methane (CH4) has a normal boiling point of -161.6*c at this temprature the H concentration = -8.17kj/mol. if 16.5g of liquid methane vaporize. how much energy is absorbed
To find out how much energy is absorbed when 16.5g of liquid methane vaporizes, we need to use the molar mass of methane (CH4) and the enthalpy of vaporization.
1. Calculate the number of moles of methane:
Molar mass of CH4 = 12.01g/mol + 4(1.008g/mol) = 16.04g/mol
Number of moles of CH4 = Mass / Molar mass = 16.5g / 16.04g/mol = 1.0266 mol
2. Calculate the energy absorbed:
Energy absorbed = Number of moles * Enthalpy of vaporization
Energy absorbed = 1.0266 mol * -8.17 kJ/mol = -8.38 kJ
(Note: The negative sign indicates that energy is absorbed.)
Therefore, when 16.5g of liquid methane vaporizes, approximately 8.38 kJ of energy is absorbed.
To calculate the energy absorbed when 16.5g of liquid methane vaporizes, we need to use the heat of vaporization (H), which is given as -8.17 kJ/mol.
First, we need to find the number of moles of methane that vaporize. We can use the molar mass of methane (CH4), which is 16.04 g/mol.
Number of moles = mass / molar mass
Number of moles = 16.5g / 16.04 g/mol
Number of moles = 1.027 mol
Now we can calculate the energy absorbed:
Energy absorbed = Number of moles * Heat of vaporization
Energy absorbed = 1.027 mol * -8.17 kJ/mol
Therefore, the energy absorbed when 16.5g of liquid methane vaporizes is approximately -8.38 kJ.
To calculate the amount of energy absorbed when 16.5g of liquid methane vaporizes, you can use the formula:
Energy absorbed = moles of methane × enthalpy of vaporization
First, let's determine the number of moles of methane. We know the molar mass of methane (CH4) is 16.04 g/mol. To find the moles, divide the given mass by the molar mass:
moles = mass / molar mass
moles = 16.5g / 16.04g/mol ≈ 1.03 mol
Next, we need to calculate the energy absorbed using the enthalpy of vaporization of methane. The enthalpy of vaporization (ΔHvap) for methane is given as -8.17 kJ/mol.
Energy absorbed = moles × enthalpy of vaporization
Energy absorbed = 1.03 mol × -8.17 kJ/mol
Energy absorbed ≈ -8.44 kJ
Therefore, the amount of energy absorbed when 16.5g of liquid methane vaporizes is approximately 8.44 kJ (rounded to two decimal places).