Solute s has a partition coefficient of 6.2 between water (phase 1) and hexane (phase 2). 74.0 ml solution of s in water is extracted six times with 17.0 ml of hexane. calculate the fraction of s remaining in the aqueous phase.

Explanation:

The given data is as follows.

      Partition coefficient, K = 6.2

      Volume of phase 1, = 74.0 mL

      Volume of phase 2, = 17.0 mL

So, after one extraction fraction of solute remaining is given as follows.

              q =

After 3 times extraction, fraction of S remaining is as follows.

              q =

Putting the given values into the above formula as follows.

               q =

                  =

                  =

                  = 0.0699

Thus, we can conclude that the fraction of S remaining in the aqueous phase is 0.0699.

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.