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Unformatted text preview: Version 256 – Final Exam – sparks – (50990) 1 This printout should have 44 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 1.0 points For dinitrogen monoxide, the arrangement of the atoms is NNO. In the Lewis structure using only double bonds, what are the formal charges on N, N, and O, in order from left to right? 1. 1, +1, 0 correct 2. 0, 0, 0 3. 2, +1, +1 4. 0, +1, 1 5. 0, 1, +1 Explanation: N N O b b b b b b b b 1 +1 002 1.0 points HF is a corrosive gas. At 2.0 atm and 300 K, HF occupies a 4.50 L volume. How many grams of HF are in the volume? 1. 2.74 g 2. 15.6 g 3. Not enough information is given. 4. 0.573 g 5. 4.82 g 6. 7.3 g correct Explanation: P = 2 atm V = 4 . 50 L T = 300 K Using the ideal gas law, solve for moles of HF: P V = nRT n = P V RT = (2 . 0 atm)(4 . 50 L) ( . 08206 L · atm mol · K ) (300 K) = 0 . 365586 mol HF Now we can solve for grams: ? g HF = (0 . 365586 mol HF) 20 g HF 1 mol HF = 7 . 31172 g HF 003 1.0 points What is the subshell notation and the num ber of electrons that can have the quantum numbers n = 3 , ℓ = 2? 1. 3 p ; 5 2. None of these 3. 3 p ; 10 4. 3 d ; 10 correct 5. 3 d ; 9 6. 3 s ; 2 Explanation: The notation is n ℓ where n = 1 , 2 , 3 , 4 , 5 , ··· ℓ = 0 , 1 , 2 , ··· , ( n 1) represented by a letter: ℓ value 1 2 3 4 5 letter s p d f g h m ℓ = ℓ, ( ℓ 1) , ( ℓ 2) , ··· , , ··· +( ℓ 2) , +( ℓ 1) , + ℓ and m s = ± 1 2 . To find the number of orbitals that can have the stated values of n and ℓ (and any allowed values of m ℓ and m s ), use ℓ to find the number of different values of m ℓ . Version 256 – Final Exam – sparks – (50990) 2 To find the maximum number of electrons that can have the stated values of n and ℓ (and any allowed value of m ℓ and m s ), double the number of different values of m ℓ . 004 1.0 points What is the energy of the photon of wave length 6580 ˚ A? (Planck’s constant = 6 . 626 × 10 − 27 erg · s) 1. 3 . × 10 − 15 erg 2. 3 . × 10 − 12 erg correct 3. 8 . 7 × 10 − 24 erg 4. 6 . 6 × 10 − 27 erg 5. 8 . × 10 − 13 erg Explanation: h = 6 . 626 × 10 − 27 erg · s c = 3 . × 10 8 m/s You are given λ = 6580 ˚ A , and energy is E = hc λ = 6 . 626 × 10 − 27 erg · s 6580 ˚ A × (3 . × 10 8 m / s) × 10 10 ˚ A 1 m = 3 . × 10 − 12 erg 005 1.0 points Which of the following molecules A) CH 3 COOH C) CH 3 CH 2 OH B) CH 3 OCH 3 D) CH 3 CHO are likely to form hydrogen bonds? 1. All are likely to form hydrogen bonds. 2. A and B only 3. None forms hydrogen bonds. 4. B and C only 5. Another combination of compounds 6. C and D only 7. A and C only correct 8. A and D only 9. B and D only Explanation: Only molecules with H attached to the electronegative atoms N, O, and F can hy drogen bond. Of the molecules given, only CH 3 COOH and CH 3 CH 2 OH have hydrogen attached to oxygen, so these are the only ones that can undergo hydrogen bonding.that can undergo hydrogen bonding....
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 Fall '07
 Fakhreddine/Lyon
 Atom, Electron, Chemical bond

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