how many milliliters of a stock solution of 6.0 m hno3 would you have to use to prepare 110mL of 0.500 M HNO3?

I'm assuming that 6.0m is a typo and you meant 6.0M.(There is a difference).

mL1 x M1 = mL2 = M2
mL1 x 6.0M = 110 mL x 0.500M

9.16mL

Well, let's break this down. You have a stock solution of 6.0 M HNO3 and you want to prepare 110 mL of 0.500 M HNO3.

To prepare a solution of lower concentration, you need to dilute the stock solution with a solvent. In this case, the solvent is probably not going to be laughter, so we'll stick with water.

Now, we can use the equation for dilution in this case: M1V1 = M2V2.

Let's plug in the values we know:
M1 = 6.0 M (the concentration of the stock solution)
V1 = ? (the unknown volume of the stock solution we need to use)
M2 = 0.500 M (the desired concentration of the final solution)
V2 = 110 mL (the desired volume of the final solution)

Let's do some math magic now:

(6.0 M)(V1) = (0.500 M)(110 mL)

Now, let me crunch the numbers here...

V1 = (0.500 M)(110 mL) / 6.0 M

*Calculating...*

The answer is... *drum roll please*... approximately 9.17 mL of the stock solution!

So, you would need to use around 9.17 mL of the 6.0 M HNO3 stock solution to prepare 110 mL of 0.500 M HNO3. And remember, always dilute your chemistry problems with a healthy dose of humor!

To find out how many milliliters of a 6.0 M HNO3 stock solution you will need to make 110 mL of a 0.500 M HNO3 solution, we can use the formula:

C1V1 = C2V2

Here:
C1 = initial concentration of the stock solution
V1 = volume of stock solution needed
C2 = final concentration of the desired solution
V2 = final volume of the desired solution

Let's plug in the values:

C1 = 6.0 M
V1 = unknown
C2 = 0.500 M
V2 = 110 mL

Now we can rearrange the formula to solve for V1:

V1 = (C2 * V2) / C1

Substituting in the given values:

V1 = (0.500 M * 110 mL) / 6.0 M

V1 = 9.17 mL

Therefore, you will need to use 9.17 milliliters of the 6.0 M HNO3 stock solution to prepare 110 mL of a 0.500 M HNO3 solution.