A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium:
-[Cu(CN) ₂^-]

Question

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows:

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10<sup>-3</sup> M CuNO<sub>3</sub>. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) <sub>2</sub><sup>-</sup>] A) 1.0 *10<sup>-</sup><sup>3</sup> M B) 4.0 *10<sup>-</sup><sup>12</sup> M C) 2.0*10<sup>-</sup><sup>7</sup> M D) 2.0 * 10<sup>-</sup><sup>3</sup> M E) none of these

Calculate the following concentrations at equilibrium:
-[Cu(CN) ₂^-]

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The Answer of A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10 -3 M CuNO 3 . Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN)₂^-]

A Solution Is Formed by Mixing 50