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.
Learn more about fractional distillation here:
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Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer:
This is the balanced equation:
Pb(NO₃)₂ (aq) + 2NaI (aq) → 2NaNO₃ (aq) + PbI₂ (s) ↓
Explanation:
This are the reactants:
PbNO₃
NaI
Iodide can react to Pb²⁺ to make a solid compound.
