Exam 3 MWF-solutions - Version 142 Exam 3 MWF sutclie(50965...

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Version 142 – Exam 3 MWF – sutcliffe – (50965) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. THIS EXAM IS ONLY TO BE TAKEN BY STUDENTS IN DR. SUTCLIFFE’S MWF SECTION. 001 10.0 points A gaseous mixture made from 6.00 grams of oxygen gas and 9.00 grams of chlorine gas is placed in a 15.0 L vessel at 10.0 degrees Celsius. What is the total pressure in the vessel? 1. 1.00 atm 2. 0.800 atm 3. 0.244 atm 4. 0.488 atm correct 5. 0.655 atm Explanation: 002 10.0 points Ethylene (C 2 H 4 ) burns in oxygen to produce carbon dioxide and water. What is the correct form of the balanced chemical equation that describes this reaction? 1. C 2 H 4 + 3 O 2 2 CO 2 + 2 H 2 O correct 2. 2 C 2 H 4 + O 2 2 CO 2 + H 2 O 3. C 2 H 4 + O 2 CO 2 + H 2 O 4. C 2 H 4 + 2 O 2 2 CO + 2 H 2 O 5. C 2 H 4 + 4 O 2 2 CO + 2 H 2 O 6. C 2 H 4 + 6 O 2 2 CO 2 + 2 H 2 O Explanation: 003 10.0 points A molecule has one lone pair of electrons on the central atom and three atoms bonded to the central atom. What is its electronic ar- rangement and its hybridization? 1. trigonal planar; sp 3 2. tetrahedral; sp 2 3. pyramidal; sp 3 4. angular; sp 3 5. tetrahedral; sp 3 correct 6. pyramidal; sp 2 7. trigonal planar; sp 2 Explanation: One lone pair plus 3 bonded pairs equals 4 electronic regions. This means that the central atom is tetrahedral and hybridization is sp 3 . 004 10.0 points ClF 3 is a T-shaped molecule; therefore the hybridization of the Cl is 1. dspf . 2. sp 3 d . correct 3. sp 3 d 2 . 4. sp 3 . 5. sp 2 . Explanation: 005 10.0 points How many σ and π bonds are in the molecule below? C C H N H H 1. 6 σ , 2 π 2. 6 σ , 1 π
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Version 142 – Exam 3 MWF – sutcliffe – (50965) 2 3. 7 σ , 1 π 4. 5 σ , 2 π correct 5. 5 σ , 1 π Explanation: Sigma bonds are the first bond formed be- tween any two bonded atoms. Any subse- quent bond is a pi bond. 006 10.0 points If 2 . 67 liters of ammonia react with an excess of oxygen at 500 C and 2 atm pressure, 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g) what volume of steam will be produced at the same temperature and pressure? 1. 2 . 0025 L 2. 4 . 005 L correct 3. 3 . 00375 L 4. 6 . 0075 L 5. 5 . 00625 L Explanation: T = 500 C = 773 K V NH 3 = 2 . 67 L P = 2 atm R = 0 . 08206 L · atm mol · K P V = n R T n = P V R T n NH 3 = (2 atm)(2 . 67 L) (0 . 08206 L · atm mol · K ) (773 K) = 0 . 0841841 mol NH 3 n H 2 O = 0 . 0841841 mol NH 3 × 6 mol H 2 O 4 mol H 2 O = 0 . 126276 mol H 2 O V H 2 O = n H 2 O R T P = (0 . 126276 mol)(773 K) 2 atm × 0 . 08206 L · atm mol · K = 4 . 005 L OR L H 2 O = L NH 3 × 6 L H 2 O 4 L H 2 O This can ONLY be done if both substances are gases. 007 10.0 points Consider CN + , CN and CN . Based on a molecular orbital diagram, which of these will be paramagnetic? 1. CN + 2. CN and CN 3. CN 4. CN correct 5. CN + and CN Explanation: Paramagnetic substances contain unpaired electrons. For heteronuclear diatomic molecules containing atoms adjacent on the periodic table, we can treat these like homonuclear diatomics so ’blank’ MO dia- grams suitable for elements B through N can be used here. Remember to adjust for the charge on CN + and CN . Of these, only the molecular orbital diagram of CN indicates an unpaired electron, in a
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