Temperature decreases (?)
Answer:
A/1. 10.9 mol O2
Explanation:
583 g x 1 mol SO3 x 3 mol O2 /
80.057 g mol SO3 x 2 mol SO3
- You just need to find molar mass of SO3, which is 80.057 g.
- Everything else came from formula. Further explanation...
- Always start with what they give, such as 583 g. Then find 1 mol of what is being produced, in this it is SO3. We already found this because we did molar mass above. Next. find how many moles of what they want, which is O2. Look in equation and you can see 3 mol in from of O2. Next, do the same for SO3 and you can find 3 mol in front of that. Lastly, just do the math.
- If you need a further explanation or more help on any problems I would be happy to help, just let me know.
Answer:
The temperature to the nearest 0.5°C is 98.5°C
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
