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
Compare them accordingly with the formula of E=hc/wavelength
Answer:
0.06mole
Explanation:
Given parameters:
Volume of balloon = 2.5L
T = 343.5K
P = 66.7kPa; in atm gives = 0.66atm
Unknown:
Number of moles of He = ?
Solution:
To solve this problem, we use the ideal gas equation.
It is mathematically expressed as;
Pv = nRT
where p is the pressure
v is the volume
n is the number of moles
R is the gas constant
T is the temperature
0.66 x 2.5 = n x 0.082 x 343.5
n = 0.06mole
