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anonymous

  • one year ago

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  1. Jhannybean
    • one year ago
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    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
    • one year ago
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    Is that understandable, @kaylee_crps_strong ?

  3. Jhannybean
    • one year ago
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    Yep.

  4. Jhannybean
    • one year ago
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    Oh wait

  5. Jhannybean
    • one year ago
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    It's not 2.00, it's meant to be 2.50*

  6. Jhannybean
    • one year ago
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    Typo.

  7. Jhannybean
    • one year ago
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    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.

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