Answer:
Acetic acid Ka = 1.74 × 10⁻⁵
Trichloroacetic acid Ka = 2 × 10⁻¹
Explanation:
Let's consider the acid dissociation of acetic acid.
CH₃COOH(aq) ⇄ CH₃COO⁻(aq) + H⁺(aq)
The pKa of acetic acid is 4.76. The acid dissociation constant (Ka) is:
pKa = -log Ka
- pKa = log Ka
Ka = anti log (-pKa)
Ka = anti log (-4.76)
Ka = 1.74 × 10⁻⁵
Let's consider the acid dissociation of trichloroacetic acid.
CCl₃COOH(aq) ⇄ CCl₃COO⁻(aq) + H⁺(aq)
The pKa of trichloroacetic acid is 0.7. The acid dissociation constant (Ka) is:
pKa = -log Ka
- pKa = log Ka
Ka = anti log (-pKa)
Ka = anti log (-0.7)
Ka = 2 × 10⁻¹
Answer:
(c) only Ca2+(aq) and Hg2+(aq)
Explanation:
- In the first step, hydrochloric acid (HCl) is added to the solution. In this case the equilibrium that could take place is:
Ag⁺(aq) + Cl⁻(aq) ↔ AgCl(s)
But no precipitate was formed, so Ag⁺(aq) is absent.
- By adding H₂SO₄(aq) the next equilibrium that could take place is:
Ca⁺²(aq) + SO₄⁻²(aq) ↔ CaSO₄(s)
A white precipitate was formed, so Ca⁺² is present in the solution.
- The following could take place after adding H₂S(aq):
Hg²⁺(aq) + S⁻² ↔ HgS(s)
A black precipitate formed, so Hg⁺² is present as well.
Chicken has the most calories
Answer: 0.0826mol
PV=nRT
n=PV/RT
n=(1atm)(2.1L)/(310K)(0.082057L*atm/mol*K)=0.0826mol
When light passes from one transparent object, this is called refraction
