<u>Answer:</u> The activation energy of the reaction is 124.6 kJ/mol
<u>Explanation:</u>
To calculate activation energy of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B79%5EoC%7D%7D%7BK_%7B26%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 79°C = 
= equilibrium constant at 26°C = 
= Activation energy of the reaction = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![26^oC=[26+273]K=299K](https://tex.z-dn.net/?f=26%5EoC%3D%5B26%2B273%5DK%3D299K)
= final temperature = ![79^oC=[79+273]K=352K](https://tex.z-dn.net/?f=79%5EoC%3D%5B79%2B273%5DK%3D352K)
Putting values in above equation, we get:
![\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B0.394%7D%7B2.08%5Ctimes%2010%5E%7B-4%7D%7D%29%3D%5Cfrac%7BE_a%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B299%7D-%5Cfrac%7B1%7D%7B352%7D%5D%5C%5C%5C%5CE_a%3D124595J%2Fmol%3D124.6kJ%2Fmol)
Hence, the activation energy of the reaction is 124.6 kJ/mol
From a chemistry book that I have read, it stated that
Barium Sulfate is so insoluble in water. Due to this, it is not toxic to humans
since it is not dissolved and absorbed by our body. In the other hand, Barium
Carbonate is slightly soluble in water making it toxic to us.
Answer:
Barium carbonate is soluble in plain water
Answer: Velocity
Explanation:
i just took the quiz for k12
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>
