Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid wit

He lost 1224.7 grams.

From the provided data, the unknown mixture was composed of salt, salicylic acid, and sand. It is understandable that the student suspected the presence of sand, yet scientific experimentation must verify such assumptions. The test involving salt and salicylic acid reveals that salt dissolves in water, while salicylic acid is only slightly soluble, and sand does not dissolve at all. By introducing the unknown into water, the salt would dissolve first, followed by the partial dissolution of salicylic acid. Heating the mixture could allow for the evaporation of salicylic acid, resulting in the remaining salt. If traces of sand were observed in the dissolved sample, it could suggest contamination.

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

10 mL of this solution is further diluted to 250 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

5 mL of this solution is diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 500 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

The problem provides a conversion factor---> 1 mm3= 7.0 x 10^6 RBC. Therefore, to determine the quantity of red blood cells in your sample, we must first convert Liters to cm3 using the conversion factor--> 1 mL= 1 cm3

I have shared how to resolve this issue.