8) Which one of the following compounds behaves as an acid when dissolved in wat

Question

8) Which one of the following compounds behaves as an acid when dissolved in water? 8) A) C4H10 B) RaO C) HI D) RbOH 9) What is the molarity of a potassium triodide solution, KI3(ag), if 30.00 mL of the solution is 9) required to completely react with 25.00 mL of a 0.200 M thiosulfate solution, Na25203(a)? The chemical equation for the reaction is 2 S2032- (ag)+13-(aa) - S4062 (ag)+3 I-(ag). A) 0.120 M B) 0.167 M C) 0.333 M D) 0.0833 M 10) An aqueous solution of H2S is named 10 A) hydrosulfurous acid. C) sulfuric acid. B) sulfurous acid. D) hydrosulfuric acid 11) Based on the balanced chemical equation shown below, what volume of 0.250 M K2S2O3(ag), is needed to completely react with 24.88 mL. of 0.125 M KI3(ag),according to the chemical equation: 11) 2 S2032 (ag)+ 13-(aa) B) 24.9 mL S4062(ag)+3 I-Caq). C) 99.5 mL A) 6.22 mL D) 12.4 mL

Explanation / Answer

8. C) HI (Hydroiodic acid)

9. No. of moles of thiosulfate = 0.200 mol/L x 0.025 L = 0.005 mol

From the equation 2 moles thiosulfate neutralizes 1 mol triiodide. So 0.005 mol thiosulfate neutralizes 0.005/2 = 0.0025 mol triiodide.

Molarity of triiodide = 0.0025 mol / 0.03 L = 0.0833 M (D)

10.D) Hydrosulfuric acid

11. No. of moles of potassium triiodide = 0.125 mol/L x 0.02488 L = 0.00311 mol

From the equation, 2 moles of potassium thiosulfate (K2S2O3) reacts with 1 mol of triiodide.

So 0.00311 mol triiodide needs 0.00311 x 2 = 0.00622 mol thiosulfate

Volume of thiosulfate needed = 0.00622 mol / 0.250 mol lit-1

= 0.02488 L

= 24.88 mL (D)

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.