Answer:
The molar average velocity is 0.0588 cm/s
The N₂ diffusion velocity relative to the mole average velocity is -0.1428 cm/s
The molar diffusional flux of N₂ is -3.9x10⁻³
Explanation:
Given data:
T = temperature = 265 K
O₂ = 25%
N₂ = 60%
CO₂ = 15%
vO₂ = -0.084 cm/s
vN₂ = 0.12 cm/s
vCO₂ = 0.052 cm/s
The molar average velocity is equal:
![v_{av} =(0.25*(-0.084))+(0.6*0.12)+(0.15*0.052)=0.0588cm/s](https://tex.z-dn.net/?f=v_%7Bav%7D%20%3D%280.25%2A%28-0.084%29%29%2B%280.6%2A0.12%29%2B%280.15%2A0.052%29%3D0.0588cm%2Fs)
The N₂ diffusion velocity relative to the molar average velocity is:
![v_{i} -v_{av} =-0.084-0.0588=-0.1428cm/s](https://tex.z-dn.net/?f=v_%7Bi%7D%20-v_%7Bav%7D%20%3D-0.084-0.0588%3D-0.1428cm%2Fs)
The molar diffusional flux of N₂ is:
![N_{N_{2} } =\frac{P}{RT} y_{A} (v_{i} -v_{av} )=-3.9x10^{-3}](https://tex.z-dn.net/?f=N_%7BN_%7B2%7D%20%7D%20%3D%5Cfrac%7BP%7D%7BRT%7D%20y_%7BA%7D%20%28v_%7Bi%7D%20-v_%7Bav%7D%20%29%3D-3.9x10%5E%7B-3%7D)
Answer:
These three factors are required for ionization potential or ionization energy.
Explanation:
Ionization potential refers to the amount of energy which is required for the removal of outermost electron of the atom. If the atom size is big so the outermost electron is far from the nucleus and low energy is required for its removal due to lower force of attraction between nucleus and outermost electron. If the nuclear charge is higher, so the electron is tightly held by the nucleus and require more energy for its removal. Nuclear charge means number of protons present in the nucleus.
I'm not sure, but maybe burning point...