sample of a substance with the empirical formula XCl2 weighs 0.4041 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCl by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to be 0.9424 g. The chemical reaction is XCl2 + 2 AgNO32AgCl + X(NO3)2 (a) Calculate the formula mass of XCl2. Formula mass XCl2 = g mol-1 (b) Calculate the atomic mass of X. Atomic mass X = g mol-1
sample of a substance with the empirical formula XCl2 weighs 0.4041 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCl by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to be 0.9424 g. The chemical reaction is XCl2 + 2 AgNO32AgCl + X(NO3)2 (a) Calculate the formula mass of XCl2. Formula mass XCl2 = g mol-1 (b) Calculate the atomic mass of X. Atomic mass X = g mol-1
Chemistry
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A sample of a substance with the empirical formula XCl2 weighs 0.4041 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCl by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to be 0.9424 g. The chemical reaction is XCl2 + 2 AgNO32AgCl + X(NO3)2
(a) Calculate the formula mass of XCl2.
Formula mass XCl2 = g mol-1
(b) Calculate the atomic mass of X.
Atomic mass X = g mol-1
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