Answer:
The concentration of cyclopropane after 22.0 hour is 0.0457 M.
Explanation:
Conversion of cyclopropane into propene follows first order kinetics.
The integrated rate of first order kinetic is given by :
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= Initial concentration of reactant
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= final concentration of reactant after time t
k = rate constant of the reaction
We have :
Rate constant of the reaction = k = 
![[A_o]=0.150 M](https://tex.z-dn.net/?f=%5BA_o%5D%3D0.150%20M)
t = 22.0 hour
[A] =?
![[A]=0.150 M\times e^{-5.4\times 10^{-2} hour^{-1}\times 22.hour}](https://tex.z-dn.net/?f=%5BA%5D%3D0.150%20M%5Ctimes%20e%5E%7B-5.4%5Ctimes%2010%5E%7B-2%7D%20hour%5E%7B-1%7D%5Ctimes%2022.hour%7D)
![[A]=0.0457 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0457%20M)
The concentration of cyclopropane after 22.0 hour is 0.0457 M.
<span>[H3O+] = 10^(-pH) = 10^(-4.20) = 6.3 x 10^-5 M
pOH = 14 - pH = 14 - 4.20 = 9.80
[OH-] = 10^(-pOH) = 10^(-9.80) = 1.6 x 10^-10 M
</span>
Q = M * C *ΔT
Q / <span>ΔT = M
</span>Δf - Δi = 98.4ºC - 62.2ºC = 36.2ºC
<span>
C = 1137 J / 140 * 36.2
C = 1137 / 5068
C = 0.224 J/gºC</span>
<span>For this reaction, oxidation number of Carbon in
CO would be +2 while oxidation number of carbon in CO2 would be +4 and so this
means that carbon has oxidized. Oxidation number of nitrogen in NO is +2. While
oxidation number of nitrogen in N2 is 0 so this means that nitrogen had reduced.
The reducing agent is the one which provides electrons by oxidizing itself so
in this case; CO is the reducing agent while the C in CO oxidized to produce
electrons. </span><span>I
am hoping that this answer has satisfied your query about and it will be able
to help you, and if you’d like, feel free to ask another question.</span>
Answer:
Melting butter
Explanation:
You can reverse the change of butter back to its original state but you can never reverse the rest back to there original state
