Answer:
Ksp = 1.07x10⁻²¹
Explanation:
Molar solubility is defined as moles of solute can be dissolved in 1L.
Ksp for NiS is defined as:
NiS(s) ⇄ Ni²⁺(aq) + S²⁻(aq)
Ksp = [Ni²⁺] [S²⁻]
As molar solubility is 3.27x10⁻¹¹M, concentration of [Ni²⁺] and [S²⁻] is 3.27x10⁻¹¹M for both.
Replacing:
Ksp = [3.27x10⁻¹¹M] [3.27x10⁻¹¹M]
<em>Ksp = 1.07x10⁻²¹</em>
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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
Some examples of malleable materials are gold, silver, iron, aluminum, copper and tin.
Hydrogen sulfide = hidrogen + sulfur
6.500 g
a) 0.384 g + x
=> 6.500 = 0.384 + x => x = 6.500 - 0.384 = 6.116 g
Answer: 6.116 g of sulfur must be obtained
b) this experiment demonstrate the conservation of mass.
c) Dalton's atomic model states that the atoms cannot be created, split or be destroyed, and so in a chemical reaction the atoms rearrange but the number of each type of atoms remain constant, so the mass of each type of atoms and the total mass remain constant.
