Mass of methane takne = 1.5g
moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles
mass of water = 1000 g
Initial temperature of water = 25 C
final temperature = 37 C
specific heat of water = 4.184 J /g C
1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) = 50208 Joules
2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J
3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules
This heat is released by 0.094 moles of methane
So heat released by one mole of methane =
- 622851.06 Joules = 622.85 kJ / mole
4) standard enthalpy of combustion = -882 kJ / mole
Error = (882-622.85) X 100 / 882 = 24.84 %
1. The formula for
absorbance is given as:
A = log (Io / I)
where A is absorbance, Io
is initial intensity, and I is final light intensity
log (Io / I) = A
log (Io / I) = 2
Io / I = 100
Taking the reverse which is
I / Io:
I / Io = 1 / 100
I / Io = 0.01
Therefore this means that
only 0.01 fraction of light or 1% passes through the sample.
2. What is meant by
transmittance values is actually the value of I / Io. So calculating for A:
at 10% transmittance = 0.10
A = log (Io / I)
A = log (1 / 0.10)
A = 1
at 90% transmittance = 0.90
A = log (Io / I)
A = log (1 / 0.90)
A = 0.046
So the absorbance should be
from 0.046 to 1
3. at 10% transmittance =
0.10
A = log (Io / I)
A = log (1 / 0.10)
<span>A = 1</span>
Increase the use of nuclear energy
