Suppose 35.7 g of iron(II) iodide is dissolved in 350. mL of a 0.40 M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) iodide is dissolved in it. Round your answer to 1 significant digit. M O 10 ? X $

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**Problem Statement:**

*Suppose 35.7 g of iron(II) iodide is dissolved in 350. mL of a 0.40 M aqueous solution of silver nitrate.*

*Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn’t change when the iron(II) iodide is dissolved in it.*

*Round your answer to 1 significant digit.*

**Solution Explanation:**

To solve for the final molarity of the iodide anion, follow these steps:

1. **Determine the moles of iron(II) iodide (FeI₂):**
   - Molar mass of FeI₂ = Fe (55.85 g/mol) + 2 * I (126.90 g/mol) = 309.65 g/mol
   - Moles of FeI₂ = mass (35.7 g) / molar mass (309.65 g/mol)
  
2. **Calculate the moles of iodide ions (I⁻) produced:**
   - Each mole of FeI₂ produces 2 moles of I⁻.
   - Moles of I⁻ = moles of FeI₂ * 2

3. **Assumption and Conversion:**
   - The total volume of the solution remains 350 mL (0.350 L).

4. **Determine the final molarity of the iodide ions:**
   - Final molarity (M) = moles of I⁻ / total volume in liters (L)

**Diagram or Graph Explanation:**

This problem does not include any diagrams or graphs. It only involves numerical calculations to determine the molarity of iodide anions in a given solution.

**Interactive Tools:**

On the user interface, several interactive buttons are visible:
- A box to input the answer with units for molarity (M).
- Buttons likely for submission ("X" for cancel, a reset button, and a help button).

**Important Points to Remember:**
- Always account for significant digits in the final answer.
- Pay close attention to the molar masses and conversion between grams and moles.
- Consider the assumption that volume remains constant when solutes are added to a solution.
Transcribed Image Text:**Problem Statement:** *Suppose 35.7 g of iron(II) iodide is dissolved in 350. mL of a 0.40 M aqueous solution of silver nitrate.* *Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn’t change when the iron(II) iodide is dissolved in it.* *Round your answer to 1 significant digit.* **Solution Explanation:** To solve for the final molarity of the iodide anion, follow these steps: 1. **Determine the moles of iron(II) iodide (FeI₂):** - Molar mass of FeI₂ = Fe (55.85 g/mol) + 2 * I (126.90 g/mol) = 309.65 g/mol - Moles of FeI₂ = mass (35.7 g) / molar mass (309.65 g/mol) 2. **Calculate the moles of iodide ions (I⁻) produced:** - Each mole of FeI₂ produces 2 moles of I⁻. - Moles of I⁻ = moles of FeI₂ * 2 3. **Assumption and Conversion:** - The total volume of the solution remains 350 mL (0.350 L). 4. **Determine the final molarity of the iodide ions:** - Final molarity (M) = moles of I⁻ / total volume in liters (L) **Diagram or Graph Explanation:** This problem does not include any diagrams or graphs. It only involves numerical calculations to determine the molarity of iodide anions in a given solution. **Interactive Tools:** On the user interface, several interactive buttons are visible: - A box to input the answer with units for molarity (M). - Buttons likely for submission ("X" for cancel, a reset button, and a help button). **Important Points to Remember:** - Always account for significant digits in the final answer. - Pay close attention to the molar masses and conversion between grams and moles. - Consider the assumption that volume remains constant when solutes are added to a solution.
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