Question

I need help with this question, if an aluminum cube of mass 2.45 g and each side measuring 9.68 mm is dipped in the above cylinder containing 5.00 mL water, what will be the final water level in the cylinder? the above cylinder in the question above weighs 64.55 g

Answer 1

You have to calculate the volume of the cube, LxLxL= L^3 = 9.68^3
convert the mm3 to mL and add this volume to the water volume (5.00mL) that is the final water level in the cylinder. The mass of the cylinder and the mass of the cube you dont need to solve this step of the problem