answersLogoWhite

0


Best Answer

This value is depending on the temperature; read the link below for details.

User Avatar

Wiki User

12y ago
This answer is:
User Avatar
More answers
User Avatar

AnswerBot

3mo ago

The heat of vaporization of nitrogen is 5.56 kJ/mol at its boiling point of -195.79°C. This is the energy required to change 1 mole of liquid nitrogen at its boiling point to gas at the same temperature.

This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: What is the heat of vaporization of nitrogen?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Continue Learning about Chemistry

The molar heats of sublimation and fusion of iodine are 62.3kjmol and 15.3kjmol respectively calculate the molar heat of vaporization of liquid iodide?

The molar heat of vaporization of iodine can be calculated using Hess's Law. The molar heat of sublimation is the sum of the molar heat of fusion and the molar heat of vaporization, so: 62.3 kJ/mol = 15.3 kJ/mol + x kJ/mol. Solving for x, the molar heat of vaporization is 47.0 kJ/mol.


How much heat is absorbed when 75 g of liquid carbon tetrachloride (CCl4) vaporizes into a gas?

The heat absorbed during vaporization is called the heat of vaporization. For carbon tetrachloride, the heat of vaporization is 30.5 kJ/mol. To calculate the heat absorbed when 75 g of CCl4 vaporizes, you would first convert grams to moles using the molar mass of CCl4. Then, use the heat of vaporization to calculate the total heat absorbed.


Estimate the molar heat of vaporization of a liquid whose vapor pressure doubles when the temperature is raised from 85 C to 95 C?

The molar heat of vaporization can be estimated by using the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and molar heat of vaporization. By knowing the temperature change and the corresponding increase in vapor pressure, calculations can be made to determine the molar heat of vaporization.


The heat needed to boil away one gram of a liquid at its boiling point is?

The heat needed to boil away one gram of a liquid at its boiling point is called the heat of vaporization. It is a characteristic property of each substance and represents the energy required to change one gram of the liquid to vapor at its boiling point without a change in temperature.


What is the heat of vaporization for gold?

The heat of vaporization for gold is approximately 330 kJ/mol at its boiling point of 2,700°C. This value represents the amount of energy required to transform one mole of liquid gold into vapor at constant temperature and pressure.

Related questions

Use the data to determine the heat of vaporization of nitrogen?

To determine the heat of vaporization of nitrogen, you would need the enthalpy of vaporization data for nitrogen. This value is typically around 5.57 kJ/mol at its boiling point of -195.79°C. By knowing the enthalpy of vaporization and the conditions at which nitrogen is boiling, you can calculate the heat of vaporization.


The amount of energy needed to change a material from a solid to a liquid is the heat of what?

The latent heat of evaporation


What is water and heat of vaporization?

Heat of Vaporization id the amount of heat needed to transform a liquid into a gas while not raising its temperature.


What is the exact formula of the heat of vaporization?

Heat of vaporization is the amount of heat energy required to change the state of a substance from liquid to gas.q = m·ΔHv, where q = heat energy in Joulesm = mass in gramsΔHv = heat of vaporization in J/g


What is the energy needed for a substance to change from liquid state to a gaseous state?

thermol


What is heat vaorization?

Heat vaporization, also known as heat of vaporization, is the amount of heat energy required to convert a liquid into a gas at its boiling point. This process occurs without a change in temperature. Heat of vaporization is an important property of a substance that determines its behavior during phase changes.


Which substance has the largest heat of vaporization is it F2 Cl2 Br2 O2 I2?

Iodine has the largest heat of vaporization: 41,57 kJ/mol.


The molar heats of sublimation and fusion of iodine are 62.3kjmol and 15.3kjmol respectively calculate the molar heat of vaporization of liquid iodide?

The molar heat of vaporization of iodine can be calculated using Hess's Law. The molar heat of sublimation is the sum of the molar heat of fusion and the molar heat of vaporization, so: 62.3 kJ/mol = 15.3 kJ/mol + x kJ/mol. Solving for x, the molar heat of vaporization is 47.0 kJ/mol.


How much heat is absorbed when 75 g of liquid carbon tetrachloride (CCl4) vaporizes into a gas?

The heat absorbed during vaporization is called the heat of vaporization. For carbon tetrachloride, the heat of vaporization is 30.5 kJ/mol. To calculate the heat absorbed when 75 g of CCl4 vaporizes, you would first convert grams to moles using the molar mass of CCl4. Then, use the heat of vaporization to calculate the total heat absorbed.


How is heat of vaporization related to heat of condensation?

The heat of vaporization and heat of condensation are directly related and have the same magnitude but opposite signs. The heat of vaporization is the energy required to change a substance from liquid to vapor, while the heat of condensation is the energy released when a substance changes from vapor to liquid.


Amount of energy to change a liquid to a gas is called heat of what?

Heat of Vaporization


What heat energy required to break the attraction of liquid molecules is the heat of?

vaporization