If 45.5 g of a metal that has a density of 3.65 g/mL is placed in 45.0 mL of water, what is the final volume?

The final volume of the metal and water is 54.46mL

Explanation:

Hello,

To solve this question, we'll first of all find the volume of the metal and assuming there's no loss of water by overflow in the container, we'll add the volume of the metal to the volume of the water to get the final volume.

Data;

Mass of the metal = 45.5g

Volume of the water = 45mL

Density of the metal (ρ) = 3.65g/mL

Density of the metal = mass / volume

ρ = mass / volume

Volume (v) = mass / density

Volume = 45.5 / 3.65

Volume = 12.46mL

The volume of the metal is 12.46mL.

When the metal is added to the container 45mL of water, assuming no water was lost by overflow in the container, the final volume =

Final volume = volume of metal + volume of water

Final volume = 12.46 + 45.0

Final volume = 57.46mL

The final volume of the metal and water is 57.46mL