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Semester Two Presentation Problems Set Three Period 3-AP CHEM

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Becca Trattner

on 7 December 2012

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Transcript of Semester Two Presentation Problems Set Three Period 3-AP CHEM

Problem 2 The normal melting and boiling points of xenon are -112°C and 282 torr, and its critical point is at 16.6°C and 57.6 atm. (a) Sketch the phase diagram for Xe showing the four points given and indicating the area in which each phase is stable. (b) Which is denser, Xe(s) or Xe(l)? Explain. (c) If Xe gas is cooled under an external pressure of 100 torr, will it undergo condensation or decomposition? Explain. Problem 1 1. Hydrazine (NH2NH2), hydrogen peroxide (HOOH), and water (H2O) all have exceptionally high surface tensions in comparison with other substances of comparable molecular weights. (a) Draw the Lewis Structures for these three compounds. (b) What structural property do these substances have in common, and how might that account for the high surface tensions? Problem 3 Perovskite, a mineral composed of Ca, O, and Ti, has the cubic unit cell shown in the drawing. What is the chemical formula of this mineral? Problem 4 The table shown here lists the molar heats of vaporization for several organic compounds. Use specific examples from this list to illustrate how the heat of vaporization varies with (a) molar mass, (b) molecular shape, (c) molecular polarity, (d) hydrogen bonding interactions. Explain these comparisons in terms of the nature of the intermolecular forces at work. THE END Any questions? By: Rebecca Trattner, Max Distler, and Marcela Rodriguez Period 3 PresentationProblems
Set Three 1A 1B All of these compounds form Hydrogen bonds. These intermolecular forces form a strong barrier across a surface when these molecules bond to other molecules of the same kind. The strong H-bonds that form between these molecules enhance the strength of the surface tension. 2A 2B 282 torr x (1 atm/760 torr) = 0.371 atm Xe (s) must be more dense than Xe (l) because at low temperatures and high pressures, particles have less room to move around which leads to a higher amount of collisions and therefore a higher pressure in the solid, making it more dense. 2C 100 torr x (1 atm/760 torr) = 0.132 atm Xe (g) would undergo decomposition, because at 0.132 atm, the only phase change possible is from solid to gas. All temperatures past the triple point leave Xe in the gas phase, and all those before can either exist as a solid or gas, leaving decomposition as the phase change possible instead of condensation. Perovskite Chemical Formula: CaTiO3 Why?? There are 8 Ca, each with 1/8 of the atom touching another molecule, so really there is only one 1/8 Ca touching the same molecule, and there are 8 Ca in one unit cell, so there is one Ca per molecule. Then there are 6 O, each with 1/2 touching the adjacent molecules, so similarly, there are really only 3 O touching the same molecule. The Ti is exclusive to each molecule so there is just one per molecule. Thus, the chemical formula is CaTiO3 Compound
CH3CH2CH2OH Heat of Vaporization
47.3 4A As molar mass increases, ∆Hvap increases. This is because molecules with a higher molar mass contain more electrons and intermolecular forces. This requires more energy to break the bonds. CH3CH2CH3
MM: 44.08 g/mol
∆Hvap: 19.0 CH3CHBrCH3
MM: 122.97 g/mol
∆Hvap: 31.8 4B As the molecular structure becomes more complex, more bonds are created, requiring more energy to break them all.
CH3CH2CH2CH2CH3 <---- has more bonds to break than CH3CH2CH3 4C As molecules become more polar, the heat of vaporization increases. Dipole-Dipole forces are stronger than London Dispersion forces because they have stronger intermolecular forces. Therefore, polar molecules are stronger than non-polar molecules. Stronger bonds -> harder to break them -> high ∆Hvap CH3CH2CH2Br has strong Dipole-Dipole forces 4D Molecules that have hydrogen bonding have very strong intermolecular forces. This requires more energy to break the bond which increases the heat of vaporization. CH3CH2CH2OH <- has a strong O-H bond. Oxygen is highly electronegative and creates a highly polar covalent bond.
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