Problem 59 - General Chemistry: Principles and Modern Applications

Question

The density of aluminum is \(2.70 \mathrm{g} / \mathrm{cm}^{3} .\) A square piece of aluminum foil, \(22.86 \mathrm{cm}\) on a side is found to weigh 2.568 g. What is the thickness of the foil, in millimeters?

Step-by-Step Solution

Answer

The thickness of the foil is 0.018 mm.

Step 1: Identify and Write down all Known Variables

Mass of aluminum square = 2.568 g\nDensity of aluminum = 2.70 g/cm³\nSide measurement of aluminum square = 22.86 cm\n

Step 2: Calculate the Volume of the Aluminum Square

The volume can be found using the formula of density: Density = Mass / Volume, rearranged to solve for volume: Volume = Mass / Density.\nSo plugging in the known values for mass and density gives: Volume = 2.568 g / 2.70 g/cm³ = 0.95 cm³.

Step 3: Determine the Thickness of the Foil

Since the foil is a square, the volume of the aluminum square can be expressed as Volume = side² * thickness.\nRearranging to solve for thickness, we get: thickness = Volume / side².\nSubstituting the known values gives: thickness = 0.95 cm³ / (22.86 cm)² = 0.0018 cm.\nFinally, convert the thickness into millimeters (since 1 cm = 10 mm): thickness = 0.0018 cm * 10 = 0.018 mm.

Key Concepts

Understanding Aluminum Properties

Aluminum is a lightweight and versatile metal, known for its numerous beneficial properties. One of the key characteristics is its low density, which is only 2.70 grams per cubic centimeter (g/cm³). This makes it an excellent choice for applications where reducing weight is essential. Another significant aspect is its remarkable corrosion resistance, meaning it doesn't rust easily, which prolongs the lifespan of items made from it. Moreover, aluminum is malleable, allowing it to be easily shaped into sheets, like aluminum foil, without cracking. It's also a good conductor of electricity and heat, making it ideal for electrical applications as well as heat sinks. These properties make aluminum a popular material across industries including aviation, packaging, and construction.

Steps in Volume Calculation

Calculating volume is essential when determining the amount of space an object occupies. In our exercise, we want to find the volume of a piece of aluminum foil. We can use the relationship between density, mass, and volume to find this.Given the formula for density: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]We can rearrange it to solve for volume: \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]By substituting the mass of the aluminum foil, 2.568 g, and the density of aluminum, 2.70 g/cm³, into the formula, we arrive at the volume:\[ \text{Volume} = \frac{2.568 \, \text{g}}{2.70 \, \text{g/cm}^3} \approx 0.95 \, \text{cm}^3 \]This volume indicates how much space the aluminum occupies. Knowing this is a crucial step in finding other dimensions, such as thickness.

Converting Units for Practical Use

Unit conversion is a crucial skill, particularly in scientific calculations, as it allows for precise communication and comparison. In this exercise, after calculating the thickness of the aluminum foil in centimeters, it needs to be converted to millimeters for use in real-world applications.The thickness formula derived from the volume is:\[ \text{Thickness} = \frac{\text{Volume}}{\text{Side}^2} \]By substituting, we find:\[ \text{Thickness} = \frac{0.95 \, \text{cm}^3}{(22.86 \, \text{cm})^2} = 0.0018 \, \text{cm} \]To convert centimeters to millimeters (as there are 10 millimeters in a centimeter), simply multiply by 10:\[ 0.0018 \, \text{cm} \times 10 = 0.018 \, \text{mm} \]This conversion allows the thickness to be readily understood in industries or settings that prefer measurements in millimeters.