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Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 11 - Problem 11.2
Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 11 - Problem 11.2

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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.

ISBN: 9780134414232 1274

## Solution for problem 11.2 Chapter 11

Chemistry: The Central Science | 14th Edition

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Problem 11.2

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.

(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.

Step-by-Step Solution:

Step 1 of 5) The equilibrium between a saturated solution and undissolved solute is dynamic; the process of solution and the reverse process, crystallization, occur simultaneously. In a solution in equilibrium with undissolved solute, the two processes occur at equal rates, giving a saturated solution. If there is less solute present than is needed to saturate the solution, the solution is unsaturated. When solute concentration is greater than the equilibrium concentration value, the solution is supersaturated. This is an unstable condition, and separation of some solute from the solution will occur if the process is initiated with a solute seed crystal. The amount of solute needed to form a saturated solution at any particular temperature is the solubility of that solute at that temperature.

Step 2 of 2

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