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.

Molarity = 0.0428 M = 42.8 mM

Explanation:

Step 1: Data given

Mass of nickel(II) bromide = 1.87 grams

Molar mass of nickel(II) bromide = 218.53 g/mol

Volume = 200 mL = 0.200 L

Step 2: Calculate moles of nickel(II) bromide

Moles nickel (II) bromide = mass / molar mass

Moles nickel (II) bromide = 1.87 grams / 218.53 g/mol

Moles nickel (II) bromide = 0.00856 moles

Step 3: Calculate moles nickel (II) cation

For 1 mol NiBr2 we have 1 mol Ni^2+

For 0.00856 moles NiBr2 we have 0.00856 moles Ni^2+

Step 4: Calculate final molarity of Ni^2+

Molarity = moles / volume

Molarity = 0.00856 moles / 0.200 L

Molarity = 0.0428 M = 42.8 mM

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

it would be a gas formation