Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)
Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol
Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol
<em>Answer: -1367.5 kJ/mol</em>
Answer:
A) increasing dispersion interactions
Explanation:
Polarizability allows gases containing atoms or nonpolar molecules (for example, to condense. In these gases, the most important kind of interaction produces <em>dispersion forces</em>, <em>attractive forces that arise as a result of temporary dipoles induced in atoms or molecules.</em>
<em>Dispersion forces</em>, which are also called <em>London forces</em>, usually <u>increase with molar mass because molecules with larger molar mass tend to have more electrons</u>, and <u>dispersion forces increase in strength with the number of electrons</u>. Furthermore, larger molar mass often means a bigger atom whose electron distribution is more easily disturbed because the outer electrons are less tightly held by the nuclei.
Because the noble gases are all nonpolar molecules, <u>the only attractive intermolecular forces present are the dispersion forces</u>.
The chemical could have more or less of a reaction to the other chemicals in the experiment
<span> </span> <span>V = nRT/P =
(0.875)(0.082057)(273)/(1) = 19.6 L</span>
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