Based on the type or types of intermolecular forces, predict the substance in each pair that has the higher boiling point Match
the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer Reset Help stronger dispersion forcesG Given the molecules propane (C3Hs) and n-butane (C4H10), Г- has the higher boiling point mainly due to n-butane (C,H Given the molecules diethyl ether (CH3 CH2OCH2 CHs) and 1-butanol (CH3 CH2CH2 CH2OH) higher boiling point mainly due to 1-butanel CH,CH, CH,CH,OH) diethyl ether (CH CH,OCH CH, has the propane (C, H ts greater melar mass
C4H10 has a higher boiling point due to stronger dispersion forces
1-butanol has the greater boiling point mainly due to hydrogen bonding influences
Explanation:
If we consider propane and n-butane, we discover that they are both alkanes. However, the magnitude of dispersion forces in alkanes depends on the length of the carbon chain. The greater the length of the carbon chain is greater in n-butane than in propane. The greater the chain length, the greater the magnitude of dispersion forces and the greater the boiling point. Hence, n-butane has a higher boiling point than than propane.
Also, if we compare the boiling points of 1-butanol to that of diethyl ether, we will discover that 1-butanol has a higher boiling point mainly due to intermolecular hydrogen bonding influences.
N is your Nitrogen and H is your Hydrogen. Elements are just one specific kind of atom. You have two elements with names Nitrogen and Hydrogen. Therefore, you have two elements.
Wavelength (m)Frequency (Hz)Radio> 1 x 10-1< 3 x 109Microwave1 x 10-3 - 1 x 10-13 x 109 - 3 x 1011Infrared7 x 10-7 - 1 x 10-33 x 1011 - 4 x 1014Optical4 x 10-7 - 7 x 10-74 x 1014 - 7.5 x 1014
Identify which lone pairs are participating in resonance (in aromatic rings).
One lone pair on the sulfur atom.
Explanation:
For a compound to be aromatic it has to comply with Hückel's Rule, which says that the quantity of electrons in the pi orbitals, has to be a multiple of
