Answer:
[O₃]= 8.84x10⁻⁷M
Explanation:
<u>The photodissociation of ozone by UV light is given by:</u>
O₃ + hν → O₂ + O (1)
<u>The first-order reaction of the equation (1) is:</u>
![rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20k%20%5BO_%7B3%7D%5D%20%3D%20-%20k%20%5Cfrac%7B%5CDelta%20%5BO_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%20)
(2)
<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>
<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>
![[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt}](https://tex.z-dn.net/?f=%20%5BO_%7B3%7D%5D_%7Bt%7D%20%3D%20%5BO_%7B3%7D%5D_%7B0%7D%20%5Ccdot%20e%5E%7B-kt%7D%20)
(3)
<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>
We can calculate the initial ozone concentration using equation (3):
![[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M](https://tex.z-dn.net/?f=%20%5BO_%7B3%7D%5D_%7Bt%7D%20%3D%205.0%20%5Ccdot%2010%5E%7B-3%7DM%20%5Ccdot%20e%5E%7B-%281.0%5Ccdot%2010%5E%7B-5%7Ds%5E%7B-1%7D%29%20%28%5Cfrac%7B10d%20%5Ccdot%2024h%20%5Ccdot%203600%20s%7D%7B1d%20%5Ccdot%201h%7D%29%7D%20%3D%208.84%20%5Ccdot%2010%5E%7B-7%7DM%20)
So, the ozone concentration after 10 days is 8.84x10⁻⁷M.
I hope it helps you!
They're based on the results of your testing.
False - A producer always provides food for the consumer.
C, that cells cannot be created nor destroyed.
Cells are produce by other cells. Cells can not be man-made nor destroyed.
<u>Answer:</u> The volume of concentrated solution required is 9.95 mL
<u>Explanation:</u>
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
pH = 0.70
Putting values in above equation, we get:
![0.70=-\log[H^+]](https://tex.z-dn.net/?f=0.70%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-0.70}=0.199M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-0.70%7D%3D0.199M)
1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.
Molarity of nitric acid = 0.199 M
To calculate the volume of the concentrated solution, we use the equation:
where,
are the molarity and volume of the concentrated nitric acid solution
are the molarity and volume of diluted nitric acid solution
We are given:
Putting values in above equation, we get:
Hence, the volume of concentrated solution required is 9.95 mL
