Consider the reaction of Chlorine and Fluorine: Cl2 (g) + F2 (g)  2ClF (g) with Kc = 108A 18.0 L reactor charged with 24.0 moles of ClF, 18.0 moles Cl2, and 18.0 moles F2. d) Calculate [ClF]i , [Cl2]i , and [F2] i .
e) Recalculate [ClF]i , [Cl2 ] i , and [F2 ] i if the reaction is shifted entirely to the reactants side. e) Recalculate [ClF]i , [Cl2 ] i , and [F2 ] i if the reaction is shifted entirely to the reactants side.
f) Recalculate [ClF]i , [Cl2 ] i , and [F2 ] i if the reaction is shifted entirely to the products side.
g) Set up three separate i. c.e. tables using the three different “initial” concentrations from d, e and f .
d) i. c.e table for concentrations calculated in d):
e) i. c.e table for concentrations calculated in e):
f) i. c.e. table for concentrations calculated in f):

gravity causes stares to cluster i think so

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.

pie chart because it is in percents ( % )