Elemental mercury has $ρ = 13.6 \cdot g \cdot m L^{- 1}$ $rho=13.6gmL^-1$ . What is the density expressed in $k g \cdot m^{- 3}$ $kg*m^-3$ ?

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Elemental mercury has $ρ = 13.6 \cdot g \cdot m L^{- 1}$ $rho=13.6gmL^-1$ . What is the density expressed in $k g \cdot m^{- 3}$ $kg*m^-3$ ?

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Answer 1 · 2016-02-21T18:20:48

This is dimensional analysis, and we use the $centi$ $"centi"$ $=$ $=$ $10^{- 2}$ $10^-2$ , and the $kilo$ $"kilo"$ $=$ $=$ $10^{3}$ $10^3$ prefixes.

Explanation:

$13.6$ $13.6$ $g \cdot c m^{- 3}$ $g*cm^-3$ $=$ $=$ $13.6$ $13.6$ $cancel(g)xx10^-3*kg*cancel(g^-1)xx(10^-2*m)^-3$

$=$ $13.6$ $xx$ $10^-3xx10^6xxkgxx(m)^-3$

$=$ $13.6$ $xx10^3*kg*m^-3$ .

This is a lot of mass; of course, mercury metal is pretty dense stuff. I think I am right in this analysis. If you (or someone else) finds an error, let me know.

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Elemental mercury has $ρ = 13.6 \cdot g \cdot m L^{- 1}$ $rho=13.6gmL^-1$ . What is the density expressed in $k g \cdot m^{- 3}$ $kg*m^-3$ ?

1 Answer

Explanation:

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Elemental mercury has rho=13.6*g*mL^-1ρ=13.6⋅g⋅mL−1rho=13.6*g*mL^-1. What is the density expressed in kg*m^-3kg⋅m−3kg*m^-3?

1 Answer

Explanation:

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Elemental mercury has $ρ = 13.6 \cdot g \cdot m L^{- 1}$ $rho=13.6gmL^-1$ . What is the density expressed in $k g \cdot m^{- 3}$ $kg*m^-3$ ?