Consider the equation A(aq) 2B(aq) 3C(aq) 2D(aq). In one experiment, 45.0 mL of 0.050 M A is mixed with 25.0 mL 0.100 M B. At equilibrium the concentration of C is 0.0410 M. Calculate K. g

K = 0.0396

Explanation:

Based on the reaction:

A + 2B ⇄ 3C + 2D

Where equilibrium constant, K, is:

K = [C]³[D]² / [A] [B]²

The initial concentrations of A and B are:

[A]₀ = 0.050M * (45.0mL / 70.0mL) = 0.0321M

[B]₀ = 0.100M * (25.0mL / 70.0mL) = 0.0357M

As [C] = 0.0410M, the molar concentration of D is:

0.0410M * (2mol D / 3mol C) = 0.0273M = [D]

And the concentration of A and B that reacted was:

0.0410M * (2mol B / 3mol C) = 0.0273M B

0.0410M * (1mol A / 3mol C) = 0.0137M A

Equilibrium concentration B and A:

0.0357M - 0.0273M = 0.0084M = [B]

0.0321M - 0.0137M = 0.0184M = [A]

And K is:

K = [0.0410M]³[0.0273M]² / [0.0184M] [0.0084M]²

the answer is c. nacl

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.