This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to your questions: 1) The amount of silver deposited by passing 241.25 coulombs of electricity through silver nitrate is Step 1: Write the half-reaction for the deposition of silver. Ag^+ (aq) + e^- Ag (s) From the reaction, 1 mole of electrons is required to deposit 1 mole of silver. Step 2: Calculate the moles of electrons passed using Faraday's constant (F = 96500 C/mol). Moles of electrons = Charge (Q)Faraday's constant (F) Moles of electrons = 241.25 C96500 C/mol Moles of electrons = 0.0025 mol Step 3: Calculate the moles of silver deposited. Since 1 mole of electrons deposits 1 mole of silver: Moles of Ag = Moles of electrons = 0.0025 mol Step 4: Calculate the mass of silver deposited using its molar mass (M_Ag = 108 g/mol). Mass of Ag = Moles of Ag × M_Ag Mass of Ag = 0.0025 mol × 108 g/mol Mass of Ag = 0.27 g The amount of silver deposited is 0.27 g. 2) For 2nd Order if the Con, twice the rxn rate. what will be the ans: Step 1: Write the general rate law for a second-order reaction. For a reaction that is second order with respect to a reactant A, the rate law is: Rate_1 = k[A]^2 where k is the rate constant and [A] is the concentration of reactant A. Step 2: Determine the new rate when the concentration is doubled. If the concentration of A is doubled, the new concentration is 2[A]. Rate_2 = k(2[A])^2 Rate_2 = k(4[A]^2) Rate_2 = 4k[A]^2 Step 3: Compare the new rate to the original rate. Rate_2 = 4 × (Rate_1) When the concentration of a reactant in a second-order reaction is doubled, the reaction rate will increase by a factor of 2^2 = 4. The reaction rate will be four times the original rate. That's 2 down. 3 left today — send the next one.