dsv

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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 \(\rho = \dfrac{m}{v}\) \[\frac{22.57 g}{cm^3} = \frac{m_{\text{block}}}{(1.00~cm~\times~4.00~cm~\times~2.00~cm)}\]\[m_{\text{block}} = \frac{22.57~g}{cm^3} \cdot (1.00~\times~4.00~\times~2.00)~cm^3\]
Is that understandable, @kaylee_crps_strong ?
Yep.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Oh wait
It's not 2.00, it's meant to be 2.50*
Typo.
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question