An experiment requires 0.7 M NH3(aq). The

stockroom manager estimates that 15 L of
the base is needed. What volume of 15 M
NH3(aq) will be used to prepare this amount of 0.7 M base?
Answer in units of mL

See your post above.

To determine the volume of 15 M NH3(aq) needed to prepare 0.7 M NH3(aq), we can use the following formula:

C1V1 = C2V2

Where:
C1 = concentration of the initial solution (15 M NH3(aq))
V1 = volume of the initial solution (unknown, in mL)
C2 = concentration of the final solution (0.7 M NH3(aq))
V2 = volume of the final solution (15 L)

Rearranging the formula, we have:

V1 = (C2V2) / C1

Now, let's substitute the given values:
C1 = 15 M
V2 = 15 L = 15000 mL
C2 = 0.7 M

Plugging the values into the formula:

V1 = (0.7 M * 15000 mL) / 15 M

Calculating:

V1 = 0.7 M * 1000 mL = 700 mL

Therefore, to prepare 0.7 M NH3(aq) using the 15 M NH3(aq) stock solution, you will need 700 mL of the 15 M NH3(aq) solution.