Answer:
% yield = 73.48 %
Explanation:
The fermentation reaction is:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
The percent yield of C₂H₅OH is given by:
<em>where
: is the obtained mass of C₂H₅OH = 67.7g and
: is the theoretical mass of C₂H₅OH. </em>
The theoretical mass of C₂H₅OH is calculated knowing that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH:
<em>where M: is the molar mass of C₂H₅OH = 46.068 g/mol</em>
Hence, the percent yield of C₂H₅OH is:

I hope it helps you!
A. potential energy.
It can't be any of the above questions. When the automobile burns gas it converts the potential energy into chemical energy which intern gets converted into mechanical energy and then into kinetic energy.
I would have thought it would be Chemical potential energy, which is referring to the energy between the bonds in the molecule.
Hoped this helped.
The balanced chemical reaction is expressed as:
M + F2 = MF2
To determine the moles of the element fluorine present in the product, we need to determine the moles of the product formed from the reaction and relate this value to the ratio of the elements in MF2. We do as follows:
moles MF2 produced = 0.600 mol M ( 1 mol MF2 / 1 mol M ) = 0.600 mol MF2
molar mass MF2 = 46.8 g MF2 / 0.6 mol MF2 = 78 g/mol
moles MF2 = 46.8 g ( 1 mol / 78 g ) = 0.6 mol
moles F = 0.6 mol MF2 ( 2 mol F / 1 mol MF2 ) = 1.2 moles F
Answer:
0.124 M
Explanation:
The reaction obeys second-order kinetics:
![r = k[BrO^-]^2](https://tex.z-dn.net/?f=r%20%3D%20k%5BBrO%5E-%5D%5E2)
According to the integrated second-order rate law, we may rewrite the rate law in terms of:
![\dfrac{1}{[BrO^-]_t} = kt + \dfrac{1}{[BrO^-]_o}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_t%7D%20%3D%20kt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D)
Here:
is a rate constant,
is the molarity of the reactant at time t,
is the initial molarity of the reactant.
Converting the time into seconds (since the rate constant has seconds in its units), we obtain:

Rearranging the integrated equation for the amount at time t:
![[BrO^-]_t = \dfrac{1}{kt + \dfrac{1}{[BrO^-]_o}}](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7Bkt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D%7D)
We may now substitute the data:
![[BrO^-]_t = \dfrac{1}{0.056 M^{-1}s^{-1}\cdot 60.0 s + \dfrac{1}{0.212 M}} = 0.124 M](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7B0.056%20M%5E%7B-1%7Ds%5E%7B-1%7D%5Ccdot%2060.0%20s%20%2B%20%5Cdfrac%7B1%7D%7B0.212%20M%7D%7D%20%3D%200.124%20M)