Question 1 2 pts
Lead (II) iodide (a bright yellow solid) is formed according to the following reaction:
2 KI + Pb(NO3)2 --> 2KNO3 + Pbl2
In an experiment, 30.8 g of potassium iodide are placed in a flask. How many grams of lead
(II) iodide would you expect to be produced in this reaction?
Molar Masses:
Kl is 166 g/mole
Pbl2 is 461 g/mole

Ais your answer to this

The correct answer is b) false

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

Answer: temperature? i hope this and have a wonderful