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devlian [24]
3 years ago
7

2) Calculate the percent composition of each element in Mgso,

Chemistry
1 answer:
iogann1982 [59]3 years ago
7 0

Answer:

2)

\% Mg=20.2\%\\\\\% S=26.6\%\\\\\% O=53.2\%

3)

\% Ag=93.1\%\\\\\% O=6.9\%

Explanation:

Hello!

2) In this case, since magnesium sulfate is MgSO₄, we can see how magnesium weights 24.305 g/mol, sulfur 32.06 g/mol and oxygen 64.00 g/mol as there is one atom of magnesium as well as sulfur but four oxygen atoms for a total of g/mol; thus the percent compositions are:

\% Mg=\frac{24.305}{120.36 } *100\%=20.2\%\\\\\% S=\frac{32.06}{120.36 } *100\%=26.6\%\\\\\% O=\frac{64.00}{120.36 } *100\%=53.2\%

3) In this case, although the element seems to contain Ag and O, we infer its molecular formula is Ag₂O; thus, since we have two silver atoms weighing 215.74 g/mol and one oxygen atom weighing 16.00 g/mol for a total of 231.74 g/mol, we obtain the following percent compositions:

\% Ag=\frac{215.74}{231.74} *100\%=93.1\%\\\\\% O=\frac{16.00}{231.74} *100\%=6.9\%

Best regards!

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The correct answer for the question that is being presented above is this one: "<span>0.3."

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Answer:

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The dissociation reaction of hydrogen cyanide can be given as

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