Q77E - Chemistry (OpenStax)

Question

Question:  Make the conversion indicated in each of the following:

(a) the length of a soccer field, 120 m (three significant figures), to feet

(b) the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers

(c) the area of an 8.5 × 11-inch sheet of paper in cm2

(d) the displacement volume of an automobile engine, 161 in.3, to liters

(e) the estimated mass of the atmosphere, 5.6 × 1015 tons, to kilograms

(f) the mass of a bushel of rye, 32.0 lb, to kilograms

(g) the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)

Step-by-Step Solution

Answer

1. 394  ft
2. 5.9634 km
3. 6.0 x 10² cm²
4. 2.64 L
5. 5.1   10¹⁸ kg
6. 14.5 kg
7. 324 mg

Step 1: Defining the conversion factor concept

A relationship is used to determine the required conversion of units. The magnitude does not change during the conversion of units.

An example is the conversion of ft to the inch using the conversion factor.

\(1\,\,ft\,\, = \,\,1\,ft\,\, \times \,\,\frac{{12\,\,inch}}{{ft}}\,\, = \,\,12\,\,inch\)

This is a very convenient method to convert the units.

Step 2: Converting the unit from given quantities into required quantities

1. The length of a soccer field, 120 m (three significant figures), to feet

\(\)\(\begin{array}{c}120\,\,m\,\, = \,\,120\,\,\left( {m\,\, \times \,\frac{{3.28084\,ft}}{m}} \right)\,\,\\ = \,\,393.701\,\,ft\,\,\\ = \,\,394\,\,ft\end{array}\)

       b. The height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to    

Kilometers

\(\begin{array}{c}19,565\,\,ft\,\, = \,\,19,565\,\,\left( {ft\, \times \,\frac{{0.0003048\,km}}{{ft}}} \right)\,\,\\ = \,\,5.963412\,km\\ = 5.9634km\end{array}\)

       c.  The area of an 8.5 × 11-inch sheet of paper in cm²

\(\begin{array}{c}8.5\,\, \times \,\,11\, - inch\,\, = \,\,8.5\,inch\,\, \times \,11\,inch\,\,\\ = \,\,8.5\,\, \times \,11\,i{n^2}\,\,\\ = \,\,93.5\,\,i{n^2}\,\\\, = \,\,93.5\,\left( {i{n^2}\, \times \,\frac{{6.4516\,\,c{m^2}}}{{i{n^2}}}} \right)\,\,\\ = \,\,603.2246\\ = 6.0\, \times {10^2}\,c{m^2}\end{array}\)

(d) The displacement volume of an automobile engine, 161 in.3, to liters\(\begin{array}{c}161\,\,i{n^3}\,\, = \,\,161\,\,\left( {i{n^3}\, \times \,\,\frac{{16.387\,c{m^3}}}{{i{n^3}}}} \right)\,\,\\ = \,\,2638.32\,\,c{m^3}\\ = 2.64L\end{array}\)

(e) The estimated mass of the atmosphere, 5.6 × 1015 tons, to kilograms

\(\begin{array}{c}5.6\, \times \,{10^{15}}\,tons\,\, = \,\,5.6\, \times \,{10^{15}}\,\left( {tons\, \times \,\frac{{907.185\,\,kg}}{{tons}}} \right)\,\,\\ = \,\,5080.23\, \times \,{10^{15}}\,kg\,\,\\ = \,\,5.08023\,\, \times \,\,{10^{18}}\,kg\\ = 5.1 \times {10^{18}}\,kg\end{array}\)

(f) The mass of a bushel of rye, 32.0 lb, to kilograms

\(\begin{array}{c}32\,\,lb\,\, = \,\,32\,\,\left( {lb\, \times \,\frac{{0.453592\,\,kg}}{{lb}}} \right)\,\,\\ = \,\,14.515\,\,kg\\ = 14.5kg\end{array}\)

(g) The mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)

\(\begin{array}{c}5\,grain\, = \,5\,\, \times \,0.00229\,oz\,\\\, = \,\,0.01145\,\,oz\,\\\, = \,0.01145\,\,\left( {oz\, \times \,\frac{{28349.5\,\,mg}}{{oz}}} \right)\,\\\, = \,\,324mg\end{array}\)