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Chapter 1: Problem 80

A package of aluminum foil contains \(50 \mathrm{ft}^{2}\) of foil, which weighs approximately \(8.0\) oz. Aluminum has a density of \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\). What is the approximate thickness of the foil in millimeters?

Short Answer

Expert verified
The approximate thickness of the aluminum foil is 0.0181 millimeters.

Step by step solution

01

Convert weight to grams and area to square centimeters

First, we need to convert the given weight in ounces to grams and the area of aluminum foil in square feet to square centimeters. We use the conversion factors 1 ounce = 28.35 grams, and 1 ft虏 = 929.03 cm虏. Given: Weight = 8.0 oz Area = 50 ft虏 Weight in grams: \(Weight(g) = 8.0\ oz \times 28.35\frac{g}{oz} = 226.8\ g\) Area in square centimeters: \(Area(cm^2) = 50\ ft^2 \times 929.03\frac{cm^2}{ft^2} = 46451.5\ cm^2\)
02

Calculate volume using weight and density

To find the volume of the aluminum foil, we can use the formula relating weight, density and volume: \(Volume = \frac{Weight}{Density}\) Given: Weight = 226.8 g Density of aluminum = 2.70 g/cm鲁 Volume: \(Volume\ (cm^3) = \frac{226.8\ g}{2.70\frac{g}{cm^3}} = 84\ cm^3\)
03

Calculate the thickness of the aluminum foil

Now that we have found the volume of the aluminum foil, we can find the thickness using the formula relating volume, area, and thickness: \(Volume = Area \times Thickness\) Given: Volume = 84 cm鲁 Area = 46451.5 cm虏 Isolating the thickness, we get: \(Thickness = \frac{Volume}{Area}\) Thickness: \(Thickness\ (cm) = \frac{84\ cm^3}{46451.5\ cm^2} = 0.00181\ cm\)
04

Convert thickness to millimeters

Finally, to express the thickness in millimeters, we convert it from centimeters to millimeters using the conversion factor 1 cm = 10 mm. Thickness in millimeters: \(Thickness\ (mm) = 0.00181\ cm \times 10\frac{mm}{cm} = 0.0181\ mm\) So, the approximate thickness of the aluminum foil is 0.0181 millimeters.

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Most popular questions from this chapter

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 \(\mathrm{g} / \mathrm{mL}\).

Indicate the number of significant figures in each of the following measured quantities: (a) \(3.774 \mathrm{~km}\), (b) \(205 \mathrm{~m}^{2}\), (c) \(1.700 \mathrm{~cm}\), (d) \(350.00 \mathrm{~K}\), (e) \(307.080 \mathrm{~g}\), (f) \(1.3 \times 10^{3} \mathrm{~m} / \mathrm{s}\).

A 40 -lb container of peat moss measures \(14 \times 20 \times 30\) in. A 40 -lb container of topsoil has a volume of \(1.9 \mathrm{gal}\). (a) Calculate the average densities of peat moss and topsoil in units of \(\mathrm{g} / \mathrm{cm}^{3}\). Would it be correct to say that peat moss is "lighter" than topsoil? Explain. (b) How many bags of peat moss are needed to cover an area measuring \(15.0 \mathrm{ft} \times 20.0 \mathrm{ft}\) to a depth of \(3.0\) in.?

Indicate which of the following are exact numbers: (a) the mass of a 3 by 5 -inch index card, (b) the number of ounces in a pound, (c) the volume of a cup of Seattle's Best coffee, (d) the number of inches in a mile, (e) the number of microseconds in a week, (f) the number of pages in this book.

In 2009, a team from Northwestern University and Western Washington University reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. The density of this new material is \(0.20 \mathrm{~g} / \mathrm{cm}^{3}\), and its surface area is \(1242 \mathrm{~m}^{2}\) per gram of material. (a) Calculate the volume of a 10.0-mg sample of this material. (b) Calculate the surface area for a \(10.0\)-mg sample of this material. (c) A \(10.0\)-mL sample of contaminated water had \(7.748 \mathrm{mg}\) of mercury in it. After treatment with \(10.0 \mathrm{mg}\) of the new spongy material, \(0.001 \mathrm{mg}\) of mercury remained in the contaminated water. What percentage of the mercury was removed from the water? (d) What is the final mass of the spongy material after the exposure to mercury?
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