The chemistry problem is "How would you make 500.0 mL of a 3.00 M HCl solution form a stock solution that is 6.00 M?" I have no clue how to even approach this problem.
To solve this problem, you need to understand the concept of dilution. Dilution involves adding a solvent (usually water) to a concentrated solution to obtain a solution with a lower concentration.
The formula to calculate dilutions is:
C1V1 = C2V2
where:
C1 = initial concentration of the stock solution
V1 = volume of the stock solution used
C2 = final desired concentration
V2 = final volume of the diluted solution
In this case, you are starting with a stock solution of 6.00 M HCl and you want to make a 500.0 mL solution with a concentration of 3.00 M.
Using the formula for dilution, we can plug in the known values:
(6.00 M) × V1 = (3.00 M) × (500.0 mL)
Next, solve for V1 by rearranging the equation:
V1 = (3.00 M) × (500.0 mL) / (6.00 M)
V1 = (1500.0 mL·M) / (6.00 M)
V1 = 250.0 mL
So, to make a 500.0 mL solution with a concentration of 3.00 M, you would measure 250.0 mL of the 6.00 M HCl stock solution and dilute it with sufficient water to reach a total volume of 500.0 mL.
Remember to always add the concentrated solution to the solvent while preparing a dilution. In this case, add the 250.0 mL of the stock solution to a container and then add water to reach the final volume of 500.0 mL.