How much of a 15.0 M stock solution do you need to prepare 250 ml of a 2.35 M HF solution?
.25 L

250 mL

39.2 L

39.2 mL

We need 39.2 mL of the 15.0M stock solution

Explanation:

Step 1: Data given

Initial concentration ( of stock) = 15.0 M

Final volume = 250 mL = 0.250 L

Final concentration = 2.35 M

Step 2: Calculate the volume of the stock volume

C1*V1 = C2*V2

⇒with C1 = the initial concentration (stock)= 15.0 M

⇒with V1 = the volume of the stock solution = TO BE DETERMINED

⇒with C2 = the final concentration of the HF solution = 2.35 M

⇒with V2 = the final volume of the HF solution = 0.250 L

15.0 M * V1 = 2.35 M * 0.250 L

V1 = (2.35 M * 0.250 L) / 15.0 M

V1 = 0.0392 L = 39.2 mL

We need 39.2 mL of the 15.0M stock solution

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

idk what is it? is this a question that you know the answer or u dont