,\text{mi}}\cdot\frac{1000\,\text{m}}{1\,\text{km}}\cdot\frac{1\,\text{h}}{60\,\text{min}}\amp\amp\text{Use unit ratios to make cancellations.}\\
\amp=\frac{65\,\cancelhighlight{\text{mi}}}{1\,\cancel{\text{h}}}\cdot\frac{1.609\,\secondcancelhighlight{\text{km}}}{1\,\cancelhighlight{\text{mi}}}\cdot\frac{1000\,\text{m}}{1\,\secondcancelhighlight{\text{km}}}\cdot\frac{1\,\cancel{\text{h}}}{60\,\text{min}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{65}{1}\cdot\frac{1.609}{1}\cdot\frac{1000\,\text{m}}{1}\cdot\frac{1}{60\,\text{min}}\\
\amp=\frac{65\cdot1.609\cdot1000}{60}\,\frac{\text{m}}{\text{min}}\\
\amp\approx1743\,\frac{\text{m}}{\text{min}}
\end{align*}
Notice that the last unit ratio is used to convert the hours to minutes and the hour must be placed in the numerator to cancel the hour in the original rate that was in the denominator. Also, note that this will automatically cause minutes to end up in the denominator, which is where this unit should end up so that we end up with meters per minute for our final unit.
An important thing to keep in mind, as demonstrated in the previous example, as well as the next example, is that we avoid multiplying or dividing any numbers until the end, after the final units that we want have been obtained. Stopping partway through to multiply or divide some numbers could lead to confusion and mistakes.
Checkpoint 8.2.14.
Convert 180 mg/L into g/dL, given that there are \(10\) deciliters in a liter.
Explanation.
We start by writing what we are converting, which is 180 mg/L, as a ratio. Then, we need to convert the milligrams into grams and the liters into deciliters, converting one unit at a time. We will start by converting the milligrams into grams. Then, we will convert the liters to deciliters.
\begin{equation*}
\begin{aligned}
180\,\frac{\text{mg}}{\text{L}}\amp=\frac{180\,\text{mg}}{1\,\text{L}}\amp\amp\text{Write the rate as a ratio.}\\
\amp=\frac{180\,\text{mg}}{1\,\text{L}}\cdot\frac{1\,\text{g}}{1000\,\text{mg}}\cdot\frac{1\,\text{L}}{10\,\text{dL}}\amp\amp\text{Use unit ratios to make cancellations.}\\
\amp=\frac{180\,\cancel{\text{mg}}}{1\,\cancel{\text{L}}}\cdot\frac{1\,\text{g}}{1000\,\cancel{\text{mg}}}\cdot\frac{1\,\cancel{\text{L}}}{10\,\text{dL}}\amp\amp\text{Units may now cancel.}\\
\amp=\frac{180}{1}\cdot\frac{1\,\text{g}}{1000}\cdot\frac{1}{10\,\text{dL}}\\
\amp=\frac{180}{1000\cdot10}\,\frac{\text{g}}{\text{dL}}\\
\amp\approx0.018\,\frac{\text{g}}{\text{dL}}
\end{aligned}
\end{equation*}
So for example if salt is mixed into water with a concentration of 180 mg/L, the concentration can also be described as 0.018 g/dL.
Reading Questions 8.2.5 Reading Questions
1.
Unit conversion is a lot like multiplying .
2.
If you are using a unit ratio to convert inches to feet, how do you decide whether to use \(\frac{1\,\text{ft}}{12\,\text{in}}\) or to use \(\frac{12\,\text{in}}{1\,\text{ft}}\text{?}\)
3.
If you use a power of a unit ratio to make a unit conversion, what do you need to remember?
Exercises 8.2.6 Exercises
Review and Warmup.
1.
Multiply: \(\displaystyle{ \frac{2}{7} \cdot \frac{2}{9} }\)
2.
Multiply: \(\displaystyle{ \frac{3}{10} \cdot \frac{3}{8} }\)
3.
Multiply: \(\displaystyle{ \frac{10}{7} \cdot \frac{13}{12} }\)
4.
Multiply: \(\displaystyle{ \frac{12}{5} \cdot \frac{14}{9} }\)
5.
Multiply: \(\displaystyle{5\cdot \frac{5}{6} }\)
6.
Multiply: \(\displaystyle{5\cdot \frac{5}{9} }\)
Unit Conversions.
7.
Convert \({86\ {\rm mg}}\) to grams.
8.
Convert \({2.4\ {\rm m^{2}}}\) to hectares.
9.
Convert \({6.1\ {\rm mL}}\) to cubic centimeters.
10.
Convert \({18\ {\rm mL}}\) to cubic inches.
11.
Convert \({5.15\ {\rm m}}\) to decimeters.
12.
Convert \({88.3\ {\rm m}}\) to millimeters.
13.
Convert \({361\ {\rm lb}}\) to tons.
14.
Convert \({7.37\ {\rm km^{2}}}\) to hectares.
15.
Convert \({21.5\ {\rm mi^{2}}}\) to hectares.
16.
Convert \({52\ {\rm km^{2}}}\) to square meters.
17.
Convert \({9.9\ {\rm kg}}\) to milligrams.
18.
Convert \({4.6\ {\rm km}}\) to hectometers.
19.
Convert \({84\ {\rm B}}\) to megabytes.
20.
Convert \({2.1\ {\rm kB}}\) to gigabytes.
21.
Convert \({66.8\ {\rm kB}}\) to bits.
22.
Convert \({145\ {\rm in^{3}}}\) to liters.
23.
Convert \({3.05\ {\rm mm^{2}}}\) to square meters.
24.
Convert \({95.5\ {\rm hm^{3}}}\) to cubic meters.
25.
Convert \({75\ {\rm mi^{2}}}\) to square feet.
26.
Convert \({4.4\ {\rm dm^{3}}}\) to cubic meters.
27.
Convert \({24\ {\rm in^{2}}}\) to square yards.
28.
Convert \({84\ {\rm cm^{3}}}\) to cubic meters.
29.
Convert \({6.4\ {\rm km^{2}}}\) to square miles.
30.
Convert \({36.4\ {\rm m^{3}}}\) to cubic yards.
31.
Convert \({811\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut wk}}}\) to decimeters per day.
32.
Convert \({29.7\ {\textstyle\frac{\rm\mathstrut mi}{\rm\mathstrut wk}}}\) to kilometers per day.
33.
Convert \({665\ {\textstyle\frac{\rm\mathstrut mi^{2}}{\rm\mathstrut d}}}\) to acres per hour.
34.
Convert \({1.2\ {\textstyle\frac{\rm\mathstrut mi^{2}}{\rm\mathstrut h}}}\) to acres per day.
35.
Convert \({5.9\ {\textstyle\frac{\rm\mathstrut in^{3}}{\rm\mathstrut h}}}\) to gallons per minute.
36.
Convert \({86\ {\textstyle\frac{\rm\mathstrut in^{3}}{\rm\mathstrut min}}}\) to milliliters per hour.
37.
Convert \({3.4\ {\textstyle\frac{\rm\mathstrut oz}{\rm\mathstrut min}}}\) to pounds per second.
38.
Convert \({7.1\ {\textstyle\frac{\rm\mathstrut oz}{\rm\mathstrut s}}}\) to pounds per minute.
39.
Convert \({218\ {\textstyle\frac{\rm\mathstrut kb}{\rm\mathstrut ms}}}\) to megabits per second.
40.
Convert \({7.7\ {\textstyle\frac{\rm\mathstrut B}{\rm\mathstrut d}}}\) to kilobytes per hour.
41.
Convert \({96.3\ {\textstyle\frac{\rm\mathstrut oz}{\rm\mathstrut L}}}\) to pounds per cubic centimeter.
42.
Convert \({449\ {\textstyle\frac{\rm\mathstrut oz}{\rm\mathstrut c}}}\) to pounds per pint.
Applications.
43.
Randi’s bedroom has \({95\ {\rm ft^{2}}}\) of floor. He would like to carpet the floor, but carpeting is sold by the square yard. How many square yards of carpeting will he need to get?
44.
Dawn’s bedroom has \({107\ {\rm ft^{2}}}\) of floor. She would like to carpet the floor, but carpeting is sold by the square yard. How many square yards of carpeting will she need to get?
45.
Maygen is traveling in Europe and renting a car. She is used to thinking of gasoline amounts in gallons, but in Europe it is sold in liters. After filling the gas tank, she notices it took \({32\ {\rm L}}\) of gas. How many gallons is that?
46.
Andrew is traveling in Europe and renting a car. He is used to thinking of gasoline amounts in gallons, but in Europe it is sold in liters. After filling the gas tank, he notices it took \({35\ {\rm L}}\) of gas. How many gallons is that?
47.
Kandace found a family recipe from the old country that uses \({250\ {\rm mL}}\) of soup stock. The recipe serves four, but Kandace wants to scale it up to serve eleven. And none of Kandace’s measuring devices use the metric system. How many cups of soup stock should she use?
48.
Stephanie found a family recipe from the old country that uses \({280\ {\rm mL}}\) of soup stock. The recipe serves four, but Stephanie wants to scale it up to serve nine. And none of Stephanie’s measuring devices use the metric system. How many cups of soup stock should she use?
49.
Alisa was driving at a steady speed of \({55\ {\rm mph}}\) for \(18\) minutes. How far did she travel in that time?
50.
Parnell was driving at a steady speed of \({60\ {\rm mph}}\) for \(11\) minutes. How far did he travel in that time?
51.
The algae in a pond is growing at a rate of \({0.48\ {\textstyle\frac{\rm\mathstrut kg}{\rm\mathstrut d}}}\text{.}\) How much algae is in the pond after \(4\) weeks?
52.
The algae in a pond is growing at a rate of \({0.11\ {\textstyle\frac{\rm\mathstrut kg}{\rm\mathstrut d}}}\text{.}\) How much algae is in the pond after \(17\) weeks?
53.
Stephen is downloading content at an average rate of \({23\ {\rm Mbps}}\) (megabits per second). After \(78\) minutes, how much has he downloaded? It is appropriate to express an amount of data like this in bytes, kilobytes, megabytes, gigabytes, or terabytes.
54.
Maygen is downloading content at an average rate of \({34\ {\rm Mbps}}\) (megabits per second). After \(13\) minutes, how much has she downloaded? It is appropriate to express an amount of data like this in bytes, kilobytes, megabytes, gigabytes, or terabytes.
This section is adapted from Dimensional Analysis, Converting Between Two Systems of Measurements, and Converting Rates by Wendy Lightheart, OpenStax CNX, which is licensed under CC BY 4.0
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