If 93.3 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.733 g of precipitate, what is the molarity of lead(II) ion in the original solution?

0.34 M

Explanation:

From the equation of reaction:

1 mole of lead (II) nitrate yields 1 mole of the PbI2 precipitate

mole of the precipitate = mass/molar mass

                                        = 0.733/461.01 = 0.00159 mole

If 1 mole of lead (II) nitrate gives 1 mole of PbI2, then 0.00159 mole of PbI2 will require:

   0.00159 x 1 = 0.00159 mole of lead (II) nitrate.

1 mole of lead (II) nitrate contains 2 moles of lead (II) ion

0.00159 mole of lead (II) nitrate will contain:

    0.00159 x 2 = 0.00318 mole of lead (II) ion

volume of lead (II) nitrate = 93.3 mL = 0.0933 Liter

Hence, molarity of lead (II) ion = mole/volume

     = 0.00318/0.0933

             = 0.034 M

The molarity of lead(II) ion in the original solution is 0.034 M.

Explanation:

Below is an attachment containing the solution

fire could start, buildings collapsing, and ash falling from the sky.

explanation:

this is because the buildings or plants could catch on fire and the buildings would most likely collapse. lastly ash would fall because all the burning would cause ash to fall from the sky.

hope this .