A 1.728 g sample of a Pb/Cd alloy was dissolved in acid and diluted to exactly 500 mL. Titration of a 25.00 mL aliquot at a pH = l0 in an NH4 + /NH3 buffer required 43.60 mL of standard 0.01080 M EDTA and involved both cations (Pb2+ and Cd2+). The cadmium ion was then masked by using an HCN/NaCN buffer to bring the pH of a second 25.00 mL aliquot to 10; 32.70 mL of the 0.01080 M EDTA standard was used in this titration and involved only the Pb2+. Calculate the respective percentages of lead and cadmium in the alloy.

Titration of a 25.00 mL aliquot at a pH = l0 in an NH4 + /NH3 buffer required 43.60 mL of standard 0.01080 M EDTA and involved both cations (Pb2+ and Cd2+). The cadmium ion was then masked by using an HCN/NaCN buffer to bring the pH of a second 25.00 mL aliquot to 10; 32.70 mL of the 0.01080 M EDTA standard was used in this titration and involved only the Pb2+. Calculate the respective percentages of lead and cadmium in the alloy.

idk

explanation:

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current