How many milliliters of water would be displaced by 408 grams of lead?

To find out how many milliliters of water would be displaced by 408 grams of lead, we need to use the concept of density and the principle of displacement.

Density is mass per unit volume. The density of lead is approximately 11.34 grams per cubic centimeter (g/cm³). This can be used to determine the volume of lead, using the formula:

Density (ρ) = Mass (m) / Volume (V)

Rearranging the equation to find volume:

Volume (V) = Mass (m) / Density (ρ)

Now, plug in the values for the mass of lead and its density:

V = 408 g / 11.34 g/cm³

V ≈ 35.973 cm³

Water has a density very close to 1 g/cm³ (at 4°C), which also means that 1 milliliter of water has a volume of 1 cubic centimeter (1 ml = 1 cm³), under standard conditions.

Therefore, the volume of water that would be displaced by 408 grams of lead would be approximately equal to the volume of the lead, which is 35.973 cm³.

Since the volume of water in milliliters is numerically the same as the volume in cubic centimeters, 408 grams of lead would displace approximately 35.973 milliliters of water.