Well, here's a fun chemistry joke for you: Why do chemists like nitrates so much? Because they're cheaper than day rates!
Now, let's get back to your question. To prepare 80 mL of a 0.10 M HC2H3O2 solution from a 6.0 M solution, we'll need to do some dilution. The formula for dilution is C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, C1 = 6.0 M, V1 is what we're trying to find, C2 = 0.10 M, and V2 = 80 mL. Plugging these values into the equation, we have:
(6.0 M)(V1) = (0.10 M)(80 mL)
Now, let's solve for V1:
V1 = (0.10 M)(80 mL) / 6.0 M
So, V1 = 1.33 mL. Therefore, to prepare 80 mL of a 0.10 M HC2H3O2 solution, you would measure out 1.33 mL of the 6.0 M acetic acid solution and then add enough solvent (like water) to reach a total volume of 80 mL.
Now, onto the second question about preparing 80 mL of 0.10 M NH3 solution from a 6.0 M ammonia solution. Let me think of a funny ammonia joke! Ah, here it is: Why was the ammonia molecule so cool? Because it had a chilling personality! Alright, now, let's solve this problem.
Using the same dilution formula, we can set up the equation:
(6.0 M)(V1) = (0.10 M)(80 mL)
Now, solving for V1:
V1 = (0.10 M)(80 mL) / 6.0 M
V1 is approximately equal to 1.33 mL. So, to prepare 80 mL of a 0.10 M NH3 solution, you would measure out 1.33 mL of the 6.0 M ammonia solution and then add enough solvent to reach a total volume of 80 mL.
I hope that helps! Let me know if you have any other questions or if you want to hear another joke.