# Question #6d9dd

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you will use the volume of *displaced water* to determine the volume of the stone, then use the stone's mass to determine its density.

Notice that you're starting with

So why did this happen? The stone has a volume of its own, and so you can say that the change in the level of the water was caused by the added volume of the stone.

#V_"final" = V_"water" + V_"stone"#

#V_"stone" = V_"final" - V_"water"#

#V_"stone" = "7.0 mL" = "2.0 mL" = "5.0 mL"#

You know the volume of the stone and its mass, which means that you can determine its density by using the formula

#"density" = "mass"/"volume"" "# , or#" "rho = m/V#

In your case, you would get

#rho = m/V = "25 g"/"5.0 mL" = color(green)("5.0 g/mL")#

This means that *each mililiter* of stone has a mass of

Here's how the procedure would look like - this example uses other values for the initial volume and the final volume of the cylinder.

Can you use this example to find the density of the piece of gold?