B, they belong to group 1
A: Li is period 2 and Na is period 3
C: Na has one more electron shell
D: Not related to how it is spelled
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
Answer:
![[A_t]=54.5\ g](https://tex.z-dn.net/?f=%5BA_t%5D%3D54.5%5C%20g)
Explanation:
Given that:
Half life = 14.0 days
Where, k is rate constant
So,
The rate constant, k = 0.04951 days⁻¹
Initial concentration [A₀] = 60.0 g
Time = 46.7 hrs
Considering, 1 hr = 0.041667 days
So, time = 1.9458 days
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
So,
![[A_t]=60.0\times e^{-0.04951\times 1.9458}\ g](https://tex.z-dn.net/?f=%5BA_t%5D%3D60.0%5Ctimes%20e%5E%7B-0.04951%5Ctimes%201.9458%7D%5C%20g)
The enthalpy change for the formation of hydrogen peroxide from water and oxygen is +98.2 kJ·mol⁻¹.
The negative sign tells you that the reaction <em>releases energy</em>.
To make the reaction go in the reverse direction, you must put the same amount of energy back into the system.
Thus, you simply <em>reverse the sign</em> of the enthalpy change for the forward direction.
