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CHEM 101 TEST 3 SUMMER 2001 (DELANEY) KEY


  1. For each of the compounds below, draw the Lewis structure, indicate what hybrid orbitals would be used around the central atom, using VSEPR theory predict, name and draw the molecular structure: (20 points)

    1. CH4



    2. PCl3



    3. SF6



    4. CaF2



  2. From the names write the proper formula for the compounds below or vise versa as appropriate: (5 points)

    1. ruthenium (III) nitrate - Ru(NO3)3
    2. MnI2 - manganese (II) iodide

    3. Rh2(SO3)3 - rhodium (III) sulfite

    4. As2O3 - diarsenic trioxide

    5. silver bromide - AgBr

  3. Define: (5 points)

    1. ionic bond - a bond between metal and nonmetal in which electrons are transferred. The bond is held together by the electrostatic attraction of the ions formed.

    2. covalent bond - a bond between two non-metals in which the electrons are shared.

    3. valence electrons - reactive electrons above the previous noble gas (the electrons used in bonding)

  4. To what volume must 110.0 mL of 3.950 N NaOH be diluted to in order to form a 0.2600 N NaOH solution? (10 points)

    N1V1 = N1V1, rearrange to

    (N1V1)/N1 = V1

    [(3.950 N)(110.00 mL)]/(0.2600 N) = 1671 mL

  5. What type of reactions are the following? Indicate your choice by putting the corresponding letter in front of the reaction. (5 points)

    a) addition or synthesis
    b) decomposition
    c) single displacement
    d) double displacement

    b 2 HgO(s) --------------> 2 Hg(l) + O2(g)

    a 4 RuO(s) + O2(g) ------------> 2 Ru2O3(s)

    d LiBr(aq) + AgNO3(aq) ---------------> AgBr(s) + LiNO3(aq)

    c Mg(s) + 2 HCl(aq) -----------------> MgCl2(aq) + H2(g)

    a 2 Fe(s) + O2(g) ---------------> 2 FeO(s)

  6. Use the molecular orbital scheme below to find the bond order of the proposed idatomic species below. Inidcate whether the species are diamagentic or parameagnetic.



    1. O2
      8 + 9 = 16 electrons total
      Bond order = (bonding electrons - antibonding electrons)/2 = (10-6)/2 = 4/2 = 2, paramagnetic

    2. BF
      5+9 = 14 electrons total
      Bond order = (bonding electrons - antibonding electrons)/2 = (10-4)/2 = 6/2 = 3, diamagnetic

    3. CO
      6+8 = 14 electrons total
      Bond order = (bonding electrons - antibonding electrons)/2 = (10-4)/2 = 6/2 = 3, diamagnetic

    4. C22+
      6+6-2 = 10 electrons total
      Bond order = (bonding electrons - antibonding electrons)/2 = (6-4)/2 = 2/2 = 1, diamagnetic

  7. Write the total ionic equation and then the net ionic equation for of the reaction below: (5 points)

    2 AgNO3(aq) + BaBr2(aq) ------> 2 AgBr(s)3)2(aq)

    Total ionic equation:

    2 Ag+(aq) + 2 NO3-(aq) + Ba2+(aq) + 2 Br-(aq) ------> 2 AgBr(s) + Ba2+(aq) + 2 NO3-(aq)

    Net Ioinic Equation:

    Ag+(aq) + Br-(aq) ------> AgBr(s)

  8. Interconvert between normality and molarity as appropriate: (5 points)

    1. 3.633 N H3PO4 =1.211 M H3PO4

    2. 11.00 M NaOH =11.00 N NaOH

    3. 2.840 N Ba(OH)2 =1.420 M Ba(OH)2

    4. 3.862 M H2SO4 =7.724 N H2SO4

  9. Balance the reactions below by inspection (10 points)

    1. MgCl2(aq) +2 AgNO3(aq) --> 2 AgCl(s) + Mg(NO3)2(aq)

    2. 2 NaClO3(s) --------------> 2 NaCl(s) + 3 O2(g)

    3. 2 Al(s) +3 H2SO4(aq) -----> Al2(SO4)3 (aq) +3 H2(g)

    4. 3 Ba(OH)2(aq) +2 H3PO4(aq) ------> Ba3(PO4)2(s) + 6 H2O

    5. C3H8(g) + 5 O2(g) ------->3 CO2(g) + 4 H2O(g)

  10. What is the molarity of an H2SO4 solution if a 15.00 mL portion of that solution is titrated with 39.85 mL of 1.650 N NaOH? (10 points)

    NAVA = NBVB, rearrange to:

    NA = (NBVB)/VA

    NA = [(1.650 N)(39.85 mL)]/(15.00 mL) = 4.384 N H2SO4

    M = N/#eq = (4.384 eq/L H2SO4)/(2 eq/mol) = 2.192 M H2SO4

  11. What is the limiting reactant when 10.00 g of rhodium (III) chloride react with 5.00 g of KOH to yield 3.91 g of rhodium (III) hydroxide and 5.69 g of KCl? What is the percent yield? The unbalanced reacton is shown below: (15 points)

    RhCl3 + KOH ---------> Rh(OH)3 + KCl

    Balanced equation:

    RhCl3 + 3 KOH ---------> Rh(OH)3 + 3 KCl

    FW RhCl3 = 102.90 +3(35.45) = 209.25 g/mol
    moles RhCl3 = (10.00 g)/( 209.25 g/mol) = 0.04779 mol

    FW KOH = 39.10 + 16.00 + 1.008 = 56.11 g/mol
    moles KOH = (5.00 g)/(56.11 g/mol) = 0.08911 mol

    [(3 mol KOH)/(1 mol RhCl3)](0.04779 mol RhCl3) = 0.1434 mol of KOH needed to react completely with RhCl3 present. There is not that much KOH so KOH must be the limiting reactant.

    moles of Rh(OH)3 produced thoerectically = [(1 mol Rh(OH)3)/(3 mol KOH)](0.08911 mol KOH) = 0.0297 mol Rh(OH)3

    FW Rh(OH)3 = 102.90+3(16.00)+3(1.008) = 153.92 g/mol

    g Rh(OH)3 theorectically produced = (0.0297 mol)(153.92 g/mol) = 4.57 g

    % yield = [(actual yield)/(thoerectical yield)](100) = [(3.91 g)/(4.57 g)](100) = 85.6%

    EXTRA CREDIT:
  12. What special reaction types (i.e precipitation, neutralization, combustion, gas-evolving) are the reactions in problem 10? (5 points)

    1. precipitation

    2. gas-evolving

    3. gas-evolving

    4. precipitation and neutralizatiuon

    5. combustion

  13. What is the molarity of a solution containing 16.00 g of C6H12O6 in 145.00 mL of aqueous solution? (5 points)

    FW C6H12O6 = 6(12.01)+12(1.008)+6(16.00) = 180.16 g/mol
    moles C6H12O6 = (16.00 g)/(180.16 g/mol) = 0.08881 mol
    (145.00 mL)(1 L/1000 mL) = 0.145 L
    M = (0.08881 mol)/(0.145 L) = 0.6125 M