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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 calculate the heat energy required to boil 75.25 g of ethanol, we need to use the formula: Q = m * ΔHv. Where Q is the heat energy, m is the mass of ethanol, and ΔHv is the heat of vaporization of ethanol. Substituting the given values, we get: Q = 75.25 g * 0.826 kJ/g Q = 62.1785 kJ
3 maj 2023 · The heat of vaporization is a measure of how much heat energy is required to break the intermolecular forces between the molecules of a liquid and convert it to a gas at a constant temperature. In the case of ethanol, it takes 0.826 kJ of heat energy to vaporize one gram of the liquid.
17 sie 2023 · Step 1/5 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 Step 3/5 3.
Heat of vaporization= 0.826 KJg-1. Mass of Ethanol = 68.15 g. Let suppose temperature of ethanol is already at boiling point .
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.
Answer. 3. Finally, we can convert the answer to joules: Heat energy = 0.031 kJ x 1000 J/kJ Heat energy = 31.0 J Therefore, the heat energy required to boil 37.55 g of ethanol is 31.0 J. Video Answer. 27 people are viewing now. Oops! There was an issue loading this video. View Best Match Solution Instead. Instant Answer. EXPERT VERIFIED. Step 1/3.
