Answer:
1.1 percent
Explanation:
C6H5COOH⇌C6H5COO+H
Ka = [C6H5COO-][H+]/[C6H5COOH]
From pKa of 4.2, Ka = 6.3x10^-5
6.3x10^-5 = (x)(x)/0.51-x and assuming x is small...
6.3x10^-5 = x^2/0.51
x^2 = 3.213x10^-5
x =5.67 x10^-3
(5.67x10^-3/0.51)x 100 =1.1 percent
The sun also heats up the air in the atmosphere at the equator. This air will move towards the poles & cool over time. In the process, wind that occurs due to air currents will introduce currents at the surface of the ocean. Wind & Air
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
<em>Where X is reaction coordinate</em>
<em />
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
<h3>[Cu²⁺] = 2.01x10⁻²⁶</h3>
<span>The theoretical yield for a reaction is calculated based on the limiting reagent. This allows researchers to determine how much product can actually be formed based on the reagents present at the beginning of the reaction.</span>
<span>The actual yield will never be 100 percent due to limitations.</span>
