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LenKa [72]
3 years ago
12

The molar heat of vaporization of hydrogen fluoride (HF) is 31.18 kJ/mol. How much heat is released when 100.0 g HF condenses at

its boiling point?
Chemistry
2 answers:
mash [69]3 years ago
8 0

the answer is -155.8


OLga [1]3 years ago
6 0

<u>Answer:</u> The amount of heat released is 155.9 kJ

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of HF = 100.0 g

Molar mass of HF = 20 g/mol

Putting values in above equation, we get:

\text{Moles of HF}=\frac{100.0g}{20g/mol}=5mol

To calculate the heat of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released =

n = number of moles = 5 moles

\Delta H_{rxn} = enthalpy change of the reaction = 31.18 kJ/mol

Putting values in above equation, we get:

31.18kJ/mol=\frac{q}{5mol}\\\\q=(31.18kJ/mol\times 5mol)=155.9kJ

Hence, the amount of heat released is 155.9 kJ

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The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

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From the balanced equation above,

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MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

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