Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
The answer to the problem is 4.5 kilometers. you can solve this problem by cross multiplying
The answer is na sodium, bromine
Answer:
Explanation:
Given: Entropy of surrounding: ΔSsurr = ?
Temperature: T= 355 K
The change in enthalpy of reaction: ΔH = -114 kJ
Pressure: P = constant
As we know, ΔH = -114 kJ ⇒ negative
Therefore, the given reaction is an exothermic reaction
Therefore, Entropy of surrounding at <em>constant pressure</em> is given by,
<u><em>In the given reaction:</em></u>
2NO(g) + O₂(g) → 2NO₂(g)
As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.
As we know, <em>for a spontaneous process, that the total entropy should be positive.</em>
<u>Therefore, at the given temperature,</u>
- if
then the given reaction is spontaneous
- if
then the given reaction is non-spontaneous
