Chem 161-2013 exam I + solutions

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Rutgers University *

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Chemistry

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Dec 6, 2023

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Chemistry 161 Exam I October 2, 2013 Student Name (Print): _____TAVSS______________ The exam booklet has 25 questions for credit and one additional question to check the color of your exam booklet. Please answer all 26 questions on the OpScan sheet. There is no penalty for guessing. If you do not find the exact answer, bubble in the best choice. A proctor will check your picture ID, OpScan form, signature and calculator during the exam. The use of calculators with permanent memories (graphing calculators), cell phones, pagers, PDAs or other electronic devices other than a basic scientific calculator is expressly forbidden. The last page of the booklet contains a periodic table along with other useful data. The use of any other notes or information on this test or taking extra time to bubble in your answers will be considered a violation of the Academic Honesty provisions of the student code. Exam scores will be posted as soon as possible. On the Opscan Form: (Use a #2 pencil or darker) 1. Sign your name across the top of the form. 2. Code the following information ( blacken circles) Your Name (Last Name First) Your Ru ID Number [Start under Box A and continue to Box I ] Your Recitation Section Number in K & L [For sections 01-09, bubble 0 under box K ] Your Exam Form Number under box P Your EXAM FORM is Chem161f13e1v1 1
Your EXAM FORM is 1 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Significant Figures/Precision/Accuracy A homogeneous mixture was prepared by mixing 0.15g A, 112.3g B and 24.55g C. The volume is found to be 85.80 cm 3 . What is the density of the mixture to correct number of significant digits? A. 1.5967 g/cm 3 B. 1.59674 g/cm 3 C. 1.6 g/cm 3 D. 1.60 g/cm 3 E . 1.597 g/cm 3 0.15g 112.3g 24.55g 137.00g The rule for addition-subtraction is to consider the smallest number of digits to the right of the decimal point. Since 112.3 has only one digit to the right of the decimal point, then the sum, 137.00, can only have one digit to the right of the decimal point, i.e., 137.0, which has four significant figures. Nevertheless, don’t round off in the middle, so the next process should use the full 137.00, but we should keep in mind that this number really has only four significant figures. D = g/cm 3 D = 137.00g/85.80cm 3 = 1.59674g/cm 3 Since 137.00 has four significant figures and 85.80 has four significant figures then the final answer must have only four significant figures. Rounding off gives 1.597g/cm 3 . E Chem161f13e1v1 2
2 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Units/unit conversions/dimensional analysis Which of the following is not an SI base unit? A. kilogram B. mole C . liter D. Kelvin E. second The SI base units are: meter, kilogram, second, Kelvin, ampere, mol and candela. Liters are derived SI units. meter → cm → cm 3 → mL → L C Chem161f13e1v1 3
3 Chem 161-2013 Exam I B&O Chapter 4 – Periodic Trends of the Elements Filed in: H&P Chapter 8 – Electron Configurations & Periodic Table Trends Periodic table trends Which of the following would have the highest 2nd ionization energy? A. Ba B. Ga C. Sr D . K E. Ca D Chem161f13e1v1 4
The first ionization energy is M → M + + 1e - . The second ionization energy is M + → M 2+ + 1e - Therefore, plot M + , and mentally consider the relative energy required to remove another electron to form M 2+ . 20 Ca + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 56 Ba + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 1 19 K + = 1s 2 2s 2 2p 6 3s 2 3p 6 38 Sr + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 56 Ba + 31 Ga + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 20 Ca + 19 K + 38 Sr + 31 Ga+ K + has its valence electrons in the third principal shell. The other four ions have their valence electrons in higher principal shells. This means that the K + ion has the highest ionization energy, in that the valence electrons are held relatively strongly by the protons in the nucleus because the valence shell is close to the nucleus. Now look at the effective nuclear charge of the ions. The effective nuclear charge on K + is far greater than the effective nuclear charge on any of the other ions, adding further strength to the conclusion that K + has the highest second ionization energy. This conclusion is different than what one would expect Chem161f13e1v1 5
regarding the generalization that atoms and ions in the lower left-hand corner of the periodic table have low ionization energies. Chem161f13e1v1 6
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