In the process: H2O(s, - 10^∘C, 1 atm)⟶ H2O(l, 10^∘C, 1 atm) CP for ice = 9 cal deg^-1 mol^-1, CP for H2O = 18 cal deg^-1 mol^-1 . Latent heat of
4 \mathrm{~L} \) - atm is equal to: a. \( 40.50 \mathrm{cal} \) ... - YouTube
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SOLVED: Convert 0.121 kcal to L atm; (1 L atm = 101.3 J; 1 cal = 4.184 J) Select one: a. 5.00 L atm b. 4.998 L atm c. 4.99 x 10^-3 atm d. 5.00 x 10^-3 L atm e. 4.99 L atm
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