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.
Solution: All the statements are true.
21. Which of the following is a primary alcohol? a. CH3CH2CHO b. CH3CH(OH)CH3 c. CH3C(O)CH3 d. CH3CH2CH2OH e. CH3OCH3
Solution: only b and d are alcohols, and b is a secondary alcohol and d is a primary alcohol. A primary alcohol is one that is attached to a primary carbon—a carbon atom singly bonded to only one other carbon atom.
22. Shown below are three possible Lewis structures for nitrous oxide, N2O
1. N NO 2. NNNO3. NO
The valid Lewis structure(s) is (are): a. 1 b. 2 c. 3 d. 1 and 2 e. 2 and 3
Solution: The Lewis structure for this molecule must show a total of 16 electrons (shared and unshared). In addition, the octet rule needs to be satisfied for each atom in the structure. Structure 1 has 18 electrons, and the central nitrogen has 10 electrons. Structure 2 has the correct number of electrons, but the central nitrogen has 10 electrons. Structure 3 is a valid structure.
-3 23. How many moles of hydroxide ion are present in 0.300 L of 5.0 x 10 M Ba(OH)2 (M = 171.3 g/mol) solution? a. 0.0015 mol b. 0.0030 mol c. 0.0050 mol d. 0.0060 mol e. 0.0075 mol
5.00 x 10 -3 mol Ba(OH) 2 mol OH- Solution : mol OH- = 2 x 0.300 L soln x = 0.0030 mol L soln 1 mol Ba(OH)2
24. When an atom becomes an anion it: a. loses electrons b. loses protons c. gains neutrons d. gains electrons e. gains protons
Solution: atoms become ions by gain or loss of electrons. An anion is a negatively charged ion. An atom acquires a negative charge by gaining one or more electrons.
25. Which of the following is not an electrolyte in water? a. ethanol (ethyl alcohol) b. HCl c. copper sulfate d. NaBr e. acetic acid
Solution: all ionic substances and molecular substances that are either acids or bases are electrolytes in water. Ethanol is a molecular substance but not an acid or a base.
26. The formula of the amino acid alanine is C3H7NO2. Three moles of alanine contains moles C, moles H, moles N, and moles O. a. 9, 21, 3, 6 b. 6, 21, 3, 6 c. 6, 10, 3, 5 d. 3, 6, 9, 21 e. 9, 21, 3, 5
Solution: multiply each subscript by the three mols of alanine. Hence, 9 mol C, 21 mol H, 3 mol N, and 6 mol O.
27. The ground state electron configuration of magnesium is: a. 1s22p63d4 b. 1p22s10 c. 1s42s42p4 d. [Ar]4s23d5 e. 1s22s22p63s2
Solution: magnesium possesses 12 electrons. Following the rules for writing electron configurations its configuration corresponds to e.
+ - 28. When [H3O ] = [OH ]: a. pH > 7 b. pH < 7 c. the solution is neutral d. the solution is acidic e. the solution is basic
+ - Solution: by definition, a solution is neutral when [H3O ] = [OH ]
29. Which arrangement lists the elements in correct order with respect to increasing first ionization energy? a. C, Si, Li, Ne b. Li, Si, C, Ne c. Li, Si, Ne, C d. Li, C, Si, Ne e. Ne, C, Si, Li
Solution: The first ionization energy increases moving left to right across a period and it also increases from bottom to top within a group. Applying these two principles, Li has the lowest first ionization energy and Ne has the highest first ionization energy. The ionization energy of Si is smaller than C. Hence the correct order is given in b.
30. The atomic number of an element is 69. From this information it can be concluded that there are 69 in the neutral atom a. electrons b. neutrons c. protons d. electrons and neutrons e. electrons and protons
Solution: The atomic number corresponds to the number of protons in an atom. Atoms are neutral, and hence, there must also be an equal number of electrons (answer e).
31. From its position in the periodic table, manganese is most likely a: a. gas b. low-melting solid c. high melting metal d. liquid e. metalloid
Solution: manganese (Z = 25) is a first row transition metal. Neighboring first row transition metals (Cr, Fe, Co, etc) are high melting as is manganese.
32. One explanation for the chemical similarity of elements in a group is that they possess the same number of electrons a. core b. inner c. subshell d. unpaired e. valence
Solution: elements in a group possess the same number of valence electrons. And valence electrons determine the chemical properties of an element.
33. Treatment of propene with bromine in carbon tetrachloride yields: a. 1-bromopropane b. 2-bromopropane c. cyclopropane d. 1,2-dibromopropane e. 1,1,2,2-tetrabromopropane
Solution: The reaction is CH3CH2=CH2 + Br2 CH3CH2BrCH2Br. The product is 1,2-dibromopropane.
34. Which of the following 0.10 M aqueous solutions has the highest boiling point? a. Al(NO3)3 b. MgCl2 c. Li2SO4
d. NaBr e. (NH4)2SO4
Solution: Boiling point elevation is a colligative property—a property that depends only on the number of dissolved solute particles. All listed solutes are ionic and each separates into its component ions in water. Although the solute concentrations are the same for each solution, it is the total ion concentration that is the determining factor. 3+ - Each unit of Al(NO3)3 is the source of four ions (1 Al and 3 NO3 ). The total ion concentration of a 0.10 M Al(NO3)3 solution is 0.4 M. By a similar analysis, the ion concentrations of the other solutions are 0.30 M, 0.30 M, 0.20 M, and 0.30 M respectively.
35. The concentration of CaCO3 in a sample of hard water is 0.0215 g/L. This concentration expressed in grains per gallon (gpg) is: (1 grain = 0.0648 g; 1 gallon = 3.78 L) a. 3.69 x 10-4 gpg b. 0.00527 gpg c. 0.0878 gpg d. 0.332 gpg e. 1.25 gpg
0.0215 g æ 3.78 L öæ 1 grain ö Solution : gpg = ç ÷ç ÷ = 1.25 grains per gallon (gpg) L è 1 gal øè 0.0648 g ø
36. The phase change CO2 (s) CO2 (g) is called: a. boiling b. fusion c. melting d. sublimation e. vaporization
Solution: sublimation, by definition.
37. The specific heat of water is 4.18 J/g-oC, and the specific heat of copper is 0.382 J/g- oC. Water heat compared to copper when equal masses of water and copper both initially at 75oC cool to 25oC. a. absorbs less b. releases less c. absorbs more d. releases more e. absorbs the same amount of
Solution: When a sample cools it releases (loses) heat. Since the specific heat of water is greater than the specific heat of copper and equal masses of both are being compared, more heat will be released by the water as it cools.
38. Which of the following substances possesses polar bonds, but is a non-polar molecule? a. CO2
b. Cl2 c. HF d. H2O e. NaCl
Solution: A molecule is polar if it possesses a polar bond(s) and has an overall non-zero dipole moment. Both C-O bonds in CO2 are polar. The associated bond dipoles (vector) for these bonds are equal, but because CO2 is linear, they point in opposite directions and cancel. Hence, the overall dipole moment of CO2 is zero making it a non-polar molecule.
39. Consider the following table:
Acid Formula Ka -5 acetic acid HC2H3O2 1.8 x 10 -2 Chlorous acid HClO2 1.1 x 10 Hydrofluoric acid HF 6.8 x 10-4 Hypobromous acid HBrO 2.5 x 10-9 -1 iodic acid HIO3 1.7 x 10
If 0.1 M solutions of each acid listed in the preceding table are prepared, which solution would have the lowest pH? a. HC2H3O2 b. HClO2 c. HF d. HBrO e. HIO3
Solution: All the listed acids are monoprotic weak acids. Thus the Ka values can be directly compared. The larger the value of Ka the greater the extent of dissociation and + + + the higher [H3O ] (or [H ]) is at equilibrium. By definition pH = -log[H ]. And pH decreases as [H+] increases. The acid solution having the lowest pH corresponds to the acid with the largest Ka.
40. What are the approximate bond angles about a carbon atom in the alkyne C2H2? a. 60o b. 90o c. 109.5o d. 120o e. 180o
Solution: Alkynes are characterized by a carbon-carbon triple bond. C2H2 (acetylene) is the simplest alkyne. Its Lewis structure is H-C≡C-H. According to VSEPR rules, the geometry about an atom is determined by the number of substituents attached to it. Both carbons have two substituents and no unshared electron pairs (AX2 in VSEPR). The predicted geometry is linear with a bond angle of 180o.
o Alternatively, both carbons are sp hybridized, and the two hybrid orbitals are 180 apart.