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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.
What is the min no of weighing operations required to measure 31 kg of rice if only one stone of 1 kg is available?
32 weighing operations are required to measure 31 kg of rice with a 1 kg stone. Start by weighing 1 kg, then 2 kg, 3 kg, and so on up to 31 kg. Each weighing operation will add 1 kg to the total.
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 of copper Use the table below and this equation?
To calculate the energy required to vaporize 2 kg of copper, you need to use the specific latent heat of vaporization for copper, which is approximately 300,000 J/kg. The total energy (Q) can be calculated using the equation Q = m * L, where m is the mass (2 kg) and L is the latent heat of vaporization (300,000 J/kg). Thus, Q = 2 kg * 300,000 J/kg = 600,000 J. Therefore, 600,000 joules of energy is required to vaporize 2 kg of copper.
How many nail in kg 100 sq MTr required?
100 and 150 kg of nails is what is needed for the 100 square metre.
What is the criteria for Jr KG admission in Bombay Scottish School Bombay and which website for the same with all details?
The criteria for Jr KG admission in Bombay Scottish School Bombay and which website for the same with all details is the infinitycourses official website.
What is the minimum age require for admission in primary school in India?
DOE notice on Minimum age for admission in pre school-nursery and kG pre primary in Delhi 1) Nursery/Pre School age to be 3 as on 31st March 2011. 2) KG/Pre Primary age to be 4 as on 31st March 2011.
Jerry weighed 155 lbs on admission He has lost 4.5 kg since admission Today he weighs kg?
65.807 kg
When is the admission form given for KG Class at auxilium convent school in Dum Dum kolkata?
when the school starts from a new session then nursery will have admission form
What kg of piston required to crush 31gm tin can?
31 gm to kg
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.
What is the min no of weighing operations required to measure 31 kg of rice if only one stone of 1 kg is available?
32 weighing operations are required to measure 31 kg of rice with a 1 kg stone. Start by weighing 1 kg, then 2 kg, 3 kg, and so on up to 31 kg. Each weighing operation will add 1 kg to the total.
How much energy is required to vaporize 1.5 kg of copper?
1650kj
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 heat is required to convert 1 kg of ice to liquid?
lf = 3.35 x 105 J kg-1 This much amount of heat required to convert 1 kg of ice to liquid Mani.Ra
How much energy is required to melt 1.5 kg of gold?
The heat of fusion for gold is 64.4 kJ/mol. To convert this to energy required to melt 1.5 kg of gold, we need to calculate the number of moles in 1.5 kg of gold (1.5 kg of gold is approximately 0.047 moles). Then, the energy required would be approximately 3.03 kJ.
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.
