Mass of vaporized triethylamine : 2.606 g
<h3>Further explanation</h3>
Given
0.5 L triethylamine
P = 18.5 psi
T = 25 °C
Required
mass of vaporized triethylamine
Solution
Conversion :
P 18.5 psi = 1,26 atm
T = 25 +273 = 298 K
Ideal gas law :
PV=nRT
n = PV/RT
Input the value :
n = (1.26 atm x 0.5 L) /(0.08205 x 298)
n = 0.0258
MW triethylamine = 101 g/mol
Mass triethylamine :
= n x MW
= 0.0258 x 101 g/mol
= 2.606 g
Chose 3 you know this because you will need heat..
Answer:
-pneumonoultramicroscopicsilicovolcanoconiosis
-pneumonoultramicroscopicsilicovolcanoconiosis
Explanation:
Number of moles of HCl =

=

= 0.763 mole
From balanced equation:
4 moles of HCl give 2 moles Cl2
0.763 mole HCl give ?? mole Cl2
by cross multiplication :
= 0.38 mole Cl2
mass of Cl2 = moles x molar mass = 0.38 x 70.9 = 27 g