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.

Related questions
See all questions in Unit Conversion
Impact of this question
3130 views around the world
You can reuse this answer
Creative Commons License