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>
Answer:
5.8 L
Step-by-step explanation:
This looks like a case where we can use the <em>Combined Gas Law</em> to calculate the temperature.
p₁V₁/T₁ = p₂V₂/T₂ Multiply both sides by T₂
p₁V₁T₂/T₁ = p₂V₂ Divide each side by V₂
V₂ = V₁ × p₁/p₂ × T₂/T₁
=====
<em>Data</em>:
p₁ = 5.6 atm
V₁ = 20 L
T₁ = 35 °C = 308.15 K
p₂ = 23 atm
V₂ = ?
T₂ = 95 °C = 368.15 K
=====
<em>Calculation:
</em>
V₂ = 20 × 5.6/23 × 368.15/308.15
V₂ = 20 × 0.243 × 1.19
V₂ = 5.8 L
Answer:
There will be 16 electrons in O2-
Answer:
a
The expected value is = 39 units
The standard deviation is = 0.1212 unites
b
The probability is = 0.2047 units
Explanation:
The explanation of this answer is shown on the first uploaded image
