Amines are derivatives of
Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;
Primary Amines: In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.
Secondary Amines: In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.
Tertiary Amines: In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.
Below are three isomers of tertiary amines with molecular formula
C₅H₁₃N.
Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential
![P_w=P_p+P_s](https://tex.z-dn.net/?f=P_w%3DP_p%2BP_s)
![P_w=P_p+(-iCRT)](https://tex.z-dn.net/?f=P_w%3DP_p%2B%28-iCRT%29)
We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose= ![P_w](https://tex.z-dn.net/?f=P_w)
(At standard temperature)
![P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar](https://tex.z-dn.net/?f=P_s%3D-iCRT%3D-%5Ctimes%201%5Ctimes%208.314%5Ctimes%2010%5E%7B-2%7Dbar%20L%2Fmol%20K%5Ctimes%20273.15%20K%3D-3.406%20bar)
![P_w=0 bar+(-3.406 ) bar](https://tex.z-dn.net/?f=P_w%3D0%20bar%2B%28-3.406%20%29%20bar)
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Answer:
![1s^{2} 2s^{2} 2p^{4}](https://tex.z-dn.net/?f=1s%5E%7B2%7D%202s%5E%7B2%7D%202p%5E%7B4%7D)
Explanation:
Oxygen has eight eletrons and six valence electrons, giving it the electron configuration of
.
Answer:
272 m/s
Explanation:
The boy jogs= 250 m
time he takes 110 seconds
average speed = 250/110
272 m/s