Question

dsv

Answer 1

The density of \(\sf 1.00~\times~10^3~g\)of osmium is \(\sf~\rho = \dfrac{22.57~g}{cm^3}\). They want to know what the mass would be if you had the given volume of osmium.

I think we can solve it by using the formula \(\rho = \dfrac{m}{v}\)

\[\frac{22.57 g}{cm^3} = \frac{m_{\text{block}}}{(1.00~cm~\times~4.00~cm~\times~2.00~cm)}\]\[m_{\text{block}} = \frac{22.57~g}{cm^3} \cdot (1.00~\times~4.00~\times~2.00)~cm^3\]

Answer 2

Is that understandable, @kaylee_crps_strong ?

Answer 3

Yep.

Answer 4

Oh wait

Answer 5

It's not 2.00, it's meant to be 2.50*

Answer 6

Typo.

Answer 7

mass of the block after my calculation turns out to be 226. g with the proper number of s.f remember, the cm\(^3\) cancels out.