Question

A graduated cylinder contains 20.5 mL of water. What is the new water level after 35.2 g of silver metal with a density of 10.5 g/mL is submerged in water?

Answer 1

Answer : The volume of new water level in graduated cylinder is 23.85 mL.

Explanation :

To calculate volume of a silver metal, we use the equation:

$\text{Density of silver metal}=\frac{\text{Mass of silver metal}}{\text{Volume of silver metal}}$

We are given:

Density of silver metal = 10.5 g/mL

Mass of silver metal = 35.2 g

Putting values in above equation, we get:

$10.5g/mL=\frac{35.2g}{\text{Volume of silver metal}}\\\\\text{Volume of silver metal}=3.35mL$

Now we have to calculate the volume of new water level in graduated cylinder.

Volume of water in graduated cylinder = 20.5 mL

Volume of silver metal = 3.35 mL

Volume of new water level = Volume of water in graduated cylinder + Volume of silver metal

Volume of new water level = 20.5 mL + 3.35 mL

Volume of new water level = 23.85 mL

Therefore, the volume of new water level in graduated cylinder is 23.85 mL.

Answer 2

 V = 24.0 mL + (35.2 g)(mL/10.5g) = I think i'm not all that sure but I think its this.