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An organic mixture weighing 2.416 grams contains inert materials and benzoic acid. The benzoic acid is extracted into 100.0 mL of 0.1513M NaOH. After extraction, the excess base is titrated with .01513 M HCl, with 41.67mL required to reach the equivalence point. What is the mass percent of benzoic acid in the original mixture?
What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under the following conditions?
Initially the stopcocks are closed. One of the flasks contains 2.0 of , the second 3.7 of , and the third, 0.10 of ethanol. The vapor pressure of at 308 is 42 and that of ethanol is 102 . The stopcocks are then opened and the contents mix freely.
explain why it is appropriate to group a polyatomic ion in parentheses in a chemical formula, if more than one of that ion is present in the formula.
assume you dissolve .297 g of the weak acid benzoic acid in enough water to make 105 ml of solution and then titrate
Calculate the concentration of FeSCN^2+ after 25 mL of 0.002 M Iron nitrate Fe(NO3)3 is combined with 15.0 mL of 0.200 M
A student mixes four reagents together, thinking that the solution will neutralize each other. The solutions mixed together are 53.6 mL of 0.129 M hydrochloric acid, 104.1 mL of 0.215 M of nitric acid, 501.5 mL of 0.0126 M calcium hydroxide
calculate the density of a solid cube that measures 4.79 cm on each side and has a mass of
Calculate the molar solubility of Fe(OH)2 in a buffer solution where the pH has been fixed at the indicated values. Ksp = 7.9 x 10-16. When pH is 7.8
A 1.50 mole sample of an ideal gas at 28.5 degrees C expands isothermally from an initial volume of 22.5dm^3 to a final volume of 75.5dm^3
A sample of H2 gas (2.0 L) at 3.5 atm was combined with 1.5 L of N2 gas at 2.6 atm pressure at a constant temperature of 25 °C into a 7.0 L flask
An alternate method for determining rate law is to moniter concentration of each reactant continuously over span of rxn. explain no more in 1 paragraph how rate law is derived from data obtained.
Calculate the potential E for the following cell: Fe(s)| Fe(NO3)2 (0.050M || Cu(NO3)2 (0.020 M) | Cu(s) Cell reaction: Cu+2 + Fes ? Cus + Fe+2 Fe+2 + 2e- ? Fe(S) Eo = -0.44 V Cu+2 + 2e- ? Cu(s) Eo = 0.339V
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