Question

Some Units conversion/ density problems

1. A gold refinery produces a gold ingot weighing 125 pounds. If the gold is drawn into wire whose diameter is 9.25 mm, how many feet of gold wire can be obtained from the ingot? The density of gold is 19.32 grams/cm3 .

2.A 15.0 cm long cylindrical glass tube, sealed at one end, is filled with ethanol. The mass of ethanol needed to fill the tube is found to be 11.86 grams. The density of ethanol is 0.789 grams/mL. Calculate the diameter of the tube in inches.

3. A 40 pound bag of peat moss needs to be spread onto a flower bed that measures 48 inches by 36 inches. Calculate the depth of the peat moss in meters. The density of peat moss is 0.132 grams/cm3 .

Show your work neatly and methodically. Be sure you work supports your answer.

Answer 1

I will answer Question 1 and Question 2 to give you an idea of how to solve these kind of questions. Try to solve question 3 by yourself. If you can't, then post it in a new question thread.

1. Let's assume the wire is like a cylinder, and volume of a cylinder is:

V = pi*r²*h from here, we solve for h:

h = V / pi*r²

We have the diameter, which we can obtain radius. pi = 3.14. Now it's time to determine the volume of gold.

d = m/V ---> V = m/d

m = 125 pounds * 453.59 g / pound = 56698.75 g

V = 56698.75 / 19.32 = 2934.71 cm³

r = D/2 = 9.25 mm * 1x10^-1 cm/mm / 2 = 0.4625 cm

h = 2934.71 / 3.14 * (0.4625)² = 4369.302 cm

h = 4369.302 cm * 1x10^-2 m/cm * 3.281 ft/m = 143.36 ft

2. The same thing as question 1, the same expression, only that this time we solve for r:

r = (V / pi*h)^1/2

V = 11.86 / 0.789 = 15.03 mL * 1 cm³/mL = 15.03 cm³

r = (15.03 / 3.14*15)^1/2 = 0.56 cm

D = 2*0.56 = 1.12 cm * 1 in / 2.54 cm = 0.4409 in.

Hope this helps