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Get Full Access to Atkins' Physical Chemistry - 11 Edition - Chapter 9c - Problem D9c.3
Get Full Access to Atkins' Physical Chemistry - 11 Edition - Chapter 9c - Problem D9c.3

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ISBN: 9780198769866 2042

## Solution for problem D9C.3 Chapter 9C

Atkins' Physical Chemistry | 11th Edition

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Atkins' Physical Chemistry | 11th Edition

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Problem D9C.3

What is the justification for treating s and p atomic orbital contributions to molecular orbitals separately?

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Example: which solvent will more likely dissolve the listed species, CHCl 3 (polar) or C6H6(nonpolar) A) C 10 22- C6H6because it is nonpolar B) H 2 – CHCl b3cause it is polar C) HF – CHCl 3ecause it is polar Miscible = able to dissolve completely Fractional Crystallization- the separation of a mixture of substances into pure components on the basis of their differing solubilities Henry’s Law- the solubility of a gas in a liquid in directly proportional to the partial pressure of the gas over the liquid, C = kP. C = concentration (M) of the dissolved gas, P = pressure of gas over the solution, k = constant for each gas-liquid that depends only on the temperature Example: the partial pressure of carbon dioxide in a sealed soda can is approximately 3.8atm and the solubility of carbon dioxide in the soda at this pressure is .13mol/L. Calculate the new solubility of carbon dioxide when the can is opened. The partial pressure of carbon dioxide in the atmosphere is 3.1 x 10 atm. C C 1 C 2 =k = (.13mol/L)/3.8atm = x/3.1x10^-4 atm 3.8x = P P1 P2 4.03 x 10^-5 x = 1.1 x 10^-5 mol/L Concentration Units: 1) Molarity (M) = mol solute/ L solution Disadvantages : 1. Volume varies with temperature 2. Volumes are not always additive 2) Molality (m) = mol solute / mass solvent (kg) Example: calculate the molality of a solution prepared by adding 3.5g of NaCl to 100 mL of water m = mol NaCl / kg H20 1m

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