Question

How many millimoles of Cr2O72- ions and K1+ ions are present in 483. mL of a 0.0750 M K2Cr2O7?1. mmolCr2O72-

2. mmol K1+

Answer 1

72.45 mmol K⁺ and 36.225 mmol Cr₂O₇²⁻

Step-by-step explanation:

It is possible to obtain millimoles of a substantce by the multiplication of molarity times milliliters of the solution.

In the solution that is 0.0750M K₂Cr₂O₇:

K⁺ = 0.0750M * 2 = 0.150M

Cr₂O₇²⁻ = 0.0750M

The millimoles of these ions are:

Millimoles K⁺:

483mL * 0.150M = 72.45 mmol K⁺

Millimoles Cr₂O₇²⁻:

483mL * 0.075M = 36.225 mmol Cr₂O₇²⁻