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
T½=18.72days
therefore t¾=18.72+½ of 18.72
we have 18.72+9.36=28.08days
Answer: 94.07%
Explanation:
Percentage yield can be calculated by the formula
%yield = Experimental yield/Theoretical yield x100
Experimental yield = 7.93g
Theoretical yield = 8.43
%yield = Experimental yield/Theoretical yield x100
%yield = 7.93/8.43 x 100 = 94.07%
The equation is as follow,
<span> HBr </span>₍aq₎ + H₂O ₍l₎ →
Solution:
HBr being strong acid with Ka value of 1.0 × 10⁹. When HBr is added to water, water acts as a base and HBr acts as a acid. Water picks the proton (H⁺) from HBr and converts into Conjugate acid (H₃O⁺) ahile HBr is converted into Conjugate Base (Br⁻) after loosing proton. The equation for this reaction is as follow,
HBr ₍aq₎ + H₂O ₍l₎ → H₃O⁺ ₍aq₎ + Br⁻ ₍aq₎
