Evaporation condensation transpiration precipitation
Emily, come to papa
<span>Mixing an acid and a base results in neutralization, but the results are potentially dangerous. No matter which acid or base is used, the resulting solution is water and varying types of salt. The process of neutralization often involves the substances heating up when they come together. If the solution heats up too much or too fast, a violent explosion or the creation of harmful or flammable gases is a possibility. This occurs when the chemicals are mixed too quickly, the acid and base are too strong or if there is no available salt to be made in the solution.</span>
Number of moles = volume / (molar volume)
Molar volume at stp = 22.4 dm^3
Volume = no of moles × molar volume
= 0.987 × 22.4
= 22.1088 dm^3
= 22108.8 cm^3
Hope it helped!
(35/18.02)(6.02kJ)=11.6926 kiloJoules
Answer:
Equilibrium concentrations of the gases are



Explanation:
We are given that for the equilibrium

Temperature, 
Initial concentration of



We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of



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



By solving we get

Now, equilibrium concentration of gases


