Practical work refers to the art of conducting experiments in order to answer certain research questions.
<h3>What is practical work?</h3>
In science, practical work refers to the art of conducting experiments in order to answer certain research questions. This could occur in a laboratory under controlled conditions or in the field.
In the physical sciences, most of the practical work is conducted in the laboratory under controlled conditions. However, some experiments in the biological sciences and most experiments in the social sciences are conducted outside the laboratory.
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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:
a. pH = 2 b. pH = 3 c. pH = 1 d. Unanswerable
Explanation:
pH = -log[H+] OR pH = -log{H3O+]
and inversely
pOH = -log[OH-]
1. Determine what substance you are working with, (acid/base)
2. Determine whether or not that acid or base is strong or weak.
a. 1.0 x 10^-2M HCl
HCl is a strong acid, therefore it will dissociate completely into H+ and Cl- with all ions going to the H+, therefore, the concentration of HCl and concentration of H+ are going to be equal, meaning we simply take the negative logarithm of the concentration of HCl and that would equal pH
pH = -log[H+]
pH = -log(1.0x10^-2)
pH = 2
b. 1.0 x 10^-3M HNO3
HNO3 like part a, is a strong acid, therefore it would simply require you to take the negative logarithm of the concentration of the compound itself, to find its pH.
pH = -log[H+]
pH = -log(1.0 x 10^-3)
pH = 3
c. 1.0 x 10^-1M HI
Like the previous parts, HI is a strong acid
pH = -log[H+]
pH = -log(0.10)
pH = 1
d. HB isn't an element, nor is it a compound so that would be unanswerable.
Answer:
Explanation:
Using the necessary reagents to faciliate the synthesis of the organic compounds as shown in the attached file.
