Question #d7ebc

1 Answer
Jul 11, 2017

Answer:

#0.25# #"mol O"_2#

#1.5 xx 10^23# #"molecules O"_2#

Explanation:

NOTE: I'll have to make the assumption that the conditions are *standard temperature and pressure*.

If this is true then

One mole of an (ideal) gas at standard temperature and pressure conditions occupies a volume of #22.41# #"L"#.

We can use this to convert from liters of #"O"_2# to moles (treating oxygen as an ideal gas). We need to convert from milliliters to liters:

#5600cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = color(red)(5.6# #color(red)("L"#

Using the above mole-liter conversion:

#5.6cancel("L")((1color(white)(l)"mol O"_2)/(22.41cancel("L O"_2))) = color(blue)(0.25# #color(blue)("mol O"_2#

Now, we can use Avogadro's number (#6.022xx10^23#) to convert from moles to individual units (molecules) of oxygen gas:

#color(blue)(0.25)cancel(color(blue)("mol O"_2))((6.022xx10^23color(white)(l)"molecules O"_2)/(1cancel("mol O"_2)))#

#= color(green)(1.5xx10^23# #color(green)("molecules O"_2#