The amount of silver deposited by passing 241.25 coulombs of electricity through silver nitrate is (Molar mass of Ag=108g/mol, F=96500C/mol. For 2nd order if the concentration twice the rate. what will be the ans)?

$$\text{Ag}^+ (\text{aq}) + \text{e}^- \to \text{Ag} (\text{s})$$

From the reaction, 1 mole of electrons is required to deposit 1 mole of silver.

$$\text{Moles of electrons} = \frac{\text{Charge (Q)}}{\text{Faraday's constant (F)}}$$ $$\text{Moles of electrons} = \frac{241.25 \text{ C}}{96500 \text{ C/mol}}$$ $$\text{Moles of electrons} = 0.0025 \text{ mol}$$

Since 1 mole of electrons deposits 1 mole of silver:

$$\text{Moles of Ag} = \text{Moles of electrons} = 0.0025 \text{ mol}$$

$$\text{Mass of Ag} = \text{Moles of Ag} \times M_{\text{Ag}}$$ $$\text{Mass of Ag} = 0.0025 \text{ mol} \times 108 \text{ g/mol}$$ $$\text{Mass of Ag} = 0.27 \text{ g}$$

The amount of silver deposited is $\boxed{\text{0.27 g}}$.

2) For 2nd Order if the Con, twice the rxn rate. what will be the ans:

For a reaction that is second order with respect to a reactant A, the rate law is:

$$\text{Rate}_1 = k[\text{A}]^2$$

where $k$ is the rate constant and $[\text{A}]$ is the concentration of reactant A.

If the concentration of A is doubled, the new concentration is $2[\text{A}]$.

$$\text{Rate}_2 = k(2[\text{A}])^2$$ $$\text{Rate}_2 = k(4[\text{A}]^2)$$ $$\text{Rate}_2 = 4k[\text{A}]^2$$

$$\text{Rate}_2 = 4 \times (\text{Rate}_1)$$

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.