Wie würden Sie # g / (cm ^ 3) # in # mL # konvertieren?

Wenn du meintest #"g/mL"#, then it requires little effort. #"1 cm"^3# is known to be equivalent to #"1 mL"#, so you're done.

Wenn du meintest #"mL"#, it's not the same units back, so it's not a proper conversion; you're inputting a Dichte and getting back a volume. Regardless, you would do this:

#"g"/"cm"^3# #stackrel("reciprocate"" ")(->)# #"cm"^3/"g"#

#"cm"^3/cancel"g" xx cancel"g" = "cm"^3 = "mL"#

So, you would take a mass #m# in #"g"# and divide it by a density #rho# in #"g/cm"^3# or #"g/mL"#, giving you the volume #V# in #"mL"#:

#rho = m/V#

#=> color(blue)(V = m/rho)#

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