Answer:
2KOH(aq) + NiSO₂(aq) → K₂SO₄(aq) + NiOH₂(s)
Explanation:
This reaction is an example of a <em>double-replacement reaction </em>where the cations of two compounds exchange with its anions. In the reaction:
KOH(aq) + NiSO₄(aq)
There are produced K₂SO₄ and NiOH₂ salts (The last one is insoluble, its state is (s) but K₂SO₄ is very soluble, its state is (aq). The unbalanced reaction is:
KOH(aq) + NiSO₄(aq) → K₂SO₄(aq) + NiOH₂(s)
To balance the potassiums:
<h3>
2KOH(aq) + NiSO₂(aq) → K₂SO₄(aq) + NiOH₂(s)</h3>
And now, the reaction is balanced
Answer:
k = 2,04x10⁻⁵
Explanation:
The equilibrium of acetic acid (CH₃COOH) in water is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺.
And the equilibrium constant is defined as:
k = [CH₃COO⁻] [H⁺] / [CH₃COOH] <em>(1)</em>
The equiibrium concentration of each specie if the solution of acetic acid is 0,05M is:
[CH₃COOH] = 0,05M - x
[CH₃COO⁻] = x
[H⁺] = x
<em>-Where x is the degree of reaction progress-</em>
As the pH is 3, [H⁺] = 1x10⁻³M. That means x = 1x10⁻³M
Replacing in (1):
k = (1x10⁻³)² / 0,05 - 1x10⁻³
k = 1x10⁻⁶ / 0,049
<em>k = 2,04x10⁻⁵</em>
<em></em>
I hope it helps!
Answer:
2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.
Explanation:
∵ pH = - log[H₃O⁺]
∴ 4.6 = - log[H₃O⁺].
∴ log[H₃O⁺] = - 4.6.
<em>∴ [H₃O⁺] = 2.51 x 10⁻⁵.</em>
<em></em>
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
[H₃O⁺] = 2.51 x 10⁻⁵ M.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] </em>= 10⁻¹⁴/(2.51 x 10⁻⁵ M) = <em>3.98 × 10⁻¹⁰ M ≅ 4.0 × 10⁻¹⁰ M.</em>
<em>So, the right choice is: 2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.</em>
Answer:
PBr5 is the formula of phosphorus pentabromide.
hope it helps!