Answer:
HD video is the best way too much for a TV now
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:
Equilibrium concentrations of the gases are
![H_2S=0.596M](https://tex.z-dn.net/?f=H_2S%3D0.596M)
![H_2=0.004 M](https://tex.z-dn.net/?f=H_2%3D0.004%20M)
![S_2=0.002 M](https://tex.z-dn.net/?f=S_2%3D0.002%20M)
Explanation:
We are given that for the equilibrium
![2H_2S\rightleftharpoons 2H_2(g)+S_2(g)](https://tex.z-dn.net/?f=2H_2S%5Crightleftharpoons%202H_2%28g%29%2BS_2%28g%29)
Temperature, ![T=700^{\circ}C](https://tex.z-dn.net/?f=T%3D700%5E%7B%5Ccirc%7DC)
Initial concentration of
![H_2S=0.30M](https://tex.z-dn.net/?f=H_2S%3D0.30M)
![H_2=0.30 M](https://tex.z-dn.net/?f=H_2%3D0.30%20M)
![S_2=0.150 M](https://tex.z-dn.net/?f=S_2%3D0.150%20M)
We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of
![H_2S=0.30+2x](https://tex.z-dn.net/?f=H_2S%3D0.30%2B2x)
![H_2=0.30-2x](https://tex.z-dn.net/?f=H_2%3D0.30-2x)
![S_2=0.150-x](https://tex.z-dn.net/?f=S_2%3D0.150-x)
At equilibrium
Equilibrium constant
![K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bproduct%7D%7BReactant%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Substitute the values
![9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}](https://tex.z-dn.net/?f=9%5Ctimes%2010%5E%7B-8%7D%3D%5Cfrac%7B%280.30-2x%29%5E2%280.150-x%29%7D%7B%280.30%2B2x%29%5E2%7D)
![9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}](https://tex.z-dn.net/?f=9%5Ctimes%2010%5E%7B-8%7D%3D%5Cfrac%7B%280.30-2x%29%5E2%280.150-x%29%7D%7B%280.30%2B2x%29%5E2%7D)
![9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}](https://tex.z-dn.net/?f=9%5Ctimes%2010%5E%7B-8%7D%3D%5Cfrac%7B%280.30-2x%29%5E2%280.150-x%29%7D%7B%280.30%2B2x%29%5E2%7D)
By solving we get
![x\approx 0.148](https://tex.z-dn.net/?f=x%5Capprox%200.148)
Now, equilibrium concentration of gases
![H_2S=0.30+2(0.148)=0.596M](https://tex.z-dn.net/?f=H_2S%3D0.30%2B2%280.148%29%3D0.596M)
![H_2=0.30-2(0.148)=0.004 M](https://tex.z-dn.net/?f=H_2%3D0.30-2%280.148%29%3D0.004%20M)
![S_2=0.150-0.148=0.002 M](https://tex.z-dn.net/?f=S_2%3D0.150-0.148%3D0.002%20M)
Answer:
because of catenation of carbon.
Explanation:
Catenation is the binding of an element to its self through covalent bonds to form chain or ring molecules. carbon is able to form continuous links with other carbon atoms which is the reason for the existence of a large number of organic compounds.
Reactivity of non-metals depend on their ability to gain electrons. So, smaller is the size of a non-metal more readily it will attract electrons because then nucleus will be more closer to valence shell. ... Hence, Br is the non-metal which will be more reactive than At.