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 flow rate of a tube is the volume of fluid flowing through the tube per unit time. The flowrate is proportional to the product of the velocity of the fluid through the tube, and the cross-sectional area of the tube.
That is
Q = AV
where
A is the area of the tube
V is the velocity of the tube
The cross-sectional area of the tube is proportional to the radius of the tube. From the above equation, we can deduce that if the velocity of the fluid flowing through the tube is held constant, the flowrate of the fluid through the tube will increase with an increase in the radius of the tube, and it will decrease with a decrease in the radius of the tube.
Protons:
- Have a mass
- Positively charged
- Found inside the nucleus of an atom
Electrons:
- Have a mass. (9.10938188×10−31 kilograms), though this can sometimes be considered negligible due to how small that actually is. Barely factored into atomic mass
- Negatively charged
- Found outside the nucleus in the electron shell
Neutrons:
- Have a mass
- Neutral (no charge)
- Found inside the nucleus of an atom
Atom A:
- 1 proton
- 0 Neutrons
- 1 electron
- Atomic mass of 1
- Atomic number of 1
Atom B:
- 8 Protons
- 10 Neutrons
- 8 electrons
- Atomic mass of 18
- Atomic number of 8
Atomic mass includes the number of protons and neutrons in the nucleus. Atomic number is the number of protons, as this is what defines what type of element the atom is.
3.0e23 atoms Ne
"E" means 10^
Then we multiply it by a mole of Ne. By the definetion of a mole, it is always 6.022e23 atoms of an element.
So now, we do this:
3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne)
After that, we use molar mass. A mole of Neon is equal, in terms of grams, to its avg. atomic mass. This goes true for any element.
It ends up like this:
3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne) x (20.1797 g Ne / 1 mol Ne)
Now cancel out the "atoms Ne" and "1 mol Ne"
You end up with a grand total of...
*plugs everything into a calculator*
10.05298... g Ne.
We need to round to 2 sig. figs. (3.0) so now it's....
10 g Ne.
Note that this method can only be used for converting atoms of an element to mass in grams.
Source(s):
A periodic table for the atomic mass of neon.
A chemistry textboook
A chemistry class.
