4.If 15.00 mL of 3.00 M potassium iodide is needed to reach the equivalence point with 10.00 mL of lead (Il) nitrate, determine the molarity of the lead (Il) nitrate solution

2.25 M

Explanation:

The reaction that takes place is:

2KI + Pb(NO₂)₃ → PbI₂ + 2KNO₃

First we calculate how many potassium iodide moles reacted, using the given volume and concentration:

15.00 mL * 3.00 M = 45 mmol KI

Then we convert 45 mmoles of KI into mmoles of Pb(NO₂)₃, using the stoichiometric coefficients of the balanced reaction:

45 mmol KI * = 22.5 mmol Pb(NO₂)₃

Finally we calculate the molarity of the Pb(NO₂)₃ solution, using the calculated number of moles and given volume:

22.5 mmol Pb(NO₂)₃ / 10.00 mL = 2.25 M

Sorry i dont understand the question can you rephase it?

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.