Question

I am needing help finding the answers the first. I am just having a bad brain fart and cant figure it out to save my life.

please show how you calculated it and found what you did.

Thanks!!

Data and Calculations Step 3 Initial concentrations [Pb2+ ] initial = Moles of Pb2+ ion = Moles of Clion = Step 4 Is there any precipitate? Record the total volume of HCI added until Total volume of HCI added: LOML the precipitate forms. ml Step 4 -ml Calculate new concentrations by taking into account the total volume of the solution. Total volume of solution (Pb(NO3)2 + HCI) = 3.0 mL Moles of Pb2+ ion = Moles of Clion = [Pb2+] = [CH] =

Answer 1

Step-3 2.00 ml of 0.30M PLNO3 is takew i a estube me ans 0.30 M 0.30 males P(o)is tresent In 1000 ml solution 0.30 Mmolex 2 m Solutin So, in 2 ml 1o00 m 0.0006 mdles 1 male Pb(N) 24 1 mde Ph and 2mdes no means So, Initial concentratin 2t Pb ini tial =.30 M Moles of Pht ions=0.000 mole Soluhion meams 2.00 ml of 0,30 M HCI 100 m Soluhim contain0.3 mdes He Solution Comtain 0-3 nde x2 m! 2 mt 0.0006 mdas HC So males of ci im O.0006 mdes