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Chapter 1: Problem 83

A lead sphere has a mass of \(1.20 \times 10^{4} \mathrm{~g}\), and its volume is \(1.05 \times 10^{3} \mathrm{~cm}^{3} .\) Calculate the density of lead.

Short Answer

Expert verified
The density of lead is approximately 11.43 g/cm³.

Step by step solution

01

Understanding Density Formula

The formula to calculate density \( \rho \) is given by \( \rho = \frac{m}{V} \), where \( m \) is the mass and \( V \) is the volume of the object.
02

Identify Given Values

From the exercise, we have the mass \( m = 1.20 \times 10^{4} \) g and the volume \( V = 1.05 \times 10^{3} \) cm³.
03

Substitute Values into the Formula

Using the formula \( \rho = \frac{m}{V} \), substitute \( m = 1.20 \times 10^{4} \) g and \( V = 1.05 \times 10^{3} \) cm³ into the equation: \[ \rho = \frac{1.20 \times 10^{4} \, \mathrm{g}}{1.05 \times 10^{3} \, \mathrm{cm}^{3}} \].
04

Calculate the Density

Perform the division: \( \rho = \frac{1.20 \times 10^{4}}{1.05 \times 10^{3}} \approx 11.43 \) g/cm³.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass and Volume Relationship
Mass and volume are fundamental properties of matter that describe how much stuff is present in an object and how much space it takes up. Understanding the relationship between mass and volume is crucial in various scientific calculations, including those involving density.

**Mass** is a measure of the amount of matter in an object. It's usually measured in grams (g) or kilograms (kg).

**Volume**, on the other hand, is a measure of the space that an object occupies. It's commonly measured in cubic centimeters (cm³) or liters (L).

The relationship between mass and volume tells us about the concentration of matter in a given space, which is where the concept of density comes in. By dividing the mass of an object by its volume, we can determine its density, which gives us an indication of how tightly packed the matter is within that space.
  • Mass = Total amount of matter
  • Volume = Space occupied by matter
  • Density = Mass divided by Volume
Formula for Density
The formula for calculating density is one of the simplest yet most important equations in physics. It's given by:
\[\rho = \frac{m}{V}\]
Where:
  • \(\rho\) represents density
  • \(m\) is the mass of the object
  • \(V\) is the volume of the object

To find the density of an object, you simply need to divide its mass by its volume. This formula is essential for understanding how different materials compare in terms of how fast they sink or float, their compactness, and various practical applications. In our example of the lead sphere, we used this formula to determine the density by substituting the given mass and volume. By solving \(\rho = \frac{1.20 \times 10^{4} \, \mathrm{g}}{1.05 \times 10^{3} \, \mathrm{cm}^{3}}\), we found the density to be around 11.43 g/cm³.
Density of Lead
Lead is a dense, heavy metal that has many uses due to its high density. Its high density makes it very effective for applications where weight and stability are important, like in radiation shielding and in making weights.

The density of lead can be calculated by using its mass and volume to see how compact the material is. In the exercise, the mass of lead was given as \(1.20 \times 10^{4} \, \mathrm{g}\) and its volume as \(1.05 \times 10^{3} \, \mathrm{cm}^{3}\). By plugging these values into the density formula, the density of lead was evaluated to be approximately 11.43 g/cm³.

This value signifies that lead is much denser than many other materials, meaning it has more mass packed into each cubic centimeter. This property is leveraged across diverse fields including physics and engineering for specific purposes.
  • Used for protection against radiation due to its density
  • Ideal for making precise weights owing to its compact nature
  • Commonly found in batteries, construction, and more
Understanding the density of lead helps in choosing the right material for the right application.

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Most popular questions from this chapter

Lithium is the least dense metal known (density = \(\left.0.53 \mathrm{~g} / \mathrm{cm}^{3}\right)\). What is the volume occupied by \(3.15 \times 10^{3} \mathrm{~g}\) of lithium?

Suppose that a new temperature scale has been devised on which the melting point of ethanol \(\left(-117.3^{\circ} \mathrm{C}\right)\) and the boiling point of ethanol \(\left(78.3^{\circ} \mathrm{C}\right)\) are taken as \(0^{\circ} \mathrm{S}\) and \(100^{\circ} \mathrm{S},\) respectively, where \(\mathrm{S}\) is the symbol for the new temperature scale. Derive an equation relating a reading on this scale to a reading on the Celsius scale. What would this thermometer read at \(25^{\circ} \mathrm{C} ?\)

Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of 1 ppm of fluorine is sufficient for the purpose ( 1 ppm means one part per million, or \(1 \mathrm{~g}\) of fluorine per 1 million g of water). The compound normally chosen for fluoridation is sodium fluoride, which is also added to some toothpastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 150 gal. What percent of the sodium fluoride is "wasted" if each person uses only 6.0 \(\mathrm{L}\) of water a day for drinking and cooking (sodium fluoride is 45.0 percent fluorine by mass; \(1 \mathrm{gal}=3.79 \mathrm{~L} ;\) 1 year \(=365\) days; 1 ton \(=2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{~g}\); density of water \(=1.0 \mathrm{~g} / \mathrm{mL}\) )?

Classify each of the following statements as a hypothesis, law, or theory. (a) Beethoven's contribution to music would have been much greater if he had married. (b) An autumn leaf gravitates toward the ground because there is an attractive force between the leaf and Earth. (c) All matter is composed of very small particles called atoms.

A \(37.2-\mathrm{g}\) sample of lead \((\mathrm{Pb})\) pellets at \(20^{\circ} \mathrm{C}\) is mixed with a 62.7 -g sample of lead pellets at the same temperature. What are the final mass, temperature, and density of the combined sample? The density of \(\mathrm{Pb}\) at \(20^{\circ} \mathrm{C}\) is \(11.35 \mathrm{~g} / \mathrm{cm}^{3}\).
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