Name______KEY______Sec______

Chem 1151 June 1, 2006Exam 195 pts

This exam covers Chapters 1-3, and Sections 1-5 of Chapter 4. There are 6 pages plus a copy of the Periodic Table. You may use your calculator.

N = 6.022E+231 in = 2.54 cm1 lb = 453.6 g

1 L = 1.057 qt = 1000 cm3

Multiple Choice (24 point =12 questions @ 2 points)Circle your multiple choice answer.

  1. 4301. mL is equivalent to
  1. 4.069 qts

**B.4.546 qts(4301. mL)(1L/1000ml)(1.057 qt/1L) = 4.546157

C.4.546E-3 qts

D.4069. qts

E.4546. qts

  1. Using the rules of significant figures, calculate the following:

(4.0021- 0.004) x 2.4961 = 3.9981 x 2.4961 = 9.979657 = 9.980

  1. 10.
  2. 9.97965
  3. 9.98

**D.9.980

E.9.98

  1. 30. mL of a 1.80M solution of H2SO4 (aq) has _____mols of H+.

**A.0.11 mols = (1.80M) (0.030L) (2H+/1 acid) = 0.108 mols

B.3.6 mols

C.54. mols

D.0.054 mols

E.1.1E+2 mols

  1. Which of the following has 45 neutrons, 35 protons and 36 electrons?
  1. 8035Hg
  2. 10345Rh -

**C.8035Br -Z = 35, A = 45+35=80 and charge = -1

D.7335Lu –

E.8135Tl -

  1. All of the following are ionic compounds except
  1. MgCl2

**B.N2O

C.Rb2SO4

D.(NH4)2S

E.CuO

  1. What set of compounds illustrates the Law of Multiple Proportions?
  1. F2, Cl2, Br2
  2. SO2, NO2, CO2,
  3. CH4, CH3Cl, CH2Cl2,
  4. H2, H2S, H2O

**E.C2H4, C2H2, C2H6This law applied when two or more elements more two or more different cmps.

  1. The element in Group IVA and the fourth period is
  1. Si

**B.Ge

C.Sn

D.Ti

E.Zr

  1. 12.8 g of an unknown compound was found to contain 0.256 mol of compound. What is the identity of the unknown?
  1. C2H4O
  2. CO2

**C.CH3Cl12.8g/0.256 mol = 50 g/mol

D.C2H6

E.CH4

  1. The empirical formula for the yellow dye, apigenin, is C3H2O. Which of the following is a possible molecular formula for apigenin?
  1. C2H4O2
  2. C9H20
  3. C5H10O5
  4. C3H8O

**E.C15H10O5

  1. Pb(NO3)2 (aq) + 2KI  2KNO3(aq) + PbI2 (s) is an example of a ______reaction.
  1. neutralization

**B.precipitation

C.redox

D.combustion

E.decomposition

11.All of the following are strong acids except

**A.HF

B.HCl

C.HBr

D.H2SO4

E.HClO4

12.The net ionic equation for

NiCl2(aq) + Na2S(aq)  NiS(s) + 2NaCl(aq)

A.NiCl2(aq) + Na2S(aq)  NiS(s) + 2NaCl(aq)

B.2Na+(aq) + 2Cl-(aq)  2NaCl(aq)

C.Ni+(aq) + 2Cl-(aq) + 2Na+(aq) + S-(aq)

 NiS(s) + 2NaCl(aq)

D.Ni+(aq) + S-(aq)  NiS(s)

**E.None of the above

  1. Problems (44)Show all work for full or partial credit; if you do not show your work, no credit will be given.

1.(12)Fill in the table as noted.

Name (2 pts each) / Formula (2)
A / Mercury(I) oxide / Hg2O
B / Ammonium phosphate / (NH4)3PO4
C / Diphosphorus pentasulfide / P2S5
D / Aluminum sulfate / (Al)2(SO4)3
E / Copper(II) sulfide / CuS
F / Nitrogen dioxide / NO2

2.(6)a.How many mL of a 0.55 M CdCl2aqueous solution contain 10.0 g cadmium chloride?

M*V = # mols

(10.0g/183.31g/mol)/0.55M = V = 0.099186L or 99. mL

b.Consider this balanced chemical equation

CdCl2(aq) + (NH4)2S(aq)  CdS(s) + 2NH4Cl(aq)

How many mL of 0.300M ammonium sulfide are required to react with 225. mL 0.500M cadmium chloride?

Calculate # mol CdCl2(aq) and equate it to # mol (NH4)2S(aq) since mole ratio between the two reactants is 1/1.

(225 mL * 0.500M)/0.300M = 375 mL

3.(12)Consider this balanced chemical equation

CH4 + 2Cl2  CH2Cl2 + 2HCl

a.If 50.0 g methane and sufficient chlorine reacted, how many grams of HCl would be produced?

(50.0g/16.04g/mol) (2 mol HCl/1mol methane) (36.45g/mol) = 227.24 g HCl = 227. g HCl maximum

b.If this reaction had a % yield of 65%, how many grams of HCl would actually be produced?

227.244 g * 0.65 = 148 g HCl actual

c.If 50.0 g methane and 350. g of chlorine, how many grams of HCl would be produced? Assume 100% yield.

Calculate mol methane: (50.0g/16.04g/mol) = 3.12

Calculate mol chlorine: (350/70.90g/mol) = 4.94

Actual ratio of chlorine/methane = 1.58

Stoichiometric ratio = 2 > 1.58 so there is too little chlorine. Cl2 = LR

(350/70.90g/mol) (2 mol HCl/2 mol chlorine) (36.45g/mol) = 180 g HCl

d.Given the data in part c, how many grams of the excess reagent would remain at the end of the reaction?

Methane = excess reagent so calculate how much methane reacts with 350 g chlorine.

(350/70.90g/mol) (1 mol methane/2 mol chlorine) 16.04 g/mol = 39.6 g methane will react.

We have 50.0 g methane, so the left over methane = 50.0-39.6 g = 10.4 g methane.

4.(10) Determine if an aqueous solution of each salt is soluble or insoluble. Place an “X” in the correct box.

Salt / Soluble / Insoluble
a. Ba (NO3)2 / X
b. AgCl / X
c. (NH4)2CO3 / X
d. K2SO4 / X
e. PbSO4 / X

5.(6) Adrenaline was found to contain 56.79% C, 6.56% H, 28.37% O and 8.28% N. What is the empirical formula for this compound?

Assume 100 g cmp

C(56.79g/12.01 g/mol) = 4.7286 mol C

H(6.56g/1.01g/mol) = 6.495 mol H

O(28.37g/16.00g/mol) = 1.7731 mol O

N(8.28g/14.00g/mol) = 0.5914 mol N

Divide by 0.5914 to get whole number mol ratios

C7.9952 or 8

H10.98 or 11

O2.99799 or 3

N1

Empirical Formula:C8H11O3N

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