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user100 [1]
2 years ago
11

If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?

Chemistry
1 answer:
photoshop1234 [79]2 years ago
7 0
The combustion reaction is as expressed,

                CxHy + O2 --> CO2 + H2O

The mass fraction of carbon in CO2 is 3/11. Hence,
       mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.

Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g. 

     moles of C = 0.86 g C / 12 g = 0.0713
     moles of H = 0.14 g H / 1 g  = 0.14

The empirical formula for the hydrocarbon is therefore, CH₂.
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When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for
balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-

Whereas the half reactions are:

Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O

Next, we exchange the transferred electrons:

1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow  4N^{4+}O^{2-}_2+4H_2O

Afterwards, we add them to obtain:

Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O

By adding and subtracting common terms we obtain:

Sn^0+4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O

Finally, by removing the oxidation states we have:

Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

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3 years ago
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Explanation:

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8 0
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2 years ago
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Answer: Percent composition by element

Element Symbol Mass Percent

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