Consider the equilibrium for the vapor pressure of br(): bry() = bry(8) k = 0.29 if a 1.0 l container has 1.00 bar of bra(g) and 3.45 mol of br: (i), what will happen? a. the excess bra(s) will condense to form brz(1) until equilibrium is established. b. the excess br(s) will evaporate to form brz(l) until equilibrium is established. c. the rate of condensation will be equal to the rate of evaporation since the mixture is at dynamic equilibrium and no change will be observed d. the excess bra(1) will evaporate to form bra(g) until equilibrium is established. e. the excess brz) will condense to form br(l) until equilibrium is established.

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.

c.

explanation:

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