Answer:
HCl < CH₃COOH < NH₃ < NaOH
Explanation:
Given compounds:
Acetic acid: CH₃COOH
Ammonia; NH₃
Hydrochloric acid: HCl
Sodium hydroxide: NaOH
All the solutions are of the same molarity which is 0.1M. We need to see how these compounds dissociate to form solutions in order to establish their pH value:
For Acetic acid;
CH₃COOH + H₂O ⇄ H₃O⁺ + CH₃COO⁻
Acetic acid is a weak acid and it ionizes slightly in solutions. It would have a pH close to 7
For Ammonia;
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Ammonia is a weak base and it ionizes slightly in solutions. It sets up an equilibrium in the process. It's would be slightly above 7
For HCl:
HCl + H₂O → H₃O⁺ + Cl⁻
HCl is a strong acid and ionizes completely in solutions. It has a very low pH
For NaOH:
NaOH → Na⁺ + OH⁻
NaOH ionizes also completely in solutions and it breaks down into sodium and hydroxide ions. It is a strong base and it would have a high PH value.
HCl < CH₃COOH < NH₃ < NaOH
This is the trend of increasing pH
The worlds largest tidal range is 16.3 meters and it occurs in Bay of Fundy, Canada
Answer:
NH3(g) + H2O(1) → NH4+(aq) + OH (aq)
HF(aq) + H2O(1) → H3O+(aq) + F (aq)
Explanation:
Acid-base reactions are chemical reactions involving acids and bases. Acids tend to ionize/dissociate in water, a property which determines their strength. Ionization of an acid refers to the acid losing its hydrogen ion (H+) in water solution. An acid ionizes or dissociates to form a conjugate base.
A strong acid is so because it ionizes completely in water i.e. loses all its hydrogen ion (H+) while a weak acid partially ionizes in water.
In the chemical reactions;
1) NH3(g) + H2O(1) → NH4+(aq) + OH (aq)
H20 loses its hydrogen ion (H+) in this reaction to form an anion (OH-). Hence, water (H20) is an acid in this case which ionizes to form a conjugate base (OH-). This is an example of ionization of acid.
2) HF(aq) + H2O(1) → H3O+(aq) + F (aq)
Hydrogen fluoride (HF) loses its hydrogen ion (H+) in the presence of water to form anion (F-). The HF is the acid while F- is it's conjugate base. Thus, an example of ionization of acid
The answer is false, oxygen
Answer:
6
Explanation:
Li has an oxydation number of 1 2*1 = 2
O has an oxidation number of -2 4(-2) = -8
S has an oxydation number of x 1 * x = x
The oxidation number on the molecule is 0.
So here is the equation
2 - 8 + x = 0 Combine like terms
-6 + x = 0 Add 6 to both sides
-6 + 6 + x = 6 Combine like terms
x = 6
Sulphur in this case has an oxidation number of 6
