Answer:
shell and tube type heat exchanger
Explanation:
for evaporation the shell and tube type heat exchanger is best suited.
- in the plate heat exchanger there is gaskets in between every part so this part become weak part in heat echanger and there is possibilities of leakage through this part, there is no such problem in shell and tube type.
- the plate type cant be used when there is high temperature and high pressure drop but shell and tube type can be used
- in evaporation there the liquids change into vapors due to which there is sudden change in pressure and in which plate type is not used because there is chances of leakage
Answer: I am actually studying about Stars, so I got you.
3. As the temperature of a star Increases, it's luminosity increases.
As the temperature of a star decreases, it's luminosity decreases.
4. Hot and Bright. The bigger the star, the hotter it gets is from what I learned.
Answer: 
for the reaction is -186.75 J/K
Explanation:
Change in entropy () for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So = ![[3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol]](https://tex.z-dn.net/?f=%5B3%5Ctimes%2031.8%20J%2FK.mol%5D%2B%5B2%5Ctimes%20188.825%20J%2FK.mol%5D-%5B2%5Ctimes%20205.79%20J%2FK.mol%5D-%5B1%5Ctimes%20248.22%20J%2FK.mol%5D)
= -186.75 J/K
<u>Answer:</u> The pH value of the solution is 10 and the solution is basic in nature.
<u>Explanation:</u>
To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=1\times 10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1%5Ctimes%2010%5E%7B-4%7DM)
Putting values in above equation, we get:
To calculate pH of the solution, we use the equation:
There are three types of solution: acidic, basic and neutral
To determine the type of solution, we look at the pH values.
- The pH range of acidic solution is 0 to 6.9
- The pH range of basic solution is 7.1 to 14
- The pH of neutral solution is 7.
As, the pH of the solution is 10 and is lying in the range of basic solution, so the solution is basic in nature.
