Question

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.

Answer 1

To coat the small ball bearing with a thickness of 0.200 mm, approximately 0.297 mg of gold is required.

Step-by-step explanation:

To determine the amount of gold required to coat a small ball bearing, we need to calculate the volume of the coating first. The volume of the coating can be calculated using the formula for the volume of a cylinder, which is V = πr²h. Given that the diameter of the ball bearing is 1.00 mm and the thickness of the coating is 0.200 mm, we can calculate the radius and the volume. The radius (r) is half of the diameter, so r = 0.50 mm. Converting the radius to cm, we get r = 0.005 cm. Plugging the values into the volume formula, we get V = π(0.005 cm)²(0.200 cm) = 0.0000157 cm³.

Now, we can use the density of gold (19.3 g/cm³) to calculate the mass of the coating. The formula for density is density = mass/volume, so we can rearrange it to solve for mass: mass = density × volume. Substituting the given values, we get mass = (19.3 g/cm³)(0.0000157 cm³) = 0.000297 g. To convert the mass to milligrams and report the answer to 3 significant figures, we need to multiply by 1000 and round to 3 decimal places. The final answer is 0.297 mg.

Answer 2

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.