Please advise. I have tried these many times and don't feel confident in my answers Calculate the morality of the OH- in the 10.00ml sample using your answer for #1Answer: 0.0064Calculate the molarity of the Ca2+ in the 10.00ml sample.Answer: 0.00000013Calculate the Ksp for Ca(OH)2 in the solution at 82.9 °C(See equation 2)Your answer should have 4 sig figsAnswer: 4.036e-5 A student prepares a solution for titration by adding a 10.00 ml aliquot of a solutionsaturated with Ca(OH)2 at 82.9°C to 25.0 ml of deionized water. This solution is titrated witha standardized 0.01252 M HCI solution. The equivalence point is observed after 17.94 mL ofthe HCl solution has been added.use this data to answer the questions below.Calculate the moles of OH- in the 10.00ml sample.(Hint, moles of OH- at the equivalence point = moles HCI)Your answer should have 4 sig figs.Enter answers in scientific notation using E, as 1.234 E-2 for example.Answer: 8.984e-3Calculate the moles of Ca2* in the 10.00ml sample.(See equation 1 in the procedure)Your answer should have 4 sig figs.Answer:0.000032

We are authorized to answer three subparts at a time, since you have not mentioned which part you are looking for, so we are answering the first three subparts, please repost your question separately for the remaining subpart.The volume of 0.01252 M HCl used for titration is 17.94 mL. The volume of solution titrated against HCl solution is 35 mL.The reaction that takes place between Ca(OH)2 and HCl is expressed as, CaOH2+2HCl→CaCl2+2H2O The above reaction shows that two mole of HCl reacts with one mole of Ca(OH)2. The number of moles of OH- present in 1 mol of Ca(OH)2 is 2. At equivalence point equal number of moles of HCl and OH- reacts. The number of moles of HCl is calculated as, nHCl=0.01252 mol L-1×17.941000L=0.0002246 mol Therefore, the number of moles of OH- present in 10.00 mL of solution is 2.246×10-4 mol.The number of moles of Ca2+ present in 1 mol of Ca(OH)2 is 1. The number of moles of Ca2+ is calculated as, nCa2+=nOH-2=2.246×10-4 mol2=1.123×10-4 molThe molarity of OH- in 10.00 mL of solution is calculated as, MOH-=2.246×10-4 mol10.00 L×1000=2.246×10-2 MAnswer: The number of moles of OH- and Ca2+ present in 10.00 mL of solution is 2.246×10-4 mol and 1.123×10-4 mol, respectively. The molarity of OH- in 10.00 mL of solution is 2.246×10-2 M.