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

1 Answer
anor277
Feb 21, 2016

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

Explanation:

#13.6# #g*cm^-3# #=# #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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