Answer:
The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
Explanation:
Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.

Molar heat of molar vaporization of Trichlorofluoromethane = 24.8 kJ/mol
Temperature at which Trichlorofluoromethan boils , T= 296.95 K
The molar entropy of the evaporation of Trichlorofluoromethan :

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
Answer:
Option D = 0.2 Kj
Explanation:
Given data:
Mass of diethyl ether = 1.0 g
Hvap = 15.7 Kj / mol
Heat absorbed = ?
Solution:
Q = mass × Hvap / molar mass
Q = 1.0 g × 15.7 Kj / mol / 74.12 g/mol
Q = 15.7 Kj / 74.12
Q = 0.212 KJ
Answer:
<h2>Ammonia Gas</h2>
Explanation:
It result in formation of ammonia gas.
N2 + 3H2 ---<u>iron</u><u>/</u><u>molybdenum</u><u>/</u><u>high</u><u> </u>temp/pres--- > 2 NH3
It forms ammonia gas.
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Alter means change or cause something in character