15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O

Transcription

1 WYSE Academic Challenge Sectinal Chemistry Exam 2008 SOLUTION SET 1. Crrect answer: B. Use PV = nrt t get: PV = nrt 2. Crrect answer: A. (2.18 atm)(25.0 L) = n( L atm/ml K)(23+273) n = 2.24 ml Assume 100.0g g hydrgen is 8.16 ml H (8.23/1.008) g silicn ( ) is 3.27 ml Si (91.77/28.09). The rati Si:H is 1:2.5 (8.16/3.27 = 2.50). Multiplying each by 2 t get a whle number becmes a rati 2:5. The empirical rmula is therere Si 2 H 5. The mlar mass the empirical rmula is g/ml, which ges int the mlar mass the mlecular rmula 2 times (122.44/61.22). The mlecular rmula is therere Si 4 H Crrect answer: B. Use PV = nrt t get: V1 V2 = T1 T2 (1.30 L) V2 = ( )K ( )K V = L 4. Crrect answer: E. 5. Crrect answer: C. 2 The balanced equatin is 2Na S O ( aq) + I ( aq) Na S O ( aq) + 2NaI( aq) sum the ceicients is therere = 6. The 6. Crrect answer: A. Sample 3 has the least amunt space r the number atms present, resulting in a higher number cllisins with the walls the cntainer. 7. Crrect answer: E. The balanced equatin is 4Cr + 3O 2 2Cr 2 O 3, and the slutin is: ml Cr 2 ml Cr2O g Cr2O g Cr = 21.9 g Cr O g Cr 4 ml Cr ml Cr O 8. Crrect answer: A

2 9. Crrect answer: D. HCN, NH 4 +, and NO 2 bey the ctet rule. SF 6 cntains 12 electrns arund the central sulur atm. 10. Crrect answer: B. NO 2 exhibits resnance. 11. Crrect answer: B. Slutin #4 cntains 28 mles ins ( ). Slutin #3 cntains 36 mles ins ( ). Slutin #2 cntains 9.0 mles ins ( ). Slutin #1 cntains 6.0 mles ins ( ). 12. Crrect answer: C. Slutin #1 cntains 3.0 ml HCl ( ). Slutin #4 cntains 14.0 ml HCl ( ). The ttal number mles HCl in slutin is 17.0 ml ( ). The ttal vlume in slutin is 5.0 L. Therere, the new cncentratin is 3.40 M (17.0/5.0). 13. Crrect answer: B. There are 12 ml in Slutin #3 ( ). The cncentratin Slutin #2 is 2.25 M. Therere, a ttal vlume 5.3 L is needed (12/2.25). There are already 3.0 L in Slutin #3, s nly 2.3 L water wuld need t be added ( ). 14. Crrect answer: A. The greater the electrnegativity dierence between tw bnded atms, the mre plar the bnd is. The greatest dierence exists between C and O. 15. Crrect answer: B. The slutin is: 1 ml KHC H O 1 ml NaHCO g NaHCO 8.0 g KHC H O = 3.6 g NaHCO g KHC4H4O6 1 ml KHC4H4O6 ml NaHCO3 16. Crrect answer: D. Chice I wuld shit the psitin t the right t use up the nitrgen gas. Chice II wuld shit the psitin t the let t create mre mles gas and restre the pressure (reactant side has mre mles gas). Chice III wuld shit the psitin t the let t create mre hydrgen gas. Chice IV wuld cause n shit because yu are adding an inert gas. 17. Crrect answer: B. The balanced equatin is: 3CH 3 OH + O 2 3H 2 CO + 2H 2 O. Since yu have equal mles each, methanl must be limiting. Yu can therere make 4.0 ml rmaldehyde since the mle t mle rati is 3:3 (r 1:1) and yu are starting with 4.0 ml methanl.

3 18. Crrect answer: E. The mre nn-plar the gas, the mre ideally it behaves (mre elastic cllisins). Helium is the mst nn-plar gas because it is a nble gas and cntains the smallest amunt Lndn dispersin rces. 19. Crrect answer: E. The lwer the change in energy rm ne state t anther, the lnger the wavelength (inversely related). The energy change is the lwest rm the n = 5 t the n = 4 energy levels. 20. Crrect answer: C. Oxygen cntains 8 electrns. The 1s and 2s rbitals are illed, using up 4 electrns. There are three 2p rbitals, each cntaining ne electrn. The inal electrn must g int ne the 2p rbitals, making tw 2p rbitals each cntain ne unpaired electrn. 21. Crrect answer: E. Beaker #1 cntains 1 ml sulur atms (32.07/32.07). Beaker #2 cntains 1 ml arsenic atms (74.92/74.92). 1 ml atms = atms, therere bth beakers cntain the same number atms. 22. Crrect answer: D. Elements cnsist istpes which cntain a dierent number neutrns r elements can be in an inic rm which cntains a dierent number electrns. 23. Crrect answer: E. In general, atmic radius increases as yu mve t the let acrss a perid and dwn a clumn n the peridic table. 24. Crrect answer: D. Statements II, III, and IV are true. Statement I is alse. The pressures the gases in the tw ballns are abut the same. I pressure, temperature, and vlume are the same, then the number mles gas in bth ballns is the same. The mlar masses helium and hydrgen are dierent, causing the masses gas in the tw ballns t be dierent. 25. Crrect answer: A. In general, irst inizatin energy increases as yu mve t the right acrss a perid and up a clumn n the peridic table.

4 26. Crrect answer: E. The net inic equatin r the reactin is Ag + (aq) + Cl (aq) AgCl(s). There are ml Ag + ins ( L M) and ml Cl ins ( L M). Since Ag + and Cl react in a 1:1 rati, the Cl ins will get used up, leaving sme Ag + ins in slutin ater the reactin is cmplete. There will be ml Ag + ins letver ( ) in a ttal vlume L ( ). The resulting cncentratin ( M) is less than M. 27. Crrect answer: C. 28. Crrect answer: D. The Rman numeral is nt necessary n the calcium (it always rms a +2 charge with a nnmetal in an inic cmpund). 29. Crrect answer: A. Hydrgen is nn-plar, therere it nly cntains Lndn dispersin rces. 30. Crrect answer: C = ml. When adding, rund t the number that cntains the least number decimal places (which is 1.3). The least number decimal places is t the tenths place, therere the answer is runded t 10.8 ml (which has 3 signiicant igures). 31. Crrect answer: C. We can use the llwing ICE chart: HC 2 H 3 O 2 (aq) + H 2 O(l) H + (aq) + C 2 H 3 O - 2 (aq) Initial Change -x +x +x Equilibrium 0.83-x x x x 2 /(0.83-x) = 1.8 x 10-5 x = M = [H + ] ph = Crrect answer: B. The slutin is: -q = +q lss gain -mcδt = mcδt (c cancels) -mδt = mδ T -(50.0 g)(t C) = (100.0 g)(t C) T = 46.7 C

5 33. Crrect answer: D. The weaker the IM rces, the lwer the biling pint. Ammnia and water exhibit hydrgen bnding, which are relatively strng intermlecular rces. Lndn dispersin rces generally are weaker IM rces, with hydrgen gas being the lwest. 34. Crrect answer: C. The xidatin state N in NH 3 is +3. The xidatin state N in NO 2 is +4. The xidatin state N in NH 4 + is +5. The xidatin state N in NO 3 - is Crrect answer: A. The percent nitrgen by mass in each letter selectin is as llws: a) 82% N, b) 30% N, c) 64% N, d) 78% N, and e) 23% N. 36. Crrect answer: A. In general, electrnegativity increases as yu mve t the right acrss a perid and up a clumn n the peridic table. 37. Crrect answer: E. The charge n xygen is -2, therere the charge n metal in M must be +2. The metal in as 24 electrns, which means it must have 26 prtns (since tw electrns were taken away rm the metal in its neutral state t rm the +2 charge). Irn (Fe) cntains 26 prtns. 38. Crrect answer: D. There are ml NaOH (2.50/39.998) in 1.50 L slutin. The cncentratin is M NaOH (0.0625/1.50). Since NaOH is a strng base, it dissciates cmpletely making the cncentratin OH = M. The poh is 1.38 (-lg[0.0417]). Therere, the ph is ( ). 39. Crrect answer: D. Oxygen has a -2 charge, therere the charge n irn must be Crrect answer: E. All the ther selectins are endthermic prcesses. Reacting hydrgen and xygen gases t make water releases a lt energy and is therere exthermic.