The equation for the equivalent of Ag is:
$$\text{Eq. of Ag} = \text{Eq. of } O_2$$
Let x grams of silver be displaced.
$$\frac{x}{108} = \frac{5.6}{22.7} \times 4$$
Using the molar volume of gas at STP (22.7 L), we get:
$$x = 106.57 \, \text{g}$$
Thus, the answer is approximately 107 g.
Alternatively, using 22.4 L as the molar volume at STP:
$$\frac{x}{108} = \frac{5.6}{22.4} \times 4$$
which gives $$x = 108 \, \text{g}$$.
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is:
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]