Question

1. aluminum foil is often incorrectly termed tin foil. if the density of tin is 7.28 g/cm³, what is the thickness of a piece of tin foil that measures 5.70 cm by 4.25 cm and has a mass of 0.655 g?
2. (optional) a silver sphere has a mass of 5.492 g and a diameter of 10.0 mm. what is the density of silver metal in grams per cubic centimeter? the volume of a sphere equals 4πr³/3, where π is 3.14, and r is the radius.

Explanation:

Step1: Calculate volume of tin foil using density formula

We know that density $
ho=\frac{m}{V}$, so $V = \frac{m}{
ho}$. Given $m = 0.655$ g and $
ho=7.28$ g/cm³, then $V=\frac{0.655}{7.28}\text{ cm}^3\approx0.09\text{ cm}^3$.

Step2: Calculate thickness of tin foil

The tin - foil is in the shape of a rectangular prism with volume $V = l\times w\times h$, where $l = 5.70$ cm, $w = 4.25$ cm and $h$ is the thickness. We know $V$, $l$ and $w$, so $h=\frac{V}{l\times w}$. Substituting the values, $h=\frac{0.09}{5.70\times4.25}\text{ cm}\approx0.0037$ cm.

for second question:

Step1: Convert diameter to radius and units

The diameter $d = 10.0$ mm. The radius $r=\frac{d}{2}=5.0$ mm. Convert radius to cm: $r = 0.5$ cm.

Step2: Calculate volume of the silver sphere

The volume of a sphere $V=\frac{4}{3}\pi r^{3}$. Substituting $\pi = 3.14$ and $r = 0.5$ cm, we get $V=\frac{4}{3}\times3.14\times(0.5)^{3}\text{ cm}^3=\frac{4}{3}\times3.14\times0.125\text{ cm}^3\approx0.523$ cm³.

Step3: Calculate density of silver

We know that density $
ho=\frac{m}{V}$. Given $m = 5.492$ g and $V\approx0.523$ cm³, then $
ho=\frac{5.492}{0.523}\text{ g/cm}^3\approx10.5$ g/cm³.

Answer:

The thickness of the tin foil is approximately $0.0037$ cm.