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Taya2010 [7]
3 years ago
13

Pls help this is due tonight pls and thnks

SAT
2 answers:
umka2103 [35]3 years ago
7 0

Answer:

D. 3.3 g/mL

Explanation:

Firlakuza [10]3 years ago
4 0

Answer:

C) 3.0/mL should be correct

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Find δs o for the combustion of methane to carbon dioxide and liquid water.
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The entropy change ΔS for the combustion of methane to carbon dioxide and liquid water is -242.6 J/mol.K

<h3>What is the Entropy in a Chemical system?</h3>

In a chemical system, Entropy (ΔS) is a quantitative measure of the variety of possible ways to disperse energy inside a chemical system.

  • In an isothermal process, entropy is determined as the quantity of heat transmitted to a system.

The chemical reaction for the combustion of methane can be expressed as:

\mathbf{CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}}

The standard entropy of the combustion is expressed as:

\mathbf{\Delta S^0 =(S^0 _{CO_{2}} + 2 \times S^0 _{H_{2}O} )+(-S^0 _{CH_{4}}  -2 \times S^0 _{O_{2}} })

Replacing the standard entropies of the compounds into the above equation, we have:

\mathbf{\Delta S^0 =(213.8 \8 \ J/mol.K + 2 \times69.92 \ J/mol.K  )+(-186.16 \ J/mol.K -2 \times 205.04 \ J/mol.K})

\mathbf{\Delta S^0 =-242.6 \ J/mol.K}

Learn more about entropy here:

brainly.com/question/6364271

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