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describe the polority of the structure of the left which is most stable 8. How many grams of BaCl2 are...

7. describe the polority of the structure of the left which is most stable
8. How many grams of BaCl2 are needed to prepare 100 mL of 0.10 M Cl- (aq) ions?
9. Why is oxygen gas able to dissolve in water? (hint... watch AP daily video 3.1)
Free Response
1. Write the net ionic equation for the reaction of lead II nitrate and potassium iodide. Both reactants are in the aqueous solution, the precipitate drawing the reaction is the formation of lead II iodide.
2. Draw a particle diagram in the beaker at right with all participating species AFTER the reaction has gone to completion.
3. The student filters and dries the precipitate of PbI2 and records the data below:
Volume of Pb(NO3)2 solution: 5.00 mL
Volume of 1.0 M KI added: 10.0 mL
Mass of PbI2 precipitate collected: 0.67 g
4. Determine the moles of Pb(NO3)2 in the original 5.00 mL of solution.
4. Describe the contents of the filtrate after the precipitate has been removed. Would this solution behave like an electrolyte? Why or why not?

Answer

7. Describe the polarity of the structure of the left which is most stable

The left structure consists of a molecular arrangement that has a significant dipole moment due to the electronegativity differences among its constituent atoms. This characteristic leads to polar bonds, causing the molecule to exhibit a net dipole moment. The stability of this polar structure arises from the ability of the polar bonds to engage in dipole-dipole interactions, contributing to a lower energy state when compared to nonpolar arrangements.

8. How many grams of BaCl2 are needed to prepare 100 mL of 0.10 M Cl- (aq) ions?

To prepare a 0.10 M Cl- solution, we need to consider that each formula unit of BaCl2 produces two chloride ions (Cl-) upon dissociation. Therefore, for 0.10 M Cl- ions, the concentration of BaCl2 required is half that value:

0.10 M Cl- / 2 = 0.050 M BaCl2

Next, calculate the moles of BaCl2 needed for 100 mL (0.1 L) of this solution:

0.050 moles/L * 0.1 L = 0.005 moles of BaCl2

Now, we compute the mass of BaCl2 required using its molar mass (Ba: 137.33 g/mol, Cl: 35.45 g/mol × 2 = 70.90 g/mol, total = 208.23 g/mol):

Mass = moles × molar mass = 0.005 moles * 208.23 g/mol = 1.04115 g

Therefore, approximately 1.04 g of BaCl2 is needed.

9. Why is oxygen gas able to dissolve in water?

Oxygen gas is able to dissolve in water due to a combination of factors including its partial solubility and interactions with water molecules. While O2 is a nonpolar molecule, it can still interact with the polar water molecules through weak Van der Waals forces. Additionally, the process of dissolution allows for some oxygen molecules to occupy spaces between water molecules, facilitating its presence in the aqueous phase to maintain equilibrium. This phenomenon is also enhanced at lower temperatures, where the kinetic energy of water molecules permits more stable interactions.

Free Response

1. Write the net ionic equation for the reaction of lead II nitrate and potassium iodide.

The balanced chemical equation for the reaction is:

Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)

The net ionic equation is:

Pb2+(aq) + 2 I-(aq) → PbI2(s)

2. Draw a particle diagram in the beaker at right with all participating species AFTER the reaction has gone to completion.

After the reaction goes to completion, the particle diagram would show solid lead(II) iodide (PbI2) precipitate at the bottom of the beaker, with remaining potassium ions (K+) and nitrate ions (NO3-) uniformly dispersed in the solution above the precipitate.

3. The student filters and dries the precipitate of PbI2 and records the data below:

  • Volume of Pb(NO3)2 solution: 5.00 mL
  • Volume of 1.0 M KI added: 10.0 mL
  • Mass of PbI2 precipitate collected: 0.67 g

4. Determine the moles of Pb(NO3)2 in the original 5.00 mL of solution.

Using the stoichiometry from the balanced equation, we first recognize that 5 mL of Pb(NO3)2 solution is being assessed. Assume the original concentration of Pb(NO3)2 is x M. The moles in 5.00 mL (0.005 L) can be calculated:

Moles of Pb(NO3)2 = x * 0.005 L

We do not have the concentration of Pb(NO3)2 provided, so we cannot calculate an exact number, but this is how it would be expressed with given concentration.

5. Describe the contents of the filtrate after the precipitate has been removed. Would this solution behave like an electrolyte? Why or why not?

The contents of the filtrate after the precipitate of PbI2 has been removed would consist of the potassium ions (K+) and nitrate ions (NO3-). This solution would behave like an electrolyte because it contains ions that can move freely in the solution, allowing it to conduct electricity. The presence of these dissociated ions facilitates the conduction of electric current through the solution.

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