Chapter 11: Problem 69
For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?
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Chapter 11: Problem 69
For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?
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True or false: (a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. (b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, dipole- dipole interactions, when present, are always greater than dispersion forces.( \(\mathbf{d}\) ) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The larger the atom, the more polarizable it is.
The boiling points, surface tensions, and viscosities of water and several alcohols are as shown below:(a) From ethanol to propanol to \(n\) -butanol the boiling points, surface tensions, and viscosities all increase. What is the reason for this increase? (b) How do you explain the fact that propanol and ethylene glycol have similar molecular weights (60 versus 62 amu), yet the viscosity of ethylene glycol is more than 10 times larger than propanol? (c) How do you explain the fact that water has the highest surface tension but the lowest viscosity?
(a) Which type of intermolecular attractive force operates between all molecules? (b) Which type of intermolecular attractive force operates only between polar molecules? (c) Which type of intermolecular attractive force operates only between the hydrogen atom of a polar bond and a nearby small electronegative atom?
A watch with a liquid crystal display (LCD) does not function properly when it is exposed to low temperatures during a trip to Antarctica. Explain why the LCD might not function well at low temperature.
Butane and 2 -methylpropane, whose space-filling models are shown here, are both nonpolar and have the same molecular formula, \(\mathrm{C}_{4} \mathrm{H}_{10},\) yet butane has the higher boiling point \(\left(-0.5^{\circ} \mathrm{C}\) compared to \(-11.7^{\circ} \mathrm{C}\right) .\) Explain.
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