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Chapter 2: Problem 106

One liter is equal to \(0.264\) gallon. Suppose you have \(1.000 \times 10^{3} \mathrm{~cm}^{3}\) of water. How many gallons do you have? Use unit analysis to calculate your answer, and show your work. Treat all conversion factors as exact.

Short Answer

Expert verified
The given volume of water is \(1{,}000 \times 10^3\mathrm{~cm}^3\), and the conversion factor between liters and gallons is provided. First, we convert cubic centimeters to liters by dividing the volume by 1000, resulting in \(1{,}000\mathrm{~liters}\). Then, we multiply the volume in liters by the conversion factor (0.264) to find the equivalent volume in gallons: \(1{,}000 \times 0.264 = 264\mathrm{~gallons}\).

Step by step solution

01

Convert cubic centimeters to liters

Since 1 liter is equivalent to 1000 cubic centimeters, we need to convert the given volume in cubic centimeters to liters. We can do this by dividing the volume in cm³ by 1000. \[ \frac{1{,}000 \times 10^3 \thinspace \mathrm{cm}^3}{1\,000\thinspace \mathrm{cm}^3/\mathrm{liters}} \]
02

Simplifying the expression

Now, cancel the units (cubic centimeters) and simplify the expression: \[ \frac{1{,}000 \times 10^3 }{1\,000} = 1{,}000 \thinspace \mathrm{liters} \]
03

Convert liters to gallons

Now that we have the volume in liters, we can use the given conversion factor to convert liters to gallons. For this, multiply the volume in liters by the conversion factor (0.264 gallons per liter): \[ 1{,}000 \thinspace \mathrm{liters} \times 0.264\thinspace\mathrm{gallons}/\mathrm{liters} \]
04

Simplifying the expression

Cancel the units (liters) and simplify the expression: \[ 1{,}000 \times 0.264 = 264\thinspace\mathrm{gallons} \] Hence, 1,000x10³ cm³ of water is equal to 264 gallons.

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Most popular questions from this chapter

Calculate the amount of heat energy in joules required to heat \(50.0 \mathrm{~g}\) of each substance from \(25.0^{\circ} \mathrm{C}\) to \(37.0^{\circ} \mathrm{C}\). (specific heats shown in parentheses): (a) Iron \(\left(0.449 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) (b) Aluminum (0.901 J/g \(\cdot{ }^{\circ} \mathrm{C}\) ) (c) Mercury \(\left(0.14 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) (d) Water \(\left(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\)

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