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Final%20Exam%20Fall%202010

Final%20Exam%20Fall%202010 - Version 256 – Final Exam –...

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Unformatted text preview: Version 256 – Final Exam – sparks – (50990) 1 This print-out should have 44 questions. Multiple-choice 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 N-N-O. 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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Final%20Exam%20Fall%202010 - Version 256 – Final Exam –...

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