Calculate the density of a solid substance if a cube measuring 2.72 cm on one side has a mass of 130g.

- anonymous

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- UnkleRhaukus

density \(\varrho\), is equal to the mass \(m\), per volume \(V\)
\[\varrho = m/V\]
Calculate the volume of the cube

- anonymous

How do you calculate the volume of the cube?

- UnkleRhaukus

the volume \(V\), of a cube is equal to the length of its sides \(s\), cubed.
\[V =s^3\]

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## More answers

- anonymous

so you would cube the 2.72?

- UnkleRhaukus

you are given that the sides are \(s=2.72[\text{cm}]\)

- UnkleRhaukus

What volume does this make?

- anonymous

If you cube the 2.72 cm?

- UnkleRhaukus

yes,

- anonymous

20.123648 cm^3

- UnkleRhaukus

good, thats the volume,
now the density is
\[\varrho = m/V = \frac{130[\text g]}{20.123648 [\text{cm}^3]}=\]

- anonymous

6.460061317g/cm^3

- UnkleRhaukus

yeah thats it,
now you might want to round to 3 significant figures, (because the values given are themselves only accurate to 3sig.figs)

- anonymous

So 6.460

- UnkleRhaukus

(that is four significant figures)

- anonymous

Whoops
6.46

- UnkleRhaukus

yeah, i would leave my final answer as \(\varrho=6.46\,[\text g/\text{cm}^3]\)

- anonymous

Alright thank you. There is another question like it wanting to calculate the mass of a cube of the same substance measuring 7.51cm on one side.

- UnkleRhaukus

OK, well since the substance is the same, the density (an intrinsic quality) will remain the same

- UnkleRhaukus

rearranging\[\varrho=m/V\]
we get\[m =\varrho\cdot V\]
again the volume of a cube is \(s^3\)

- anonymous

7.51^3 = 423.564751

- UnkleRhaukus

now multiply that, by the density we found earlier

- anonymous

2736.228291

- UnkleRhaukus

and what are the units?

- anonymous

grams

- anonymous

since mass was being calculated for

- UnkleRhaukus

yeah, so what is our final answer with units (and to 3sig.figs)

- anonymous

Alright and if it wants it in scientific notation would it be 2.73 x 10^3?

- UnkleRhaukus

\[2.74 \times 10^3\,[\text g]=2.74\,[\text{kg}]\]

- anonymous

Thanks. Didn't notice the 6 there.

- anonymous

Thanks for the help!

- UnkleRhaukus

notice that the ratio of volumes \(424\,[\text{cm}^3]/20.1\,[\text{cm}^3]\approx21\), is equal to the ratio of masses \(2740\,[\text g]/130\,[\text g]\approx21\)

- UnkleRhaukus

(as we should expect for a substance of constant density )

- anonymous

I didn't notice that. That's interesting and that does make sense for it to.

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