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To calculate the heat energy required to boil 49.05 g of ethanol, you can use the given heat of vaporization and the mass of the ethanol. The calculation will utilize the formula Q = m × ΔHvap, where Q is the heat energy, m is the mass of the substance, and ΔHvap is the heat of vaporization.
12 gru 2020 · Answer: The heat energy required to boil 94.15 g of ethanol is 77.7679 kJ. Explanation: The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.
22 sie 2023 · To convert this to joules, we know that 1 kJ = 1000 joules. Therefore, the heat of vaporization for ethanol is 0.826 * 1000 = 826J/g. Now that we have converted the heat of vaporization into joules, we can multiply it by the mass of ethanol to calculate the heat energy required: 826J/g * 70.75g = 58419.5J.
To find the heat energy, we can use the formula: Heat energy (Q) = Mass (m) × Heat of vaporization (L) where Q is the heat energy, m is the mass of ethanol, and L is the heat of vaporization. Now, we can plug in the given values: Q = 54.65 g × 0.826 kJ/g Since we need the answer in joules, we need to convert kJ to J.
Heat of vaporization= 0.826 KJg-1. Mass of Ethanol = 68.15 g. Let suppose temperature of ethanol is already at boiling point .
25 sie 2023 · Explanation: To calculate the heat energy required to boil 66.45 g of ethanol, we use the given value of the heat of vaporization, which is 0.826 kJ/g. The equation for heat energy is q = mass x heat of vaporization.
The molar heat of vaporization of ethanol is 39.3 kJ/mol and the boiling point of ethanol is 78.3 degrees C. Calculate the energy required to heat 100 g of liquid ethanol from 0 degrees...
