Answer:
The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = Heat gained = ?
c = Specific heat = 
ΔT = The change in temperature = 3.08°C
Now put all the given values in the above formula, we get:
Now we have to calculate molar enthalpy of combustion of this substance :
where,
= enthalpy change = ?
q = heat gained = 8.2544kJ
n = number of moles methane = 
Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Answer:
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<span>(15.0 g) / (150.0 g) x (100 g) = 10.0 g/100 g H2O </span>
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
