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!
Answer:
The answer would be B.
Explanation:
Ocean water near areas with low evaporation has higher salinity.
if im wrong please tell me .__.
<u>Answer:</u> The correct answer is Option A.
<u>Explanation:</u>
Mole ratio is defined as the ratio between the stoichiometric coefficients of the molecules present in the chemical reaction.
For the given balanced chemical equation:
By Stoichiometry of the reaction:
3 moles of iron metal reacts with 4 moles of water to produce 1 mole of iron oxide and 4 moles of hydrogen gas.
The mole ratio of 
Hence, the correct answer is Option A.
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
