Astudent obtains 100 ml of a 0.0003321 m solution of agno 3 (silver nitrate). he labels this "solution #1". he then pipets 5 ml of solution #1 into a 50 ml volumetric flask and dilutes to the mark with water. he labels this "solution #2." he then pipets 10 ml of solution #1 into a 250 ml volumetric flask and dilutes to the mark with water. this is "solution #3". finally, the prepares "solution #4" by pipetting 18 ml of solution #3 into a 50 ml volumetric flask and dilutes to the mark with water. draw relational pictures to illustrate the relationship between the original solution and all of the diluted solutions that the student prepared. what is the concentration of silver nitrate in solution #4?

The concentration of silver nitrate in solution #4 is 4.781 × 10⁻⁷ M.

Explanation:

Let´s express the initial concentration in cientific notation:

0.0003321 M = 3.321 × 10⁻⁴ M

In the first dilution step, the student takes 5 ml and add water to 50 ml. The dilution factor is 5 ml/50 ml = 1/10. It means that the concentration is reduced ten times. Then, after the first step the concentration of the solution will be:

3.321 × 10⁻⁴ M · 1/10 = 3.321 × 10⁻⁵ M (concentration of solution #2)

In the next step, the solution is diluted by 10 ml /250 ml = 1/25. Then, the solution will be diluted 25 times. The concentration of solution #3 is:

3.321 × 10⁻⁵ M · 1/25 = 1.328 × 10⁻⁶ M

Finally, the dilution factor will be 18 ml / 50 ml = 9/25. Then, the concentration of solution #4 will be:

1.328 × 10⁻⁶ M · 9/25 = 4.781 × 10⁻⁷ M

The concentration of silver nitrate in solution #4 is 4.781 × 10⁻⁷ M.

Attached you will find a figure that illustrates what amount of the original solution is present in each diluted solution. In the first dilution, there are 5 ml of the original solution in the 50 ml solution. In the second dilution, the volume of the original solution is 1 ml in the 250 ml of solution #3. In solution #4, the volume from the original solution is 0.072 ml.

The solution #4 could be prepared taking 0.072 ml from the original solution and adding water to 50 ml, but it wouldn´t be a good practice.

for points guys

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.