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Fed [463]
3 years ago
9

A gas sample is at 25°C and 1.0 atmosphere. Which changes in temperature and pressure will cause this sample to behave more like

an ideal gas?
(1) decreased temperature and increased pressure
(2) decreased temperature and decreased pressure
(3) increased temperature and increased pressure
(4) increased temperature and decreased pressure
Chemistry
1 answer:
-BARSIC- [3]3 years ago
3 0
What's the answer. tell me please please please please please please please please please please please please 
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Which of the following statements about the combustion of glucose with oxygen to form water and carbon dioxide (C6H12O6 + 6 O2 →
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Answer:

The correct statement is:

E - The entropy of the products is greater than the entropy of the reactants.

Explanation:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

As glucose is a large molecule and then it is transformed into many molecules of water and carbon dioxide, the entropy of the system increases. If the number of molecules increases, the disorder increases.

Initial state: 7 molecules (1 glucose + 6 oxygen)

Final state: 12 molecules (6 carbon dioxide + 6 water)

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Estimate ΔG°rxn for the following reaction at 775 K. 2Hg(g) + O2(g) → 2HgO(s) ΔH°= -304.2 kJ; ΔS° = -414.2 J/K
hichkok12 [17]

Answer:

\Delta G^0_{rxn}=16.81kJ

Explanation:

Hello!

In this case, since the Gibbs free energy of any process is related with the enthalpy change, temperature and entropy change as shown below:

\Delta G^0=\Delta H^0-T\Delta S^0

For a chemical reaction it is simply modified to:

\Delta G^0_{rxn}=\Delta H^0_{rxn}-T\Delta S^0_{rxn}

Thus, since the enthalpy of reaction is given as -304.2 kJ and the entropy as -414.2 J/K (-0.4142 kJ/K), at 775 K the Gibbs free energy of reaction turns out:

\Delta G^0_{rxn}=-304kJ-775K*(-0.4142\frac{kJ}{K} )\\\\\Delta G^0_{rxn}=16.81kJ

Whose result means this is a nonspontaneous reaction.

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Answer:

+ 3

Ex planation:

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hope this helps!

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