Answer:
The correct statement is option c, that is, particles discharged in the air by volcanoes fall to the ground and enrich the soil.
Explanation:
The eruptions of volcanoes lead to the dispersion of ash over the broader regions surrounding the site of eruption. On the basis of the chemistry of the magma, the ash will be comprising different concentrations of soil nutrients. While the major elements found in the magma are oxygen and silica, the eruptions also lead to the discharging of carbon dioxide, water, hydrogen sulfide, sulfur dioxide, and hydrogen chloride.
In supplementation, the eruptions also discharge bits of rocks like pyroxene, potolivine, amphibole, feldspar that are in turn enriched with magnesium, iron, and potassium. As an outcome, the areas which comprise huge deposits of the volcanic soil are quite fertile.
The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or FF atom.
A hydrogen atom acquires a partial positive charge when it is covalently bonded to an FF atom.
A hydrogen bond is possible with only certain hydrogen-containing compounds.
Explanation:
A hydrogen bond does not occur in all hydrogen containing compounds. Hydrogen bonds only occur in those compounds where hydrogen is bonded to a highly electronegative element such as fluorine, oxygen or nitrogen.
In a hydrogen bonded specie, hydrogen acquires a partial positive charge and the electronegative element acquires a partial negative charge which extends throughout the molecule.
2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x
Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5
The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.
K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
<em>K = 0.0556</em>