How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold The density of the ore is 8.0 g/cm³ and the price of gold is $818 per troy ounce. (14.6 troy oz = 1.0 ordinary pound, called an avoirdupois pound; 1 lb = 454 g) A) 0.48 cm³ B) 220 cm³ C) 1.4 * 10³ cm³ D) 1.7 * 10³ cm³ E) 1.8 * 10⁴ cm³

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First, calculate the mass of gold needed to obtain $100.00 worth of gold. The price of gold is $818 per troy ounce. There are 14.6 troy ounces in an avoirdupois pound, and 1 pound is 454 grams. Therefore, $100.00 worth of gold is $100.00 / $818 per troy ounce = 0.1222 troy ounces. Converting troy ounces to grams (using 1 troy ounce = 31.1035 grams), 0.1222 troy ounces of gold is approximately 3.8 grams of gold. Since the ore contains only 0.22% gold by mass, the total mass of ore needed to get 3.8 grams of gold is 3.8 grams / 0.0022 = 1727 grams. Given the density of the ore is 8.0 g/cm^3, the volume of ore needed is 1727 grams / 8.0 g/cm^3 = 215.875 cm^3, which rounds to approximately 220 cm^3.

How Many Cubic Centimeters of an Ore Containing Only 0