In the store room, I have a bottle of 18 M acetic acid. I would like to make a dilute solution. I need 150.0 mL of 3.852 M acetic acid. How much of the concentrated acetic acid should I use (in mL)? How do I make this solution? Be specific.

M1V1 = M2V2

M1 = 18 M  M2 = 3.852 M

V1 = ?V2 = 150.0 mL

(18 M)(V1)     =  (3.852 M)(150.0 mL)

    18 M       18 M

V1 = 32.1 mL

First measure the mass of the compound. Next add water to the calculated mass in the beaker to dissolve it, and pour the solution into the flask you will use. Since we're only going to need 150.0 mL, find a flask that holds up to 150 mL. Rinse the beaker with water multiple times and pour the water into the flask until the solution almost makes it to the neck. Mix the solution thoroughly until it's all dissolved. Very carefully fill the flask until it gets to the 150 mL mark. Turn your flask upside down and rightside up at least 10 times carefully to make sure you mixed the solution properly. Then see the results.

Explanation:

You also taking this clc? Don.t copy word from word lol

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :