We take activities of Solids and liquids equal 1. The reason is there concentrations not at all change amount of reactant at equilibrium in the reaction. Hence we don't consider concentrations of pure solids and liquids while writing equilibrium constant.
So for above reaction, equilibrium constant (K) = [Sn^2 +] ^3 x 1 / [Fe^3+]^2 x 1. Ping me if you have any doubts.
This problem is providing us with the chemical equation depicting the production of ammonia from nitrogen and hydrogen at equilibrium and asks for the correct change when the concentration of nitrogen is increased. At the end, the answer is the forward reaction would increase to start reducing the concentration of N2.
<h3>Chemical equilibrium</h3>
In chemistry, chemical reactions not always reach a 100-% conversion when reactants get in contact in order to carry out the chemical reaction. Thus, there is a point wherein the concentrations remain the same and is called equilibrium.
In such a way, for this problem, we have the following chemical reaction at equilibrium:
Now, according to the Le Ch.atelier's principle, an increase in the concentration of any species, shifts the equilibrium away from it, which means that if we increase the concentration of nitrogen, a reactant, the forward reaction will be favored.
Thereby, the correct answer is "the forward reaction would increase to start reducing the concentration of N2".
Learn more about chemical equilibrium: brainly.com/question/26453983
Answer:
1.Handpicking,winnowing and sieving 2. distillation 3.distillation 5. winnowing 6.magnet
Well, percent ionization, simply allows you to find X which is the change in amount of substance into the final amount, it is dependent on the initial concentration of the substance present that is being dissociated.
For example, if the percent dissociation of CH3COOH is 1.5 % for an initial concentration of 0.15, then to solve for x, you would need to write percent as a decimal and solve.
% dissociation = X/Concentration of substance dissociated
0.015 = X/0.15
X = 0.0025.
Now since you know X, use the ice tables to solve for concentrations of species at equilibrium and plug in Kc expression.
FeO is Iron II Oxide. You identify that it's an ionic bond so it won't have any prefixes. By reverse criss-crossing the charges, Iron gets a 1+ charge, and Oxygen gets a 1- charge. Since you know that Oxygen has to have a 2- charge, you multiply the 1+ and 1- by 2 to fix their charges, thus giving Iron a 2+ charge and Oxygen a 2- charge.
Since Iron has a 2+ charge, it's roman numeral is II. So the answer is Iron II Oxide.
The other one is Iron III Oxide. By reverse criss-crossing the charges, Fe geta 3+ charge, and Oxygen gets a 2- charge. Since it's correct, the 3+ charge is the roman numeral. So, the answer is Iron III Oxide.
