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Chemistry
UTME / JAMB 1999
Metals, Transition Elements & Extraction
What is observed when aqueous solutions of each of tetraoxosulphate (VI) acid, potassium trioxoiodate (V) and potassium iodide are mixed together?
A
A white precipitate is formed
B
A green precipitate is formed
C
The mixture remains colourless
D
The mixture turns reddish brown
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #1380)
Chemistry
UTME / JAMB 1999
Acids, Bases, pH & Salts
Solution
The solution with the lowest pH value is
A
5 ml of (m/10) HCl
B
10 ml of (m/10) HCl
C
15 ml of (m/5) HCl
D
20 ml of (m/8) HCl
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
pH = - Log[\(H^+\)]. All we need is the concentration of the acid, and not the volume.
Option A : 5ml of \(\frac{M}{10}\)HCl; [\(H^+\)] = \(\frac{M}{10}\)= \(\frac{1}{10}\times M\) = 0.1M. pH = - Log(0.1)= 1
Option B : 10ml of \(\frac{M}{10}\)HCl;[\(H^+\)] = \(\frac{M}{10}\)= \(\frac{1}{10}\times M\) = 0.1M. pH = - Log(0.1)= 1
Option C : 15ml of \(\frac{M}{5}\)HCl; [\(H^+\)] = \(\frac{M}{5}\)= \(\frac{1}{5}\times M\) = 0.2M. pH = - Log(0.2)= 0.699
Option D : 20ml of \(\frac{M}{8}\)HCl; [\(H^+\)] = \(\frac{M}{8}\)= \(\frac{1}{8}\times M\) = 0.125M. pH = - Log(0.125)= 0.903
Among the options, option C tends to have the least pH value of 0.699
Source: Myschool Past Questions Archive (Record #1389)
Chemistry
UTME / JAMB 1999
Water, Solutions & Solubility
Solution
The solubility product of Cu(IO3)2 is 1.08 * 10-7. Assuming that neither ions react appreciable with water to form H+ and OH-, what is the solubility of this salt?
A
2.7 * 10-8 mol dm-3
B
9.0 * 10-8 mol dm-3
C
3.0 * 10-3 mol dm-3
D
9.0 * 10-3 mol dm-3
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
On dissociation, Cu(IO\(_3\))\(_2\) ⇌ Cu\(^{2+}\)(aq) + 2IO\(_3\)\(^-\)(aq).
Solubility product constant = Ksp
The formula is : Ksp = [Cu\(^{2+}\)] [IO\(_3\)\(^-\)]\(^2\)
Let the concentrations of Cu\(^{2+}\) and IO\(_3\)\(^-\) be expressed as d.
[Cu\(^{2+}\)] = d ; [ IO\(_3\)\(^-\)] = 2d (because 2IO\(_3\)\(^-\) )
Substituting these expressions into the Ksp
Ksp = [Cu\(^{2+}\)] [IO\(_3\)\(^-\)]\(^2\)
Ksp = (d) x (2d)\(^2\)
Ksp = d.4d\(^2\)
Ksp = 4d\(^3\)
Recall that Ksp = 1.08 x 10\(^{-7}\)
4d\(^3\) = 1.08 x 10\(^{-7}\)
Divide both sides by 4
d\(^3\) = \(\frac{1.08 x 10^{-7}}{4}\)
d\(^3\) = 0.000000027
d = ∛0.000 000 027
d = 0.003
d = 3 x 10\(^{-3}\) moldm\(^{-3}\)
Source: Myschool Past Questions Archive (Record #1391)
Chemistry
UTME / JAMB 1999
Chemical Energetics & Thermochemistry
The entropy and enthalpy of a system are a measure of
A
degree of disorderlinesss and heat content respectively
B
heat content and degree of disorderliness respectively
C
heat content of a system only
D
degree of disorderliness only
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #1393)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
2SO2(g) + O2(g) ↔ 2SO3(g). In the chemical reaction above, the substance that will increase the rate of production of sulphur (VI) oxide is
A
manganese (IV) oxide
B
finely divided iron
C
vanadium (V) oxide
D
nickel
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #1396)
Chemistry
UTME / JAMB 1999
Reaction Rates & Chemical Equilibrium
Solution
N\(_2\)O\(_4\)(g) ↔ 2NO\(_2\)(g). Increase in total pressure of the equilibrium reaction above will
A
produce more of NO2(g) in the mixture
B
convert all of N2O4(g) to NO2(g)
C
have no effect on the concentrations of NO2(g)and N2O4(g)
D
produce more of N2O4(g) in the mixture
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
An increase in pressure on a system at equilibrium, according to Le Chatelier's principle, shifts the equilibrium towards the side with fewer moles of gas to relieve the stress of increased pressure.
In the above reaction, the side with the lesser number of moles of gaseous species is the reactant,N\(_2\)O\(_4\) with 1 mole as against the product side with 2 moles. Hence, an increase in total pressure will favour the side with the lesser number moles. This implies more N\(_2\)O\(_4\) will be produced.
Source: Myschool Past Questions Archive (Record #1399)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
The gas that will form a white precipitate with acidified silver trioxonitrate (v) is
A
NH3
B
SO2
C
CO2
D
HCL
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #1461)
Chemistry
UTME / JAMB 1999
Oxidation, Reduction & Redox Reactions
The salt that reacts with dilute hydrochloric acid to produce a pungent smelling gas which decolourizes acidified purple potassium tetraoxomanganate (VII) solution is
A
Na2SO4
B
Na2SO3
C
Na2S
D
Na2CO3
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #1465)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
A pair of compounds that can be used to generate a gas which has physiological effect on human beings is
A
sodium trioxonitrate (V) and calcium chloride
B
sodium dioxonitrate (III) and ammonium chloride
C
sodium trioxonitrate (V) and ammonium chloride
D
sodium dioxonitrate (III) and potassium chloride
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
The reaction between NaNO\(_3\) and NH\(_4\)Cl gives NaCl and NH\(_4\)NO\(_3\).
Upon heating, NH\(_4\)NO\(_3\) yields 2 moles of H\(_2\)O and N\(_2\)O.
Source: Myschool Past Questions Archive (Record #1467)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Hydrogen is used in oxy-hydrogen flames for melting metals because it
A
evolves a lot of heat when burnt
B
combines explosively with oxygen
C
is a very light gas
D
is a rocket fuel
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #1469)
Chemistry
UTME / JAMB 1999
Electrolysis & Electrochemical Cells
Solution
What quantity of electricity will liberate 0.125 mole of oxygen molecules during the electrolysis of dilute sodium chloride solution?
[ F = 96 5000 C mol-1 ]
A
24 125 coulombs
B
48 250 coulombs
C
72 375 coulombs
D
96 500 coulombs
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
4 Faradays of electricity is required to liberate one mole of oxygen gas
∴ quantity of electricity required
= 0.125 * 4 * 96500 = 48250 coulombs
Source: Myschool Past Questions Archive (Record #1402)
Chemistry
UTME / JAMB 1999
Electrolysis & Electrochemical Cells
Solution
When a current I was passed through an electrolyte solution for 40 minutes, a mass X g of a univalent metal was deposited at the cathode. What mass of the metal will be deposited when a current 2I is passed through the solution for 10 minutes?
A
X/4 g
B
X/2 g
C
2X g
D
4X g
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
From M = zit
Where M = to the mass of the metal liberated
z = Is the electrochemical equivalent
∴ X g = 1 * 1 * 40 * 60
X g = 2400 I
Let Y represent the new mass
Y = 1*2I * 10 * 60
Y = 1200 I
Y/(X g) = 1200 I/2400 I
Y = X/2 g
Source: Myschool Past Questions Archive (Record #1427)
Chemistry
UTME / JAMB 1999
Chemical Energetics & Thermochemistry
RS(aq) + HF(aq) → RF(s) + HS(aq)
ΔH = -65.7 kj mol(-1).
From the equation above, it can be deduced that
A
the heat content of the reactants is lower than that of the products
B
the heat content of the reactants is higher than that of the products
C
the reaction is slow
D
a large amount of heat is absorbed
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #1458)
Chemistry
UTME / JAMB 1999
Electrolysis & Electrochemical Cells
Which of the following statement is true of the electrochemical series?
A
Electropositivity of metals increase down the series
B
Electropositivity of non-metals decrease down the series
C
Electronegativity of non-metals decrease down the series
D
Electropositivity of metals decreases down the series
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #1460)
Chemistry
UTME / JAMB 1999
Metals, Transition Elements & Extraction
What properties of duralumin make it more useful than its constituent metals?
A
It is heavy with a high melting point
B
It is malleable and has high density
C
It is strong and light
D
It is hard and ductile
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #1470)