Sulfur Has 4 P Electrons ( )( )( ), Two of Which Are Unpaired

Multiple Choice

1.

A / Sulfur has 4 p electrons (­¯)(­ )(­ ), two of which are unpaired.

2.

C / Ga is in column 13, which has one p electron.

3.

C / CH2=CH2. CO2 has two double bonds, and H2O and PH3 have only single bonds.

4.

B / H2O and PH3 are the only polar molecules. H2O has a greater electronegativity difference.

5.

D / PH3 has a pair of electrons around P, which pushes the H's down forming a trigonal pyramidal shape.

6.

C / KMnO4, an important oxidizing agent, is purple.

7.

D / All column 1 metals form soluble salts. KMnO4 is purple. FeCl3 is soluble but yellow.

8.

D / v = (3RT/MM)½ \ the gas with a molar mass closest to 28 would have the same velocity.

9.

B / The volumes are the same for all gases, so the one with the greatest MM would be most dense.

10.

A / Rate » v = (3RT/MM)½ \ the gas with the smallest MM would have the greatest rate.

11.

C / A precipitation reactions are:
cation + anion ® insoluble ionic compound.

12.

D / Complex ions contain a transition metal cation surrounded by ligands, which has a overall charge.

13.

B / One element shows up in two products with different oxidation states.

14.

A / Combustion is a general term for redox reaction where oxygen gas is reduced.

15.

B / Freezing occurs at the lower plateau (boiling occurs at the higher plateau.

16.

A / Boiling is occurring during second plateau, where liquid molecules are entering the gas phase.

17.

A / Isoelectric series includes elements that are within three squares of a noble gas.

18.

A / The normal melting point is » 37oC and normal boiling is » 70oC (where line crosses 1 atm).

19.

D / The oxidation number for S in H2SO4 is +6, which is the same for SO2Cl2 (S + 2(-2) + 2(-1) = 0).

20.

D / SO2(g) is 1/5 the total moles \ 1/5 the total pressure (800/5 = 160 mm Hg).

21.

B / H2(g) is produced when strong acid (HCl) is added to metal with a negative Eored.

22.

B / This is a proton exchange reaction, where NH3 receives H+ forming the conjugate acid NH4+.

23.

D / A = 235 + 1 – (141 + 3 + x) \ x = 92
Z = 92 + 0 – (55 + 0 + y) \ y = 37 \ nuclide is 9237Rb

24.

C / The mole ratio in the compound is 2 K + 1 Te + 3 O \ the formula is K2TeO3.

25.

A / DHo = DHfoC6H6 –3 DHfoC2H2 = 83 – 3(230) =

26.

B / 0.250 L x 0.10 mol/L x 250 g/mol = 6.25 g

27.

A / The rate is based on the slow step, where reactant coefficients become exponents \ Rate = k[NO]2.

28.

D / Most ionic compounds are composed of a metal cation and a non-metal anion.

29.

C / Diamonds are covalent network solids where carbon atoms are bonded covalently to each other.

30.

C / pH = 8 \ pOH = 14 – 8 = 6 and [OH-] = 1 x 10-6 M

31.

A / At common temperature and pressure, moles of gas are directly proportional to moles \ 4 L CO2 = 12 L O2.

32.

B / Molecular substances are gases, which are composed of nonmetals, but not SiO2 (covalent network).

33.

D / Indicator changes color at equivalence, which occurs during vertical section of graph \ pH = 9.

34.

A / Buffer action occurs when acid and conjugate base have equal concentration (plateau section) \ V.

35.

A / Adding solute raises boiling pt. and osmotic pressure, but lowers vapor pressure and freezing pt.

36.

B / The yellow ppt. indicates Pb2+. Dissolving the precipitate with NH3 indicates Ag+.

37.

A / Using Le Chatelier's principle, lowering the exothermic reaction temperature will shift to right.

38.

D / Na2SO4 has a van't Hoff factor of 3, where NaCl has a factor of 2 \ Na2SO4 produces more moles of ions.

39.

B / KNO3 has the greatest change in solubility between 90oC and 30oC \ a higher % precipitates at lower T.

40.

D / .1 L x .4 mol H+/L x 1 mol H2/2 mol H+ x 22.4 L/mol H2 = 0.448 L = 448 mL

41.

C / Reaction happens \ DG < 0, Temperature drops \ DH > 0, Gas is produced \ DS > 0.

42.

B / K for the reverse reaction is 1/K for the forward reaction \ (2.0 x 105)-1 = 5.0 x 10-6.

43.

D / 63.55 = 63(1 – x) + 65(x) = 63 – 63x + 65x = 63 + 2x
0.55 = 2x \ x » 0.25

44.

D / Atomic radius decreases because each element in a period gains a proton w/o increasing core electrons.

45.

A / mol ethanol = 46 g x 1 mol/46 g = 1 mol
mol H2O = 54 g x 1 mol/18 g = 3 mol \ 1/(1 + 3) = 0.25

46.

B / The effective nuclear charge for Na is lower, which results in a weaker attraction for valence electrons.

47.

C / Reaction order is determined from experiment because it depends on the slow step in a mechanism.

48.

B / Ionic compounds are soluble in polar solvents. CCl4 is nonpolar.

49.

D / Cr2O72- + 14 H+ + 6e- ® 2 Cr3+ + 7 H2O
3(H2S ® S + 2 H+ + 2 e-)
Cr2O72- + 8 H+ + 3 H2S ® 2 Cr3+ + 7 H2O + 3 S

50.

C / Adding solid acid to distilled water is necessary for titration because the acid has to be in solution.

51.

B / NO3-: N + 3(-2) = -1 \ N = +5
NO: N + (-2) = 0 \ N = +2

52.

B / C3H8 + 5 O2 ® 3 CO2 + 4 H2O

53.

D / Ea for forward and reverse reactions are equal then DE between reactants and products = 0.

54.

A / Only first order reactions have a constant half life, which is 3 days for this substance.

55.

D / Unshared pairs of electrons take up more space, which leaves less space for H and a tighter angle.

56.

A / Higher boiling pt. is due to stronger intermolecular bonds—dispersion forces, which increase with MM.

57.

D / The concentration of M and N are both doubled. M is 1st order and N is 2nd order \ rate is 2 x 4 = 8 x faster.

58.

A / N2H4: 8 g x 1 mol/32 g x 4 mol H2O/2 mol = 0.5 mol
N2O4: 92 g x 1 mol/92 g x 4 mol H2O/1 mol = 4 mol
\ 0.5 mol H2O x 18 g/mol = 9 g.

59.

A / All are diatomic (all have odor and color, only F2 and Cl2 are gases).

60.

A / .02 L x .2 mol MnO4-/L x 3 mol ClO2-/4 mol MnO4- = 0.0030

61.

C / pH 12 is 1/10 as concentrated as pH 13 \ dilute the solution to 10 x the volume (1 liter).

62.

C / All Na+, K+ and NO3- salts are soluble.

63.

C / Same number and kinds of atoms, but the Br is on the middle carbon. (A, B and D are 1-bromopropane)

64.

B / Buffer requires a weak acid and its conjugate (HCl and HBr are strong acids).

65.

D / Small K2 means very few molecules have both H+ removed.

66.

C / 4 g/2 L x 22.4 L/1 mol = 44.8 g/mol \ CO2

67.

B / Phase change involves intermolecular bonds, which are H-bonds for water.

68.

A / 0.60 M Mg2+ x 200 mL/600 mL = 0.20 M Mg

69.

C / SiO2 is covalent network, which contains covalent bonds throughout. The others are all molecular.

70.

B / .200 F x 1 mol e-/1 F x 1 mol Ni2+/2 mol e- x 58.7 g/mol = 5.87 g

71.

C / Melting occurs along the plateau.

72.

D / H+ reacts with CO32-. NH3 precipitates with Al3+ (forms OH- ppt.), SO42- precipitates with Ba2+.

73.

D / At its normal melting point DG = 0 = DH – TDS
\ DH = TDS

74.

B / CO32- + 2 H+ ® H2O + CO2(g)
CO32- + H2O D HCO3- + OH- (basic solution)

75.

A / Ksp = [Zn2+][OH-]2 = (1.0 x 10-6)(2.0 x 10-6)2 = 4.0 x 10-18

Free Response

1. a.

[H+] = 10-4.95 = 1.12 x 10-5 M

b.

Ka = [OBr-][H+]/[HOBr] = [H+]2/[HOBr]
2.3 x 10-9 = (1.8 x 10-5)2/[HOBr] \ [HOBr] = 0.14 M

c. i.

2 HOBr + Ba(OH)2 ® 2 H2O + 2OBr- + Ba2+
0.0650 L HOBr x 0.146 mol x 1 mol Ba(OH)2 x 1 L = 0.0413 L
1 L 1 mol HOBr 0.115 mol

ii.

The pH is greater than 7 because at equivalence the weak base OBr- exists, which reacts with water to generate OH-: OBr- + H2O D HBrO + OH-.

d.

[OBr-] = Ka[HOBr]/[H+] = 2.3 E-9(0.160)/(5.00 E-9) = 0.0736 M
0.125 L x 0.0736 mol/L = 9.20 x 10-3 mol NaOBr

e.

The additional oxygens around Br pull the electrons away from the H attached to O, which weakens the O-H bond and strengthens the acid.

2. a. (1)

1.50 g Zn x 1 mol Zn x 1 mol Zn2+ = 0.0275 mol Zn2+
65.4 g Zn 1 mol Zn
0.250 L Ag+ x 0.110 mol Ag+ x 1 mol Zn2+ = 0.0138 mol Zn2+
1 L 2 mol Ag+

(2)

0.0138 mol Zn2+/0.250 L = 0.0550 M Zn2+

b.

Eo = EoAg+/Ag + EoZn/Zn2+ = 0.80 V + 0.76 V = 1.56 V

c.

DGo = -nFEo = (2 mol)(96,500 J V-1 mol-1)(0.46 V) = -89,000 J

d.

E = Eo –RT/nF(lnQ)
E = 0.46 V – (8.31)(298)/(2)(96,500)ln[(.045)/.010)2] = 0.38 V

e.

The cell potential in part (d) is positive. Therefore the reaction is spontaneous under these conditions.

3. a.

C5H12 + 8 O2 ® 5 CO2 + 6 H2O

b.

2.50 g C5H12 x 1 mol C5H12/72.1 g x 5 mol CO2/1 mol = .173
V = nRT/P = (0.173)(0.0821)(298)/(785/760) = 4.10 L

c.

-243 kJ/5.00 g x 72.1 g/1 mol = -3.51 x 103 kJ/mol

d.

RateA/RateB = (MMB/MMA)½
2/1 = (72.1/MMA)½ \ MMA = 18.0 g/mol

e.

C C
| |
C–C–C–C–C C–C–C–C C–C–C
|
(Your answer should include H's) C

4. a. (1)

CuSO4•5 H2O(s) ® CuSO4(s) + 5 H2O

(2)

1 mol CuSO4•5 H2O x 90 g H2O/mol = 90 g H2O

b. (1)

Ni2+ + 4 NH3 ® Ni(NH3)42+ (coordination # 2 or 6 is ok)

(2)

Ni2+ (electron pair acceptor)

c. (1)

CH3NH2 + H+ ® CH3NH3+

(2)

The solution is basic (CH3NH2 + H2O D CH3NH3+ + OH-).

5. a.

q has units in joules, m has units in grams, c has units in J/g•oC, T has units in oC

b.

mass of HCl and NaOH solutions, initial temperature of HCl or NaOH solution, highest temperature after mixing

c. (1)

Moles of H2O = moles of HCl = moles NaOH.
__L HCl x mol HCl/1 L x 1 mol H2O/1 mol HCl = __mol H2O

(2)

q = mcDT, where m = total mass of solution
Divide -q by mol H2O from part c(1) = DHneut

d. (1)

The DT will be greater, so q increases. There are more grams of HCl and NaOH reacting so the final temperature of the mixture will be higher.

(2)

DHneut = -q/mol H2O. Since q and moles are both doubled, there is no change in DHneut.

e.

Heat lost will result in a smaller DT, which reduces q
(q = mcDT) and make DHneut less negative ( -q/mol H2O).

6. a.

Both Li and Be have their outer electrons in the same energy level. However, Be has four protons and Li has only three protons. Therefore, the effective nuclear charge experienced by the valence electrons is greater in Be than in Li, so Be has a smaller atomic radius.

b.

The 2nd electron removed from K comes from the 3rd level. The 2nd electron removed from a Ca comes from the 4th level. 3rd energy electrons are closer to the nucleus and harder to remove than 4th energy electrons.

c.

C2H4 has a double bond between carbons, whereas C2H6 has a single bond between carbons. More energy is needed to break a double bond compared to a single bond \ the carbon-to-carbon bond energy is greater.

d.

Both Cl2 and Br2 are nonpolar molecules with dispersion forces generating the intermolecular attraction, but Br2 has more electrons to be polarized \ the attraction is greater in Br2 and requires a higher boiling point.

7. a.

O3 + Cl ® O2 + ClO
ClO + O ® Cl + O2
O3 + O ® 2 O2

b.

Cl is the catalyst. It is a reactant in Step I and reappears as a product in step II.

c.

ClO is the intermediate. It is a product in step I and reappears as a reactant in step II.

d. (1)

The overall order is 1 + 1 = 2

(2)

k = rate/[O3][Cl] = M/t/M2 = M-1t-1

(3)

Step I. The reactant coefficients in Step I correspond to the exponents in the rate law equation.

8. a.

DS > 0. Gases are more disordered than solids \ there is more disorder in the products and S increases.

b.

DH > 0. More CO at the higher temperature indicates that the reaction shifts to the right with increasing temperature. This occurs for endothermic reaction.

c.

Energy

2 CO(g)

DH

C(s) + CO2(g)

Reaction Coordinate

d.

C(s) has no effect because the concentration of C doesn't change and is not included in the equilibrium expression.