Exam-1 - Rahman, Tarique – Exam 1 – Due: Sep 20 2007,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Rahman, Tarique – Exam 1 – Due: Sep 20 2007, 11:00 pm – Inst: Vandenbout 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Vandenbout 2pm c = 3 × 10 8 m/s N A = 6 . 022 × 10 23 mol- 1 h = 6 . 626 × 10- 34 J · s ν = R µ 1 n 2 1- 1 n 2 2 ¶ R = 3 . 29 × 10 15 Hz E n =- h R n 2 h R = 2 . 18 × 10- 18 J ψ n ( x ) = µ 2 L ¶ 1 2 sin ‡ nπx L · n = 1 , 2 , ··· E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , ···- ¯ h 2 2 m d 2 ψ d x 2 + V ( x ) ψ = Eψ 001 (part 1 of 1) 10 points Which set of quantum numbers could corre- spond to a 4 f orbital? 1. n = 4, ‘ = 3, m ‘ = +4 2. n = 4, ‘ = 3, m ‘ =- 3 correct 3. n = 3, ‘ = 2, m ‘ = +1 4. n = 4, ‘ = 4, m ‘ = +3 5. n = 3, ‘ = 2, m ‘ = 0 Explanation: A 4 f orbital has n = 4, ‘ = 3 and m ‘ =- 3 ,- 2 ,- 1 , , +1 , +2 , +3. 002 (part 1 of 1) 10 points Which of the following atoms has the largest radius? 1. Ar 2. H 3. Al 4. Na correct 5. P 6. F Explanation: Atomic radii increase from left to right and from top to bottom on the Periodic Table. 003 (part 1 of 1) 10 points According to the Bohr explanation of the hydrogen emission spectrum, the individual lines of this spectrum 1. depend directly on the differences in en- ergy between the energy levels the electron is permitted to have. correct 2. are spaced at equal intervals along the wavelength scale. Rahman, Tarique – Exam 1 – Due: Sep 20 2007, 11:00 pm – Inst: Vandenbout 2 3. depend directly on the absolute energies which the electron is permitted to have in the hydrogen atom. 4. are due to the different number of neu- trons present. Explanation: 004 (part 1 of 1) 10 points Balance the reaction ? Fe + ? H 2 O → ? Fe 2 O 3 + ? H 2 using the smallest possible integers. The co- efficient of H 2 is 1. 6. 2. 3. correct 3. 1. 4. 2. 5. 4. Explanation: A balanced equation has the same num- ber of each kind of atom on both sides of the equation. We find the number of each kind of atom using equation coefficients and com- position stoichiometry. For example, we find there are 6 H atoms on the reactant side: ? H atoms = 3 H 2 O × 2 H 1 H 2 O = 6 H The balanced equation is 2 Fe + 3 H 2 O → Fe 2 O 3 + 3 H 2 , and the coefficient of H 2 is 3. 005 (part 1 of 1) 10 points Consider the reaction of CaCN 2 and water to produce CaCO 3 and NH 3 : CaCN 2 + 3 H 2 O → CaCO 3 + 2 NH 3 . How much NH 3 is produced upon reaction of 65.0 g of CaCN 2 and 4.0 moles of H 2 O? 1. 6.0 mol 2. 0.8 mol 3. 1.6 mol correct 4. 2.7 mol 5. 8 mol Explanation: m CaCN 2 = 65 . 0 g n H 2 O = 4 . 0 moles We recognize this as a limiting reactant problem because the amounts of more than one reactant are given. We must determine which of these would be used up first (the limiting reactant). We find moles of CaCN 2 : ? mol CaCN 2 = 65 g CaCN 2 × 1 mol CaCN 2 80 . 11 g CaCN 2 = 0 . 811 mol CaCN 2 The balanced chemical equation shows that...
View Full Document

This note was uploaded on 10/31/2009 for the course CH 56435 taught by Professor Flowers during the Spring '08 term at University of Texas.

Page1 / 8

Exam-1 - Rahman, Tarique – Exam 1 – Due: Sep 20 2007,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online