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marissa [1.9K]
4 years ago
5

Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less s

table than those of the other halogens. Bonding theory, however, does allow one to propose structures for these acids and use formal charges for the evaluation of these structures. For a molecule of fluorous acid, the atoms are arranged as HOFO. (Note: In this oxyacid, the placement of fluorine is an exception to the rule of putting the more electronegative atom in a terminal position.) What is the formal charge on each of the atoms? Enter the formal charges in the same order as the atoms are listed. Express your answers as charges separated by comma. For example, a positive one charge would be written as +1.
Chemistry
1 answer:
inysia [295]4 years ago
5 0

Answer:

Formal Charge for H, O, F, O = 0,0,1,-1

While an octet of electrons is attained by the oxygen and fluorine atoms in this structure and the formal charges add to zero, having to assign a formal charge of +1 to such an electronegative atom as fluorine may indicate that such a structure is not likely to occur. In fact, fluorine commonly only forms one oxyacid, HOF.

Explanation:

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Answer:

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3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

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Answer:

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Explanation:

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It expresses a cube using cm
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