Answer: The activation energy Ea for this reaction is 22689.8 J/mol
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
= rate constant at temperature 
= 
= rate constant at temperature 
= 
= activation energy = ?
R= gas constant = 8.314 J/kmol
= temperature = 
= temperature = 
Putting in the values ::
![ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7B4.8%5Ctimes%2010%5E8%7D%7B2.3%5Ctimes%2010%5E8%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B649%7D%20-%20%5Cfrac%7B1%7D%7B553%7D%5D)
The activation energy Ea for this reaction is 22689.8 J/mol
Answer:
Approximately 
.
Explanation:
Make use of the molar mass data (
) to calculate the number of moles of molecules in that 
of 
:
.
Make sure that the equation for this reaction is balanced.
Coefficient of in this equation: 
.
Coefficient of 
in this equation: .
In other words, for every two moles of that this reaction consumes, two moles of would be produced.
Equivalently, for every mole of that this reaction consumes, one mole of would be produced.
Hence the ratio: 
.
Apply this ratio to find the number of moles of that this reaction would have produced:
.
Answer:
A. 20 grams of milk at 10°C
Explanation:
Since we refrigerate milk, it would be cooler than the room temperature, which standard norm is 25°C. So the milk has to be colder than the room temperature. Therefore, our answer is A.
Conversion process of non-organic carbon dioxide into organic compounds by living organisms.
Example: Photosynthesis
You must know the concentration of the acetic acid. Suppose the concentration is 0.1 M. The solution is as follows:
CH₃COOH → CH₃COO⁻ + H⁺
I 0.1 0 0
C -x +x +x
E 0.1 - x x x
Ka = (x)(x)/(0.1 - x)
1.8×10⁻⁵ = x²/(0.1 - x)
Solving for x,
x = 1.333×10⁻³ = H⁺
pH = -log[H⁺] = -log(1.333×10⁻³)
pH = 2.88
