"<span>A water molecule is formed from the hydrogen and oxygen atoms removed" is what happens in the process described in the question. The correct option among all the options that are given in the question is the second option. I hope the answer has actually come to your great help.</span>
You get a tub of warm water and tip the warm water into the cold water
Answer: The expression for equilibrium constant is ![\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as 
For a general reaction:

The equilibrium constant is written as:
![k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Chemical reaction for the formation of ammonia is:


Expression for
is:
![k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
![1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E2%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Answer:
He will decide which drink is to be served to whom, by the use of litmus paper.
Explanation:
The litmus paper is the most common indicator to determine the acidity or basicity of a solution. Blue litmus paper changes its color to red when a solution changes from basic to acidic while red litmus paper changes its color to blue when the opposite occurs (acid → basic).
First of all the litmus paper strip, pH indicator, is immersed in a solution and allowed to pass between 10 and 15 seconds while keeping the strip submerged. Afterwards it is removed, and then the strip compares the color. If the color is diffuse, there is a color scale where it is determined which solution has alkaline or acidic pH
Molar mass of ( NH₄)₃PO₄ = 14.01×3 + 1.01×12 + 30.97 + 16.00×4 = 149.12 g/mol. Mass of 0.183 mol ...