On the basis of stoichiometry, justify why 1 mL of 0.025M of Na2S2O3 is equivalent to 1.0 mg/L of DO? Calculate the DO value if your 250 mL sample consumes 18 mL of 0.014M Na2S2O3.

Boron has an average atomic mass of 10.81. one isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percent . the other isotope has a relative abundance of 80.20 percent what is the mass of that isotope? report two decimal places