Answer:
Antifreeze is whats used to keep your engine cool without freezing.
Explanation:
it keeps the engine from overheating.
It also prevents corrosion.
Here is a quote from google "Antifreeze works because the freezing and boiling points of liquids are “colligative” properties. This means they depend on the concentrations of “solutes,” or dissolved substances, in the solution. A pure solution freezes because the lower temperatures cause the molecules to slow down"
That quote is from "The Science Behind Antifreeze"
If you have any questions feel free to ask in the comments.
Isotopes are substances that have the same number of protons but differ in the number of neutrons. Hence, the pair of isotopes above should be of the same element. In the given choices, 14C is not an isotope of 14N. 206Pb is an isotope of 208 Pb. O2 and O3 differ in molecular formula but still made up of same kind of atom, hence they are allotropes, while 32S and 32S2- are not isotopes.
The transfer of energy that occurs when a force is applied over a distance is WORK.
Hope this helps!
<em>Hello there, and thank you for asking your question here on brainly.
The answer to this is Answer choice C: CaCl2
</em>Hope this helped you! ♥<em>
</em>
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!
