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 volume of the sphere is calculated using the formula \(V = \frac{4}{3} \pi (0.50\mathrm{~cm})^{3}\), which yields approximately \(0.52 \mathrm{~mL}\). Then, we calculate the density of the sphere using the formula \(Density = \frac{2.0\mathrm{~g}}{0.52 \mathrm{~mL}} \approx 3.85 \mathrm{~g} / \mathrm{mL}\). Since the density of the sphere (3.85 \(\mathrm{g/mL}\)) is greater than the density of water (1.0 \(\mathrm{g/mL}\)), the ball will sink when placed in water.

Step by step solution

01

Calculate the volume of the sphere

The formula to calculate the volume of a sphere is given: \[ V = \frac{4}{3} \pi r^{3} \] With the given radius (\(r = 0.50\mathrm{~cm}\)), we can calculate the volume: \[ V = \frac{4}{3} \pi (0.50\mathrm{~cm})^{3} \]
02

Calculate the density of the sphere

The density formula is: \[ Density = \frac{Weight}{Volume} \] Using the sphere's given weight of \(2.0\mathrm{~g}\) and the volume calculated in Step 1 (assuming you already calculated it), we can calculate the density: \[ Density = \frac{2.0\mathrm{~g}}{V} \]
03

Compare the density of the sphere to the density of water

If the calculated density of the sphere is less than \(1.0\mathrm{~g} / \mathrm{mL}\) (the density of water), the sphere will float. If it is greater than \(1.0\mathrm{~g} / \mathrm{mL}\), the sphere will sink.

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