Final answer:
The specific gravity of mercury is approximately 849.024 lb/ft^3, which is calculated by converting the known density of mercury, 13.6 times greater than water, into pounds per cubic foot.
Step-by-step explanation:
The question asks for the specific gravity of mercury in terms of pounds per cubic foot (lb/ft3). The specific gravity of a substance is the ratio of its density compared to the density of water. For mercury, its density is 13.6 times the density of water, which is about 1000 kg/m3 at 4 degrees Celsius, the temperature at which water is densest.
Since 1 kg/m3 is equivalent to approximately 0.0624 lb/ft3, the density of mercury in lb/ft3 can be found by multiplying 13.6 by 1000 kg/m3 to get its density in kg/m3 and then converting that figure to lb/ft3. This results in:
(13.6 × 1000 kg/m3) × 0.0624 lb/ft3/kg/m3 = 849.024 lb/ft3
The specific gravity of mercury expressed in lb/ft3 is therefore approximately 849.024 lb/ft3.