make up exam 2 - Version 135 Make Up Exam 2 Laude (53755) 1...

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Unformatted text preview: Version 135 Make Up Exam 2 Laude (53755) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 6.0 points A sample of water vapor is allowed to reach equilibrium in a closed container at 102 C. The equilibrium pressure is mea- sured as 1 . 2 atm. What is the molarity of the water vapor? 1. . 039 mol L 1 correct 2. not enough information is given 3. 1 . 16 mol L 1 4. . 143 mol L 1 Explanation: T = 102 C = 375 K M = mol L = n V = P RT = 1 . 2 atm . 0821 L atm mol 1 K 1 375 K = 0 . 039 mol L 1 002 6.0 points Which of the following statements regarding intermolecular forces (IMF) is/are true? I) Intermolecular forces are not affected by the distance between two molecules. II) Water molecules do not exhibit disper- sion forces. III) Bonds within a molecule are usually one or two orders of magnitude stronger than intermolecular forces. 1. I, II, and III 2. II and III 3. I only 4. II only 5. III only correct 6. I and II 7. I and III Explanation: Statement III is true; with a few no- table exceptions in the ion-ion category, IMF are much weaker than the bonds within a molecule. Statements I and II are false; all IMF are the result of Coulombic attraction, which diminishes greatly over even very short distances; all molecules have dispersion forces. 003 6.0 points The following molecules are composed of hy- drogen(s) bound to another atom, X. Rank the HX bonds in terms of decreasing polar- ity: BeH 2 , CH 4 , H 2 O , HF. 1. BeH 2 < HF < CH 4 < H 2 O 2. HF < BeH 2 < H 2 O < CH 4 3. H 2 O < HF < BeH 2 < CH 4 4. CH 4 < BeH 2 < H 2 O < HF correct 5. HF < CH 4 < H 2 O < BeH 2 Explanation: Note that all of the bonds within both BeH 2 , CH 4 , H 2 O are identical to each other and the fact that there are multiple bonds does not change the polarity of the individual bonds. The EN for BeH 2 , CH 4 , H 2 O , HF are 0.7, 0.3, 1.3, 1.8, respectively. Arranged from least to greatest: CH 4 < BeH 2 < H 2 O < HF. 004 6.0 points Ignoring resonant structures, in a molecule of the form AB 2 U x where A is a central atom, B is a bonding electron pair, and U is an un- bonded electron pair, what value of x would yield a non-polar molecule that is not hypo- valent? 1. 3 Version 135 Make Up Exam 2 Laude (53755) 2 2. 4 3. 1 4. 5. 2 correct Explanation: Having 2 identical bonded atoms and 2 non-bonding electron pairs would correspond to angular molecular geometry and be po- lar. Having 0 or 1 non-bonding electron pairs would be hypovalent (excluding reso- nant structures). Having 3 or 4 non-bonding electron pairs would correspond to linear molecular geometry and be non-polar....
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This note was uploaded on 06/06/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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make up exam 2 - Version 135 Make Up Exam 2 Laude (53755) 1...

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