The molarity of the resulting solution obtained by diluting the stock solution is 3 M
<h3>Data obtained from the question </h3>
- Molarity of stock solution (M₁) = 15 M
- Volume of stock solution (V₁) = 500 mL
- Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL
- Molarity of diluted solution (M₂) =?
<h3>How to determine the molarity of diluted solution </h3>
M₁V₁ = M₂V₂
15 × 500 = M₂ × 2500
7500 = M₂ × 2500
Divide both side by 2500
M₂ = 7500 / 2500
M₂ = 3 M
Thus, the volume of the resulting solution is 3 M
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Answer:
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Explanation:
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Answer:
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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