If 61.1 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.636 g of precipitate, what is the molarity of lead(II) ion in the original solution?

During trial 2, what allowed you to determine that aluminum was the limiting reactant? check all that apply. all of the copper dissolved. all of the aluminum dissolved. the solution turned clear. the number of grams of copper(ii) chloride used in the reaction was greater than the number of grams of aluminum. the molar ratio of copper(ii) chloride to aluminum was greater than 3: 2, the equation’s molar ratio.

Will give brainliest it is a lab from k12 here is the linkfor each metal that participated in a chemical change, write the type of metal it is, based on your examination of the periodic table. type your answer here. (score for question 3: of 5 points) were there any metallic compounds that did not react with either the acid or the base? write the type of metal, based on your examination of the periodic table. type your answer here. (score for question 4: of 5 points) make a general statement about the reactivity of the metals in this experiment. type your answer here.

Achemistry student needs to standardize a fresh solution of sodium hydroxide. he carefully weighs out of oxalic acid , a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in of distilled water. the student then titrates the oxalic acid solution with his sodium hydroxide solution. when the titration reaches the equivalence point, the student finds he has used of sodium hydroxide solution.calculate the molarity of the student's sodium hydroxide solution. be sure your answer has the correct number of significant digits.