<p>Name______KEY______Sec______Quiz 3 6/8/06 35 points</p><p>1. (3) Determine the correct oxidation number for each underlined atom. 3- PO4 _X=+5____ H2 ___0___ PbO2 _X = +4___ X +4(-2) = -3 elemental hydrogen X + 2(-2) = 0 2. (6) Consider this balanced chemical equation:</p><p>2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(ℓ) a.. What atom is oxidized? S goes from ox# = -2 to +4 b. What is the oxidizing agent? O2(g) c. What atom gains electrons? each O gains 2 electrons</p><p>3. (3) Use this portion of the Activity Series to determine which reaction will occur. Li  Li+ + e- Ca  Ca2+ + 2e- Mn Mn2+ + 2e- Cu  Cu2+ + 2e- Au  Au3+ + 3e-</p><p>A. Mn + 2Li+  2Li + Mn2+ Li is stronger RA than Mn B. Ca + Cu  Ca2+ + Cu2+ not possible to oxidize both metals **C. 2Au3+ + 3Cu  2Au + 3Cu2+ D. Mn2+ + Ca2+  Mn + Cu not possible to reduce both metals E. Ca + 2Li+  2Li + Ca2+ Li is stronger RA than Ca</p><p>4. (5) Use this balanced chemical eqn to determine the mass of vanadium (V) in - 2+ a sample of ore if 26.45 mL of 0.02250 M MnO4 was required to titrate VO .</p><p>2+ - + + 2+ 5VO (aq) + 11H2O(ℓ) + MnO4 (aq)  5V(OH)4 (aq) + 2H (aq) + Mn (aq)</p><p>- - 2+ V, M of MnO4  mol MnO4  mol VO  mol V  g V</p><p>2+ - 2+ (0.02645 L x 0.02250 M) * (5 mol VO / 1 mol MnO4 ) * (1 mol V/ 1 mol VO ) * (50.94 g/mol) = 0.152 g</p><p>5. (3) Calculate the frequency of an X-ray with a wavelength of 3.44E-9 m. </p><p>ν = c/λ = 3E+8/3.44E-9 = 8.72E+16 1/s 6. (6) Balance this redox reaction in acid.</p><p>2+ - 3+ - Mn (aq) + BiO3 (s)  Bi + MnO4 (aq)</p><p>Write half reactions, balance atoms, O, H, charge 2+ - + Mn + 4H2O  MnO4 + 8H + 5e- oxidation - + 3+ BiO3 (s) + 6H + 2e-  Bi + 3H2O reduction</p><p>Balance e- 2+ - + 2 x (Mn + 4H2O  MnO4 + 8H + 5e-) - + 3+ 5 x (BiO3 (s) + 6H + 2e-  Bi + 3H2O)</p><p>Add half rxns and cancel common species 2+ - + - + 3+ 2Mn + 8H2O + 5BiO3 (s) + 30 H + 10e-  2MnO4 + 16H + 10e- + 5Bi + 15H2O</p><p>2+ - + - 3+ 2Mn + 5BiO3 (s) + 14 H +  2MnO4 + 5Bi + 7H2O</p><p>Reactants 2 5 15 14 +13 Mn Bi O H charge Products 2 5 15 14 +13</p><p> h = 6.626E-34 Js c = 3.00E+8 m/s R = (1.097E+7)/m</p><p>7. (4) Match the equation with the description</p><p>_C___1. duality of nature A. E = hν </p><p>__B__2. electromagnetic radiation B. λ = c/ν is a wave</p><p>_D___3. atomic spectral lines C. λ = h/(mv) </p><p>2 2 __A__4. electromagnetic radiation D. (1/λ) = R(1/nf – 1/ni ) is a photon (particle)</p><p>8. (5) Calculate the wavelength of light emitted when an electron in hydrogen goes from ni = 7 to nf = 4.</p><p>1/λ= 1.097E+7 1/m (1/16 – 1/49) = 4.617E+5</p><p>λ = 2.17E-6 m or 2.17 μm</p>