How do the volatilities of methane, methyl bromide, methylene bromide, bromoform, and tetrabromomethane, relate to their dipole moments?
Both methane and methyl bromide are gases. As to the volatilities of the other, you are going to have to consult the literature.
Even in university first-year chemistry, this question would never be asked without providing the methane derivatives' boiling points. You might be asked to use the dipole moment to account for the given volatilities, but in the absence of any information, your guess is as good as mine.
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Volatility generally decreases with increasing dipole moment. In the given compounds, methane has the lowest dipole moment and highest volatility, while tetrabromomethane has the highest dipole moment and lower volatility. The trend is influenced by the intermolecular forces, with higher dipole moments leading to stronger attractive forces, thereby reducing volatility.
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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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