What makes the classical dipole moment for water different than its experimental result? I used #r = 0.958# angstroms, #A_(HOH) = 104.4776^o#, and #q_(OH) = -0.52672 a.u.#, where #a.u. = # electron-charge. My result was #1.484 D#, compared to #1.85 D#.
I wrote the following, though perhaps I did it incorrectly:
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The classical dipole moment for water is different from its experimental result due to several factors including the use of classical approximations, neglect of quantum mechanical effects, and simplifications in the model used for calculation. Additionally, experimental measurements can be influenced by various factors such as temperature, pressure, and molecular interactions, leading to discrepancies between theoretical predictions and observed values. In your specific calculation, the discrepancy may arise from the assumptions and approximations made in the classical model used, such as treating the water molecule as a rigid structure and neglecting the influence of electron distribution and molecular vibrations.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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