0.14g of methanol is burnt and 3.04kJ of heat is released. Calculate the enthalpy of combustion of methanol.

-694.8 kJ/mol ≅ -695 kJ

Explanation:

The enthalpy of combustion (ΔHc) is defined as the amount of heat that is released when 1 mol of the compound is burned. We have the mass of methanol (0.14 g) and we have to calculate the moles present in that mass. For that, we need the molecular weight of methanol, which is calculated from the molar mass of the chemical elements.

Methanol: CH₃OH

Mw(CH₃OH) = (12 g/mol x 1 C) + (1 g/mol x 3 H) + 16 g/mol O + 1 g/mol H = 32 g/mol

moles of CH₃OH = mass/Mw(CH₃OH) = 0.14 g/(32 g/mol) = 4.4 x 10⁻³ mol

As the heat is released, the change in enthalpy is a negative value: -3.04 kJ.

Finally, we divide the released heat (-3.04 kJ) into the number of moles of CH₃OH:

ΔHc= (-3.04 kJ)/(4.4 x 10⁻³ mol CH₃OH) = -694.8 kJ/mol ≅ -695 kJ

idk what the question is about

the correct answer is:

a. formation of precipitation