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rachel2005smith
28.09.2021 •
Chemistry
Which statement correctly explains why H2S is a stronger acid than water?
Sulfur (S) has a larger atomic size than oxygen (O), so H2S is a more stable, weaker base than H2O.
Sulfur (S) has a larger atomic size than oxygen (O), so H, 2, S, , is a more stable, weaker base than H, 2, O.
Sulfur (S) has a larger atomic size than oxygen (O), so H2S is a more stable, stronger acid than H2O.
Sulfur (S) has a larger atomic size than oxygen (O), so H, 2, S, , is a more stable, stronger acid than H, 2, O.
Sulfur (S) has a larger atomic size than oxygen (O), so HS– is a more stable, stronger acid than OH–.
Sulfur (S) has a larger atomic size than oxygen (O), so HS, – , is a more stable, stronger acid than OH, –, .
Sulfur (S) has a larger atomic size than oxygen (O), so HS– is a more stable, weaker base than OH–.
Sulfur (S) has a larger atomic size than oxygen (O), so HS, – , is a more stable, weaker base than OH, –, .
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Ответ:
Enthalpy of formation of strontium chloride, SrCl = -558.1 kJ/mol
Explanation:
Sr(s) > Sr(g) ΔHsublimation = +164 kJ/mol
Sr⁺(g) > Sr⁺(g) First Ionization energy, IE₁ = +549 kJ/mol
Sr⁺(g) > Sr²⁺(g) Second ionization energy, IE₂ = +1064 kJ/mol
I₂(s) > I₂(g) ΔHsublimation = +62.4 kJ/mol
I₂(g) > 2I(g) Bond dissociation energy, BE = +152.55 kJ/mol
2I(g) > 2I⁻(g) Electron affinity, EA = 2 (-295.15 kJ/mol) = -590.3 kJ/mol
Sr²⁺(g) + 2I⁻(g) > SrI₂(s) Lattice energy, LE = -1959.75 kJ/mol
Overall equation for the formation of SrCl₂ is given as:
Sr(s) + I₂(s) > SrI₂(s) ΔHformation = ?
Enthalpy of formation, ΔHformation = (ΔHsublimation of Sr(s) + IE₁ + IE₂ + ΔHsublimation of I₂(s) + BE + EA + LE)
ΔHformation = {164 + 549 + 1064 + 62.4 + 152.55 + (-590.3) + (-1959.75)} kJ/mol
Sr(s) + I₂(s) > SrI₂(s) ΔHformation = -558 kJ/mol
Therefore, enthalpy of formation of strontium chloride, SrCl = -558.1 kJ/mol