Answer:
How to convert volts to electron-volts
How to convert electrical voltage in volts (V) to energy in electron-volts (eV).
You can calculate electron-volts from volts and elementary charge or coulombs, but you can't convert volts to electron-volts since volt and electron-volt units represent different quantities.
Volts to eV calculation with elementary charge
The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electric charge Q in elementary charge or proton/electron charge (e):
E(eV) = V(V) × Q(e)
The elementary charge is the electric charge of 1 electron with the e symbol.
So
electronvolt = volt × elementary charge
or
eV = V × e
Example
What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 40 electron charges?
E = 20V × 40e = 800eV
Volts to eV calculation with coulombs
The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C) divided by 1.602176565×10-19:
E(eV) = V(V) × Q(C) / 1.602176565×10-19
So
electronvolt = volt × coulomb / 1.602176565×10-19
or
eV = V × C / 1.602176565×10-19
Example
What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 2 coulombs?
E = 20V × 2C / 1.602176565×10-19 = 2.4966×1020eV
Explanation:
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 200 = 0
Resolving the second degree equation:
= 
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
Answer:
4.83% of acetic acid in the vinegar
Explanation:
