Answer:
The rate of the reaction increased by a factor of 1012.32
Explanation:
Applying Arrhenius equation
ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)
where;
k₂/k₁ is the ratio of the rates which is the factor
Ea is the activation energy = 274 kJ/mol.
T₁ is the initial temperature = 231⁰C = 504 k
T₂ is the final temperature = 293⁰C = 566 k
R is gas constant = 8.314 J/Kmol
Substituting this values into the equation above;
ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)
ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)
ln(k₂/k₁) = 6.92
k₂/k₁ = exp(6.92)
k₂/k₁ = 1012.32
The rate of the reaction increased by 1012.32
Mass 1 + %abundance of first isotope + Mass 2 + %abundance of second isotope
/ 100
This is RAM.
Answer:
i am sorry i can not help with that there isnt enough information to work with if you can tell me what you think about the topic i can help you put together a doc
Explanation:
Answer:
Because they are extremely stable molecules, CFCs do not react easily with other chemicals in the lower atmosphere. ... Free chlorine atoms then react with ozone molecules, taking one oxygen atom to form chlorine monoxide and leaving an ordinary oxygen molecule.
Explanation:
The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.
The mass percent of lithium hydroxide can be calculated with the following equation:
(1)
Where:
(2)
We need to find the mass of LiOH.
From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.
Since mol = m/M, where M: is the molar mass and m is the mass, we have:
(3)
Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:
Solving for 
, we have:
Hence, the percent lithium hydroxide is (eq 1):
Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.
Learn more about mass percent here:
I hope it helps you!
