A sample of a substance with the empirical formula XCl3 weighs 0.4393 g. When it is...
A sample of a substance with the empirical formula XCl3 weighs 0.4393 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 1.1926 g. The chemical reaction is
XCl3 + 3 AgNO33AgCl + X(NO3)3
(a) Calculate the formula mass of XCl3.
Formula mass XCl3 =
g mol-1
(b) Calculate the atomic mass of X.
Atomic mass X =
g mol-1
Homework Answers
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A sample of a substance with the empirical formula XCl3 weighs
0.4393 g. When it is...
A sample of a substance with the empirical formula XCl3 weighs 0.5183 g. When it is...
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A sample of a substance with the empirical formula XCl3 weighs 0.5183 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCI by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to he 1.4071 g. The chemical reaction is XCl3 +3 AgNO3 3AgCI + X(NO3)3 (a) Calculate the formula mass of XCl3. Formula mass XCI3- g mol (b) Calculate the atomic mass of X Atomic mass X-...
A sample of a substance with the empirical formula XC1_2 weighs 0.5785 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 1.2794 g. The chemical reaction is XCI_2 + 2 AgNO_3 rightarrow 2 AgCI + X(NO_3)_2 Calculate the formula mass of XCI_2. Formula mass XCI_2 = g mol^-1 Calculate the atomic mass of X. Atomic...
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