How do we explain the normal boiling points of #"ethanol"# (#78.4# #""^@C#), #"methanol"# (#64.7# #""^@C#), #"ethane"# (#-89# #""^@C#), #"ethyl acetate"#, (#77.1# #""^@C#), and #"methyl acetate"# (#56.9# #""^@C#)?
Because the esters are not capable of effective hydrogen bonding.
On the other hand, methanol is INSOLUBLE in petroleum ether, whereas ETHANOL is SOLUBLE. Why should this be so?
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The normal boiling points are influenced by intermolecular forces. Ethanol and ethyl acetate have hydrogen bonding, leading to higher boiling points. Methanol and methyl acetate exhibit weaker hydrogen bonding, resulting in lower boiling points. Ethane lacks significant intermolecular forces, leading to a lower boiling point.
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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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