Temperature is an example of a categorical variable a quantitative variable either a quantitative or categorical variable neithe
1 answer:
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Explanation:
The given data is as follows.
Solvent 1 = benzene, Solvent 2 = water
= 2.7,
= 100 mL
= 10 mL, weight of compound = 1 g
Extract = 3
Therefore, calculate the fraction remaining as follows.
=
=
=
Hence, weight of compound to be extracted = weight of compound - fraction remaining
= 1 -
= 0.00001
or, =
Thus, we can conclude that weight of compound that could be extracted is .
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value =
Molar mass of pentane = 72 g/mol
Fuel value =
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
Answer:
mole fraction of C6H6 = 0.613 atm
Explanation:
The equation for this reaction is :
Initial P₁ P₂ 0
Final 0 P₂ -P₁/2 P₁
After completion of the reaction;
P₁ + P₂ = 1.21 atm ----- (1)
P₂ - P₁/2 + P₁ = 0.839 atm
P₂ + P₁/2 = 0.839 atm ----- (2)
Subtracting (2) from (1); we have:
P₁/2 = 0.371
P₁ = 0.742 atm
From(1)
P₁ + P₂ = 1.21 atm
0.742 atm + P₂ = 1.21 atm
P₂ = 1.21 atm - 0.742 atm
P₂ = 0.468 atm
Thus, the partial pressure of C6H6 = 0.742 atm
∴
Partial pressure = Total pressure × mole fraction of C6H6
mole fraction of C6H6 = Partial pressure / Total pressure
mole fraction of C6H6 = 0.742 atm / 1.21 atm
mole fraction of C6H6 = 0.613 atm