WYSE ‘Academic Challenge’ 2002 Regional Solutions Key 1. Consider the following liquids and their respective densities: pentane, d = 0.626 g/mL; carbon tetrachloride, d = 1.88 g/mL; diiodomethane, d = 3.33 g/mL; mercury, d = 13.7 g/mL. If equal masses of each liquid are compared, which liquid occupies the largest volume? a. pentane b. carbon tetrachloride c. diiodomethane d. mercury e. all occupy the same volume Solution: Of the listed liquids, pentane has the smallest density, and it will occupy the largest volume (volume = mass/density). 2. A substance X melts at 10oC and boils at 108oC. This substance is soluble in water and its aqueous solutions do not conduct electricity. This substance is best classified as a(n): a. acid b. ionic compound c. metal d. molecular compound e. network covalent compound Solution: The physical data are consistent with a molecular compound. Metals and network covalent substances are not water soluble; most are solids at and above room temperature. Acids and ionic compounds are electrolytes in water. 3. Which property of a sample of water remains the same when it changes from a solid (ice) to a gas? a. density b. mass c. enthalpy d. volume e. all remain the same Solution: all but mass change as a consequence of this phase change. Volume increases dramatically and density decreases. A significant amount of heat must be absorbed to transform ice to a gas. 4. 3.54 x 10-4 is the same as: a. 0.0000354 b. 0.000354 c. 0.0354 d. 354 e. 35,400 Solution: to express this number in expanded notation move the decimal place four places to the left or divide it by 10000. 5. The following electronic transitions occur when lithium atoms are sprayed into a hot flame. The various steps are numbered for identification. (Note: 4s 3p means an electronic transition from a 4s to a 3p orbital.) 2s I 2p II 3d III 3p IV 4s V 2p Which of these steps would result in the emission of light? a. step III b. III and V c. I, II, and V d. III, IV, and V e. all of the steps Solution: light emission can occur when an electron moves from an orbital of higher energy to one of lower energy. In a many electron atom, orbital energies are determined by both the principle quantum number, n, and the azimuthal quantum number, l. Orbital energies increase with increasing n and l. Hence, the transitions 3d to 3p and 4s to 2p result in the emission of light. 6. The prefix equivalent for 10-3 is: a. deci, d b. kilo, k c. micro, µ d. milli, m e. pico, p Solution: the prefix equivalent for 10-3 is milli. 7. A sample of a copper oxide is transferred to a test tube having a mass of 24.738 g. The combined mass of the test tube and sample is 25.899 g. After heating the tube in a methane atmosphere the copper oxide is converted to copper. The mass of the test tube and copper is 25.765 g. What is the mass percent of copper in the copper oxide? a. 4.25% b. 13.4% c. 79.6% d. 88.5% e. 99.5% mass Cu in sample (25.765 g - 24.738 g) Solution : %Cu = x100 = x100 = 88.5% mass of sample 25.899 g - 24.738 g 8. Which formula is incorrect? a. NaCO3 b. BaSO4 c. Ca(OH)2 d. NH4NO3 e. KI Solution: The sum of charges in an ionic substance is zero—principle of electroneutrality. Sodium ion has a charge of +1 and carbonate ion has a charge of –2. The formula NaCO3 is incorrect. The correct formula is Na2CO3. 9. The correct name for Al2S3 is: a. aluminum sulfite b. aluminum sulfide c. aluminum bisulfite d. dialuminum trisulfur e. sulfur aluminate Solution: The component ions of an ionic substance are included in its name in the sequence cation(s) then anion(s). Metal cations having only one common oxidation state retain the element name (aluminum in this case). Non-metal anions are named by adding the suffix –ide to the stem of the element name (sulfide in this case). 10. What information about a molecule does an empirical (simplest) formula provide? a. the relative number and kind of atoms that make up the molecule. b. the three dimensional arrangement of atoms in the molecule. c. the atom connectivity within the molecule. d. the number of shared and unshared electron pairs in the molecule. e. all of the above. Solution: by definition, an empirical formula is the simplest whole number ratio of atoms in a substance. 11. Consider the following ions: Si2+, N3-, Mg+, Ar2+, H-, P+, Se2-, Br2+. Which ions possess a closed-shell (noble gas) electron configuration? a. Si2+, Ar2+, H- b. N3-, H-, Se2- c. N3-, P+, Br2+ d. Mg+, P+, Se2- e. Si2+, Mg+, Br2+ Solution: the number of electrons associated with the listed ions are: Si2+ (12), N3- (10), Mg+ (11), Ar2+ (16), H- (2), P+ (14), Se2- (36), Br2+ (33). N3-, H-, and Se2- have a closed- shell configuration. 12. The stoichiometric reaction between an acid and a base is called: a. disproportionation b. hybridization c. neutralization d. oxidation e. reduction Solution: neutralization, by definition. 13. Which set of coefficients balances the following unbalanced equation: CH4 (g) C3H8 (g) + H2 (g) a. 3, 1, 1 b. 3, 2, 1 c. 3, 1, 2 d. 6, 2, 2 e. 6, 2, 6 Solution: 3CH4 1C3H8 + 2H2 14. Ethanol (C2H6O, M = 46.1 g/mol) is produced by the fermentation of glucose (C6H12O6, M = 180.2 g/mol). When 20.0 g of glucose is fermented, 10.1 g of ethanol is produced. What is the mole ratio of ethanol to glucose (mol ethanol/mol glucose) in this reaction? a. 0.010 b. 0.50 c. 2.0 d. 8.0 e. 10 Solution: from the given information there is 0.219 mol ethanol (10.1 g/46.1 g/mol) and 0.111 mol glucose (20.0 g/180.2 g/mol). The calculated mole ratio is1.97. 15. What mass of barium chloride (M = 208.3 g/mol) is required to completely react with 10.0 g of aluminum sulfate (M = 342.2 g/mol): 3BaCl2 (aq) + Al2(SO4)3 3BaSO4 (s) + 2AlCl3 (aq) a. 2.03 g b. 6.06 g c. 15.2 g d. 18.3 g e. 49.3 g æ 1 mol Al SO öæ 3 mol BaCl öæ 208.3 g BaCl ö = ç 2 4 ÷ç 2 ÷ç 2 ÷ = Solution : mBaCl2 10.0 g Al2SO4 ç ÷ç ÷ç ÷ 18.3 g è 342.2 g Al2SO 4 øè1 mol Al2SO4 øè molBaCl2 ø - 16. The oxidation number of C in C2H3O2 is: a. 2 b. 1 c. 0 d. -1 e. –2 Solution: The oxidation numbers of H and O are +1 and –2 respectively. In an ion the sum of oxidation numbers equals the charge: -1 = 3(+1) + 2(-2) + 2(oxidation number of C). Hence, the oxidation number of C is 0. 17. A cylinder in an automobile engine has a volume of 0.600 L. It takes in air at a pressure of 0.950 atm and a temperature of 30oC, and compresses it to a volume of 75.0 mL at 85oC. What is the air pressure in the cylinder after compression? a. 8.98 x 103 atm b. 21.5 atm c. 8.98 atm d. 0.140 atm e. 8.98 x 10-3 atm P V P V P (0.0750L) 1 1 = 2 2 = (0.950 atm)(0.600 L) = 2 Þ = Solution : P2 8.98 atm T1 T2 303.2 K 358.2 K 18. A 0.100 L sample of carbon monoxide, CO, is at a pressure of 0.905 atm and a temperature of 565oC. What mass of CO does this represent? a. 0.00131 g b. 0.0368 g c. 0.0546 g d. 21.3 g e. 174 g Solution: find moles of CO by appropriate substitution of the given information into the ideal gas law (= 0.001315 mol). Use the found moles of CO and the molar mass of CO (28.01 g/mol) to determine the corresponding mass of CO (= 0.0368 g). 19. Sodium azide (NaN3, M = 65.0 g/mol) is the propellant used to inflate an automobile airbag in a collision. When detonated, it decomposes into sodium and nitrogen, and the liberated nitrogen rapidly inflates the airbag: 2NaN3 (s) 2Na (s) + 3N2 (g) What volume of nitrogen measured at STP is produced by the decomposition of 1.30 2 x 10 g of NaN3? a. 22.4 L b. 28.9L c. 33.6 L d. 44.8 L e. 67.2 L 2 Solution: 1.30 x 10 g NaN3 corresponds to 2.00 mol NaN3. This produces 3 mol N2 when decomposed. At STP, 1 mol of gas occupies 22.4 L. Hence, 3 mol N2 occupies 67.2 L at STP. 20. Which of the following statements is true about a gas in a container at a temperature of 25oC and a pressure of 1 atm? a. the pressure exerted by the gas results from gas particles colliding with the container wall. b. the space inside the container is mostly empty. c. the gas can be compressed. d. the density of the gas is significantly smaller than its density in the solid or liquid states. e. all of the above.