An aqueous solution of hydroiodic acid is standardized by titration with a 0.168 M solution of sodium hydroxide. If 18.7 mL of base are required to neutralize 29.6 mL of the acid, what is the molarity of the hydroiodic acid solution

0.106M

Explanation:

Equation of reaction

NaOH + HI ===》 NaI + H₂O

If 0.168 moles of NaOH = 1000cm³

X moles = 18.7 cm³

X = (18.7 * 0.168) / 1000 = 0.00314 moles

From equation of reaction,

1 moles of NaOH = 1 mole of HI

0.00314 moles of NaOH = 0.00314 moles of HI

Therefore,

0.00314 moles of HI = 29.6cm³

X moles = 1000cm³

X = (0.00314 * 1000) / 29.6 = 0.106M

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.

yes

explanation:

search chemistry final b exam in google and a site called quizlet will pop up