Co carbon monoxide
sorry friend i don't know other ones
Answer: a) E= 6.63x10^-19J
E= 3.97×10^2KJ/mol
b) E = 3.31×10^-19J
E= 18.8×10^4 KJ/mol
C) E = 1.32×10^-33J
E= 8.01×10^-10KJ/mol
Explanation:
a) E = h ×f
h= planks constant= 6.626×10^-34
E=(6.626×10^-34)×(1.0×10^15)
E=6.63×10^-19J
1mole =6.02×10^23
E=( 6.63×10^-19)×(6.02×10^23)
E=3.97×10^2KJ/mol
b) E =(6.626×10^-34)/(1.0×10^15)
E=3.13×10^-19J
E= 3.13×10^-19) ×(6.02×10^23)
E= 18.8×10^3KJ/MOL
c) E= (6.626×10^-34) /0.5
E= 1.33×10^-33J
E= (1.33×10^-33) ×(6.02×10^23)
E= 8.01×10^-10KJ/mol
Answer:
A ball thrown into the air has the most potential energy when it has reached the highest point above the ground before it begins descending. If we consider the vertical motion only beginning when the ball leaves the thrower’s hand, the ball is exchanging kinetic energy for (gravitational) potential energy. When all of the kinetic energy has been transformed, the ball begins falling, and exchanges it’s potential energy back into kinetic energy. If you ignore air resistance, the ball will land with as much energy as it began with.
This is more chemistry. But it is a process called fractional distillation, and it basically separates the long chained hydrocarbons from the short chained hydrocarbons through separation dependant on the boiling point of the crude oil.
Answer:The 2s orbital different than the 1s orbital because the 2s orbital extends farther from the nucleus than the 1s.
As we move away from the nucleus, the values of the principal quantum number (n) continues to increase.
As the principal quantum number (n) increases, the orbital becomes farther away from the nucleus.
The higher energy orbital are always larger than the orbitals closer to the nucleus.
Hence, the 2s orbital different than the 1s orbital because the 2s orbital extends farther from the nucleus than the 1s.
Explanation: