The answer to this question would be: 2.36 mol
To answer this question, you need to know the molecular weight of copper. Molecular weight determines how much the weight of 1 mol of a molecule has. Copper molecular weight about 63.5g/mol. Then, the amount of mol in 150g copper should be: 150g / (63.5g/mol)= 2.36 mol
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.
Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:
54.3 mm = 5.43 cm
Now, we determine the number of gold atoms that will be present in this:
5.43 / 1 x 10⁻⁹
There will be 5.43 x 10⁹ atoms
We now determine the number of moles this is by:
one mole = 6.02 x 10²³ atoms
Moles = 5.43 x 10⁹ / 6.02 x 10²³
Moles = 9.01 x 10⁻¹⁵ moles
The molar mass of gold is 197 g/mol
The mass is 9.01 x 10⁻¹⁵ * 197
The mass of the strand is 1.76 x 10⁻¹² grams
I don't exactly know but probably so.
