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)

216.0cm³ of ore must be processed

Explanation:

First, we need to know how many grams of gold we need to obtain $100.00

Mass of gold to obtain $100.00:

$100.00 gold * (1 troy oz / $818) = 0.122 troy oz

0.122 troy oz * (1.0lb / 14.6troy oz) = 8.37x10⁻³ lb

8.37x10⁻³ lb * (454g / 1lb) =

3.80g of gold we need to obtain $100.00

Now, the ore contain 0.22g of gold in 100g and the density of the ore is 8.0g/cm³. We can solve the cubic centimeters to obtain $100.00:

Cubic centimeters of ore:

3.80g Gold * (100g Ore / 0.22g Gold) = 1727.9g of Ore

1727.9g Ore * (1cm³ / 8.0g Ore) =

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.