an aluminum sphere has a mass of 22.7 g.\npart a\nfind the radius of the sphere. (the density of aluminum is…

an aluminum sphere has a mass of 22.7 g.\npart a\nfind the radius of the sphere. (the density of aluminum is 2.7 g/cm³, and the volume of a sphere is given by the equation v = 4/3πr³.) express your answer to two significant figures and include the appropriate units.

an aluminum sphere has a mass of 22.7 g.\npart a\nfind the radius of the sphere. (the density of aluminum is 2.7 g/cm³, and the volume of a sphere is given by the equation v = 4/3πr³.) express your answer to two significant figures and include the appropriate units.

Answer

Explanation:

Step1: Calculate the volume of the sphere

We know that density $\rho=\frac{m}{V}$, so $V = \frac{m}{\rho}$. Given $m = 22.7\ g$ and $\rho=2.7\ g/cm^{3}$, then $V=\frac{22.7}{2.7}\ cm^{3}\approx8.407\ cm^{3}$.

Step2: Use the volume - radius formula for a sphere

The volume formula of a sphere is $V=\frac{4}{3}\pi r^{3}$. We need to solve for $r$. Rearranging the formula for $r$, we get $r^{3}=\frac{3V}{4\pi}$. Substituting $V = 8.407\ cm^{3}$ into the formula, $r^{3}=\frac{3\times8.407}{4\pi}\ cm^{3}$. Then $r^{3}=\frac{25.221}{4\pi}\ cm^{3}\approx2.004\ cm^{3}$.

Step3: Find the radius

Taking the cube - root of $r^{3}$, $r=\sqrt[3]{2.004}\ cm\approx1.26\ cm$. Rounding to two significant figures, $r = 1.3\ cm$.

Answer:

$r = 1.3\ cm$