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.
Metallic solids or metallic structures experience metallic bonds which are the forces of attractions between the sea of electrons and the nucleus of the metallic atoms. They share a network of highly delocalized electrons.
I therefore think that the packing efficiency decreases as the number of nearest neighbors decreases.
Answer:
The molar mass of the gas is 36.25 g/mol.
Explanation:
- To solve this problem, we can use the mathematical relation:
ν = 
Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,
R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,
T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,
M is the molar mass of the gas in <em>(Kg/mol)</em>.
ν = 
(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,
<em>by squaring the two sides:</em>
(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,
∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.
<em>∴ The molar mass of the gas is 36.25 g/mol.</em>
<span>At 100 feet, the diver is under about 4 atmospheres pressure. If she is free diving, her lungs will be compressed to about 1/4 their size on the surface (with some movement of the major abdominal organs). If she is scuba diving, the air which she is breathing is also at 4 atmospheres and there is no problem. (The non-gas spaces in the body are not-compressible and are unaffected.) The only problems she has to concern herself with are the beginnings to nitrogen narcosis and the nitrogen which is dissolving (Henry's law) into her body tissues. On the way up, she also has to remember that the air in her lungs will expand by a factor of 4 and she better exhale! Hope this helps you</span>