Math Problem Statement
Identify the oxidizing and reducing agents in the given reactions and calculate the molarity of a diluted solution.
Solution
Solution for the Questions:
b) Identify the oxidizing and reducing agents:
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Reaction (i):
- Oxidation: Carbon in increases oxidation state from to in . Therefore, is the reducing agent.
- Reduction: Iodine in decreases oxidation state from to in . Therefore, is the oxidizing agent.
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Reaction (ii):
- Oxidation: Iron increases oxidation state from to . Therefore, is the reducing agent.
- Reduction: Oxygen in decreases oxidation state from to in . Therefore, is the oxidizing agent.
c) Molarity of the resulting solution:
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Step 1: Molar mass of :
- Sodium (Na): 22.99 g/mol
- Fluorine (F): 18.998 g/mol
- Total:
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Step 2: Calculate moles of :
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Step 3: Molarity of the original solution:
- Volume = 1 L
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Step 4: Calculate moles in 180.0 mL of the solution:
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Step 5: Molarity of the diluted solution:
- Total volume after dilution = 500 mL = 0.500 L
Final Answers:
-
b (i):
- Oxidizing agent:
- Reducing agent:
-
b (ii):
- Oxidizing agent:
- Reducing agent:
-
c:
- Molarity of the resulting solution:
Do you want further explanations or clarifications?
Related Questions:
- What is the difference between oxidizing and reducing agents in chemical reactions?
- How do you identify the oxidation states of elements in a compound?
- What is the significance of molarity in chemical reactions?
- How would the molarity change if the solution were further diluted?
- What practical applications involve identifying oxidizing and reducing agents?
Tip: Always confirm oxidation states by calculating for each element systematically to avoid errors.
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Math Problem Analysis
Mathematical Concepts
Oxidation-Reduction Reactions
Solution Chemistry
Stoichiometry
Formulas
Molar mass = Sum of atomic masses of elements in a compound
Moles = Mass / Molar Mass
Molarity = Moles of Solute / Volume of Solution
Dilution Formula: M1V1 = M2V2
Theorems
Law of Conservation of Mass
Suitable Grade Level
Grades 11-12