Starting at 50 degree C, how much heat (in kJ) is used to completely vaporize 83

Question

Starting at 50 degree C, how much heat (in kJ) is used to completely vaporize 83.75 g of 2-propanol? Show all calculations below, and clearly label each step with the process that is occurring (The specific heat capacity of the liquid is 2.56 J/g degree C and the specific heat capacity of the gas is 1.49 J/g. degree C). How much heat is required to vaporize one mole of 2-propanol at its boiling point? Consider the molecules of 2-propanone and 2-propanel(MM = 60.094 g/mol). Fill in the blanks to match each molecule with its heating curve.

Explanation / Answer

3.Use the equation H = mLv or = mHv, where Lv is the latent heat of vaporization and m is mass in g or moles depending on the units of the value of Lv.

heat = H = mLv = (1 mole)/(46 KJ/mol) = 46 KJ/mol

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