Add 7 water atom to the right hand side to adjust the quantity of oxygen. Increase Cr(+3) by two to adjust the quantity of Cr. Duplicate Cl-by two to adjust the quantity of chlorine molecules.
Cr2O7[2-](aq) +2 Cl[-](aq) < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
Presently adjust that charges.
you have - 4 charges on the left hand side, while +18 charges on the right hand side, there for include 14H+ the left hand side to adjust the charges
Cr2O7[2-](aq) +2 Cl[-](aq)+14H+ < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
take note of that the oxidation number of hydrogen in water is +1
Answer:
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Explanation:
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
