The closer to the top the metal is in the list, the more active the metal is and the stronger a reducing agent the metal is. When two different metals are involved in a redox reaction, the metal higher in the list will be oxidized and give up electrons that will reduce the cation of the less active metal.
Answer:
0.34 M
Explanation:
I assume that the compound is PbCl2.
One mole of PbCl2 contains one mol of Pb+2 and 2 moles of Cl-
Molarity (M)= moles (n) /Volume (V)
Moles Pb2+ = M x V = 0.17 V
Moles Cl- = moles Pb2+ x (2 moles Cl-/1 mole Pb2+) = 0.17 V x 2 = 0.34 V
M Cl- = moles Cl-/V = 0.34V/V = 0.34 M
0.025 is how many moles of oxygen to react with NH3
