T₁ = 50,14 K.
p₁ = 258,9 torr.
T₂ = 161,2 K.
p₂ = 277,5 torr.
R = 8,314 J/K·mol.
Using Clausius-Clapeyron equation:
ln(p₁/p₂) = - ΔHvap/R · (1/T₁ - 1/T₂).
ln(258,9 torr/277,5 torr) = -ΔHvap/8,314 J/K·mol · (1/50,14 K - 1/161,2 K).
-0,069 = -ΔHvap/8,314 J/K·mol · (0,0199 1/K - 0,0062 1/K).
0,0137·ΔHvap = 0,573 J/mol.
ΔHvap = 41,82 J.

5 0

3 years ago

CAN SOMEONE HELP PLEASE!!!!

miv72 [106K]

Answer: The Hubble

Explanation:

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3 years ago

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Calculate the molarity of the acid solution.<br>H2SO4(aq) with a pH of 2.

Lena [83]

Concentration "molarity" of H₂SO₄ in this solution:

5 × 10⁻³ mol / dm³.

<h3>Explanation</h3>

What's the concentration of H⁺ ions in this solution?

$[\text{H}^{+}] = 10^{-\text{pH}}$ ,

where $[\text{H}^{+}]$ is in the unit mol / dm³.

$\text{pH} = 2$

$[\text{H}^{+}] = 10^{-2} \;\text{mol}\cdot\text{dm}^{-3}$ .

What's the concentration "molarity" of H₂SO₄ in this solution?

Sulfuric acid H₂SO₄ is a strong acid. Note the subscript "2". Each mole of this acid dissolves in water to produce two moles of H⁺ ions. It takes only $\dfrac{10^{-2}}{2} = 5 \times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ of H₂SO₄ to produce twice as much H⁺ ions.

As a result, the <em>molarity</em> of H₂SO₄ is 5 × 10⁻³ mol / dm³ or 0.005 M.

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3 years ago