Exam 3 Chem 111 Fa23 vA KEY

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Colorado State University, Fort Collins *

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Chemistry

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Jun 13, 2024

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1 CHEM 111 – Exam 3 Thursday, October 26, 2023 Version A Name (printed): _______________________________________ CSU ID: ______________________ For this exam, you will need a pencil, a TI-30X IIS, TI-30Xa, or a TI-36X Pro calculator (without the cover), and your CSU ID card. NO OTHER MATERIALS are allowed on your desk during the exam. Calculators may not be shared. Having a cell phone, smart watch or other electronic device on your person will be considered a violation of CSU’s Academic Integrity Policies. HONOR PLEDGE : “I have not given, received, or used any unauthorized assistance .” Signature: ________________________________________________ Choose the single best answer to each multiple-choice question. When you have finished taking the exam, the cover sheet and the answer sheet must be turned in. Make sure to include your CSU ID number and version letter on your answer sheet. You will be graded solely on the answers that you report on your answer sheet, so be sure to bubble in your exam version and answers carefully. This exam has 35 equally weighted multiple-choice questions. There are 12 pages including this page. Good luck! Conversion Factors for SI and Other Common Units Mass 1 kg = 2.205 pounds (lb); 1 lb = 0.4536 kg = 453.6 g 1 g = 0.03527 ounce (oz); 1 oz = 28.35 g Length 1 m = 1.094 yards (yd); 1 yd = 0.9144 m (exactly) 1 m = 39.37 inches (in); 1 foot (ft) = 0.3048 m (exactly) 1 in = 2.54 cm (exactly); 1 mi = 1.609 km 1 mi = 5280 ft (exactly) Volume 1 L = 0.2642 gallon (gal); 1 gal = 3.785 L 1 L = 1.057 quarts (qt); 1 qt = 0.9464 L Select Electronegativity Values: H (2.1), C (2.5), N (3.0), O (3.5), F (4.0), S (2.5), Cl (3.0), P (2.1) Electron Domain and Molecular Geometries: Linear Trigonal Planar (bent) Tetrahedral (trigonal pyramidal, bent) Trigonal Bipyramidal (seesaw, T-shaped, linear) Octahedral (square pyramidal, square planar)

2 Constants and Conversion F act ors h = 6.626 ×10 -34 J ∙ s c = 2.9979 × 10 8 m/s Avogadro’s number = 6.022 × 10 23 T (in K) = 273.15 + T (in  C) 1 mL = 1 cm 3 (exactly) The Electromagnetic Spectrum Equations d = m/V V (sphere) = 4/3  r 3 V (cube) = length  width  height 𝐸 = −2.179 × 10 −18 𝐽 ( 1 𝑛 2 )  = c E photon = h ν = ℎ? 𝜆 𝐸 ∝ 𝑞 1 𝑞 2 ? Δ𝐸 = −2.179 × 10 −18 𝐽 ( 1 𝑛 𝑓 2 − 1 𝑛 𝑖 2 )

3 You have version A of the exam. Please bubble A on your answer sheet. 1. Arrange aluminum, nitrogen, and phosphorus in order of predicted increasing electronegativity. A. N < P < Al B. P < N < Al C. Al < P < N D. P < Al < N E. None of these orders is correct. 2. A nonpolar covalent (i.e., pure covalent) bond would form between .... A. Na – Cl B. H – Cl C. Br – Br D. Li – Br E. Na – Br 3. Which of these choices is the best Lewis structure for ozone, O 3 ? A. B. C. D. E.

4 4. What is the mass percent sodium in sodium carbonate? A. 43.4% B. 11.3% C. 45.3% D. 27.7% E. 21.7% 5. 50.00 g of a compound contains 38.87 g iron and 11.14 g oxygen. What is the empirical formula of the compound? A. Fe 2 O B. FeO C. Fe 2 O 2 D. Fe 2 O 3 E. Iron (II) oxide 6. Which of the following is a molecular formula for a compound with an empirical formula of CH 2 O and a molar mass of 150. g/mol? A. C 7 H 18 O 3 B. C 5 H 10 O 5 C. C 3 H 6 O 3 D. CH 2 O E. None of the answers is correct. 7. Which of these covalent bonds is the most polar (i.e., highest percent ionic character)? A. Al – I B. Si – I C. Al – Cl D. Si – Cl E. Si – P

5 8. The electronegativity of an element is a measure of . . . A. the ability of a bonded atom to draw shared electrons to itself. B. the amount of energy required to remove an electron from atoms. C. the amount of energy released when atoms gain an electron. D. the distance between electrons. E. the number of valence electrons. 9. How many lone pairs of electrons need to be added to complete this Lewis structure? A. 5 B. 8 C. 6 D. 1 E. 14 10. What is the formal charge of the sulfur atom in the Lewis structure below? A. 0 B. +1 C. – 1 D. +2 E. – 2 11. Consider the two possible Lewis structures for formaldehyde CH 2 O. Considering formal charges, which structure is to be preferred and why ? A. The left structure because the total of formal charges is zero B. The left structure because the total of formal charges is -1 C. The left structure because all formal charges are zero D. The right structure because the total of formal charges is +1 E. The right structure because all formal charges are zero

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