The density of gold is 19.3 g/cc. how many milligrams of gold are required to coat a small ball bearing of diameter 1.00 mm? the thickness of the coating is to be 0.200 mm. report answer to 3 sig figs.

Solution: Density of gold is 19.3 g/cc or 19.3 g/cm^{3}. Diameter of small ball is 1.0 mm. Thus, radius will be 0.5 mm. Since, 1 mm =0.1 cm thus, 0.5 mm =0.05 cm.

First calculate the volume of small ball from formula of volume of sphere as follows:

V=\frac{4}{3}\pi r^{3}

V=\frac{4}{3}(3.14)(0.05 cm)^{3}=5.02\times 10^{-4}cm^{3}

Now, if a layer of gold covered the small ball with thickness 0.200 mm or 0.02 cm, a sphere with radius more than the small ball is formed. The radius of bigger sphere will be sum of radius of small ball and the thickness of gold layer. Thus, R=0.05+0.02=0.07 cm

Calculate volume of bigger sphere:

V_{1}=\frac{4}{3}\pi R^{3}

V_{1}=\frac{4}{3}(3.14)(0.07 cm)^{3}=1.43\times 10^{-3}cm^{3}

now, from the figure attached, volume of gold used can be calculated as follows:

V_{gold}=V_{1}-V

=1.43\times 10^{-3}cm^{3}-5.02\times 10^{-4}cm^{3}

=9.28\times 10^{-4}cm^{3}

Mass of gold can be calculated from volume and density as follows:

m=d×V=19.3 g/cm^{3}×9.28\times 10^{-4}cm^{3}=0.0179 g

Since, 1 g=10^{3}mg

Thus,

Mass of gold will be 17.9 mg. This is up to three significant figures.

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of  rain  or snow. this dry air produces a  rain shadow.