This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.
<h3>pH calculations</h3>
In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:

Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:

Which conserves as much decimals as significant figures in the molarity.
Learn more about pH calculations: brainly.com/question/1195974
The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at approximately 273.1575 K (0.0075 °C; 32.0135 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).[4][5] At that point, it is possible to change all of the substance to ice, water, or vapor by making arbitrarily small changes in pressure and temperature. And the temperature is called Triple point
<u>Answer:</u>
When the temperature increases, the molecules of the gas gain energy. Therefore, they move faster.
This causes the molecules to hit the walls of the container more frequently and with greater force. Hence the pressure inside the container increases.
Explanation:
The given data is as follows.
Volume of lake =
= 
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake = 
=
mg
=
kg
Flow rate of river is 50 
Volume of water in 1 day = 
=
liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are
or 
Flow rate of sewage = 
Volume of sewage water in 1 day =
liter
Concentration of sewage = 300 mg/L
Total amount of pollutants =
or 
Therefore, total concentration of lake after 1 day = 
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence,
= 
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.