Suppose of nickel(II) bromide is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) bromide is dissolved in it. Round your answer to significant digits.

The question is incomplete, the complete question is;

Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) bromide is dissolved in it. Be sure your answer has the correct number of significant digits. The answer is needed in M.

0.0428 M

Explanation:

The equation of the reaction is;

NiBr2 + K2CO3 → NiCO3 + 2KBr

Number of moles of NiBr2 = mass/molar mass = 1.87 g/218.53 g/mol =

8.557 *  10^-3 moles

Now, we know that the Ni^2+ comes from the NiBr2. Also, the concentration of Ni^2+ in NiBr2 = concentration of Ni^2+ in NiCO3

So,

Concentration of Ni^2+ = 8.557 *  10^-3 moles/0.2 L = 0.0428 M

a:

by the combination of an unlimited number of elements