Question #945e2

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Answer 1 · 2017-06-02T00:05:30

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

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.