Illinois Smith the explorer extraordinaire find a solid gold statue on one of his expeditions into the jungles of South America he measures its volume by determining the amount of water it will displace and find that is displaces 276.3 mL of water how much is the gold in the statue worth if the current price of gold is $972.40 per oz and the density of gold is 19.32 g/cm^3
First, we need to calculate the mass of the gold statue by multiplying the volume of water displaced by the density of gold:
Volume of gold = Volume of water displaced = 276.3 mL = 276.3 cm^3
Density = 19.32 g/cm^3
Mass = Volume x Density
Mass = 276.3 cm^3 x 19.32 g/cm^3
Mass = 5334.516 g
Now, we need to convert the mass from grams to ounces:
1 oz = 28.35 g
5334.516 g / 28.35 g/oz = 188 oz (approximately)
Finally, we can calculate the value of the gold in the statue:
Value = Mass x Price per ounce
Value = 188 oz x $972.40/oz
Value = $182,903.20
Therefore, the gold in the statue is worth approximately $182,903.20.