Answer:
Molecular formula => C₃H₈O₃
Explanation:
From the question given above, the following data were obtained:
Carbon (C) = 39.12%
Hydrogen (H) = 8.75%
Oxygen (O) = 51.12%
Molar mass of compound = 92.09 g/mol
Molecular formula =?
Next, we shall determine the empirical formula of the compound. This can be obtained as follow:
C = 39.12%
H = 8.75%
O = 51.12%
Divide by their molar mass
C = 39.12 / 12 = 3.26
H = 8.75 / 1 = 8.75
O = 51.12 / 16 = 3.195
Divide by the smallest
C = 3.26 / 3.195 = 1
H = 8.75 / 3.195 = 2.7
O = 3.195 / 3.195 = 1
Thus, the empirical formula is CH₂.₇O
Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:
Empirical formula = CH₂.₇O
Molar mass of compound = 92.09 g/mol
Molecular formula =?
Molecular formula = Empirical formula × n
Molecular formula = [CH₂.₇O]ₙ
92.09 = [12 + (2.7×1) + 16] × n
92.09 = 30.7n
Divide both side by 30.7
n = 92.09 / 30.7
n = 3
Molecular formula = [CH₂.₇O]ₙ
Molecular formula = [CH₂.₇O]₃
Molecular formula = C₃H₈O₃
Answer:
the first line
Explanation:
A rarefaction is a region in a longitudinal wave where the particles are furthest apart. ... The region where the medium is compressed is known as a compression and the region where the medium is spread out is known as a rarefaction.
use this picture as a reference:
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
Saponification is a process that involves conversion of fat or oil into soap and alcohol by the action of heat in the presence of aqueous alkali
