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Unformatted text preview: thus the volume of nitrogen gas produced: V = m r V = 345.1 g .99757 gmL-1 V = 345.9 mL = .3459 L The temperature of the nitrogen gas was assumed to be the same as that of the water in the flask. The number of moles of nitrogen, PV = nRT (.9464)(.3459) = n (.08206)(30.8 + 273) n = .01313 mol N 2 is equal to the number of moles of the unknown nitrate due to the molar ratio in the equation. The molar mass of the unknown compound, 1.194 g .01313 mol = 90.94 g/mol was used to determine the molar mass of the unknown metal by subtracting the molar mass of NO 2 from it. 90.94 46.01 = 44.93 g/mol Conclusion: Knowing the unknown metal must be an alkaline metal, the metal was determined to be potassium (which has a molar mass of 39.10 g/mol) with an error of 14.91% | 44.93- 39.10 | 39.10 100% = 14.91%...
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