Question

Suppose 1.35g of iron(II) chloride is dissolved in 250.mL of a
29.0mM aqueous solution of silver nitrate.Calculate the final
molarity of chloride anion in the solution. You can assume the
volume of the solution doesn't change when the iron(II) chloride is
dissolved in it.Round your answer to 3 significant
digits.

Answer #1

**Number of moles of FeCl2 = 1.35/126.75 = 0.01065
moles**

**number of moles of AgNO3 = 29 m(moles/L) * 250 mL = 7.25
mmoles**

**FeCl2 + 2 AgNO3 -------> 2 AgCl +
Fe(NO3)2**

**2 moles of AgNO3 require 1 moles of FeCl2**

**7.25 mmoles AgNo3 requires 7.25/2 * 10^-3moles of
FeCl2**

**
= 0.003625 moles**

**Number of moles remaining = 0.01065 - 0.003625 =
0.007025 moles**

**Concentration of Cl- ion or FeCl2 solution =
0.007025*1000/250**

**Concentration of Cl- ions or FeCl2 solution = 0.0281
mol/L**

Suppose 0.938g of iron(II) chloride is dissolved in 300.mL of a
46.0mM aqueous solution of silver nitrate. Calculate the final
molarity of chloride anion in the solution. You can assume the
volume of the solution doesn't change when the iron(II) chloride is
dissolved in it. Be sure your answer has the correct number of
significant digits.

Suppose
2.55g
of iron(II) chloride is dissolved in
50.mL
of a
0.30M
aqueous solution of silver nitrate.
Calculate the final molarity of iron(II) cation in the solution.
You can assume the volume of the solution doesn't change when the
iron(II) chloride is dissolved in it.
Be sure your answer has the correct number of significant
digits.

Suppose 1.06g of iron(II) bromide is dissolved in 200.mL of a
52.0mM aqueous solution of silver nitrate. Calculate the final
molarity of bromide anion in the solution. You can assume the
volume of the solution doesn't change when the iron(II) bromide is
dissolved in it. Be sure your answer has the correct number of
significant digits.

Suppose 3.10g of sodium iodide is dissolved in 250.mL of a
73.0mM aqueous solution of silver nitrate. Calculate the final
molarity of iodide anion in the solution. You can assume the volume
of the solution doesn't change when the sodium iodide is dissolved
in it. Round your answer to 2 significant digits.

Suppose
0.332g
of zinc chloride is dissolved in 50.mL
of a
48.0mM
aqueous solution of potassium carbonate.
Calculate the final molarity of chloride anion in the solution.
You can assume the volume of the solution doesn't change when the
zinc chloride is dissolved in it.
Be sure your answer has the correct number of significant
digits.

Suppose 0.829g of ammonium iodide is dissolved in 250.mL of a
30.0mM aqueous solution of potassium carbonate.
Calculate the final molarity of iodide anion in the solution.
You can assume the volume of the solution doesn't change when the
ammonium iodide is dissolved in it.
Round your answer to 3 significant digits.

Suppose
2.10g
of lead(II) acetate is dissolved in
350.mL
of a
65.0mM
aqueous solution of ammonium sulfate.
Calculate the final molarity of acetate anion in the solution.
You can assume the volume of the solution doesn't change when the
lead(II) acetate is dissolved in it.
Be sure your answer has the correct number of significant
digits.

Suppose 2.27g of lead(II) nitrate is dissolved in 300.mL of a
52.0mM aqueous solution of ammonium sulfate. Calculate the final
molarity of nitrate anion in the solution. You can assume the
volume of the solution doesn't change when the lead(II) nitrate is
dissolved in it.
Be sure your answer has the correct number of significant
digits.

Suppose 35.7g of potassium iodide is dissolved in 350.mL of a
0.50 M aqueous solution of silver nitrate. Calculate the final
molarity of iodide anion in the solution. You can assume the volume
of the solution doesn't change when the potassium iodide is
dissolved in it. Be sure your answer has the correct number of
significant digits.

Suppose
1.87g
of ammonium iodide is dissolved in
300.mL
of a
45.0mM
aqueous solution of potassium carbonate.
Calculate the final molarity of ammonium cation in the solution.
You can assume the volume of the solution doesn't change when the
ammonium iodide is dissolved in it.
Round your answer to
3
significant digits.

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