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Chapter 1: Problem 9

When you convert units, how do you decide which part of the conversion factor is in the numerator and which is in the denominator? [Section 1.6]

Short Answer

Expert verified
To decide which part of the conversion factor goes in the numerator and which goes in the denominator, first identify the undesired unit and the desired unit. Write the given value as a fraction over 1. Then, multiply the given value by the conversion factor in such a way that the undesired unit cancels out. This will ensure the correct placement of the conversion factor and yield the desired unit in your result.

Step by step solution

01

(Step 1: Understand Conversion Factors)

A conversion factor is a ratio or a fraction that represents the relationship between two different units. It allows us to convert a quantity from one unit to another by multiplying or dividing it by a conversion factor.
02

(Step 2: Determine the Desired and Undesired Units)

Identify the initial unit of the given quantity that you want to convert (the undesired unit) and the unit you want to convert it to (the desired unit). For example, if you want to convert 3 miles to kilometers, the undesired unit is miles and the desired unit is kilometers.
03

(Step 3: Set up the Conversion Factor)

Identify the conversion factor that relates the desired and undesired units. The conversion factor should be written as a fraction where the numerator and the denominator must represent equivalent quantities in different units. In the above example, we know that 1 mile is approximately equal to 1.60934 kilometers. Thus, the conversion factors can be written as either \(\frac{1\, mile}{1.60934\, km}\) or \(\frac{1.60934\, km}{1\, mile}\).
04

(Step 4: Determine the Placement of the Conversion Factor)

Write the given value as a fraction over 1 (in our example, \(\frac{3\, miles}{1}\)), and then multiply the given value by the conversion factor in such a way that the undesired unit cancels out. This will ensure the correct placement of the conversion factor. In our example, to cancel the miles unit, we need to choose the conversion factor that has miles in the denominator. So, we multiply the given value by \(\frac{1.60934\, km}{1\, mile}\).
05

(Step 5: Perform the Calculation)

Multiply the given value by the selected conversion factor, which will cancel out the undesired unit and yield the desired unit. In our example, \(\frac{3\, miles}{1} \times \frac{1.60934\, km}{1\, mile} = 3 \times 1.60934\, km = 4.82802\, km\). So, 3 miles is approximately equal to 4.82802 kilometers. In conclusion, when converting units using conversion factors, the key is to choose the correct placement of the conversion factor in order to cancel out the undesired unit while obtaining the desired unit. Follow these steps to ensure that you place the conversion factor in the correct position for your calculations.

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Most popular questions from this chapter

Use appropriate metric prefixes to write the following measurements without use of exponents: (a) \(2.3 \times 10^{-10} \mathrm{~L}\), (b) \(4.7 \times 10^{-6} \mathrm{~g}\), (c) \(1.85 \times 10^{-12} \mathrm{~m}\), (d) \(16.7 \times 10^{6} \mathrm{~s}\), (e) \(15.7 \times 10^{3} \mathrm{~g}\), (f) \(1.34 \times 10^{-3} \mathrm{~m}\), (g) \(1.84 \times 10^{2} \mathrm{~cm}\)

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains \(1.50 \mathrm{~g}\) of carbon and \(2.00 \mathrm{~g}\) of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains \(6.35 \mathrm{~g}\) of carbon. How many grams of oxygen does it contain? Which law are you assuming in answering this question?

Carry out the following conversions: (a) \(0.105\) in. to \(\mathrm{mm}\), (b) \(0.650 \mathrm{qt}\) to \(\mathrm{mL}\), (c) \(8.75 \mu \mathrm{m} / \mathrm{s}\) to \(\mathrm{km} / \mathrm{hr}\), (d) \(1.955 \mathrm{~m}^{3}\) to \(\mathrm{yd}^{3}\), (e) \(\$ 3.99 / \mathrm{lb}\) to dollars per kg, (f) \(8.75 \mathrm{lb} / \mathrm{ft}^{3}\) to \(\mathrm{g} / \mathrm{mL}\).

A \(32.65-\mathrm{g}\) sample of a solid is placed in a flask. Toluene, in which the solid is insoluble, is added to the flask so that the total volume of solid and liquid together is \(50.00 \mathrm{~mL}\). The solid and toluene together weigh \(58.58 \mathrm{~g}\). The density of toluene at the temperature of the experiment is \(0.864 \mathrm{~g} / \mathrm{mL}\). What is the density of the solid?

(a) The diameter of Earth at the equator is \(7926.381 \mathrm{mi}\). Round this number to three significant figures and express it in standard exponential notation. (b) The circumference of Earth through the poles is \(40,008 \mathrm{~km}\). Round this number to four significant figures and express it in standard exponential notation.
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