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Chapter 1: Problem 66

For a material to float on the surface of water, the material must have a density less than that of water \((1.0 \mathrm{~g} / \mathrm{mL})\) and must not react with the water or dissolve in it. A spherical ball has a radius of \(0.50 \mathrm{~cm}\) and weighs \(2.0 \mathrm{~g} .\) Will this ball float or sink when placed in water? (Note: Volume of a sphere \(=\frac{4}{3} \pi r^{3} .\) )

Short Answer

Expert verified
The spherical ball has a volume of approximately \(1.57 \mathrm{cm^3}\) and a density of approximately \(1.27 \mathrm{g/cm^3}\). Since the density of the sphere is greater than the density of water \((1.0 \mathrm{g/cm^3})\), the sphere will sink when placed in water.

Step by step solution

01

Find the Volume of the Sphere

We know the radius of the spherical ball (r) is 0.50 cm, and the formula to calculate the volume of a sphere is \(\frac{4}{3} \pi r^3\). Substitute the given value of radius in the formula to find the volume: \[V = \frac{4}{3} \pi (0.50)^3\]
02

Calculate the Volume

Now, let's compute the volume: \[V = \frac{4}{3} \pi (0.125)\] \[V = \frac{1}{2} \pi\] \[V \approx 1.57 \mathrm{cm^3}\] So, the volume of the sphere is approximately 1.57 cm鲁.
03

Calculate the Density of the Sphere

Now that we have the volume of the sphere, we can find its density. The density is the mass of an object divided by its volume. The mass of the sphere (m) is 2.0 g, and the volume (V) is 1.57 cm鲁. Use the formula: \[\rho = \frac{m}{V}\] Substitute the given values: \[\rho = \frac{2.0}{1.57}\]
04

Compute the Density

Now, let's compute the density: \[\rho \approx 1.27 \mathrm{g/cm^3}\] So, the density of the sphere is approximately 1.27 g/cm鲁.
05

Determine if the Sphere Will Float or Sink

Now that we have the density, compare it to the density of water (1.0 g/mL or g/cm鲁). If the density of the sphere is less than the density of water, it will float; otherwise, it will sink. Since the density of the sphere (1.27 g/cm鲁) is greater than the density of water, the sphere will sink when placed in water.

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Key Concepts

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

Density of Water
Water is often used as a standard reference for density. Its density is typically about 1.0 g/mL, which is the same as 1.0 g/cm鲁. This value is crucial because it helps us determine whether objects will float or sink. Materials with densities less than this value float, while those with higher densities sink. This fundamental concept is essential for understanding buoyancy and other related topics.
Volume of a Sphere
To find the volume of a sphere, we use the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. For our spherical ball with a radius of 0.50 cm, we plug the value into the equation:
  • \( V = \frac{4}{3} \pi (0.50)^3 \)
  • \( V \approx 1.57 \; \text{cm}^3 \)
This tells us how much space the ball occupies, which is vital for further calculations, like determining its density.
Determining Buoyancy
Buoyancy is the force that makes objects float. It depends on the density of both the object and the liquid. To determine if an object will float, compare its density to the liquid's density. If the object's density is lower, it will float due to the upward buoyant force. Otherwise, it will sink. This concept helps in understanding why certain materials float in water while others don't.
Floating and Sinking
When it comes to floating and sinking, density is the key factor. The rule of thumb is:
  • If an object is less dense than the liquid, it floats.
  • If an object is more dense, it sinks.
For the spherical ball with a density of 1.27 g/cm鲁, it is denser than water's 1.0 g/cm鲁. Thus, it will sink when placed in water. Understanding this principle can explain the behavior of different materials in various liquids.
Mass-to-Volume Ratio
The mass-to-volume ratio, or density, is a measure of how much mass is in a given volume. It's calculated using the formula \( \rho = \frac{m}{V} \). For our sphere:
  • Mass (m) = 2.0 g
  • Volume (V) = 1.57 cm鲁
  • Density \( \rho = \frac{2.0}{1.57} \approx 1.27 \; \text{g/cm}^3 \)
This calculation is critical to determine whether the sphere will float or sink. It shows how compact or spread out the particles in an object are, affecting its interaction with liquids.

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Most popular questions from this chapter

Make molecular-level (microscopic) drawings for each of the following. a. Show the differences between a gaseous mixture that is a homogeneous mixture of two different compounds, and a gaseous mixture that is a homogeneous mixture of a compound and an element. b. Show the differences among a gaseous element, a liquid element, and a solid element.

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Suppose a teaspoon of magnesium filings and a teaspoon of powdered sulfur are placed together in a metal beaker. Would this constitute a mixture or a pure substance? Suppose the magnesium filings and sulfur are heated so that they react with each other, forming magnesium sulfide. Would this still be a "mixture"? Why or why not?

Many times errors are expressed in terms of percentage. The percent error is the absolute value of the difference of the true value and the experimental value, divided by the true value, and multiplied by \(100 .\) Percent error \(=\frac{\mid \text { true value }-\text { experimental value } \mid}{\text { true value }} \times 100\) Calculate the percent error for the following measurements. a. The density of an aluminum block determined in an experiment was \(2.64 \mathrm{~g} / \mathrm{cm}^{3}\). (True value \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\).) b. The experimental determination of iron in iron ore was \(16.48 \% .\) (True value \(16.12 \% .)\) c. A balance measured the mass of a \(1.000-\mathrm{g}\) standard as \(0.9981 \mathrm{~g}\)

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