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Chapter 1: Problem 39

Aluminum is a lightweight metal (density \(=2.70\) \(\mathrm{g} / \mathrm{cm}^{3}\) ) used in aircraft construction, high-voltage transmission lines, and foils. What is its density in \(\mathrm{kg} / \mathrm{m}^{3} ?\)

Short Answer

Expert verified
The density of aluminum in kilograms per cubic meter is 2700 kg/m^3.

Step by step solution

01

Conversion of g to kg

We know that 1 g equals 0.001 kg. Since the density of aluminum is 2.70 g/cm^3, this converts to 2.70 * 0.001 kg/cm^3 = 0.0027 kg/cm^3.
02

Conversion from cm^3 to m^3

In order to convert from cubic centimeters to cubic meters, we must recall that 1 m equals 100 cm. Therefore, (100 cm)^3 = 1,000,000 cm^3 in 1 m^3. Applying this conversion, the density from step 1 becomes 0.0027 * 1,000,000 kg/m^3.
03

Final Calculation

After multiplying 0.0027 kg/cm^3 by 1,000,000 cm^3/m^3, we get a final density of 2700 kg/m^3.

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Most popular questions from this chapter

The price of gold on a certain day in 2004 was \(\$ 315\) per troy ounce. How much did \(1.00 \mathrm{~g}\) of gold cost that day? \((1\) troy ounce \(=31.03 \mathrm{~g} .)\)

A bank teller is asked to assemble "one-dollar" sets of coins for his clients. Each set is made of three quarters, one nickel, and two dimes. The masses of the coins are: quarter: \(5.645 \mathrm{~g}\); nickel: \(4.967 \mathrm{~g}\); dime: \(2.316 \mathrm{~g}\). What is the maximum number of sets that can be assembled from \(33.871 \mathrm{~kg}\) of quarters, \(10.432 \mathrm{~kg}\) of nickels, and \(7.990 \mathrm{~kg}\) of dimes? What is the total mass (in g) of this collection of coins?

The total volume of seawater is \(1.5 \times 10^{21} \mathrm{~L}\). Assume that seawater contains 3.1 percent sodium chloride by mass and that its density is \(1.03 \mathrm{~g} / \mathrm{mL}\). Calculate the total mass of sodium chloride in kilograms and in tons. \((1\) ton \(=2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{~g} .)\).

Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of 1 ppm of fluorine is sufficient for the purpose. (1 ppm means \(1 \mathrm{~g}\) of fluorine per 1 million g of water.) The compound normally chosen for fluoridation is sodium fluoride, which is also added to some toothpastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 150 gallons. What percent of the sodium fluoride is "wasted" if each person uses only \(6.0 \mathrm{~L}\) of water a day for drinking and cooking? (Sodium fluoride is 45.0 percent fluorine by mass. 1 gallon \(=3.79 \mathrm{~L} ; 1\) year \(=365\) days; density of water \(=1.0 \mathrm{~g} / \mathrm{mL} .)\)

Write the numbers for these prefixes: (a) mega-,(b) kilo-, (c) deci-, (d) centi-, (e) milli-, (f) micro-, (g) nano-, (h) pico-.
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