Fractional distillation is used for the refining of crude petroleum.
<h3>How are the components of
crude petroleum separated out?</h3>
Fractional distillation is the procedure used to separate crude oil's numerous constituents.
- A mixture is divided into several components, known as fractions, using fractional distillation.
- A combination of hydrocarbons makes up crude oil. The crude oil evaporates, and in the fractionating column, its vapors condense at various temperatures.
- The hydrocarbon molecules in each percent have a comparable number of carbon atoms and a comparable range of boiling points.
- The mixture is placed above a tall fractionating column that has multiple condensers coming off at various heights.
- The bottom of the column is warm, while the top is cool. High boiling point compounds condense at the bottom, whereas low boiling point substances condense as they ascend.
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Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.
This question provides us –
- Weight of
is = 47 g - Volume, V = 375 mL
__________________________________________
- Molar Mass of
–
<u>Using formula</u> –
- Henceforth, Molarity of the solution is = 1.7M
___________________________________________
The correct answer would be the first option. Material A having a smaller latent heat of fusion would mean that it will take only less energy to phase change into the liquid phase. Latent of heat of fusion is the amount of energy needed of a substance to phase change from solid to liquid or liquid to solid.
Answer:
Mass= 2.77g
Explanation:
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g
