A 5.0 gram sample of lead and a 3.2 g sample of iron are placed into 367 mL of water. What will be the new volume level of water in units of mL? The density of lead is 11.34 g/cc and the density of iron is 7.874 g/mL. Round your answer to three significant figures. Do not enter "mL" as part of your answer.

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Density=mass / volume ⇒ volume=mass / density

1) We compute the volume of 5 g of lead

dа ta:

Density = 11.34 g / cm³

mass = 5 g

1 ml=1 cm³

volume = 5 g / (11.34 g / cc) = 0.441 cm³=0.441 ml

2) we compute the volume of 3.2 g of iron.

dа ta:

density = 7.874 g / ml

mass=3.2 g

volume=3.2 g / (7.874 g/ml) = 0.406 ml

3) we compute the total volume (lead + iron) of the metals.

Total volume of the metals = 0.441 ml + 0.406 ml=0.847 ml

4) we compute the new volume level of water:

New volume level of water=367 ml + 0.847 ml=367.847 ml

Answer: The new volume level of water in units of ml is 367.847