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

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

(b) which contains higher levels of vitamin b, oranges or spinach?

explanation:

you can eliminate a, c, and d because they are questions based on opinions.