11) What two new compounds form if the solutions. AgNO_(3) and NH_(4)OH, were mixed. Circle the new formed compound that would precipitate (insoluble) 12) What is the molarity of the following solution? a) 3.5 mol KCl in 355 mL.of solution b) 12.38 CaF_(2) 350 mL of Solution

11) When the solutions $AgNO_{3}$ and $NH_{4}OH$ are mixed, the following reaction occurs:<br /><br />$AgNO_{3} + 2NH_{4}OH \rightarrow Ag(NH_{3})_{2}OH + NH_{4}NO_{3}$<br /><br />The two new compounds formed are $Ag(NH_{3})_{2}OH$ (silver ammonia complex) and $NH_{4}NO_{3}$ (ammonium nitrate). Among these, $Ag(NH_{3})_{2}OH$ is the compound that would precipitate (insoluble).<br /><br />12) To calculate the molarity of the given solutions, we need to use the formula:<br /><br />Molarity (M) = moles of solute / volume of solution in liters<br /><br />a) For the solution with 3.5 mol $KCl$ in 355 mL of solution:<br /><br />Molarity = 3.5 mol / 0.355 L = 9.86 M<br /><br />b) For the solution with 12.38 g $CaF_{2}$ in 350 mL of solution:<br /><br />First, we need to convert the mass of $CaF_{2}$ to moles using its molar mass (78.08 g/mol):<br /><br />Moles of $CaF_{2}$ = 12.38 g / 78.08 g/mol = 0.158 mol<br /><br />Then, we can calculate the molarity:<br /><br />Molarity = 0.158 mol / 0.350 L = 0.452 M