Answer:
120.575 kJ is the activation energy for the souring process.
Explanation:
The formula for an activation energy is given as:
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at
= ![40k](https://tex.z-dn.net/?f=40k)
= rate constant at
= ![k](https://tex.z-dn.net/?f=k)
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = ![25^oC=273+25=298 K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298%20K)
= final temperature = ![4^oC=273+4=277 K](https://tex.z-dn.net/?f=4%5EoC%3D273%2B4%3D277%20K)
Now put all the given values in this formula, we get:l
![\log (\frac{k}{40k})=\frac{Ea}{2.303\times 8.314 J/mol K}[\frac{1}{298K}-\frac{1}{277 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7Bk%7D%7B40k%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B298K%7D-%5Cfrac%7B1%7D%7B277%20K%7D%5D)
![E_a=120,575.61J=120.575 kJ](https://tex.z-dn.net/?f=E_a%3D120%2C575.61J%3D120.575%20kJ)
120.575 kJ is the activation energy for the souring process.
The number of atoms present in carbon, specifically Carbon-12, is the number of atoms present in 1 gram of the substance. One moles of the substance weighs approximately 12.01 g/mole. The unit suggests that every mole of carbon weighs 12.01 grams.
A. Protons neutrons and electrons.
Haha those three make up a simple Atom.
Answer:
Higher the frequency, the higher the energy
Explanation:
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