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  1. Final answer: The heat energy required to boil 49.05 g of ethanol is 40,527.3 joules, calculated by multiplying the mass of ethanol with the heat of vaporization and converting kJ to J. Explanation: 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.

  2. 22 sie 2023 · The heat of vaporization for ethanol is given as 0.826 kJ/g. 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:

  3. 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. Thus, multiplying the given mass of ethanol by its heat of vaporization: q = 66.45 g x 0.826 kJ/g = 54.907 kJ.

  4. 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.

  5. Heat of vaporization= 0.826 KJg-1. Mass of Ethanol = 68.15 g. Let suppose temperature of ethanol is already at boiling point .

  6. EXPERT VERIFIED. Step 1/3. 1. First, we need to convert the given mass of ethanol from grams to kilograms, as the heat of vaporization is given in kJ per gram. 37.55 g = 0.03755 kg. Step 2/3. 2. Next, we can use the formula: Heat energy = mass x heat of vaporization Heat energy = 0.03755 kg x 0.826 kJ/g Heat energy = 0.031 kJ. Answer. 3.

  7. 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.

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