Answer:
Oil is extracted by three general methods: rendering, used with animal products and oleaginous fruits; mechanical pressing, for oil-bearing seeds and nuts; and extracting with volatile solvents, employed in large-scale operations for a more complete extraction than is possible with pressing.
Explanation:
Mole fraction of Oxygen=0.381
Mole fraction of Oxygen= (range of moles of oxygen) ÷(general moles)
also, mole fraction of oxygen = (partial stress of oxygen) ÷ (total strain)
consequently , mole fraction of Oxygen= (2.31 atm)÷(2.31 atm + 3.75 atm)
= 0.381
The mole fraction may be calculated by means of dividing the variety of moles of 1 element of a solution by the entire quantity of moles of all the additives of a solution. It is cited that the sum of the mole fraction of all of the components inside the solution should be identical to 1.
Mole fraction is a unit of awareness. in the solution, the relative amount of solute and solvents are measured by way of the mole fraction and it's far represented through “X.” The mole fraction is the variety of moles of a selected aspect inside the answer divided by way of the entire range of moles in the given answer.
Mole fraction is the ratio between the moles of a constituent and the sum of moles of all ingredients in a mixture. Mass fraction is the ratio between the mass of a constituent and the full mass of a mixture.
The question is incomplete. Please read below to find the missing content.
Assuming that only the listed gases are present, what would the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.
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The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
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a. They are both normally found as gases in the atmosphere. TRUE
That is correct, the oxygen and nitrogen are found in large quantities in the air around us.
b. They can be either liquids or gases. TRUE
Under certain temperatures any gas will transform into a liquid.
c.They turn from gas to liquid at the same temperature. FALSE
Oxygen it will pass into a liquid at -183 °C while nitrogen pass into a liquid at -195.8 °C.
d.They can be changed from gases to liquids by heating them. FALSE
The gases change to liquids by cooling them.