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Answer :
Sure, let's work through the problem step-by-step:
1. Find the molar mass of ammonia (NH₃):
- Ammonia (NH₃) consists of 1 nitrogen atom and 3 hydrogen atoms.
- The atomic mass of nitrogen (N) is approximately 14.01 g/mol.
- The atomic mass of hydrogen (H) is approximately 1.01 g/mol.
[tex]\[ \text{Molar mass of NH}_3 = 14.01 \, \text{g/mol for N} + 3 \times 1.01 \, \text{g/mol for H} = 17.03 \, \text{g/mol} \][/tex]
2. Calculate the number of moles in 50.0 g of ammonia:
Use the formula:
[tex]\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{Number of moles of NH}_3 = \frac{50.0 \, \text{g}}{17.03 \, \text{g/mol}} \approx 2.94 \, \text{moles} \][/tex]
3. Calculate the heat released during condensation:
The standard enthalpy of vaporization (ΔHvap) is given as 23.3 kJ/mol. Since condensation is the process of a gas turning into a liquid, it is the reverse of vaporization. Thus, the enthalpy change is negative for condensation.
[tex]\[ \text{Heat released} = - (\Delta H_{vap} \times \text{moles of NH}_3) \][/tex]
[tex]\[ \text{Heat released} = -(23.3 \, \text{kJ/mol} \times 2.94 \, \text{moles}) \approx -68.4 \, \text{kJ} \][/tex]
4. Select the correct answer:
The amount of heat released when 50.0 g of ammonia is condensed at [tex]\(-33^{\circ}C\)[/tex] is approximately [tex]\(-68.4 \, \text{kJ}\)[/tex].
Therefore, the correct answer is A. -68.4 kJ.
1. Find the molar mass of ammonia (NH₃):
- Ammonia (NH₃) consists of 1 nitrogen atom and 3 hydrogen atoms.
- The atomic mass of nitrogen (N) is approximately 14.01 g/mol.
- The atomic mass of hydrogen (H) is approximately 1.01 g/mol.
[tex]\[ \text{Molar mass of NH}_3 = 14.01 \, \text{g/mol for N} + 3 \times 1.01 \, \text{g/mol for H} = 17.03 \, \text{g/mol} \][/tex]
2. Calculate the number of moles in 50.0 g of ammonia:
Use the formula:
[tex]\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{Number of moles of NH}_3 = \frac{50.0 \, \text{g}}{17.03 \, \text{g/mol}} \approx 2.94 \, \text{moles} \][/tex]
3. Calculate the heat released during condensation:
The standard enthalpy of vaporization (ΔHvap) is given as 23.3 kJ/mol. Since condensation is the process of a gas turning into a liquid, it is the reverse of vaporization. Thus, the enthalpy change is negative for condensation.
[tex]\[ \text{Heat released} = - (\Delta H_{vap} \times \text{moles of NH}_3) \][/tex]
[tex]\[ \text{Heat released} = -(23.3 \, \text{kJ/mol} \times 2.94 \, \text{moles}) \approx -68.4 \, \text{kJ} \][/tex]
4. Select the correct answer:
The amount of heat released when 50.0 g of ammonia is condensed at [tex]\(-33^{\circ}C\)[/tex] is approximately [tex]\(-68.4 \, \text{kJ}\)[/tex].
Therefore, the correct answer is A. -68.4 kJ.
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