1. Dr. Nielsen finds an old bottle of nitric acid (HNO3) in the stockroom. The label is very faded, but she thinks it says that the concentration of the nitric acid is 9.780 M. Since acids of unknown molarity
are
not useful for experiments, Dr. Nielsen decides to titrate the nitric acid with barium hydroxide
(Ba(OH)2) to determine the actual concentration of the acid.

2HNO3 (aq) + Ba(OH)2 (aq) → H20 (l) + Ba(NO3)2 (aq)

a. Dr. Nielsen uses 25.00 mL of HNO3 in the titration. If the actual concentration of the HNO3 is
9.780 M, what volume (in mL) of 5.062 M Ba(OH)2 would be required to completely react with
the HNO3?
ra

b. Dr. Nielsen finds that 10.20 mL of 5.062 M Ba(OH)2 are required to completely react with the
HNO3. If Dr. Nielsen used 25.00 mL of HNO3 in the titration, what is the actual concentration of
the HNO3?

most historians agree that gabriel mouton, the vicar of st. paul’s church in lyons, france, is the “founding father” of the metric system. he proposed a decimal system of measurement in 1670. mouton based it on the length of one minute of arc of a great circle of the earth (now called a nautical mile, 1852 meters). he also proposed the swing-length of a pendulum with a frequency of one beat per second as the unit of length (about 25 cm). a pendulum beating with this length would have been fairly easy to produce, thus facilitating the widespread distribution of uniform standards. over the years, his work was revised, improved, and extended by a number of french scientists.

the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally