A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
dsv
anonymous
 one year ago
dsv

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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 \(\rho = \dfrac{m}{v}\) \[\frac{22.57 g}{cm^3} = \frac{m_{\text{block}}}{(1.00~cm~\times~4.00~cm~\times~2.00~cm)}\]\[m_{\text{block}} = \frac{22.57~g}{cm^3} \cdot (1.00~\times~4.00~\times~2.00)~cm^3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that understandable, @kaylee_crps_strong ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's not 2.00, it's meant to be 2.50*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0mass 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.