Question

Question #945e2

Answer 1

`"1.34 g cm"^(-3)`

Explanation:

The idea here si that the volume of the rock will be equal to the volume of water it displaces when added to the graduated cylinder.

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The volume of the water goes from

`"25.0 mL " -> " 36.2 mL"`

after the rock is added, so you can say that the volume of the rock will be equal to

`V_"rock" = "36.2 mL" - "25.0 mL"`

`V_"rock" = "11.2 mL"`

Now, in order to find the density of the rock, you must determine the mass of exactly one unit of volume the rock would occupy.

You know that a volume of `"11.2 mL"` corresponds to `"15.0 g"`, so you can say that `"1m L"` will correspond to

`1 color(red)(cancel(color(black)("mL"))) * "15.0 g"/(11.2color(red)(cancel(color(black)("mL")))) = "1.339 g"`

Now, you should know that

`"1 mL" = "1 cm"^3`

This means that a volume of `"1 cm"^3` corresponds to `"1.339 g"`. You can thus say that the density of the rock is equal to

`color(darkgreen)(ul(color(black)("density = 1.34 g cm"^(-3))))`

The answer is rounded to three sig figs.