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Chemistry
UTME / JAMB 2010
Acids, Bases, and Salts
Solution
The colour of methyl orange in alkaline medium is?
A
yellow
B
pink
C
orange
D
red
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Methyl orange has the property to color alkaline and neutral water yellow.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4481)
Chemistry
UTME / JAMB 2010
Periodic Table & Periodicity
Moving from left to right across a period, the general rise in the first ionization energy can be attributed to the?
A
decrease in nuclear charge
B
increase in nuclear charge
C
decrease in screening effect
D
increase in screening effect
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: WAEC / JAMB / NECO Classroom Archive (Record #4467)
Chemistry
UTME / JAMB 2010
Periodic Table & Periodicity
Solution
The stability of the noble gases is due to the fact that they?
A
have no electron in their outermost shells
B
have duplet or octet electron configurations
C
belong to group zero of the periodic table
D
are volatile in nature
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Noble / Rare gases belong to group 0 because they have no bonding electrons in their outermost shell to satisfy the octet rule for stability.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4469)
Chemistry
UTME / JAMB 2010
Periodic Table & Periodicity
Solution
Elements in the same period in the periodic table have the same?
A
number of shells
B
atomic number
C
chemical properties
D
physical properties
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Elements within the same group have the same number of electrons in their valence (outermost) shells, and they have similar electron configurations. They exhibit similar chemical properties.
But elements in the same period can determined by the same number of shells.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4471)
Chemistry
UTME / JAMB 2010
Periodic Table & Periodicity
Solution
A noble gas with a high power of fog penetration used in aerodrome beacons is?
A
krypton
B
argon
C
helium
D
neon
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
The noble gas with a high power of fog penetration used in aerodrome beacons is krypton. Krypton's light can penetrate dense fog better than other noble gases, making it ideal for airport approach lights where visibility in foggy conditions is crucial.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4473)
Chemistry
UTME / JAMB 2010
Air and Air Pollution
Solution
When iron is exposed to moist air, it gradually rusts. This is due to the formation of?
A
hydrate iron (III) oxide
B
anhydrous iron (III) oxide
C
anhydrous iron (II) oxide
D
hydrate iron (II) oxide
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Rusting is an oxidation reaction. The iron reacts with water and oxygen to form hydrated iron(III) oxide, which we see as rust.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4495)
Chemistry
UTME / JAMB 2010
Water and Solubility Concept
Solution
Permanent hardness of water can be removed by?
A
filtration
B
adding slaked lime
C
adding caustic soda
D
boiling
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
- this hardness cannot be removed by boiling. We can remove this hardness by treating the water with washing soda, caustic soda and permutit
Source: WAEC / JAMB / NECO Classroom Archive (Record #4474)
Chemistry
UTME / JAMB 2010
Water and Solubility Concept
Solution
Calculate the solubility in mol dm-3 of 40g of CuSO4 dissolved in 100g of water at 120°C?
[Cu = 64, S = 32, O = 16]
A
4.00
B
2.50
C
0.40
D
0.25
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Solubility = \(\frac{\text{number of moles of solution}}{\text{vol of solution}}\)
100g of water = 100 cm\(^3\) = 0.1 dm\(^3\).
CuSO\(_4\) = 160 g/mol
No of moles of CuSO\(_4\) in 40g = \(\frac{40}{160}\)
= 0.25 mol.
Solubility = \(\frac{0.25}{0.1}\)
= 2.5 moldm\(^{-3}\).
Source: WAEC / JAMB / NECO Classroom Archive (Record #4476)
Chemistry
UTME / JAMB 2010
Quantitative and Qualitative Analyses
Solution
Calculate the volume of 0.5 mol dm\(^{-3}\) H\(_2\)SO\(_4\) that is neutralized by 25 cm\(^3\) of 0.1 mol dm\(^3\) NaOH?
A
5.0 cm3
B
2.5 cm3
C
0.4 cm3
D
0.1 cm3
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Firstly, write a balanced equation for the chemical reaction: H\(_2\)SO\(_4\) + 2NaOH → Na\(_2\)SO\(_4\) + 2H\(_2\)O
C\(_A\) = 0.5 mol dm\(^{-3}\) V\(_A\) = ? C\(_B\) = 0.1 mol dm\(^{-3}\) V\(_B\) = 25 cm\(^3\)
N\(_A\) = 1 N\(_B\) = 2
\(\frac{{C_A}{V_A}}{{C_B}{V_B}}\) = \(\frac{N_A}{N_B}\)
V\(_A\) = \(\frac{{C_B}{V_B}{N_A}}{{C_A}{N_B}}\)
V\(_A\) = \(\frac{{0.1}\times{25}\times{1}}{{0.5}\times{2}}\)
V\(_A\) = 2.5 cm\(^3\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #4480)
Chemistry
UTME / JAMB 2010
Quantitative and Qualitative Analyses
Solution
To a solution of an unknown compound, a little dilute tetraoxosulphate (VI) acid was added with some freshly prepared iron (II) tetraoxosulphate (VI) solution. The brown ring observed after the addition of a stream of concentrated tetraoxosulphate (VI) acid confirmed the presence of?
A
CO\(^{2-}_{3}\)
B
Cl-
C
SO\(_{3}^{2-}\)
D
NO\(^{2-}_{3}\)
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
NO\(_{3}^{-}\) is confirmed, then the brown ring is due to the formation of FESO\(_4\).NO
Source: WAEC / JAMB / NECO Classroom Archive (Record #4492)
Chemistry
UTME / JAMB 2010
Quantitative and Qualitative Analyses
Solution
A compound gives an orange-red color to non-luminous flame. This compound is likely to contain?
A
Na+
B
Ca2+
C
Fe3+
D
Fe2+
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
- Calcium when introduced into the flame, it will produce brick red / red-orange colour.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4496)
Chemistry
UTME / JAMB 2010
Electrochemistry & Electrolysis
Iron can be protected from corrosion by coating the surface with?
A
gold
B
silver
C
copper
D
zinc
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: WAEC / JAMB / NECO Classroom Archive (Record #4485)
Chemistry
UTME / JAMB 2010
Electrochemistry & Electrolysis
Solution
What quantity of aluminum is deposited when a current of 10A is passed through a solution of an aluminum salt for 1930s?
[Al = 27, F = 96500 C mol -1]
A
0.2 g
B
1.8 g
C
5.4 g
D
14.2 g
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Q = It = 10 x 1930 ( given)
27g → 3F( since aluminum is trivalent)
27g → 3 x 96500
x → 19300
x = \(\frac{27 \times 19300}{3 \times 96500}\)
x = 1.8g
Source: WAEC / JAMB / NECO Classroom Archive (Record #4486)
Chemistry
UTME / JAMB 2010
Types of Reactions
Solution
6\(AgNO_{4(aq)} + PH_{3(g)} + 3H_2O_{(l)}\) → 6\(Ag_{(s)} + H_3PO_{3(g)} + 6HNO_{3(aq)}\)
In the above reaction, the reducing agent is?
A
HNO3(aq)
B
H2O(l)
C
PH3(g)
D
AgNO3(aq)
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Phosphorus(iii)hydride is a strong reducing agent which reduces solution of silver or copper(ii)salts to the metals or phosphides.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4483)
Chemistry
UTME / JAMB 2010
Chemical Energetics & Thermochemistry
Solution
In which of the following is the entropy change positive?
A
Thermal dissociation of ammonium chloride
B
Reaction between an acid and a base
C
Addition of concentrated acid to water
D
Dissolution of sodium metal in water
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Thermal dissociation of ammonium chloride is an endothermic reaction as heat is absorbed for a positive entropy, with more products than reactants.
Source: WAEC / JAMB / NECO Classroom Archive (Record #4487)