What volume of 0.215 M HCl is required to neutralize 50.0 mL of
0.800 M NaOH?

186 mL HCl

Explanation:

M1V1 = M2V2

M1 = 0.215 M HCl

V1 = ?

M2 = 0.800 M NaOH

V2 = 50.0 mL

Solve for V1 --> V1 = M2V2/M1

V1 = (0.800 M)(50.0 mL) / (0.215 M) = 186 mL HCl

can you list the answers so i can answer

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.