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Final Exam: General Chemistry 2 Fall, 2011 (1/2)

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33 views2 pages

Final Exam: General Chemistry 2 Fall, 2011 (1/2)

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siakarachi0521
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General Chemistry 2 Final Exam Fall, 2011 (1/2)

감독자인
계열/학부/학과: 학번: 고유번호: 성명:
수업요일 및 담당교수명: ( , )요일 ( )교 자필서명:

주 * 답은 지정된 자리에 반드시 영문(English)으로 기입할 것. * 시험 시간은 60분임. 400점 만점 (full score = 400 pt).

* 고유번호 및 담당교수명을 반드시 기입할 것. * 7, 8, 9, 10, 11번 문제는 풀이 과정을 반드시 적을 것.

항 * 문제지는 1장(2면)임. 연습지 및 영한사전 사용 불가. * 계산기 사용 무방하나 빌려 쓰지는 못함.

1. Write ‘T’ for true statement or ‘F’ for false statement in ( ). ( D) Which of the followings is the correct Nernst equation at
[5 pt for right answer, -2 pt for wrong, and 0 pt for no answer] 25o C for the cell reaction:
+
( T) Buffer solutions can be prepared by a weak acid plus a 2MnO4 + 16H3O + 5Zn 2Mn2+ + 24H2O + 5Zn2+ ?
2+ 2 2+ 5 2 + 16
soluble ionic salt of the weak acid. [ Q=[Mn ] [Zn ] /[MnO4 ] [H3O ] ]
( F) The chemically equivalent end point for a titration of a strong (A) E = E0 (0.0592/2) log (Q[H2O]24/[Zn]5)
cell cell
acid with a weak base means the point when the pH of 0
solution reaches 7.0. (B) Ecell = E cell (0.0592/2) log Q
( T) Fractional precipitation is a method of precipitating a specific (C) Ecell = E0cell + (0.0592/2) log Q
ion from solution while leaving others in solution by using 0
(D) Ecell = E cell (0.0592/10) log Q
the solubility products of salts. 0
(E) Ecell = E cell (0.0592/14) logQ
( F) The oxidation state of cobalt in K[Co(NH3)2(CN)4] is +2.
( F) To construct the galvanic cells, electrochemical reactions ( D) Which of the following base pairs forms two hydrogen
on both electrodes should be different from each other. bonds?
( T) The neutral Co(NH3)3Cl3 complex has a geometric isomer.
( T) The d orbital splitting patterns of central transition metal ions (A) Adenine Guanine
are dependent on geometries of coordination complexes. (B) Guanine Cytosine
( F) The coordination number is the number of ligands in (C) Cytosine Thymine
coordination complexes. (D) Adenine Thymine
( T) Octane has a higher boiling point than butane. (E) None of the above
( T) A polymer is a large molecule that consists of a
high-molecular weight chain of small molecules.
2. Choose the correct answer and write it in ( ). [9 pt ea]
( B) When salts derived from ______ acids and ______ bases are
dissolved in water, the resulting solution is always acidic. ( A) Usually, the ligand names of neutral molecules are same with
(A) strong; strong (B) strong; weak (C) weak; strong the original names. However, there are some exceptions. In the
(D) weak; weak (E) no way to determine without Ka and Kb following neutral molecules, which one has a different name as
( E) The following titration curve could describe the titration of a ligand in coordination complexes.
a solution of ______ by addition of a solution of ______. (A) nitrogen oxide (B) triphenylphosphine (C) benzene
(A) HCl; KOH (D) trimethylamine (E) pyridine
(B) KOH; HCl
pH
( C) Which of the following functional groups is not present in
(C) NaOH; KOH
7 Strychnine?
(D) KOH; CH3COOH
N (A) amine
0 (E) CH3COOH; KOH
vol. KOH added (B) alkene
H
( C) Which indicator would be the most suitable for the previous (C) aldehyde
N
titration? H (D) amide
(A) methyl orange (B) methyl red (C) phenolphthalein O O (E) ether
(D) methyl violet (E) None Strychnine

( A) Triglycerides are high molecular mass fats (or oils) which can 3. Write the IUPAC name or the systematic name. [8 pt each]
be saponified (or hydrolyzed) in basic solution to give
______ and ______. (1) K3[Fe(CN)6] ( potassium hexacyanoferrate(III) )
(A) soap; glycerol (B) fatty acid; ester (C) PET; SBR (2) [Fe(H2O)6]Cl3 ( hexaaquairon(III) chloride )
(D) tri; glyceride (E) protein; carbohydrate
CH 3
( A) What is an expected geometry of metal complexes having CH 2 OH
2 2
metals with sp d hybrid orbital? (3) ( 4-methyl-2-hexanol )
H 3C CH CH2 HC CH 3
(A) square pyramidal (B) tetrahedral (C) square planar
(D) trigonal bipyramidal (E) octahedral CH3
Cl CH
( E) Which of the followings is not an isomer of C6H14?
CH2
(A) hexane (B) 2-methylpentane CH2
H 3C H 2 C CH CH2 CH CH 3
(C) 3-methylpentane (D) 2,2-dimethylbutane (4) (2-chloro-4-ethyl-6-methyloctane)
CH3
(E) 2-cyclopropylpropane

1 2 3 4 5 6 7 8 9 10 11 Total
50 90 32 40 18 70 20 20 20 20 20 400
General Chemistry 2 Final Exam Fall, 2011 (2/2)
4. Write the structural formula of the following compounds. 8. A lead storage battery was charged at a steady current of
[10 pt each] 5.00 A for 4.00 hours:
(1) trans-3-methyl-3-heptene
CH 3
2PbSO4(s) + 2H2O(l) Pb(s) + PbO2(s) + 2H2SO4(aq)
H2
H3C C C CH 3 What masses (in g) of Pb were formed? [20 pt]
C C C [ Atomic weight for Pb = 207.2 g/mol, F = 96500 C/mol ]
H2 H2
H
(2) 4,5-dichloro-2-pentyne
Cl PbSO4 + 2e Pb + SO42 (n=2) [A]=[C/s]
Cl C ? C = (5.00 C/s)(4.00 hr)(3600 s/hr) = 7.20×104 C
C H C
H2 C ? g Pb = (7.20×104 C / 96500 C/mol)
CH 3
×(1 mol Pb /2 mol e )×207.2 g/mol= 77.3 g
(3) 3-bromobutanoic acid
Br O
C C
H3 C H C OH
H2 9. Calculate the mass (in g) of solid ammonium chloride, NH4Cl,
(4) 1-amino-3-ethylcyclohexane that must be used to prepare 1.00 L of a buffer solution of
o
H2 pH=9.15 at 25 C. In the buffer solution, the equilibrium
C concentration of aqueous ammonia is 0.100 M. [20 pt]
H2 C CH 2
[ Kb for NH3 = 1.80×10 5, M.W. of NH4Cl = 53.49 g/mol ]
H3C C CH
C H C NH2
H2 H2
5
5. For the octahedral complex ions [Fe(CN)6]3 and [Fe(H2O)6]3+ : pH=9.15 pOH=4.85 and Kb=1.80×10 pKb=4.745
+
Write the electron configuration in t2g and eg orbitals NH4Cl NH4 + Cl
[for example, t2g5eg2] and the number of unpaired electrons, and x x x (100% dissociation)
indicate high-spin or low-spin. (Atomic number of Fe is 26). NH3 + H2O NH4+ + OH
3
[Fe(CN)6] : configuration ( t2g5 ) [3 pt each] Kb = [NH4+][OH ]/[NH3]
number of unpaired electrons ( 1 )
pOH = pKb + log{[NH4+]/[NH3]}
( low )-spin complex
4.85 = 4.745 + log{x/0.100}
[Fe(H2O)6]3+ : configuration ( t2g3eg2 )
number of unpaired electrons ( 5 ) x = 100.105×0.100 = 0.127 mol
( high )-spin complex Mass of NH4Cl = 0.127 mol × 53.49 g/mol = 6.79 g
6. Write the structural formula of a major product. [14 pt each]
NO2
conc.
H2 SO4 10. What is the minimum concentration (in M) of AgNO3 needed
+ HNO3 product 18 6
(1) 50 °C ( ) for the precipitation of Ag3PO4 (Ksp = 1.8×10 ) in 1.0×10 M
Na3PO4 solution? [20 pt]
O
K2 Cr 2O 7
CH3OH product C Na3PO4 3Na+ + PO43 (100% dissociation)
(2) dil. H2 SO4 ( H H ) 3 6
so that [PO4 ] = 1.0×10 M
+ 3
i) KMnO4, OH -, heat O
Ag3PO4 3Ag + PO4
CH 3 product C
(3) ii) HCl (aq) ( OH
) 3
Ksp = [Ag+]3[PO4 ] = [Ag+]3(1.0×10 6) = 1.8×10 18

Solving, [Ag+] = (1.8×10 12 1/3 4


) = 1.22×10 M = [AgNO3]
Br 2 +
[AgNO3 Ag + NO3 (100% dissociation)]
product
(4) ( Br Br )
11. A concentrated, strong acid is added to a 0.0150-mol solid
NaOH
product
sample of Fe(OH)2 placed in 1.00 L of water at 25 oC.
Cl
(5) ( ) At what value of pH will the dissolution of the hydroxide be
complete? (Assume negligible volume change.) [20 pt]
7. For CdS(s) Cd2+(aq) + S2 (aq): 15
2+
[ Ksp for Fe(OH)2 = 7.90×10 ]
Cd (aq) + 2e Cd(s) E0 = 0.403 V
CdS(s) + 2e Cd(s) + S2 (aq) E0 = 1.21 V
Calculate the Ksp value for CdS(s) at 25
o
C. [20 pt] Fe(OH)2(s) Fe2+(aq) + 2OH (aq)
[ R = 8.314 J/mol⦁K, F = 96500 C/mol ] 2+ 2
Ksp = [Fe ][OH ]
0 When the dissolution of Fe(OH)2(s) is complete, [Fe2+]=0.0150 M.
E = 1.21 ( 0.403) = 0.81V n=2
15
/0.0150)1/2 = 7.26×10 7
logKsp = nFE0/(2.303RT) (or lnKsp = nFE0/RT ) [OH ] = (7.90×10 M
7
= [2×96500×( 0.81)]/[2.303×8.314×298] = 27.4 pOH = log(7.26×10 ) = 6.14
Ksp = 10 27.4 = 3.98×10 28 (=4.0×10 28)
pH = 14 6.14 = 7.86

copyright - Department of Chemistry, Sungkyunkwan University®

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