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 :
= 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 =
t = 22.0 hour
[A] =?
The concentration of cyclopropane after 22.0 hour is 0.0457 M.
There are:
3.41 moles of C
4.54 moles of H
3.40 moles of O.
Why?
To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.
Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.
We have that:
To know the percent of each element, we need to to the following:
So, we know that for the 100 grams of the compound, there are:
40.92 grams of C
4.58 grams of H
54.49 grams of O
We know the molecular masses of each element:
Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:
Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.
Have a nice day!