Boris recently synthesized an explosive compound he named badenoughium. the molecular formula for bdenoughium is h5c3n3o9. the balance chemical equation for the ignition of badenoughium is: 4 h5c3n3o9 (l) ? 12 co2 (g) + 10 h2o (g) + o2 (g) + 6 n2 (g) a. when badenoughium is ignited it has a ? h o rxn of - 5678 kj. i. determine the ? h o f for the compound badenoughium. ii. determine the ? h o rxn for when 1.25 grams of badnenoughium is ignited. b. boris ignites 1.25 grams of his compound in a 500.00 ml at a temperature of 100.0 oc. i. determine the partial pressure for each of the gases produced when 1.25 grams of badenoughium is ignited. ii. determine the total pressure in the reaction vessel after 1.25 grams of badenoughium has been igntited.

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.