The question is incomplete, the complete question is;
The heat of vaporization ΔHv of acetic acid HCH3CO2 is 41.0 /kJmol. Calculate the change in entropy ΔS when 954.g of acetic acid condenses at 118.1°C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits.
Answer:
-1.67 JK-1
Explanation:
Since Heat of vaporization of acetic acid = 41.0 kJ/mol
Therefore:
Heat of condensation of acetic acid = -41.0 kJ/mol
Mass of acetic acid = 954 g
Temperature of condensation = 118.1 °C or 391.1 K
Number of moles of acetic acid = 954 g/60g/mol = 15.9 moles
Heat evolved during condensation = 15.9 moles * -41.0 kJ/mol = -651.9 KJ
Entropy change (ΔS) = Heat evolved/ Temperature = -651.9 KJ/391.1 K
Entropy change (ΔS) = -1.67 JK-1
H3O+ + OH- = 2 H2O
H3O+ is acid (proton donor)
OH- is base (proton receptor)
<u>Lithium Iodide</u><u>:</u>
~formed by the reaction of hydroxide with hydroiodic acid
Hope this helped you, have a good day bro cya)
Answer:
Lithium selenide Lithium selenide (Li2Se) 12136-60-6 Dilithium selenide lithium selenidolithium
Molecular Weight 92.9 g/mol
Dates Modify 2020-11-15 Create 2005-08-08
