A50.0-ml sample of a 1.50 m naoh solution is titrated with a 2.90 m hcl solution. what will be the final volume (sample + titrant) of solution when the naoh has been completely neutralized by the hcl?

In order to proceed calculations for a titration process, it is important that we know the reaction involved. For this case, it would be:

HCl + NaOH = NaCl + H2O

To determine the final volume of the sample and the titrant added, we need to know how much of the titrant or the HCl solution was added into the sample. To do this, we use the given values in the problem and the reaction to relate the substances. We do as follows:

1.50 M NaOH ( .05 L solution ) ( 1 mol HCl / 1 mol NaOH ) = 0.075 mol HCl
0.075 mol HCl / 2.90 M HCl  = 0.0259 L HCl solution or 25.9 mL of the HCl solution was added

Therefore, the total volume of the sample and the titrant would be 50.0 + 25.90 = 75.90 mL.

This is what i found on google “when one phosphate group is removed by breaking a phosphoanhydride bond in a process called hydrolysis, energy is released, and atp is converted to adenosine diphosphate (adp). likewise, energy is also released when a phosphate is removed from adp to form adenosine monophosphate (amp).” i hope this you .