Solve for the dimensional analysis problem that is: A small cube of aluminum measures 17.5 mm on a side and weighs 12.25 g. What is the density of aluminum in kg/L?

The volume of the cube can be calculated as the cube of its side length, which is (17.5 mm)^3 = 5359.375 mm^3. Since 1 mL is equal to 1 cm^3, and there are 10 mm in 1 cm, we can convert the volume of the cube to mL by dividing by 10^3, which gives us 5378.125 mm^3 / (10^3) = 5.359375 mL. The density of the aluminum cube in g/mL can be calculated as its mass divided by its volume, which is 12.25 g / 5.359375 mL ≈ 2.2857 g/mL To convert this to kg/L, we need to multiply by the conversion factor between grams and kilograms (1000 g = 1 kg) and between milliliters and liters (1000 mL = 1 L). This gives us 2.276 g/mL * (1 kg / 1000 g) * (1000 mL / 1 L) = 2.2857 kg/L. Therefore, the density of aluminum in this case is approximately 2.2857 kg/L