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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 · Calculate the total heat required (in joules) to convert 3.95 grams liquid ethanol (C2H5OH) at 25.0 °C to gas at 95.0 °C, given that the boiling point of ethanol is 78.5 °C, heat of vaporization is 40.5 kJ/mol, the specific heat of liquid ethanol is 2.45 J/g-K, and the specific heat of gaseous ethanol is 1.43 J/g-K
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.
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.
Question: The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 60.85 g of ethanol. heat energy: J. Show transcribed image text. There are 2 steps to solve this one.
VIDEO ANSWER: The heat of Vaporization for Ethanol is zero point 825. Kill the jewel program. We have to calculate the amount of heat and jewels which is needed to boil 27 points. The amount of heat is the same as the amount of vaporization time
17 sie 2023 · 1. We know the heat of vaporization for ethanol is 0.826 kJ/g. This means that it takes 0.826 kJ of energy to vaporize 1 gram of ethanol. Step 2/5. 2. We need to find the heat energy required to boil 37.95 g of ethanol. To do this, we can use the following formula: Heat energy = Heat of vaporization x Mass of substance.
