Asked Oct 15, 2019

Hello, this is one of the problems that I cannot figure out on my homework. Thank you for your help. 

Suppose 1.85g of copper(II) nitrate is dissolved in 100.mL of a 60.0mM aqueous solution of sodium chromate.

Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) nitrate is dissolved in it.

Be sure your answer has the correct number of significant digits.


Expert Answer

Step 1

The number of moles of sodium chromate (nsc) is calculated using equation (1) in which C is the concentration of sodium chromate and V is the volume of solution.


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п, ..... 1) C ="sc V n 60.0 mM 100 mL п. SC 60.0x 103 M 100x 10 L 6.00x 103 mol n SC

Step 2

The number of moles of copper nitrate (ncn) is calculated using equation (2) in which m is the mass of solute and Mm is the molar mass of solute.


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т .. (2) п. en М, 1.85 g п. 187.56 g/mol cn 3 mol =9.86x 10

Step 3

The reaction between copper nitrate and...


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Cu(NO,), (aq)+Na,CrO, (aq) CuCrO, (s)+2NaNO, (aq)


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