I believe so, looks like it
Answer:
Depth and location affect ocean water’s temperature.
Explanation:
The main source of heat for the oceans is solar radiation. That is, water is basically heated by the radiation of the Sun, which transmits energy to the surface. The ocean absorbs this energy and stores it. Seawater has high caloric capacity. This means that more energy and more time is needed to change or increase the water temperature, compared to the air temperature. Similarly, once the ocean heats up, it takes a long time for the water to completely release or lose that heat.
The temperature decreases to greater depth, because the amount of solar radiation is reduced. On the contrary, it is greater where there is greater energy or heat content.
The closer a place is to the equator, the solar energy will affect more vertically and with more intensity on it, so the warmer the temperatures will be. The further that point of the equator is found, the solar energy will reach it with a smaller angle. And if the point is near the poles, the sun's rays will arrive at a very small angle. This causes the temperature of the water of the oceans to vary depending on the earth's latitude, being higher in areas close to the equator and the tropics, and colder the closer to the poles or the further away from the temperate zones.
Answer:
Final pH of the solution: 2.79.
Explanation:
What's in the solution after mixing?
,
where
is the concentration of the solute,
is the number of moles of the solute, and
is the volume of the solution.
.
Acetic (ethanoic) acid:
.
.
Hydrochloric acid HCl:
.
.
HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be 
.
The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be 
. 
.
.
![\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K_a%20%3D%20%5Cfrac%7B%5B%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E%7B-%7D%5C%3B%28aq%29%5D%5Ccdot%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5C%3B%28aq%29%5D%7D%7B%5B%5Ctext%7BCH%7D_3%5Ctext%7BCOOH%7D%5C%3B%28aq%29%5D%7D%20%3D%20%5Cfrac%7Bx%5Ccdot%28x%20%2B%205.00%5Ctimes%2010%5E%7B-4%7D%29%7D%7B0.100%20-%20x%7D)
.
Rewrite as a quadratic equation and solve for 
:
.
The pH of a solution depends on its H⁺ concentration.
At equilibrium
![[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5C%3B%28aq%29%5D%20%3D%205.00%5Ctimes%2010%5E%7B-4%7D%5C%3B%5Ctext%7BM%7D%20%2B%20x%5C%3B%5Ctext%7BM%7D%20%3D%200.00161%5C%3B%5Ctext%7BM%7D)
.
![\text{pH} = -\log{[\text{H}^{+}]} = 2.79](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20-%5Clog%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%7D%20%3D%202.79)
.
