An aqueous solution is made by dissolving 26.4 grams of barium sulfide in 349 grams of water. The molality of barium sulfide in the solution is
m.

Molality = 0.43 m

Explanation:

Given data:

Mass of barium sulfide = 25.4 g

Mass of water = 349 g (349 g/1000 = 0.349 Kg)

Molality of water = ?

Solution:

Number of moles of barium sulfide:

Number of moles = mass/molar mass

Number of moles = 25.4 g/169.39 g/mol

Number of moles = 0.15 mol

Molality:

Molality = number of moles o f solute/ Kg of solvent

Molality = 0.15 mol / 0.349 Kg

Molality = 0.43 m

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.