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How much energy is required to vaporize 2 kg of gold with this equation q ml vapor?

To vaporize gold, we need to consider its molar enthalpy of vaporization. The molar enthalpy of vaporization of gold is approximately 334 kJ/mol. Since the molar mass of gold is about 197 g/mol, vaporizing 2 kg (2000 g) of gold requires: ( q = \frac{2000 , \text{g}}{197 , \text{g/mol}} \times 334 , \text{kJ/mol} \approx 3,385 , \text{kJ}. ) Thus, approximately 3,385 kJ of energy is required to vaporize 2 kg of gold.


How much energy is required to vaporize 1.5 kg of aluminum (refer to table of latent heat values.)?

To calculate the energy required to vaporize 1.5 kg of aluminum, we need to use the latent heat of vaporization for aluminum, which is approximately 10,900 J/kg. The energy required can be calculated using the formula: Energy = mass × latent heat of vaporization. Thus, for 1.5 kg of aluminum: Energy = 1.5 kg × 10,900 J/kg = 16,350 J. Therefore, 16,350 joules of energy is required to vaporize 1.5 kg of aluminum.


How much energy is required to vaporize 2 kg if aluminum?

To calculate the energy required to vaporize 2 kg of aluminum, we use the heat of vaporization of aluminum, which is approximately 10,900 J/kg. Therefore, the energy required is 2 kg × 10,900 J/kg = 21,800 J, or 21.8 kJ. This is the amount of energy needed to convert 2 kg of aluminum from a liquid to a vapor at its boiling point.


How much energy is required to vaporize 2kg of aluminum?

To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10.5 MJ/kg. For 2 kg of aluminum, the energy required would be about 21 MJ (megajoules). This calculation assumes that the aluminum is already at its melting point and that no heat losses occur during the process.


How much energy is needed to vaporize 2 kg of aluminum?

To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10,700 kJ/kg. For 2 kg of aluminum, the total energy required would be 2 kg × 10,700 kJ/kg, equating to about 21,400 kJ. Therefore, around 21.4 million joules of energy is needed to vaporize 2 kg of aluminum.

Related Questions

How much energy is required to vaporize 2 kg of gold Use the table below and this equation?

The heat of vaporization of gold is 158 kJ/kg. To find the total energy required to vaporize 2 kg of gold, you can use the equation: Energy = mass * heat of vaporization. Substitute the values to get: Energy = 2 kg * 158 kJ/kg = 316 kJ. Therefore, 316 kJ of energy is required to vaporize 2 kg of gold.


How much energy is needed to vaporize 2 kg of gold?

The energy required to vaporize a material can be calculated using its heat of vaporization. For gold, the heat of vaporization is approximately 330 kJ/mol. Since gold has a molar mass of 196.97 g/mol, 2 kg of gold is equal to 10.15 moles. Therefore, the energy needed to vaporize 2 kg of gold is approximately 3.35 MJ.


How much is required to vaporize 2 kg of copper?

9460 kJ


How much energy of is required to vaporize 1.5 kg of aluminum?

1650kj


How much energy is required to vaporize 1.5 of aluminum?

1650kj


How much energy is required to vaporized 2 kg of copper?

9460 kJ


How much energy is required to vaporize 1.5 kg of aluminum (refer to table of latent heat values.)?

To calculate the energy required to vaporize 1.5 kg of aluminum, we need to use the latent heat of vaporization for aluminum, which is approximately 10,900 J/kg. The energy required can be calculated using the formula: Energy = mass × latent heat of vaporization. Thus, for 1.5 kg of aluminum: Energy = 1.5 kg × 10,900 J/kg = 16,350 J. Therefore, 16,350 joules of energy is required to vaporize 1.5 kg of aluminum.


How much energy is required to vaporize 1.5 kg of aluminum apex answer.com?

The energy required to vaporize 1.5 kg of aluminum can be calculated using the formula: energy = mass * heat of vaporization. The heat of vaporization for aluminum is around 10,000 J/g. So, the energy required would be 1.5 kg * 10,000 J/g = 15,000,000 J or 15,000 kJ.


How much energy is required to vaporize 2 kg if aluminum?

To calculate the energy required to vaporize 2 kg of aluminum, we use the heat of vaporization of aluminum, which is approximately 10,900 J/kg. Therefore, the energy required is 2 kg × 10,900 J/kg = 21,800 J, or 21.8 kJ. This is the amount of energy needed to convert 2 kg of aluminum from a liquid to a vapor at its boiling point.


How much energy is required to vaporize 1.5 kg of copper?

1650kj


How much energy is required to vaporize 2kg of aluminum?

To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10.5 MJ/kg. For 2 kg of aluminum, the energy required would be about 21 MJ (megajoules). This calculation assumes that the aluminum is already at its melting point and that no heat losses occur during the process.


How much energy is needed to vaporize 2 kg of aluminum?

To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10,700 kJ/kg. For 2 kg of aluminum, the total energy required would be 2 kg × 10,700 kJ/kg, equating to about 21,400 kJ. Therefore, around 21.4 million joules of energy is needed to vaporize 2 kg of aluminum.