Exam 2 CEM 151 October, 18, 2006 Name Section PID Multiple choice (3 points each). 1. The quantum number is most responsible for defining the shape of an orbital. a. spin b. azimuthal c. Ψ d. magnetic e. n f. m s 2. There are orbitals in the third shell. a. 25 b. 4 c. 9 d. 16 e. 1 f. 8 g. 12 3. The azimuthal quantum number is 3 in orbitals. a. s b. p c. d d. f e. a f. m s 4. Electrons in the subshell of tin experience the lowest effective nuclear charge. a. 1s b. 3p c. 3d d. 5s e. 5p f. 4f 5. [Ar]4s 2 3d 10 4p 3 is the electron configuration for a(n) atom. a. As b. V
c. P d. Sb e. Sn f. Ge 6. The ground state electron configuration for a Cd atom is: a. [Ar]4s 2 4d 9 b. [Kr]5s 2 4d 10 c. [Kr]5s 2 3d 9 d. [Xe]5s 2 5d 10 e. [Kr]5s 2 3d 10 f. [Xe]5s 1 5d 11 7. How many unpaired electrons are in an Arsenic (As) atom? a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 g. 6 8. How many valence electrons are there in Ni? a. 2 b. 3 c. 4 d. 6 e. 8 f. 9 g. 10 9. What are the valence electrons in Zn? a. 4s and 3d b. 4s c. 3d d. 4p e. 3p f. 5s 10. Suppose that the spin quantum number m s had four possible values instead of two. How many elements would then be in the first two rows of the periodic table? a. 8 b. 10 c. 14 d. 20 e. 28
f. 4 11. Predict the atomic number of the next Noble gas (below Rn). a. 126 b. 112 c. 118 d. 106 e. 96 f. 136 12. Which of the following electrons would you predict to feel the highest effective nuclear charge. a. The 3s electrons in Mg b. The 3p electrons in Al c. The 3p electrons in Cl d. The 3p electrons in P e. The 4s electrons in Ca. f. The 4s electrons in K. 13. Which of the following are correctly listed in the order of size (smallest to largest)? a. K, Ca, Cu, Ga, Ge, As, Se b. Se, As, Ge, Ga, Cu, Ca, K c. Cu, Se, As, Ge, Ga, Ca, K d. Cu, K, Ca, Ga, Ge, As, Se e. Se, As, Ge, Ga, Ca, K, Cu f. Cl, S, P, Mg, Si, Al 14. Which of the following ions has the smallest radius? a. Li + b. Be + c. K + d. Mg + e. B 3+ f. Al 2+ g. F - h. Cl - 15. Which of the following are in the correct order of first ionization energy (most positive to least positive or negative). a. Na > F > O > C > Be > Li b. Na > Li > Be > C > O > F c. Li > Be > C > O > F > Na d. F > O > C > Be > Li > Na
16. Which of the following has the largest second ionization energy? a. Ca b. K c. Ga d. Ge e. Se f. Rb 17. Which of the following elements has the most negative electron affinity? a. O b. K c. Na d. S e. Se f. As g. P 18. Chlorine is much more likely to exist as a ion than sodium. This is because. a. positive, chlorine is bigger than sodium b. negative, chlorine is bigger than sodium c. positive, chlorine has the higher electron affinity d. negative, chlorine has the higher ionization potential e. negative, chlorine has the greater electron affinity f. negative, chlorine has both, the greater electron affinity and the higher ionization energy. 19. The electron configuration belonging to the atom with the most negative electron affinity is a. 1s 2 2s 2 sp 6 3s 1 b. 1s 2 2s 2 sp 6 3s 2 c. 1s 2 2s 2 sp 6 3s 2 3p 1 d. 1s 2 2s 2 sp 6 3s 2 3p 4 e. 1s 2 2s 2 sp 6 3s 2 3p 5
Problems, show work ON THIS SHEET. Use additional paper if necessary (8 points each) 20. Given the data: N 2 (g) + O 2 (g) ----> 2NO(g) 2NO(g) + O 2 (g) -----> 2NO 2 2N 2 O(g) -----> 2N 2 (g) + O 2 (g) ΔH =185.2 kj ΔH = -113.1 kj ΔH = -163.2 kj Use Hess's law to calculate ΔH for the reaction N 2 O(g) +NO 2 (g) -----> 3NO(g) ΔH? N 2 O(g) -----> N 2 (g) + 1/2O 2 (g) NO 2 -----> NO(g) + 1/2O 2 (g) N 2 (g) + O 2 (g) ----> 2NO(g) ΔH = -163.2/2 = -81.6 kj ΔH = 113.2/2 = 56.6 kj ΔH =185.2 kj N 2 O(g) +NO 2 (g) -----> 3NO(g) ΔH = 160.2 21. Under constant-volume conditions, the heat of combustion of glucose (C 6 H 12 O 6 ) is 15.57 kj/g. A 7.8500 g sample of glucose is burned in a bomb calorimeter. The temperature increased from 18.67 C to 24.62 C. a. What is the total heat capacity of the calorimeter? b. If the size of the glucose sample had been exactly twice as large, what would the final temperature of the calorimeter have been? a. q = 15.57 kj/g(7.8500 g) = (heat capacity)(δt) = (heat capacity)(24.62-18.67 C) = 122.22 kj = (heat capacity)(5.94 C) (heat capacity) = 122.22 kj/5.94 C = 20.57 kj/ C b. 122.22(2) = (heat capacity)(x - 18.67) x = 30.55 C
22. Consider the following reaction: CH 3 OH(g) -----> CO(g) + 2H 2 (g) ΔH = 90.7 kj a. Is heat absorbed or evolved in the course of this reaction? b. Calculate the amount of heat transferred when 62 g of CH 3 OH(g) is decomposed by this reaction at constant pressure. c. For a given sample of CH 3 OH, the enthalpy change on reaction is 36.7 kj. How many grams of hydrogen gas are produced? a. Absorbed b. moles CH 3 OH: 62 g/32 g/mol = 1.93 moles 1.93 moles (90.7 kj/mole CH 3 OH) = 175.7 kj = 180 kj. c. 90.7 kj/2 moles H 2 = 45.35 kj/mole H 2 36.7 kj/45.35 kj/mole = 0.809 mole H 2 (2 g/mole) = 1.62 g 23. How many grams of methane (CH 4 (g)) must be combusted to heat 2.4 kg of water 25 C to 85 C, assuming a standard combustion reaction? CH 4(g) + 2O 2(g) -----> 2H 2 O(g or l acceptable) + CO 2 (g) ΔH form -74.8 0-241.8-393.5 2(-241.8) + (-393.5) - [-74.8 +0) = -802.3 kj q = 2400 g (4.184 j/molk)(60) = 602496 J = 802.3 kj/mol(x mole methane) x = 0.751 moles(16 g/mol) = 12 g. (answer from 11-12 if you showed you did it right) 24. Carotenoids, present in all organisms capable of photosynthesis, extend the range of light absorbed by the organism. They exhibit maximal capacity for absorption of light in the range of 440-470 nm. Calculate the energy in kilojoules represented by absorption of an Avogadro's number of photons of wavelength 455 nm. c = λν ν = 3.0 x 10 8 m/sec/455 x 10-9 m/= 6.5934 x 10 14 E = hνn A = (6.626 x 10-34 J-s/photon)(6.5934 x 10 14 sec -1 )(6.0221 x 10 23 photons) = 263 x 10 3 J