What is the equilibrium concentration of amide ion (NH₂^-)in liquid ammonia at 25°C? ("am" = dissolved in ammonia) 2NH₃(l) NH₄⁺(am)+ NH₂^-(am); K_c = 1.8 × 10^-24 at 25°C A)2.6 × 10^-12 M B)3.6 × 10^-24 M C)1.3 × 10^-12 M D)1.8 × 10^-24 M E)9.0 × 10^-25 M

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The Answer of What is the equilibrium concentration of amide ion (NH₂^-)in liquid ammonia at 25°C? ("am" = dissolved in ammonia) 2NH₃(l) NH₄⁺(am)+ NH₂^-(am); K_c = 1.8 × 10^-24 at 25°C A)2.6 × 10^-12 M B)3.6 × 10^-24 M C)1.3 × 10^-12 M D)1.8 × 10^-24 M E)9.0 × 10^-25 M

What Is the Equilibrium Concentration of Amide Ion (NH2-)in Liquid

What Is the Equilibrium Concentration of Amide Ion (NH₂^-)in Liquid