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:
will form bonds between atoms. for more details are in the pic
Answer:
30 moles
Explanation:
Water is H2O, meaning there is 2 Hydrogen atoms and 1 Oxygen atom. Oxygen is O2, because it is a diatomic molecule. (Hydrogen is also a diatomic molecule, so H2)
The equation, balanced, would have to be: 2H2 + O2 -----> 2H2O
I multiply 15 moles O2 by the molar ratio of (hydrogen/oxygen)
15 mol. O2 * (2 mol. H2/1 mol O2) = 30 moles of water