The balanced chemical equation for the combustion of butane is:
Δ
= Σ
Δ
-Σ
Δ
= ![[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]](https://tex.z-dn.net/?f=%5B%7B8%2A%28-393.5kJ%2Fmol%29%7D%2B%7B10%2A%28-241.82kJ%2Fmol%29%7D%5D-%5B%7B2%2A%28-125.6kJ%2Fmol%29%7D%2B13%2A%280%20kJ%2Fmol%29%7D%5D)
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:
Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol
Answer:
Molarity = 54.50 M
Explanation:
Molarity is defined as the number of moles of a solute in 1 liter of solution. It has mol/l unit.
Molar mass of glucose, C6H12O6 = (6*12) + (1*12) + (16*6)
= 72 + 12 + 96 = 180g/mol
Density of glucose solution = 1.03g/ml
Density in g/l = 1.03g/ml * 1ml/10^-3l
= 1030g/l
10.5% mass of glucose = 10.5/100 * 180
= 18.9g/mol
Molarity = density (g/l)/molar mass (g/mol)
= 1030/18.9
= 54.50 M
The answer is B) the second group
