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Chemistry 1010
WInter (Loader) Version 2
Test # 1
[50 Marks Total]
Name: _____________________
January, 30th 2003
MUN #: ______________________
Marking Scheme
QUESTIONS
Part A
Part B
Part C
VALUE
16
23
11
MARK
Total
Part A: Multiple choice questions. Circle the letter for the one correct answer.
[Each Question = 2 marks]
1.
The symbol for the element silicon is
(a) Si
(b) Sn
(c) S
(d) Na
(e) K
2.
According to your text book exactly 1 m = 39.37 inches. The Weather
Underground reports that on January 26th 2002 the snow depth in St. John's
was 32.7 inches. What was the depth of snow on this date in centimetres
given to the correct number of significant digits?
(a) 0.8306 cm
(b) 1.20 x 103 cm
(c) 1.29 x 1026 cm
(d) 8.31 x 10–4 cm
(e) 8.31 x 101 cm
3.
How many significant digits given when the volume of a liquid is stated as
0.00330200 L?
(a) 8
(b) 7
(c) 6
(d) 5
(e) 4
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CHEMISTRY 1010
4.
Table salt is sold in 1.0 kg packets. If table salt is pure sodium chloride,
NaCl(s), how many moles of sodium chloride is contained in a 1.0 kg packet?
(a) 0.058 mol
(b) 1.7 x 10–5 mol
(c) 1.7 x 10–2 mol
(d) 6.2 mol
(e) 17 mol
5.
Magnesium has atomic mass 24.305 and oxygen has atomic mass 15.9994.
How many grams of magesium are required in order to contain the same
number of moles of atoms as 10.0 g of oxygen?
(a) 38.9 g
(b) 15.2 g
(c) 10.0 g
(d) 6.58 g
(e) 0.152 g
6.
Which one of the following is a homogeneous mixture?
(a) bread
(b) carbon dioxide
(c) copper(II) sulfate solution
(d) oil and water
(e) salt and sand mixture
7.
The molar mass of lithium sulfate, Li2SO4, is
(a) 39.536 g mol–1
(b) 55.006 g mol–1
(c) 100.866 g mol–1
(d) 109.946 g mol–1
(e) 135.071 g mol–1
8.
Given the balanced equation for the reaction between phosphorus and
chlorine:
2 P(s) + 5 Cl2(g) → 2 PCl5(g)
If 1.00 mole of PCl5(g) was produced in a reaction, how many moles of
chlorine gas, Cl2(g), was used in the reaction?
(a) 5.00 moles
(b) 2.50 moles
(c) 2.00 moles
(d) 10.0 moles
(e) 0.400 moles
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CHEMISTRY 1010
Part B
Short Answer Questions
(23 Marks)
[12]
B1.
Complete the following table
Name
[3]
[4]
calcium chloride
CaCl2
iron(II) sulfate
FeSO4
hydrochloric acid
HCl(aq)
chlorine dioxide
ClO2
sulfur tetrachloride_____
SCl4
calcium nitrate__________
Ca(NO3)2
B2.
B3.
Give the charge on the following ions
carbonate
-2_______________________
nitrate
-1_______________________
ammonium
+1______________________
Fill in the following table
Cation
formula or symbol (as
appropriate)
[4]
Chemical Formula
Anion
formula or symbol (as
appropriate)
Compound formed
by combining the
two ions.
Cr3+
SO42–
Cr2(SO4)3
NH4+
S2–
(NH4)2S
B4.
Fill in the following table
Formula
Cation
formula or symbol (as
appropriate) and charge
Anion
formula or symbol (as
appropriate) and charge
K2CrO4
K+
CrO 2−
4
Ca3(PO4)2
Ca2+
PO 3−
4
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CHEMISTRY 1010
Part C
Long Answer Question
(11 Marks)
[7] C1.
Mercury(II) oxide, HgO (molar mass = 216.6 g mol–1), easily decomposes
when heated to give mercury and oxygen gas.
HgO(s) → Hg(l) + O2(g) (unbalanced)
A. Write the balanced equation for the reaction.
2 HgO(s) → 2 Hg(l) + O2(g)
B. If 20.00 g of HgO are heated how many mole HgO is this?
This is less than 1 mole
20.00 g
moles of HgO = 216.6 g mol −1 = 0.092336 mol
Ans: 0.09234 mol
C. How many mole of oxygen gas can be produced in the reaction if all
20.00 g of the HgO decomposed?
Divide the equation by 2
1 HgO(s) → 1 Hg(l) + ½ O2(g)
For each mole of HgO decomposed ½ moles of O2 is produced so the amount
produce in the reaction = ½ x 0.092336 mol = 0.046168 mol.
Ans: 0.04617 mol
D. Calculate the mass of mercury produced when 20.00 g HgO is
decomposed?
For each mole of HgO decomposed 1 mole of Hg is produced so the amount of
Hg produced = 1 x 0.092336 mol
The mass of Hg produced is the number of moles times the molar mass of Hg =
0.092336 mol x 200.59 g mol-1 = 18.522 g
Ans: 18.52 g
[4] C2.
The element lead occurs combined with various other elements in nature.
One of the lead containing minerals is matlockite which is composed of
lead(II) chloride, PbCl2(s). Calculate the mass of lead, Pb, that could be
extracted from 110 kg of matlockite?
1 PbCl2(s)→ 1 Pb (not an equation)
The molar mass of lead(II) chloride = 207.2 + (2 x 35.4527) g mol-1 = 278.1054 g
mol-1 and the molar mass of lead = 207.2 g mol-1of Pb
So, the mass of lead =
207.2 g mol −1 of Pb
110 kg of PbCl2(s) x 278.1054g mol −1 of PbCl (s) = 81.955 kg
2
Ans: 82.0 kg
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