I believe it’s A..but I’m not quite sure.
a metal with a large number of free-flowing electrons most likely have good conductivity
Answer:
T = 215.33 °C
Explanation:
The activation energy is given by the Arrhenius equation:

<u>Where:</u>
k: is the rate constant
A: is the frequency factor
Ea: is the activation energy
R: is the gas constant = 8.314 J/(K*mol)
T: is the temperature
We have for the uncatalyzed reaction:
Ea₁ = 70 kJ/mol
And for the catalyzed reaction:
Ea₂ = 42 kJ/mol
T₂ = 20 °C = 293 K
The frequency factor A is constant and the initial concentrations are the same.
Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

(1)
By solving equation (1) for T₁ we have:
Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.
I hope it helps you!
The rate law equation for Ozone reaction
r=k[O][O₂]
<h3>Further e
xplanation</h3>
Given
Reaction of Ozone :.
O(g) + O2(g) → O3(g)
Required
the rate law equation
Solution
The rate law is a chemical equation that shows the relationship between reaction rate and the concentration / pressure of the reactants
For reaction
aA + bB ⇒ C + D
The rate law can be formulated:
![\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cboxed%7B%5Cbold%7Br~%3D~k.%5BA%5D%5Ea%5BB%5D%5Eb%7D%7D%7D)
where
r = reaction rate, M / s
k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹
a = reaction order to A
b = reaction order to B
[A] = [B] = concentration of substances
So for Ozone reaction, the rate law (first orde for both O and O₂) :
![\tt \boxed{\bold{r=k[O][O_2]}}](https://tex.z-dn.net/?f=%5Ctt%20%5Cboxed%7B%5Cbold%7Br%3Dk%5BO%5D%5BO_2%5D%7D%7D)