There are many examples of chemical equilibrium all around you. One example is a bottle of fizzy cooldrink. In the bottle there is carbon dioxide (CO2) dissolved in the liquid. There is also CO2 gas in the space between the liquid and the cap
3 Common States of Matter:
1. Solid - particles are motionless and stick together very closely.
2. Liquid - particles are moving slowly without pattern.
3. Gas - Particles are moving rapidly again without pattern.
Answer:
- <u>Yes,</u> <em>all titrations of a strong base with a strong acid have the same pH at the equivalence point.</em>
This <u>pH is 7.</u>
Explanation:
<em>Strong acids</em> and <em>strong bases</em> ionize completely in aqueous solutions. The ionization of strong acids produce hydronium ions, H₃O⁺, and the ionization of strong bases produce hydroxide ions, OH⁻.
Since the ionization of strong acids and bases progress until completion, there is not reverse reaction.
The definition of pH is pH = - log [H₃O⁺]. Acids have low pH (below 7, and greater than 0) and bases have high pH (above 7 and less than 14). Neutral solutions have pH = 7.
Acid-base titrations are a method to determine the concentration of an acid from the known concentration of a base, or the concentraion of a base from the known concentration of an acid.
The<em> equivalence point</em> of the titration is the point at which the the number of moles of hydronium ions and hydroxide ions are equal.
Then, at that point, the hydronium and hydroxide ions will be in the stoichiometric proportion to form a neutral solution, i.e. the pH of the solution wiill be 7.
<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
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)