Answer:
mining of clay limestone and then heated to a certain temperature of 1450⁰ in a cement kiln
Answer:
HOAc is stronger acid than HClO
ClO⁻ is stronger conjugate base than OAc⁻
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Explanation:
Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵
=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺
Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸
=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺
HOAc delivers more H⁺ than HClO and is more acidic.
Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Answer:
The pKa of the conjugate acid is 17.7
Explanation:
If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,
Kₐ = [OH⁻]/[H₂O]
Now, we determine the equivalent pKa
pKa = -log[ka]
pKa = -log[100]
pKa = -2
Removal of hydrogen from water is reversible as shown below;
H₂O ⇄ OH⁻ + H⁺
15.7 -2
This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];
pKa of the conjugate acid = 15.7 - (-2) = 17.7
The pKa of the conjugate acid is 17.7
Answer:
the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.
Explanation:
V = 75 mL = 0,075 L = 0,075 dm³
C = 2.1M
n = ?
---------------
C = n/V
n = C×V
n = 2.1×0,075
n = 0,1575 mol
--------
mKCl: 39+35.5 = 74,5 g/mol
74,5g --------- 1 mol
Xg ------------- 0,1575 mol
X = 74,5×0,1575
X = 11,73375g KCl
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