You are a meteorologist measuring the changing air pressure due to an approaching weather front. What instrument would you use t
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Answer:
The required volume of hexane is 0.66245 Liters.
Explanation:
Volume of octane = v=1.0 L=
Density of octane= d =
Mass of octane ,m=
Moles of octane =
Mole percentage of Hexane = 45%
Mole percentage of octane = 100% - 45% = 55%
Total moles = 11.212 mol
Moles of hexane :
Moles of hexane = 5.0454 mol
Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g
Density of the hexane,D =
Volume of hexane = V
(1 cm^3= 0.001 L)
The required volume of hexane is 0.66245 Liters.