1211 Exam Iii Fall09 Wanswers 3c4u5e

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Clayton State University Examination III Principles of Chemistry (CHEM 1211) Fall 2009 November 17, 2009 The following exam is worth 100 points Each question is worth 10 points Please………………………….. Make sure there are seven pages in your exam copy Use non-erasable PEN Write your name on all exam sheets Sign the upper right-hand corner of the first sheet (signature is taken as a statement of honest and independent work) State appropriate units with each numerical answer Show all work and clearly indicate the final answers using boxes

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Question 1 (a) Name two elements and two compounds that exist as gases at room temperature. Elements: He, Ar, Ne, H2, O2, F2, Cl2 etc. Compounds: CO2, SO2, CH4, CO etc. (b) State Boyle’s law and Charles’s law in the form of equations. Indicate the conditions under which each law is applicable. Sketch graphs to illustrate the laws. Boyle’s law: P1V1 = P2V2 at constant T and n V

V

1/P

P

Charles’s law: V1/T1 = V2/T2 at constant P and n V

V

T, K

- 273

T, oC

Question 2 (a) A gas occupying a volume of 725 mL at a pressure of 0.970 atm is allowed to expand at a constant temperature until the pressure becomes 0.541 atm. What is its final volume (in mL)? P1V1 = P2V2

V2 

P1 V1  0.970 atm  0.725 L    1.30 L 1.30 x 10 3 mL P2  0.541 atm 

(b) At 46 oC a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-tenth (0.10) of the original value at the same temperature? V2 = 0.1V1 or 10V2 = V1

P2 

P1 V1  5.3 atm 10V2    53 atm V2  V2 

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Question 3 (a) The volume of a gas at STP is 488 mL. Calculate its volume at 22.5 atm and 150 oC. T1 = 273 K, P1 = 1.00 atm, V1 = 0.488 L, T2 = 423 K, P2 = 22.5 atm, V2 = ?

P1 V1 P2 V2  T1 T2

V2 

P1 V1T2 1.00 atm  0.488 L  423 K    0.0336 L or 33.6 mL T1 P2  273 K  22.5 atm 

(b) A sample of air contains only nitrogen and oxygen gases whose partial pressures are 0.80 atm and 0.20 atm, respectively. Calculate the total pressure and the mole fractions of the gases. PT = 0.80 + 0.20 = 1.00 atm x1 = 0.80/1.00 = 0.80 x2 = 0.20/1.00 = 0.20

Question 4 (a) Dry ice is solid carbon dioxide (CO2). A 0.00114-mol sample of dry ice is placed in an evacuated vessel of volume 4.6 L at 30 oC. Calculate the pressure inside the vessel after all the dry ice has been converted to CO2 gas. PV = nRT P

 0.00114 mol  0.08206 L atm/mol K  303 K   6.2 x 10 3 atm nRT  V 4.6 L

(b) What is the molar mass of a gas if a 0.165-g sample occupies 34.8 mL at a pressure of 1.22 atm and a temperature of 50 oC? d = m/V = 0.165 g/0.0348 L = 4.74 g/L M

dRT  4.74 g/L  0.08206 L atm/mol K  323 K    103 g/mol P 1.22 atm

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Question 5 Some commercial drain cleaners contain two components: sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: 2NaOH(aq) + 2Al(s) + 6H2O(l) → 2NaAl(OH)4(aq) + 3H2(g) The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of H2 formed at STP if 3.12 g of Al is treated with an excess of NaOH.

mol H 2  3.12 g Al x

3 mol H 2 1 mol Al x  0.173 mol H 2 26.98 g Al 2 mol Al

PV = nRT V

 0.173 mol  0.08206 L atm/mol K  273 K   3.89 L nRT  P 1.00 atm

Question 6 (a) What is the wavelength (in nanometers) of radiation of frequency 2.20 x 109 Hz? 10 9 nm c 3.00 x 10 8 m/s λ   0.136 m x  1.36 x 10 8 nm 9 ν 1m 2.20 x 10 1/s

(b) How long will it take a radio wave to travel from the planet Venus to Earth (Average distance from Venus to Earth = 28 million miles, 1 mile = 1.61 km)?

28 x 10 6 miles x

1.61 km 10 3 m 1s x x  150 seconds 1 mile 1 km 3.00 x 10 8 m

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Question 7 In each part, a set of quantum numbers is given. If the set is an allowed combination of quantum numbers, give the maximum number of electrons and the number of orbitals. If the combination of quantum numbers is not allowed, state why. a) n = 2, ms = +1/2 2s, 2p: 4 orbitals and 4 electrons b) n = 4, ml = +1 4p, 4d, 4f: 3 orbitals and 6 electrons c) n = 3, l = 2 3d: 5 orbitals and 10 electrons d) n = 2, l = 0, ms = -1/2 2s: 1 orbital and 1 electron e) n = 4, l = 3, ml = -4, ms = +1/2 invalid Question 8 (a) The ground state electron configurations listed below are incorrect. Write the correct electron configurations and state the numbers of valence electrons and core electrons for each. Al: 1s22s22p43s23p3 1s22s22p63s23p1

valence = 3

core = 10

K: 1s22s22p43s33p64s14p1 1s22s22p63s23p64s1

valence = 1

core = 18

(b) Write the complete electron configuration and draw the orbital diagram for the sulfide ion. S21s22s22p63s23p6

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Question 9 (a) How many photons of a carbon dioxide laser light at 10,060 nm would be required to produce 0.500 J of energy? E = nhc/λ n





 0.500 J  10,060 x 10 9 m  2.53 x 1019 photons Eλ  hc 6.626 x 10 34 Js 3.00 x 10 8 m/s







(b) What is the wavelength of the line in the spectrum of the hydrogen atom that arises from transition of the electron from the orbit with n = 3 to the orbit with n = 1?

 1 1 1  1  1  R H  2  2   1.097 x 10 7 m 1  2  2   9.751 x 10 6 m 1 λ n2  1   3  n1

λ

1  1.026 x 10 7 m 9.751 x 10 6

Question 10 (a) What are isoelectronic series? Atoms and ions with the same number of electrons and configuration (ii) Write five ions or atoms that are isoelectronic with Kr. Arrange them in order of increasing size. Sr2+, Rb+, Br-, Se2-, As3(iii) Indicate which species has the highest ionization energy: Mg, Ba, Ca, Be Be (iv) State which member of the following elements has the smallest electron affinity (least negative value): Mg, S, Al, P Mg

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Bonus The active ingredients of an antacid tablet contained only magnesium hydroxide and aluminum hydroxide. Complete neutralization of a sample of these active ingredients required the addition of hydrochloric acid to form chloride salts. Write balanced chemical equations for the reactions between the active ingredients and hydrochloric acid. Mg(OH)2 + 2HCl → MgCl2 + 2H2O Al(OH)3 + 3HCl → MgCl3 + 3H2O

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