For each of your NCount ACCEPTABLE trials enter the values you calculated for the molarity of your HCl solution in the same sequence corresponding to the equivalence point volumes. You should report the molarities to 4 significant figures, e. g. 0.2314 M.

The question is incomplete, here is the complete question:

For each of your 2 ACCEPTABLE trials enter the values you calculated for the molarity of your HCl solution in the same sequence corresponding to the endpoint volumes.  

Endpoint (in mL):

Trial 1: 13 mL

Trial 2: 12.75 mL

Other info:

10.00 mL of HCl solution in beaker for each trial. 0.15 M NaOH solution in burrett for titration

For Trial 1: The molarity of HCl for trial 1 is 0.195 M

For Trial 2: The molarity of HCl for trial 2 is 0.191 M

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

     ......(1)

where,

are the n-factor, molarity and volume of acid which is HCl

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in equation 1, we get:

Hence, the molarity of HCl for trial 1 is 0.195 M

We are given:

Putting values in equation 1, we get:

Hence, the molarity of HCl for trial 2 is 0.191 M

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

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