Question

A rectangular block has dimensions \(2.9 \mathrm{~cm} \times 3.5 \mathrm{~cm} \times 10.0 \mathrm{~cm}\). The mass of the block is \(615.0 \mathrm{~g}\). What are the volume and density of the block?

Short answer

The volume of the rectangular block is \(V = 101.5 \mathrm{~cm^3}\), and its density is approximately \(6.06 \, \frac{\mathrm{g}}{\mathrm{cm^3}}\).

Step-by-step solution

Find the volume of the block

To find the volume of the rectangular block, simply multiply the length, width, and height: \(V = l \times w \times h\) where V is the volume, l is the length, w is the width, and h is the height of the block. Using the given dimensions: \(l = 2.9 \mathrm{~cm}\) \(w = 3.5 \mathrm{~cm}\) \(h = 10.0 \mathrm{~cm}\) Plug the values into the formula: \(V = 2.9 \times 3.5 \times 10.0\)

Calculate the volume

Now, compute the product: \(V = 101.5 \mathrm{~cm^3}\) The volume of the block is 101.5 cubic centimeters.

Find the density of the block

To find the density, we will divide the mass of the block by its volume. The formula for density is: \(density = \frac{mass}{volume}\) We are given the mass of the block as 615.0 grams. Now plug the mass and volume values into the formula: \(density = \frac{615.0}{101.5}\)

Calculate the density

Compute the division: \(density \approx 6.06 \, \frac{\mathrm{g}}{\mathrm{cm^3}}\) The density of the block is approximately 6.06 grams per cubic centimeter.

Summary

To summarize, the volume of the rectangular block is 101.5 cubic centimeters, and its density is approximately 6.06 grams per cubic centimeter.

Key concepts

Volume of a Rectangular Block

Understanding the volume of a rectangular block is fundamental in various fields of science, especially in chemistry where it's used to calculate concentrations and densities. To find the volume, you need to know the length, width, and height of the block, measurable in units such as centimeters. The volume is the amount of space occupied by the block and is calculated by the simple formula:

The volume, V, is given by: \[V = length \times width \times height\] For a block with given dimensions in the exercise, the calculation would involve: \[V = (2.9 \text{ cm}) \times (3.5 \text{ cm}) \times (10.0 \text{ cm})\] When you work out the calculation, the volume, in this case, is found to be \(101.5 \text{ cm}^3\). This volume calculation is essential because it provides the basis for further calculations, such as determining the density of the block.

Density Formula

The concept of density is ubiquitous in chemistry as it allows us to compare different materials regardless of size or shape. Density is defined as the mass of an object divided by its volume and is typically expressed in units like grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³).

The formula is: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] In practical use, if you have a substance with a high density, more mass is packed into a given volume, whereas a low-density substance is lighter for the same volume. For the block in the exercise, with a mass of \(615.0 \text{ g}\) and volume of \(101.5 \text{ cm}^3\), the density is calculated by dividing the mass by the volume, resulting in approximately \(6.06 \text{ g/cm}^3\). This indicates how compact the material of the block is.

Mass to Volume Ratio

The mass to volume ratio is essentially another term for density but may be more specifically used in contexts where we're interested in how much mass is present in a given volume. It's a critical concept in chemistry when dealing with solutions, mixtures, and even pure substances.

For solid objects like the rectangular block in our exercise, this ratio signifies how densely packed the particles are inside the volume. The mass to volume ratio for the block is calculated by dividing the mass in grams by the volume in cubic centimeters. For our block weighing \(615.0 \text{ g}\) with a volume of \(101.5 \text{ cm}^3\), the mass to volume ratio, or density, would be the same value found using the density formula, \(6.06 \text{ g/cm}^3\). This ratio is essential in numerous scientific endeavors, including material science and engineering, where materials are chosen based on their density for specific applications.