Compute the concentration of HCl for all three trials using the formula molarity = mol / L. Trial 1 (20.61 mL HCl): M Trial 2 (20.06 mL HCl): M Trial 3 (19.67 mL HCl): M

Trial 1 (20.61 mL HCl):  

⇒ 0.121 M

Trial 2 (20.06 mL HCl):  

⇒ 0.125 M

Trial 3 (19.67 mL HCl):  

⇒ 0.122 M

the answer for the next question, Compute the average concentration (molarity) of HCl for all three trials, is 0.123 M.

Explanation:

Trial 1 (20.61 mL HCl):  

⇒ 0.121 M

Trial 2 (20.06 mL HCl):  

⇒ 0.125 M

Trial 3 (19.67 mL HCl):  

⇒ 0.122 M

0.123 M.

Explanation:

the answer to your question is c

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.