Daniel C. Harris: 7-4, 8,10,12,16
7-4) A 1.000g sample of unknown analyzed by Reaction 7-2 gave 2.500g of bis(dimethylglyoximate) nickel(II). Find the wt% of Ni in the unknown.
7-8) The man in the vat. Once upon a time, a workman at a dye factory fell into a vat containing a hot concentrated mixture of sulfuric and nitric acids, and he dissolved! Because nobody witnessed the accident, it was necessary to prove that he fell in so that the man’s wife could collect his insurance money. The man weighed 70kg, and a human body contains about 6.3 parts per thousand phosphorus. The acid in the vat was analyzed for phosphorus to see if it contained a dissolved human.
a) The vat had 8.00 x 10 raise to 3L of liquid, and 100.0mL were analyzed. If the man did fall into the vat, what is the expected quantity of phosphorus in 100.0mL?
b) The 100.0mL sample was treated with a molybdate reagent that precipitates ammonium phosphomolybdate, (NH4)3 [P(Mo12O40)] x 12H2O. This substance was dried at 110 degree C to remove waters of hydration and heated to 400 degree C until it reached a constant composition corresponding to the formula P2O5 x 24MoO3, which weighed 0.3718g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.0331g of P2O5 x 24MoO3 (FM 3596.46) was produced. This blank determination gives the amount of phosphorus in the starting reagents. The P2O5 x 24MoO3 that could have come from the dissolved man is therefore 0.3718 – 0.0331 = 0.3387g. How much phosphorus was present in the 100.0mL sample? Is this quantity consistent with a dissolved man?
7-10) Finely ground mineral (0.6324g) was dissolved in 25mL of boiling 4M HCl and diluted with 175mL H2O containing two drops of methyl red indicator. The solution was heated to 100 degree C, and 50mL of warm solution containing 2.0g (NH4)2 C2O4 were slowly added to precipitate CaC2O4. Then 6M NH3 was added until the indicator changed from red to yellow,... [continues]
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