Answer:
chlorine is therefore an anion
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>
acquires a positive charge
electrically (neutral)
Balanced chemical reaction: 2S + 3O₂ → 2SO₃.
1) Answer is: oxygen is limiting reactant.
n(S) = 3 mol; amount of sulfur.
n(O₂) = 4 mol; amount of oxygen.
From balanced chemical reaction: n(S) : n(O₂) = 2 : 3.
3 mol : n(O₂) = 2 : 3.
n(O₂) = (3 · 3 mol) ÷ 2.
n(O₂) = 4.5 mol; limiting reactant, because there is only 4 mol of oxygen.
2) Answer is: sulfur(S) is limiting reactant.
n(S) = 3 mol.
n(O₂) = 5 mol.
From balanced chemical reaction: n(S) : n(O₂) = 2 : 3.
n(S) : 5 mol = 2 : 3.
n(S) = 10 mol ÷ 3.
n(S) = 3.33 mol; there is only 3 mol of sulfur, so it is not enough.
3) Answer is: oxygen (O₂) is limiting reactant.
n(S) = 3 mol.
n(O₂) = 3 mol.
From balanced chemical reaction: n(S) : n(O₂) = 2 : 3.
3 mol : n(O₂) = 2 : 3.
n(O₂) = (3 · 3 mol) ÷ 2.
n(O₂) = 4.5 mol; limiting reactant, because there is only 3 mol of oxygen.
Answer:
turgor pressure can be done in a lab or a self test.
turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.
Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.
Explanation:
- salt is bad for turgor pressure.
- Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
- The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.
