1. Calculate partial pressure of NH₃ if 8.50 g NH₃, 12.8 g O₂ and 8.80 g CO₂ exert 3.00 atm.  (8)
     2. Calculate density of CO₂ if 2.50 g exert pressure of 1.20 atm at 25.0° C.   (8)
     3. A sample of gas at 3.50 atm occupies 235 mL. Calculate volume at 2.50 atm.    (8)
     4. If O₂ effuses at 64 mL/min, calculate rate of effusion for CO₂ at same temperature.   (8)
     5. Calculate volume of CO₂ produced at STP when 15.0 g CaCO_3(s) allowed to react with excess HCl: (8)
            CaCO_3(s) + 2HCl_(aq) CaCl_2(aq) + H₂O_(l) + CO_2(g)
     6. Draw Lewis structures with correct geometry (show lone pairs on central atom).    (15)
          If polar, indicate poles using delta (+/-).
          (a) BHF₂     (b) SeF₄      (c) SiF₄      (d) GeF₂     (e) SF₆
     7. Provide valence bond sketch including lone pair nonbonding orbitals.   (16)
         (a) H₂C=CH-CH₃      (b) H₃C-CN     (c) H-N=N-O-H     (d) NO₃^-1
     8. Identify hybrid orbital state of underlined. Use boxes/arrows to illustrate electrons in hybrid orbitals.  (15)
         (a) PI₅     (b) NO₂^-1       (c) SCl₆        (d) H-CN    (e) H₂C=N-H
     9. Provide brief explanation:        (15)
        (a) Propane (C₃H₈) has C-C-C bond angle of 109.5° while propene (C₃H₆) has bond angle of 120°.
        (b) The N-N bond length shorter in N₂H₂ than in N₂H₄.
        (c) Phosphorous forms PCl₃ and PCl₅ while nitrogen forms only NCl₃.
        (d) H₂Se has bond angle of 90° while H₂O has bond angle of 105°.
        (e) The N-O bond for NO₂^-1 not single or double bond.

     Answers
     1.  mol NH₃ = 8.50/17.0 = 0.500   mol O₂ = 12.8/32.0 = 0.400   mol CO₂ = 8.80/44.0 = 0.200
                 P_NH3 = (0.500/1.10)(3.00 atm) = 1.36 atm
     2.  d = (P)(MM)/RT = [(1.20 atm)(44.0 g/mol)]/[(0.0821 atm L/mol K)(298 K)] = 2.16 g/L
     3.  P₁V₁ = P₂V₂    (3.50 atm)(235 mL) = (2.50 atm)(V₂)    V₂ = 329 mL
     4.  (m_CO2/m_O2) = (m_O2/m_CO2)^½  (m_CO2/64 mL/min) = (32/44)^½      m_CO2 = 54.6 mL/min
     5.   mol CaCO₃ = mol CO₂ = 15.0/100 = 0.150          V = nRT/P = (0.150)(0.0821)(273)/(1.00) = 3.36 L
     

     9.   (a) Middle carbon in propane sp³ while middle carbon in propene sp².
            (b) N-N overlap for N₂H₄ between sp³ & sp³ while overlap for N₂H₂ between sp² & sp².
            (c) P can utilize 3d orbital to form sp³d hybrid orbitals; N has no 2d orbital to utilize.
            (d) Se uses p-orbitals to bond with H; O in sp³ hybrid state.
            (e) Resonace theory utilizes hybrid involving N-O & N=O bonds.