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if the density of a certain spherical atomic nucleus is 1.0x10^14 g cm^-3 and its mass is 2.0x10^-23 g, what is it radius in cm?

Verified Answer :

spinions

Density as you should know is equal to mass/volume. Solve for the volume.

Now, you know that you are dealing with a sphere. You know the volume of a sphere is V= 4/3pi r ^3.

You know the volume so just solve for r

IthaloAbreu

Answer:

r = \(3.63x10^{-13}\) cm

Explanation:

The density (d) is the mass divided by the volume, so:

d = m/V

\(1.0x10^{14} = \frac{2.0x10^{-23}}{V}\)

\(V = \frac{2.0x10^{-23}}{1.0x10^{14}}\)

V = \(2.0x10^{-37} cm^3\)

The volume of a sphere is

\(V = \frac{4xpixr^3}{3}\)

For pi = 3.14

\(2.0x10^{-37} = \frac{4x3.14xr^3}{3}\)

\(12.56r^3 = 6.0x10^{-37}\)

\(r^3 = 4.78x10^{-38}\)

\(r = \sqrt[3]{4.78x10^{-38}}\)

r = \(3.63x10^{-13}\) cm

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