Problem 109 - Introductory Chemistry Atoms First

Question

You measure one edge of a cube using a meterstick marked in centimeters. Unfortunately, the edge is longer than \(1 \mathrm{~m}\). You mark the \(1-\mathrm{m}\) point on the cube edge with a pen and then, using a \(15-\mathrm{cm}\) ruler marked in millimeters, measure the remaining distance to be \(1.40 \mathrm{~cm}\). (a) What is the length of the edge in centimeters? (b) What is the volume of the cube in cubic centimeters? (Remember, the lengths of all edges of a cube are equal.) Watch your significant figures. Use scientific notation if you have to. (c) The cube has a mass of \(111 \mathrm{~kg} .\) What is its density in grams per milliliter? Watch your significant figures.

Step-by-Step Solution

Answer

The length of an edge of the cube is \(101.40 \mathrm{~cm}\), its volume is approximately \(1.046 \times 10^6 \mathrm{~cm^3}\), and its density is approximately \(0.106 \mathrm{~g/cm^3}\).

Step 1: Find the Length of the Cube Edge in Centimeters

First, we need to find the total length of the cube edge. Since the edge is longer than \(1 \mathrm{~m}\), we have to combine the measured lengths to find the total length. The \(1 \mathrm{~m}\) mark was made with a meterstick and is given in meters, and then an additional \(1.40 \mathrm{~cm}\) was measured with a ruler. To add these lengths, we need to convert the \(1 \mathrm{~m}\) to centimeters and then add it to the remaining \(1.40 \mathrm{~cm}\). Recall that \(1 \mathrm{~m} = 100 \mathrm{~cm}\). So, Total length = \(1 \mathrm{~m} + 1.40 \mathrm{~cm} = 100 \mathrm{~cm} + 1.40 \mathrm{~cm} = 101.40 \mathrm{~cm}\) The length of an edge of the cube is \(101.40 \mathrm{~cm}\).

Step 2: Calculate the Volume of the Cube in Cubic Centimeters

The volume of a cube is calculated by raising the length of an edge to the power of 3. So, considering all the edges of the cube are equal, we can perform the calculation: Volume = (Edge length)\(^3= (101.40 \mathrm{~cm})^3\) Make sure to keep track of significant figures in your calculations. In this case, we have four significant figures. Volume \(\approx 1.046 \times 10^6 \mathrm{~cm^3}\) The volume of the cube is \(1.046 \times 10^6 \mathrm{~cm^3}\).

Step 3: Calculate the Density of the Cube

To find the density of the cube, we'll need to convert the known mass from kilograms to grams and then divide it by the volume of the cube. Remember that \(1 \mathrm{~kg} = 1000 \mathrm{~g}\). The given mass of the cube is \(111 \mathrm{~kg}\). Mass in grams = \(111 \mathrm{~kg} \times 1000 \mathrm{~g/kg} = 111,000 \mathrm{~g}\) Now we can calculate the density by dividing the mass in grams by the volume in cubic centimeters. Density = \(\frac{\text{Mass in grams}}{\text{Volume in cm}^3}=\frac{111,000 \mathrm{~g}}{1.046 \times 10^6 \mathrm{~cm^3}} \approx 0.106 \mathrm{~g/cm^3}\) The density of the cube is \(0.106 \mathrm{~g/cm^3}\).

Key Concepts

Scientific Notation

Scientific notation is a method to express very large or very small numbers in a compact form. It's written as the product of a number between 1 and 10 and a power of 10. For example, the notation \(4.18 \times 10^6\) signifies the number 4,180,000.

When dealing with cube volume calculations in scientific notation, it allows us to easily keep track of significant figures and makes calculations with very large or very small numbers more manageable. For instance, the volume of our cube \(1.046 \times 10^6 \mathrm{cm^3}\) shows the cube's size in a straightforward, readable manner that immediately indicates the scale of its volume.

Significant Figures

Significant figures are the digits in a number that contribute to its precision. They include all the nonzero digits, any zeros between significant digits, and trailing zeros in the decimal portion. In measurements, we use significant figures to indicate the precision of the measured value.

For example, in the edge length of the cube, \(101.40 \mathrm{cm}\), using four significant figures assures that we preserve the precision of our measurement when calculating the volume. This discipline in precision reflects our original measurement's accuracy and influences how we round our final answers, such as the density we computed to be \(0.106 \mathrm{g/cm^3}\), which also has four significant figures.

Unit Conversion

Unit conversion involves changing a measurement from one unit to another without altering its value. In our cube exercise, we converted meters to centimeters to maintain a consistent unit type for our calculations. Specifically, since \(1 \mathrm{m} = 100 \mathrm{cm}\), we converted the meter part of the measurement into centimeters before adding it to the remainder of the edge measured in centimeters.

Unit conversion is crucial because it ensures that all measurements are in the same units and thus can be accurately calculated together. Incorrect or inconsistent units can lead to incorrect calculations and results, so attentiveness during conversion is essential to accurate problem-solving.

Density Formula

The density formula is a fundamental concept in physics and engineering that defines the relationship between mass and volume, expressed as density (\(\rho\)) equals mass (m) divided by volume (V). In mathematical terms, it's \(\rho = \frac{m}{V}\).

Applying the formula, we calculated the cube's density by converting its mass from kilograms to grams and then dividing by its volume in cubic centimeters. Understanding the density formula is critical as it serves as the measurement of how much 'stuff' is packed into a given volume and it is widely used in various fields of science and industry. Our cube's density of \(0.106 \mathrm{g/cm^3}\) is a measure of its compactness.