Final answer:
By converting the given dimensions and density to consistent units, we calculate that there are approximately 1,666,705.6 pounds of dry soil in the top 6 inches of an acre of soil with a bulk density of 1.3 g/cm^3.
Step-by-step explanation:
The question involves a calculation to find the amount of dry soil in the top 6 inches of an acre using the given bulk density and converting the units to pounds. To solve this, we first convert the depth from inches to feet and the bulk density from grams per cubic centimeter to pounds per cubic foot, then calculate the volume of soil and use the density to find the mass in pounds.
First, convert 6 inches to feet: 6 inches = 0.5 feet. Then convert the bulk density: 1.3 g/cm3 is equal to 1.3 g/cm3 * (1/454 lb/g) * (2.54 cm/in)3 * (12 in/ft)3 = approximately 76.52 lb/ft3. Next, calculate the volume of soil covering an acre at 6 inches deep: 43,560 ft2 * 0.5 ft = 21,780 ft3.
Finally, multiply the volume by the bulk density to find the mass: 21,780 ft3 * 76.52 lb/ft3 = 1,666,705.6 lb. Therefore, there are approximately 1,666,705.6 pounds of dry soil in the top 6 inches of an acre of soil with a bulk density of 1.3 g/cm3.