Gdetermine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solution with an excess of ag ions. the ksp of agbr is 5.0 × 10–13.

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ + .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains ×1000 = 14.411 gm-mole. Thus the solubility product of AgBr = []×.

Or, 5.0× = (the given value of solubility product of AgBr is 5.0× and the charge of the both ions are same).

Thus S = (5.00×) = 7.071× g/mL.

Thus the concentration of Br or HBr is 7.071× g/mL.

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).

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2