Question

Titanium is used in airplane bodies because it is strong and light. It has a density of \(4.55 \mathrm{~g} / \mathrm{cm}^{3}\). If a cylinder of titanium is \(7.75 \mathrm{~cm}\) long and has a mass of \(153.2 \mathrm{~g}\), calculate the diameter of the cylinder. \(\left(V=\pi r^{2} h\right.\), where \(V\) is the volume of the cylinder, \(r\) is its radius, and \(h\) is the height.)

Short answer

The approximate diameter of the cylindrical titanium piece is 2.39 cm.

Step-by-step solution

Calculate the volume of the cylinder

To find the volume, we can use the formula: \(Volume = \frac{Mass}{Density}\). We have the mass as \(153.2 \mathrm{~g}\) and the density as \(4.55 \mathrm{~g} / \mathrm{cm}^{3}\). Plugging in these values, we get: \(Volume = \frac{153.2}{4.55} = 33.6483516 \mathrm{~cm}^{3}\)

Use the volume formula to find the radius

We will use the formula for the volume of a cylinder, \(V = \pi r^{2} h\), where \(V\) is the volume, \(r\) is the radius, and \(h\) is the height. We have the volume \(V = 33.6483516 \mathrm{~cm}^{3}\) and height \(h = 7.75 \mathrm{~cm}\). Plugging these into the formula and solving for \(r\): \(33.6483516 = \pi r^{2}(7.75)\) Now, divide both sides by \(\pi\) and the height: \(r^{2} = \frac{33.6483516}{\pi \times 7.75} = 1.42590957\) Take the square root to find the radius: \(r = \sqrt{1.42590957} = 1.19495623 \mathrm{~cm}\)

Calculate the diameter from the radius

To find the diameter, we simply multiply the radius by 2: \(Diameter = 2 \times 1.19495623 = 2.38991247 \mathrm{~cm}\) So, the diameter of the cylindrical titanium piece is approximately \(2.39 \mathrm{~cm}\).

Key concepts

Mass and Volume

Understanding mass and volume is crucial, not only in the realm of chemistry, but in various scientific applications. Mass refers to the amount of matter in an object and is typically measured in grams (g) or kilograms (kg). Volume, on the other hand, measures the amount of space an object occupies and is often expressed in cubic centimeters (\text{cm}^3) or liters (L).

In the context of the exercise, mass and volume are inherently linked through the concept of density, which is the mass per unit volume of a substance. So, to conceptualize density, imagine you have two blocks of equal size - one made of lead and the other of styrofoam. The lead block has a higher mass in the same volume, meaning it has a higher density. This relationship can be mathematically expressed by the formula:\[ Density = \frac{Mass}{Volume} \]

• Mass of the cylinder: 153.2 g
• Volume of the cylinder: Calculated using mass and known density.

Cylindrical Volume Calculation

Cylindrical shapes are common in geometry and come into play in various practical applications, including chemistry and engineering. The volume of a cylinder can be calculated using the formula:\[ V = \pi r^2 h \]
where:\[ V \]is the volume, \[ r \]is the radius, and \[ h \]is the height of the cylinder.

In our exercise, the volume of the cylinder was determined so that the radius could be inferred. Once the radius is calculated, it’s multiplied by two to get the diameter, which represents the width of the cylinder across its center. This type of geometric calculation is indispensable when working with cylindrical objects, whether you're figuring out the amount of metal needed for a pipe or the volume of a chemical in a test tube.

• Volume given by mass/density: 33.6483516 \text{cm}^3
• Height of cylinder: 7.75 cm
• Formula rearranged to solve for radius, then diameter.

Geometry in Chemistry

The use of geometry in chemistry is pervasive, as it helps to understand the spatial arrangement of molecules and the design of chemical equipment, such as reactors and storage containers. In our textbook problem, the principles of geometry allow us to determine an unknown dimension of a cylinder - its diameter - by applying volume formulas.

The geometric understanding goes hand-in-hand with chemical concepts to ensure precise calculations. For example, knowing the cylinder’s volume helps to ascertain the amount of substance needed or produced in a reaction. Also, the shape and size of molecules or crystalline structures can be described by geometric figures, which are essential in predicting molecular interactions and reactions.

• Applied cylindrical volume calculation to a real-world substance (titanium).
• Combination of geometric formula with density property.
• Understanding geometrical dimensions aids in material and space estimation in chemical processes.