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Fall (Loader) Version 1
Chemistry 1010
Test # 3
[48 Marks Total]
Name: _____________________ Lab. Slot ____
November 17, 2004
MUN #: ______________________
Show your full calculations for Part B and give numerical answers to the correct number
of significant digits.
Marking Scheme
QUESTIONS
VALUE
MARK
Part A
16
Part B
34
Total
Part A
[Total Marks 16]
Multiple choice questions
Circle the letter for the one correct answer. [Each Question = 2 marks]
1.
2.
, atom in the ground
state is:
(a)
1
(b)
2
(c)
3
(d)
4
(e)
5
The following mixtures are made as listed in the table below:
Mixture
A
B
C
D
E
F
Mixture (1 mL each of 0.1 mol L-1 solutions)
NaCl(aq)
K2SO4(aq)
BaCl2(aq)
AgNO3(aq)
K2SO4(aq)
NaOH(aq)
Pb(NO3)2(aq)
KI(aq)
CuSO4(aq)
NaOH(aq)
BaCl2(aq)
NH4NO3(aq)
The three mixtures that give precipitates are
3.
(a)
B, C and D
(b)
B, D and E
(c)
A, C and F
(d)
B, D and F
(e)
D, E and F
The elements with symbols
(a)
F < Cl < Se < Ge < K
(b)
K < F < Ge < Cl < Se
(c)
Ge < K < F < Cl < Se
(d)
K < Se < Ge < Cl < F
(e)
K < Ge < Se < Cl < F
-1-
CHEMISTRY 1010
4.
5.
6.
7.
8.
What volume of 0.237 mol L–1 NaOH is required to neutralize 35.86 mL of 0.280
mol L–1 sufuric acid in the reaction
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
(a)
11.8 mL
(b)
47.4 mL
(c)
71.7 mL
(d)
84.7 mL
(e)
60.7 mL
When 3.455 g of magnesium chloride hexahydrate, MgCl2·6H2O(s), (molar mass,
203.30 g.mol–1) is heated strongly in a crucible to constant weight, a residue
consisting of 0.6849 g a white solid remains. Based on this result alone the most
likely formula for the residue is
(a)
MgH2 (molar mass = 26.320 g.mol–1)
(b)
MgO (molar mass = 40.304 g.mol–1)
(c)
MgOCl (molar mass = 75.757 g.mol–1)
(d)
MgCl2 (molar mass = 95.210 g.mol–1)
(e)
MgCl2·H2O (molar mass = 113.226 g.mol–1)
What is the theoretical yield of vanadium, V, when 40.0 g of V2O5 (molar mass =
181.9 g mol–1) reacts with 40.0 g of calcium (molar mass = 40.078 g mol–1)
according to the equation below?
V2O5 + 5 Ca → 2 V + 5 CaO
(a)
5.6 g
(b)
11.2 g
(c)
20.3 g
(d)
22.4 g
(e)
40.0 g
Which of the following molecules
(i) AlCl3; (ii) PCl3; (iii) PCl5; (iv) SiCl4
contain an atom for which the octet law (rule) is not satisfied?
(a)
(i) only
(b)
(i) and (ii)
(c)
(iii) and (iv)
(d)
(i) and (iii)
(e)
(iii) only
Which of the following species is isoelectronic with the Cl − ion?
(a)
Ne
(b)
Se2-
(c)
Na+
(d)
S2-
(e)
K2+
-2-
CHEMISTRY 1010
Section B
(32 Marks)
B1
Balance the following oxidation-reduction reaction,
−
2−
S 2 O 2−
3 (aq) + NO 3 (aq) t S 4 O 6 (aq) + NO 2 (g)
done in acid solution.
−
2−
4 H + (aq) + 2 S 2 O 2−
3 (aq) + 2 NO 3 (aq) t S 4 O 6 (aq) + 2 NO 2 (g) + 2 H 2 O(l)
[7]
B2.
(a)
[2]
(b)
Write ground state electron configurations for the following:
(eg. Na+ 1s2 2s22p6)
Ca2+
1s2 2s22p6 3s23p6 _______________________________
Cr
1s2 2s22p6 3s23p63d10 4s1______________________________
In writing the electron configuration and box diagram (valence
electrons only) for sulfur in the ground state, a student has written:
S, electron configuration is 1s2 2s22p6 3s23p4
box diagram
↑
↓
↑
↓
↑
↓
The box diagram is incorrect. Explain what rule(s) it violates and
draw a the correct box diagram.
The corrected box diagram is
↑
↑
↑
↑
↓
↓
The original diagram violates Hund's Rule of maximum multiplicity.
The p-orbital have the same energy (are degenerate) and are
therefore singly occupied before spin pairing takes place.
[3]
-3-
CHEMISTRY 1010
B3
(a)
Complete the following table:
Lewis electron
dot structure.
Sketch of the
shape
H2S
PCl5
CO2
H S H
Cl
Cl
Cl P Cl
Cl
O C O
H
Cl
S
Cl
H
Cl
P
Cl
Cl
O
Name of the
shape
bent
trigonal bipyramidal
linear
Approximate
bond angles
<109.5o
180, 120 and 90o
180o
C
O
[9]
(b)
The element selenium has the electron configuration 4s24p4 for its valence
electrons. Write down the likely formula(s) for selenium in its
oxide(s) Selenium is in the same group as sulfur so would be expected
to form similar oxides. SeO2 and SeO3 are therefore likely and possibly
SO.
chloride(s)
Would you expect the oxides to be acidic, basic or amphoteric? Why?
The oxides would be acidic.
[3]
B4
SeCl2_and SeCl4___________________________
A 0.5000 g sample of iron ore is dissolved in sulfuric acid so that all of the iron is
converted to iron(II) sulfate and present as Fe2+(aq). In a titration the solution
containing the Fe2+(aq) ion requires 31.50 mL of 0.0310 mol L–1 potassium
permanganate, KMnO4(aq), for reaction:
5 Fe 2+ (aq) + MnO −4 (aq) + 8 H + (aq) d Mn 2+ (aq) + 5 Fe 3+ (aq) + 4 H 2 O(l)
(f)
Calculate the number of moles of iron present in the ore sample.
(This is a titration) Moles of KMnO4 used = 30.70 x 10-3 L x 0.0310 mol L–1
= 9.765 x 10-4 mol.
Sime 5 mol of Fe2+ react with 1 mole of MnO −4 , the number of moles of
Fe2+ = 5 x 9.765 x 10-4 mol = 4.883 x 10-3 mol.
Ans: 4.88 x 10-3 mol
[5]
(g)
Calculate the percentage by mass of iron present in the ore sample.
Mass of iron present in the ore sample = 4.883 x 10-3 mol x 55.8475 g
mol–1
= 0.2727 g
So, the percentage yield =
0.2727 g
0.5000 g
[3]
Ans: 54.5%
-4-
x 100% =54.54%