Answer: The average potential energy of the PIB is 0 irrespective of the wave function.
Explanation:
⟨H⟩=⟨KE⟩+⟨V⟩
the nn quantum number
⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )
the average kinetic energy of the wavefunction is dependent on
⟨V⟩=∫sin(kx)0sin(kx)dx=0
The average potential energy of the PIB is 0 irrespective of the wave function.
⟨H⟩=⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )
Start studying Pressure - Volume Relationships in Gases (Boyle's Law). Learn vocabulary, terms ...Select<span> all that apply. V2 = k/P2 V2 = P1V1/P2 ... What </span>two variables<span> are </span>held constant when testing Boyle's Law in a manometer<span>? Temperature hope this helps
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I had a dream once more like a nightmare,about a zombie apocalypse.
The answer is critical mass with no doub t
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, in terms of pressures, the rate becomes:
Thus, the rate of change for the partial pressure of ammonia turns out:
![r_{NH_3}=2*(-r_{N_2H_4})\\r_{NH_3}=2*[-(-70torr/h)]\\r_{NH_3}=140torr/h](https://tex.z-dn.net/?f=r_%7BNH_3%7D%3D2%2A%28-r_%7BN_2H_4%7D%29%5C%5Cr_%7BNH_3%7D%3D2%2A%5B-%28-70torr%2Fh%29%5D%5C%5Cr_%7BNH_3%7D%3D140torr%2Fh)
The rate of decrease of partial pressure of urea is taken negative as it is a reactant whereas ammonia a product which has 2 as its stoichiometric coefficient.
Best regards.
