A chemical bond is what holds atoms together in molecules.
The balanced equation for the burning of N-butane is;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
stoi. ratio 2 : 13 : 8 : 10
molar mass 58 32 44 18
moles X 13 X/2 8X/2 10X/2
105.86 688.10 423.45 529.31
produced mass 18632.8 g 9527.6 g
moles of n-butane = X = 6.14 x 10³ g / 58 g mol⁻¹
hence other moles of compounds can be calculated.
Mass = molar mass * moles
mass fraction = mass / total mass
hence the mass fraction of
CO₂ = 18632.8 / 6.14 x 10³ = 3.03
H₂O = 9527.6 / 6.14 x 10³ = 1.55
the required mass of air = 688.10 x 32 = 22019.2 g = 2.20 kg
Answer:
The heat capacity for the sample is 0.913 J/°C
Explanation:
This is the formula for heat capacity that help us to solve this:
Q / (Final T° - Initial T°) = c . m
where m is mass and c, the specific heat of the substance
27.4 J / (80°C - 50°C) = c . 6.2 g
[27.4 J / (80°C - 50°C)] / 6.2 g = c
27.4 J / 30°C . 1/6.2g = c
0.147 J/g°C = c
Therefore, the heat capacity is 0.913 J/°C
Delta Go = -RTlnKeq
delta Go = 5.95 kJ/mole = 5.95 X 1000 = 5950 J/mole ( 1 kj = 1000 J )
putting the values and finding Keq
5950 = -8.314 X 298 X ln Keq
ln Keq = -5950 / 2477.572 = -2.4015
Keq = e^-2.402 = 0.0905
suppose the equilibrium reaction is :-
chair 1 <--------> chair 2
now as Keq is less than 1 ....so chair 1 will be more stable
Keq = [chair2]/[chair 1 ] = 0.0905
this means that [chair 2] ~ 0.0905 and [chair 1] ~ 1
[total] = [chair 2] + [chair 1] ~ 1 + 0.0905= 1.0905
percentage of chair 1 = [chair 1] / [total] = 1 / 1.0905 X 100 = 91.70 %
