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)
= Initial concentration of reactant
= 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.
Answer:
6.67 moles
Explanation:
Given that:-
Moles of hydrogen gas produced = 10.0 moles
According the reaction shown below:-

3 moles of hydrogen gas are produced when 2 moles of aluminium undergoes reaction.
Also,
1 mole of hydrogen gas are produced when
moles of aluminium undergoes reaction.
So,
10.0 moles of hydrogen gas are produced when
moles of aluminium undergoes reaction.
<u>Moles of Al needed =
moles = 6.67 moles</u>
The replacing of sodium hydroxide with potassium hydroxide
(KOH) to the reaction will least affect the organic product that forms.
Potassium hydroxide is an
inorganic compound with the formula KOH, and is commonly called caustic potash. Along with sodium hydroxide, this colorless
solid is a prototypical strong base.