Question

Three spheres of equal size are composed of aluminum (density \(\left.=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\), silver (density \(\left.=10.49 \mathrm{~g} / \mathrm{cm}^{3}\right)\), and nickel (density \(\left.=8.90 \mathrm{~g} / \mathrm{cm}^{3}\right)\). List the spheres from lightest to heaviest.

Short answer

The spheres can be listed from lightest to heaviest based on their densities as follows: Aluminum sphere (lightest, density \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\)), Nickel sphere (density \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\)), and Silver sphere (heaviest, density \(10.49 \mathrm{~g} / \mathrm{cm}^{3}\)).

Step-by-step solution

Understand the problem

We are given three spheres of equal size (volume), composed of three different materials (aluminum, silver, and nickel). Each of these materials has a different density, which will lead to different weights for the spheres. We need to list the spheres from lightest to heaviest based on each sphere's density.

Express mass in terms of density and volume

The mass of an object can be calculated using its density and volume. The formula for mass is given by: \(m = \rho V\) where \(m\) is the mass, \(\rho\) is the density, and \(V\) is the volume. Since all the spheres have the same volume, we can compare their masses using only their densities.

Compare the densities of the spheres

We are given the densities of aluminum (\(2.70 \mathrm{~g} / \mathrm{cm}^{3}\)), silver (\(10.49 \mathrm{~g} / \mathrm{cm}^{3}\)), and nickel (\(8.90 \mathrm{~g} / \mathrm{cm}^{3}\)). Comparing these values: Aluminum: \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\)
Silver: \(10.49 \mathrm{~g} / \mathrm{cm}^{3}\)
Nickel: \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\) We can see that the aluminum sphere has the lowest density, followed by the nickel sphere, and then the silver sphere.

List the spheres from lightest to heaviest

Using the densities and considering that all the spheres have the same volume, we can determine the order of the spheres from lightest to heaviest: 1. Aluminum sphere (lightest) 2. Nickel sphere 3. Silver sphere (heaviest)

Key concepts

Materials Density Comparison

Understanding the concept of density is essential in material science and physics. Density is defined as mass per unit volume and expressed mathematically as \(\rho = \frac{m}{V}\), where \(\rho\) is the density, \(m\) is the mass, and \(V\) is the volume. When comparing materials based on density, you can determine which ones are heavier for the same volume.

The textbook example involves comparing aluminum, silver, and nickel's densities. Given that the density of aluminum is \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\), silver is \(10.49 \mathrm{~g} / \mathrm{cm}^{3}\), and nickel is \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\), we can tell, without knowing the exact volume of the spheres, that silver is the densest and therefore heaviest for the same volume, while aluminum is the least dense and lightest.

In practical terms, this affects a wide range of applications, from construction materials to jewelry design, as the choice of material often depends on its density relative to strength, appearance, or cost.

Mass-Volume Calculation

Mass-volume calculation is a handy skill in various scientific disciplines. It allows us to calculate the mass of an object if its volume and density are known. The formula to calculate mass is \(m = \rho V\), where \(m\) represents the mass, \(\rho\) stands for density, and \(V\) is the volume.

In our textbook problem, all spheres have equal volumes, so their masses solely depend on their densities. This simplifies the mass calculation process, effectively turning it into a comparison of densities to determine which sphere is lightest or heaviest.

Why is Mass-Volume Calculation Important?

Understanding this concept is crucial for predicting an object's weight, which in turn affects transport costs, material selection, and structural integrity in engineering projects. For instance, engineers must take the mass of construction materials into account when designing buildings to ensure that the structure can support the weight.

Chemistry Problem Solving

Chemistry problem solving often involves a mix of conceptual understanding and quantitative calculations. To solve these types of problems effectively, it's important to break them down into manageable steps. Our solution followed this approach by first understanding what was asked, then using the appropriate formulas to find a solution.

Whether dealing with reactions, concentrations, or densities, the process entails identifying known and unknown variables, selecting the right equations, and manipulating these equations to solve for the unknowns.

Identifying Key Strategy

For the given problem, the key strategy was to use the given densities and a known volume to compare the masses of the spheres. By conceptualizing the relationship between mass, volume, and density, we could solve the problem effectively without the need for actual mass measurements.

In real-world applications, such skills are vital in laboratory work, pharmaceutical dosing, or material selection in manufacturing, demonstrating how chemistry problem solving transcends the classroom to impact various aspects of science and industry.