There is 8oz for each cup
Answer:

Explanation:
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In this case, since the precipitation of the lead iodide is related to the iodide ion in solution, if we make react lead (II) nitrate with an iodide-containing salt, a possible chemical reaction would be:

In such a way, since 15.71 mL of a 0.5770-M solution of lead (II) nitrate precipitates out lead (II) iodide, we can first compute the moles of lead (II) nitrate in the solution:

Next, since there is a 1:2 mole ratio between lead (II) nitrate and iodide ions, we compute the moles of those ions:

Finally, since the mixing of the two solutions produce a final volume of 40.71 mL (0.04071 L), the resulting concentration (molarity) of the iodide ions in the student's unknown turns out:

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There will be three moles of ions.
Because in an aqueous solution, HCl will break apart into <span>H<span>+ and Cl- ions.
</span></span>HCl -> <span>H+</span><span> + </span><span>C<span>l<span>−
</span></span></span>1.5 HCl -> 1.5 H+ + 1.5 Cl−<span>
So the aqueous solution will have 1.5 moles of hydrated hydrogen ions and 1.5 moles of hydrated chloride ions. So, a total of 3 moles of ions are present in the aqueous solution of 1.5 moles of HCl. </span>