Explanation:
-plants that can absorb copper ions are grown on soil with low grade copper ores.
-the plants are burned and the copper compounds are within the Ash.
-copper ions can be leached from the Ash by adding sulphuric acid, this makes a solution of copper sulphate
-the displacement of scrap iron makes pure copper metals.
Answer:
The concentration of H₃PO₄ will increase.
Explanation:
H₃PO₄(aq) + H₂O(l) ⇄ H₂PO₄⁻(aq) + H₃O⁺(aq)
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
If we add more H₂PO₄⁻, the position of equilibrium will move to the left to get rid of the added H₂PO₄⁻.
The concentration of H₃PO₄ will increase.
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
