Chem 1151Exam 2June 12, 2008100 points

Name_____KEY______Sec______

R =0.0821 (L-atm)/(mol-K) = 8.314 J/(mol-K)

1 atm = 760 mmHg = 760 Torr urms = 3RT/M

dRT = MP1 L-atm = 101.3 J

I.Multiple Choice (36 points)

1.Which pair of substances will produce a precipitate when aqueous solutions of each are mixed?

A.RbBr and Mg(NO3)2

B.HI and KOH

C.AlCl3 and CuSO4

D.NH4NO3and Na2S

**E.K2CO3 and FeBr3

2.The oxidation numbers of the underlined element in H2PtCl6, IO3-, and OsO4, respectively are

A.+4, -1, +8

**B.+4, +5, +8

C.+2, -1, +2

D.0, +6, +8

E.+6, +7, 0

3.Which chemical equation is incorrectly labeled?

A.HBr(aq) + KOH(aq)  H2O(ℓ) + KBr(aq)acid-base neutralization

B.Hg2(NO3)2(aq) + 2KI(aq)  2KNO3(aq) Hg2I2(s)precipitation

**C.2NaNO3(aq) + CuCl2(aq)  2NaCl(aq) + Cu(NO3)2(aq) redox

D.2C2H6(g) + 7O2(g)  6H2O(ℓ) + 4CO2(g)combustion

E.2KClO3(s)  2KCl(s) + 3O2(g) decomposition

4.All of the following are strong acids except

**A.H2SO3

B.HNO3

C.HI

D.HClO4

E.HBr

5.Which statement is true about this unbalanced half-reaction?

Cr2O72-(aq) Cr3+(aq)

A.Cr is oxidized.

B.Cr2O72-(aq) is the reducing agent.

**C.Cr gains electrons.

D.Cr loses electrons.

E.The oxidation number of oxygen inCr2O72-(aq) is 0 (zero).

6.Consider four identical 1.0 L flasks containing the following gases each at 298K and 1 atm pressure. For which gas do the molecules have the highest root mean square velocity?

  1. SO2

**B.H2

C.NH3

D.O2

E.Same for all gases

7.Equal volumes of an unknown gas and nitrogen (N2) gas take 5.37 min and 3.55 min, respectively, to effuse through a small hole. Calculate the molar mass of the unknown gas.

  1. 12.2 g/mol
  2. 18.5
  3. 42.4
  4. **64.1
  5. 128

8.Which of the following laws about gases isincorrect?

A.In a mixture of gases, the total pressure of the gases equals the sum of the individual pressures of each gas.

B.At fixed P, the volume of a gas is directly proportional to its temperature in K.

C.At fixed T, the volume of a gas is inversely proportional to its pressure.

D.The average translational kinetic energy of the molecules in a gas is directly proportional to its temperature in K.

E.**At fixed T and P, the volume of a gas is inversely proportional to the number of moles of the gas.

9.Which diagram is consistent with Charles’ Law?

**A.B.

V vs T linear increasingV vs (1/T) linear, inc

C.D.

nRT vs T constantPV vs T constant

10.An important experimental method of determining enthalpy values is

A.spectroscopy

**B.calorimetry

C.gas expansion

D.solution dilution

E.titration

  1. A gas absorbs 133 J of heat and does 55 J of work. What is ΔE?
  1. 188 J
  2. -188 J
  3. **78 J
  4. -78 J
  5. 55 J
  1. Consider this thermochemical equation

4Fe(s) + 3O2(g)  2Fe2O3(s)ΔH = -1652 kJ

When 1.25 mol Fe burns in excess oxygen, how many kJ of heat is given off?

  1. -2065 kJ
  2. -1652 kJ
  3. -1322 kJ
  4. ** -516 kJ
  5. 0 kJ

II.Problems (Show all work for partial credit. INCLUDE UNITS WITH YOUR ANSWER)

1.(7) A sample of carbon dioxide occupies 25.3 L at 325K and 0.95 atm.

  1. If the temperature increases to 375K and the pressure increases to 1.00 atm, what is the new volume of the gas?

Note that the # moles of gas stays constant

P1 V1/(R T1) = n1 = n2 = P2 V2/(R T2) or

V2 = V1 (P1/P2) (T2/T1) = 27.7 L

  1. What is the density of this gas at STP?

D = MP/(RT) = (44 g/mol) (1.00 atm)/(0.0821 x 273) = 1.96 g/L

2.(5) An unknown metal of mass 17.5 g initially at 450 K was placed in 100.0 g water initially at 300 K. The temperature of the water increased to 305 K. What is the specific heat of the metal? The specific heat of water is 4.18 J/(g-K).

q = s m ΔT

Also, the heat lost by the metal = heat the warms water

heat lost by metal = s 17.5 g (450-305)

= 4.18 (100.0 g) (305-300) = heat gained by water

s = 2090/2537.5 = 0.824 J/(K g) or 0.824 J/(oC g)

3.(5) What volume of 0.250M NaOH (M = 40.0 g/mol) is needed to react completely with 6.45 g iodine (M = 253.8 g/mol) according to this equation:

3I2(s) + 6NaOH(aq)  5NaI(aq) + NaIO3(aq) + 3H2O(l)

g I2 mol I2 mol NaOH  vol NaOH

[(6.45 g/253.8g/mol) x (6 mol NaOH/3 mol I2)]/ 0.250 M = V = 0.203 L = 203 mL

  1. (5) At 323K and 800 torr, 0.20 mol HgO decomposed to give 2.00 L oxygen gas. What is the % yield of this reaction?

2HgO(s)  2Hg(l) + O2(g)

# mol O2 produced = PV/RT = (800/760) (2.00L)/0.0821 (323K) = 0.0794 mol O2 = actual yield,

But if you start with 0.20 mol HgO, you should get 0.10 mol O2 = theoretical yield.

So % yield = (0.0794/0.10) x 100 = 79.4%

  1. (7) Balance the following redox reaction in acid.

OX#0 +4, -2+2+3

Zn(s) + VO2+(aq)  Zn2+(aq) + V3+(aq)

Oxidation ½ Rxn: Zn  Zn2+ + 2 e-

Reduction ½ Rxn:

[2H+ 1 e- + VO2+(aq)  V3+(aq) + H2O] x 2

Add rxns and cancel electrons

Zn + 4H+ + 2VO2+(aq)  Zn2+ + 2V3+(aq) + 2H2O

Check: 1 Zn, 2 V, 2 O, 4 H and +8 charge on both sides

  1. What is the reducing agent?__Zn_____
  1. What atom gains electrons?__V_____

6. (5)If 1.53 L bromine (M=159.8 g/mol) and 2.18 L ammonia (M=17.0 g/mol) react according to the following equation, what is the maximum number of liters of nitrogen that can be produced?

3Br2(l) + 8NH3(g)  6NH4Br(s) + N2(g)

Use Avogadro’s Law which states that V is proportional to n at constant P and T.

This is a LR problem

1.35 L Br2 (8 L NH3/3 L Br2) = 4.08 L NH3 needed

2.18 L NH3 on hand so NH3 = LR

2.18 L NH3 (1 L N2 / 8 L NH3 ) = 0.273 L H2 produced

7.(10)Short answers

  1. When a gas expands, work = w is [ 0, = 0, > 0]
  1. Circle the insoluble salt: Na3PO4PbSO4MgCl2
  1. Circle the strong base: Fe(OH)2Sr(OH)2Fe(OH)3
  1. In the van der Waals model of a gas, molecular volume [is, is not] negligible.
  1. If the mass of the gas doubles, its urms [increases, stays the same, decreases].

Name______Sec______

  1. ______(36)
  2. ______(44)

Total______(80) x 1.25 = ______(100)

+ Extra Credit______(5)

Total______(105)

EXTRA CREDIT(5) 2KClO3(s)  2KCl(s) + 3O2(g) describes the decomposition of potassium chlorate. Oxygen was collected over water at 22 oC at a total pressure of 749 torr. The volume of gas collected was 0.650 L and the vapor pressure of water at 22 oC is 21 torr.

  1. Calculate the partial pressure (in torr) of oxygen collected.

P (oxygen) = 749 – 21 = 728 torr

  1. Calculate the number moles of oxygen collected.

n(oxygen) = P(oxygen) V/RT = 0.026 mol

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