Balancing A Redox Reaction

Step 1: Concept
Equivalent wt. = .
Step 2: Analysis
In oxidation by , becomes , a change of 1.
Step 3: Conclusion
Equivalent mass = .
Final Answer: (A)

Step 1: Analysis
The 1 and 2 resistors are in series, forming a 3 branch.
Step 2: Parallel
This branch is in parallel with the 3 resistor across the 3 V battery.
Step 3: Calculation
Current in the 3 branch is A.
Final Answer: (b)

Step 1: Determine the equivalent weight of the oxidizing agent.
The oxidizing agent in this reaction is , and its equivalent weight is determined by the amount of oxygen it can provide in the reaction. The reaction shows that 1 mole of is reduced to 1 mole of sulfur (S), and 2 moles of are converted to 2 moles of , with oxygen being released. The equivalent weight of is the molar mass divided by the number of electrons transferred in the reaction. Since 2 electrons are involved, the equivalent weight of is half of its molar mass.

Step 2: Find the choice that has twice the equivalent weight.
The molar mass of is 64 g/mol, so the equivalent weight is: The choice with twice the equivalent weight is , which corresponds to option (4).

Step 3: Conclusion.
The correct answer is , which is option (4).

Step 1: Understanding the Concept:
Graham's law of effusion: . Also, .
Step 2: Detailed Explanation:
Partial pressures: , .
.
Mole ratio effusing out = . Composition of He = (not asked). The ratio itself is 0.88.
Step 3: Final Answer:
Thus, .

Concept: Normality relation:

Step 1: FAS normality.
Equivalent weight of FAS = 392

Step 2: Apply titration formula.


Step 3: Convert to grams/L.
Equivalent weight of KMnO (acidic) = 31.6