Answer:
Products are favored.
Explanation:
The acid-base reaction of CH₃COOH (acid) with NH₃ (base) produce:
CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = ?
It is possible to know Kr of the reaction by the sum of acidic dissociations of the half-reactions. That is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵
NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹
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CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = 1.8x10⁻⁵×1.8x10⁹ = <em>3.2x10⁴</em>
<em> </em>
As Kr is defined as:
Kr = [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]
And Kr is > 1
[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],
showing <em>products are favored</em>.
Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1
Answer: 1, 3, and 4
Explanation: i just did it
Adding an atom will increase the repulsion between existing atoms and lone pairs. Added atom will result in bond pair-bond pair and bond pair-lone pair repulsion. The magnitude of the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion. The added atom will change the electron geometry and bring about a distortion in the molecular geometry.