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Chapter 1: Problem 44

In the determination of the density of a rectangular metal bar, a student made the following measurements: length, \(8.53 \mathrm{~cm}\); width, \(2.4 \mathrm{~cm}\); height, \(1.0 \mathrm{~cm}\); mass, 52.7064 g. Calculate the density of the metal to the correct number of significant figures.

Short Answer

Expert verified
The density of the metal is \(2.57 \mathrm{g/cm}^3\)

Step by step solution

01

Calculate the Volume

Start out the process by determining the volume of the rectangular metal bar. This can be calculated using the formula for the volume of a rectangular prism: Volume \(=\) Length \(\times\) Width \(\times\) Height. Substituting the given values, you get: Volume \(=\) \(8.53 \mathrm{cm} \times 2.4 \mathrm{cm} \times 1.0 \mathrm{cm} = 20.472 \mathrm{cm}^3\)
02

Calculate the Density

Next, you calculate the density by dividing the mass by the volume obtained. Density \(=\) Mass/volume. Substituting the given and calculated values, you get: Density \(=\) \(52.7064 \mathrm{g}/20.472 \mathrm{cm}^3 = 2.57 \mathrm{g/cm}^3\)
03

Determine the Significant Figures

The answer should be rounded off to three significant figures since the width has only two significant figures. However, in determining the significant figures for multiplication or division, one should look for the fewest number of significant figures in the given displacement measurements. Hence, the density calculated should be rounded off to three significant figures.

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Most popular questions from this chapter

A sheet of aluminum (Al) foil has a total area of \(1.000 \mathrm{ft}^{2}\) and a mass of \(3.636 \mathrm{~g}\). What is the thickness of the foil in millimeters? (Density of \(\left.\mathrm{Al}=2.699 \mathrm{~g} / \mathrm{cm}^{3} .\right)\)

Express these numbers in scientific notation: (a) 0.000000027 , (b) 356 (c) 0.096 .

What is the number of significant figures in each of these measured quantities? (a) 4867 miles, (b) \(56 \mathrm{~mL}\) (c) 60,104 tons, (d) \(2900 \mathrm{~g}\).

A 1.0 -mL volume of seawater contains about \(4.0 \times\) \(10^{-12} \mathrm{~g}\) of gold. The total volume of ocean water is \(1.5 \times 10^{21} \mathrm{~L} .\) Calculate the total amount of gold in grams that is present in seawater and its worth in dollars, assuming that the price of gold is \(\$ 350\) an ounce. With so much gold out there, why hasn't someone become rich by mining gold from the ocean?

Mercury is the only metal that is a liquid at room temperature. Its density is \(13.6 \mathrm{~g} / \mathrm{mL}\). How many grams of mercury will occupy a volume of \(95.8 \mathrm{~mL} ?\)
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