Question

Butane (C4H10) has a heat of vaporization of 22.44 kJ>mol and a normal boiling point of -0.4 x1fC. A 250.0 mL sealed flask contains 0.55 g of butane at -22 x1fC. How much liquid butane is present? If the butane is warmed to 25 x1fC, how much liquid butane is present?

Name: Butane (C4H10) has a heat of vaporization of 22.44 kJ>mol and a normal boiling point of -0.4 x1fC. A 250.0 mL sealed flask contains 0.55 g of butane at -22 x1fC. How much liquid butane is present? If the butane is warmed to 25 x1fC, how much liquid butane is present?
Uploaded: 2021-11-14T14:07:35-08:00
Duration: 2 min 20 s
Channel: Hitendra Singh
Description: Butane (C4H10) has a heat of vaporization of 22.44 kJ>mol and a normal boiling point of -0.4 x1fC. A 250.0 mL sealed flask contains 0.55 g of butane at -22 x1fC. How much liquid butane is present? If the butane is warmed to 25 x1fC, how much liquid butane is present?

          Butane (C4H10) has a heat of vaporization of 22.44 kJ>mol and a normal boiling point of -0.4 x1fC. A 250.0 mL sealed flask contains 0.55 g of butane at -22 x1fC. How much liquid butane is present? If the butane is warmed to 25 x1fC, how much liquid butane is present?

Added by Kimberly M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Hitendra Singh

11/14/2021

Step 1

Given: Mass of butane = 0.55 g Molar mass of butane (C4H10) = 58.12 g/mol Number of moles = Mass / Molar mass Number of moles = 0.55 g / 58.12 g/mol Number of moles = 0.0095 mol Show more…

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Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 °C. A 250.0 mL sealed flask contains 0.55 g of butane at -22 °C. How much liquid butane is present? If the butane is warmed to 25 °C, how much liquid butane is present?