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The Molar Heat Of Vaporization For Methane, Ch4, Is 8.53 Kj/Mol. How Much Energy Is Absorbed When 54.8 G Of Methane Vaporizes At Its Boiling Point? Use Q Equals N Delta H.. 6.42 Kj 29.1 Kj 137 Kj 467 Kj. The molar heat of vaporization for methane, ch_4 , is 8.53 kj/mol. 【solved】click here to get an answer to your question :
The molar heat of vaporization for methane, ch_4 , is 8.53 kj/mol. Number of moles = 54.8 g / 16.05 g/mol ≈ 3.42 mol We will use the formula q = nδh, where:.
How much energy is absorbed when 54.8 g of methane vaporizes at its boiling point? The molar heat of vaporization for methane, ch_(4) is 8 8.53kj/mol how much energy is absorbed when 54.8 g of methane. To solve this problem, we need to calculate the amount of energy absorbed when 54.8 grams of methane vaporize at its boiling point.
【solved】click here to get an answer to your question : The amount of energy absorbed when 54.8 g of methane vaporizes at its boiling point is approximately 29.1 kj \[ \text{molar mass of ch}_4 = 12.01 \, \text{g/mol (c)} + 4 \times 1.008 \, \text{g/mol (h)} = 16.042 \, \text{g/mol} \] given the mass of methane is 54.8 g, the number of.
Molar mass of ch₄ = 12.01 g/mol (c) + 4(1.01 g/mol) = 16.05 g/mol. The molar heat of vaporization for methane, ch4, is 8.53 kj/mol. Number of moles = 54.8 g / 16.05 g/mol ≈ 3.42 mol
1 calculate the number of moles of methane in 54.8 g.
