Answer:
(D) (CH3CH2)2NH
Explanation:
In order to decide which base is strongest we need to calculate its PKb
PKb = -log [Kb]
A large Kb value and small PKb value gives the strongest base
Compound Kb PKb
(A) C6H5NH2 - 4 x 10^-10 9.349
(B) NH3 1.76x 10^-5 4.754
(C) CH3NH2 4.4x 10^-4 3.357
(D) (CH3CH2)2NH 8.6x 10^-4 3.066
(E) C5H5N 1.7x10^-9 8.77
Clearly (CH3CH2)2NH is the strongest base.
The question is incomplete, here is the complete question:
A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.
<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
We are given:
Given mass of liquid bromine = 766. g
Molar mass of liquid bromine, 
= 159.8 g/mol
Putting values in above equation, we get:
Hence, the amount of liquid bromine produced is 4.79 moles.
The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
