That would be the cobalt(11) Chloride Hexahydrate
A chemical reaction (signs)
- rusting
- change in base of chemical
- for example lets say u mix two chemicals, and then it becomes a different new chemical (it changed from the inside)
a physical
- a physical reaction is outer looks not inside.
- it changes on the outside, like changing a color
The given question is incomplete. The complete question is as follows.
The enzyme urease catalyzes the reaction of urea, (), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of at . In the presence of the enzyme in water, the reaction proceeds with a rate constant of at .
If the rate of the catalyzed reaction were the same at as it is at , what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions?
Express your answer using two significant figures.
Explanation:
The reaction equation is as follows.
Urea + Water
Hence, it is given that,
without enzyme: Rate = at
with enzyme: Rate = at
Rate = at
It is known that,
ln \frac{K_{2}}{K_{1}} = \frac{-E_{a}}{R}{\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
and, ln K =
Let us assume that collision factor (A) is same for both the reactions.
Hence,
=
= 63672.8 J/mol
= 63.67 kJ/mol (as 1 kJ = 1000 J)
Thus, we can conclude that the difference in the activation energy between the catalyzed and uncatalyzed reactions is 63.67 kJ/mol.
Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.