## anonymous one year ago dsv

1. Jhannybean

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$

2. Jhannybean

Is that understandable, @kaylee_crps_strong ?

3. Jhannybean

Yep.

4. Jhannybean

Oh wait

5. Jhannybean

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

6. Jhannybean

Typo.

7. Jhannybean

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.