Answer:
7628 y
Explanation:
Carbon-14 is radioactive and it follows the first-order kinetics for a radioactive decay. The first-order kinetics may be described by the following integrated rate law:
![ln(\frac{[A]_t}{[A]_o})=-kt](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%29%3D-kt)
Here:
is the mass, moles, molarity or percentage of the material left at some time of interest t;
is the mass, moles, molarity or percentage of the material initially, we know that initially we expect to have 100 % of carbon-14 before it starts to decay;
is the rate constant;
is time.
The equation becomes:
![ln(\frac{[A]_t}{[A]_o})=-\frac{ln(2)}{T_{\frac{1}{2}}}t](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%29%3D-%5Cfrac%7Bln%282%29%7D%7BT_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7Dt)
Given:
![\frac{[A]_t}{[A]_o} = \frac{40.0 %}{100.0 %}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%20%3D%20%5Cfrac%7B40.0%20%25%7D%7B100.0%20%25%7D)

Solve for time:
![t = -\frac{ln(\frac{[A]_t}{[A]_o})\cdot T_{\frac{1}{2}}}{ln(2)}](https://tex.z-dn.net/?f=t%20%3D%20-%5Cfrac%7Bln%28%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%29%5Ccdot%20T_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Bln%282%29%7D)
In this case:

Actual question from source:-
A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette. A blank had an absorbance of 0.029. The absorbance of an unknown solution of compound A was 0.375. Find the concentration of A in the unknown.
Answer:
Molar absorptivity of compound A = 
Explanation:
According to the Lambert's Beer law:-
Where, A is the absorbance
l is the path length
is the molar absorptivity
c is the concentration.
Given that:-
c = 
Path length = 1.000 cm
Absorbance observed = 0.624
Absorbance blank = 0.029
A = 0.624 - 0.029 = 0.595
So, applying the values in the Lambert Beer's law as shown below:-

<u>Molar absorptivity of compound A =
</u>
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!
236 I DONT KNOW I AM TOTALLY THAT IS NOT THE ANSWER