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:
Aluminium + oxygen = almininum oxide
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1/2 - 3/7 = 7/14 - 6/14 = 1/14
According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.
the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent
Where
xB = mole fraction of solute=?
p = 22.8 torr
mole fraction is ratio of moles of solute and total moles of solute and solvent
moles of solvent = mass / molar mass = 500 /18 = 27.78 moles
putting the values
mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams
