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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 ft² = 929.03 cm². Given: Weight = 8.0 oz Area = 50 ft² 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/cm³ 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 cm³ Area = 46451.5 cm² 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

Gold can be hammered into extremely thin sheets called gold leaf. An architect wants to cover a \(100 \mathrm{ft} \times 82 \mathrm{ft}\) ceiling with gold leaf that is five-millionths of an inch thick. The density of gold is \(19.32 \mathrm{~g} / \mathrm{cm}^{3}\), and gold costs \(\$ 1654\) per troy ounce \((1\) troy ounce \(=31.1034768 \mathrm{~g})\). How much will it cost the architect to buy the necessary gold?

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.

Suppose you decide to define your own temperature scale with units of \(\mathrm{O}\), using the freezing point \(\left(13{ }^{\circ} \mathrm{C}\right)\) and boiling point \(\left(360^{\circ} \mathrm{C}\right)\) of oleic acid, the main component of olive oil. If you set the freezing point of oleic acid as \(0^{\circ} \mathrm{O}\) and the boiling point as \(100^{\circ} \mathrm{O}\), what is the freezing point of water on this new scale?

Perform the following conversions: (a) \(5.00\) days to \(\mathrm{s}\), (b) \(0.0550 \mathrm{mi}\) to \(\mathrm{m}\), (c) \(\$ 1.89 / \mathrm{gal}\) to dollars per liter, (d) \(0.510 \mathrm{in} . / \mathrm{ms}\) to \(\mathrm{km} / \mathrm{hr}\), (e) \(22.50 \mathrm{gal} / \mathrm{min}\) to \(\mathrm{L} / \mathrm{s}\), (f) \(0.02500 \mathrm{ft}^{3}\) to \(\mathrm{cm}^{3}\).

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A \(25.0\)-mL portion of the liquid had a mass of \(21.95 \mathrm{~g}\). A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as \(0.8787 \mathrm{~g} / \mathrm{mL}\). Is the calculated density in agreement with the tabulated value? (b) An experiment requires \(15.0 \mathrm{~g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is \(0.7781 \mathrm{~g} / \mathrm{mL}\). What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of \(5.0 \mathrm{~cm}\). What is the mass of the sphere if lead has a density of \(11.34 \mathrm{~g} / \mathrm{cm}^{3}\) ? (The volume of a sphere is \((4 / 3) \pi r^{3}\), where \(r\) is the radius.)

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