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An element X has a dibrom ide with the empirical formula XBr2 and a dichloride with the empirical formula XCI2. The dibromide is completely converted to the dichloride when it is heated in a stream of chlorine according to the reaction $XBr2 + Cl2 → XCI2 + Br2$ When 1.500 g XBr2 is treated, 0.890 g XCI2 results. (a) Calculate the atomic mass of the element X. (b) By reference to a list of the atomic masses of the elements, identify the element X.

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Answer:

58.85 g/mol is the atomic mass of the element X.

The element X is cobalt.

Explanation:

[tex]XBr_2 + Cl_2\rightarrow XCI_2 + Br_2[/tex]

Moles of [tex]XBr_2=\frac{1.500 g}{M}[/tex] mol

According to reaction, 1 mole of [tex]XBr_2[/tex] gives 1 mole of [tex]XCl_2[/tex]. then [tex]\frac{1.500 g}{M}[/tex] mol of [tex[CBr_2[/tex] will give:

[tex]\frac{1}{1}\times \frac{1.500 g}{M}[/tex] mol=\frac{1.500 g}{M}[/tex] mol[/tex]  of [tex]XCl_2[/tex]

Mass of [tex]\frac{1.500 g}{M}[/tex] moles of [tex]XCl_2[/tex] =0.890 g (given)

Atomic mass of element X  =m

M= [tex]m+79.9 g/mol\times 2[/tex]

M'= [tex]m+35.45g/mol\times 2[/tex]

[tex]\frac{1.500 g}{M}\times M'=0.890 g[/tex]

[tex]\frac{1.500 g}{m+79.9 g/mol\times 2}\times (m+35.45g/mol\times 2)=0.890 g[/tex]

m = 58.806 g/mol

58.56 g/mol is the atomic mass of the element X.

m = 58.806 g/mol  ≈ 58.93 g/mol atomic mass of cobalt

The element X is cobalt.

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