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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.
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.
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.
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.
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 Step 3/5 3.
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.
Learn more about this topic, chemistry and related others by exploring similar questions and additional content below. Solution for The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 79.95 g of ethanol. J heat energy:
