okay so im doing pre lab questions on reaction rates

I have a chart labeled effects of varying concentration

in reaction #1 flask A contains 7.50mL of .200 M KI , none of .20M KCI, 5.00 mL of .0050 M S2O3^2-, 1 drop of starch. Flask B contains 7.5mL of .100M S2O8^2-

It says for each reaction (i just gave the 1st one) determine the concentration of KI, (NH4)2S2O8 and Na2S2O3 that will be present when solutions A and B are mixed together. the total volume for each reaction is 20.0 mL. You must show one sample calculation for each component.

I'm not sure how to do this, if someone can give me the calculations for the first one, i can figure out the rest. ThankS!

Total volume is 7.50+5.00+7.50 =20.00 (+ 1 drop but I guess they don't count that).

You have 7.50 mL of 0.200 M KI so the concn in the mixed solution is (0.00750 L x 0.200 = ?? moles and that divided by 0.0020 L) = ?? M KI. All of the others are done the same way. If you are comfortable with millimoles, you can make it much simpler by 7.50 x 0.200 = ?? millimoles and that divided by 20 mL = xx M KI. Even simpler is to treat it as a dilution problem like so.
0.200 M x (7.50 mL/20.0 mL) = ?? M KI.

To correct DrBob222 it would be .020 L instead of 0.0020 L

To determine the concentrations of KI, (NH4)2S2O8, and Na2S2O3 when solutions A and B are mixed together, you first need to calculate the moles of each component in both solutions. Then you can use the moles to calculate the new concentrations in the final mixture.

Let's start with the calculation for KI in solution A:

1. Calculate the moles of KI in solution A:
Moles of KI = concentration of KI × volume of KI
= 0.200 M × 0.00750 L
= 0.0015 mol

2. Calculate the total volume of the final mixture:
Total volume = volume of solution A + volume of solution B
= 0.00750 L + 0.00750 L
= 0.015 L

3. Calculate the concentration of KI in the final mixture:
Concentration of KI = moles of KI / total volume
= 0.0015 mol / 0.015 L
= 0.100 M

So, the concentration of KI in the final mixture when solutions A and B are mixed together is 0.100 M.

Now, you can apply the same steps to calculate the concentrations of (NH4)2S2O8 and Na2S2O3 in the final mixture.

To determine the concentration of KI, (NH4)2S2O8, and Na2S2O3 when solutions A and B are mixed, we need to use the concept of moles and stoichiometry.

Let's start by calculating the number of moles of each chemical present in flask A.

For KI:
Given: Volume = 7.50 mL, Concentration = 0.200 M
Using the formula: moles = concentration × volume
moles of KI = 0.200 mol/L × 0.00750 L = 0.0015 mol

For KCl:
Given: Volume = None, Concentration = 0.200 M (meaning there is no KCl present)
moles of KCl = 0 mol

For S2O3^2-:
Given: Volume = 5.00 mL, Concentration = 0.0050 M
moles of S2O3^2- = 0.0050 mol/L × 0.00500 L = 2.5e-5 mol

Now, let's calculate the number of moles of (NH4)2S2O8 in flask B:
Given: Volume = 7.5 mL, Concentration = 0.100 M
moles of (NH4)2S2O8 = 0.100 mol/L × 0.00750 L = 7.5e-3 mol

Since the total volume for the reaction is 20.0 mL, we have to consider the dilution that occurs when combining A and B. We can assume that the total volume remains constant after mixing.

Now, let's calculate the concentration of KI in the final mixture:
moles of KI in the final mixture = moles of KI from A = 0.0015 mol
concentration of KI = moles of KI / total volume
concentration of KI = 0.0015 mol / 0.0200 L = 0.075 M

Similarly, we can calculate the concentration of (NH4)2S2O8 and Na2S2O3 in the final mixture using the same approach.

I hope this helps you understand how to calculate the concentrations of the components in the given reaction. Repeat the same steps for the other components, and feel free to ask if you have any further questions!