Search results
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
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.
Heat of vaporization= 0.826 KJg-1. Mass of Ethanol = 68.15 g. Let suppose temperature of ethanol is already at boiling point .
29 wrz 2023 · Instant Answer. We are given the heat of vaporization for ethanol, which is 0.826 kJ/g. We are asked to calculate the heat energy required to boil 75.25 g of ethanol. Therefore, the heat energy required to boil 75.25 g of ethanol is 62.1655 kJ. Final Answer: \boxed {62.1655 \, \text {kJ}}
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
Science. Chemistry questions and answers. The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 43.65 g of ethanol, heat energy. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer.
