A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate is filtered and dried and found to have a mass of 258 mg. What mass of barium was in the original solution assume that all of the barium was precipitated out of the solution you the reaction.

Explanation:

The white precipiate that is formed when sodium sulfate, Na₂SO₄, reacts with barium ions in solution, Ba²⁺, is barium sulfate, BaSO₄.

The molar mass of BaSO₄ is 233.38 g/mol. Thus, the number of moles of BaSO₄ is:

Since 1 mole of BaSO₄ contains 1 mole of Ba, there are 0.00110549 mol of Ba²⁺ ions.

Use the atomic mass of Ba to calculate the mass in grams in the original solution:

Then, with 3 significant figures, there were 152mg of barium in the original solution.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.