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s344n2d4d5 [400]
3 years ago
11

A certain substance has a heat of vaporization of 70.83 kJ / mol. 70.83 kJ/mol. At what Kelvin temperature will the vapor pressu

re be 5.00 5.00 times higher than it was at 331 K?
Chemistry
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

The answer to the question is

The temperature at which the vapor pressure will be 5.00 times higher than it was at 331 K is 353.0797 K.

Explanation:

To solve the question, we make use of the Clausius-Clapeyron equation as follows

ln(\frac{P_2}{P_1}) = \frac{d_{vap}H}{R} (\frac{1}{T_1} -\frac{1}{T_2} )

Where P₁ = Initial pressure

P₂ = Final pressure

T₁ = Initial temperature = 331 K

T₂ = Final temperature

dvapH = ΔvapH = Heat of vaporization = 70.83 kJ / mol.

R = Universal gas constant = 8.3145. J K⁻¹ mol⁻¹

We are required to find the temperature when P₂ = 5 × P₁

Therefore we have

ln(\frac{5*P_1}{P_1}) = \frac{70.83}{8.3145} (\frac{1}{331} -\frac{1}{T_2} ) = ln(5) = 8518.853 (\frac{1}{331} -\frac{1}{T_2} ) or T₂ = \frac{1}{\frac{1}{331} -\frac{ln(5)}{8518.853} } = 353.0797 K

The vapor pressure be 5.00 times higher than it was at 331 K when the temperature is raised to 353.0797 K.

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A sample of glucose (C6H1206) contains 0.75 moles of oxygen. How many moles of hydrogen does this sample contain?
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Explanation:

Hello,

In this case, given that one mole of glucose, contains six moles of oxygen (subscript), we can also see it contains twelve moles of hydrogen (subscript), therefore, the moles of hydrogen in the sample are computed by:

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Answer:

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Explanation:

Molar mass is defined as the mass in 1 mole of the substance. It is calculated by adding the molar mass of the substituents each multiplied by the subscript they have in the formula.

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