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Physics
NECO SSCE 2001
Heat Energy & Temperature
The inside of a vacuum flask is usually coated with silver to reduce heat loss by
A
condensation
B
conduction
C
convection
D
evaporation
E
radiation
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #42852)
Physics
NECO SSCE 2001
Mechanics and Properties of Matter
What is the angular velocity of an object in a horizontal circular path of radius 1.5m with a constant speed of 12ms-1?
A
1.6 rads-1
B
4.0 rads-1
C
8.0 rads-1
D
16.0rads-1
E
32.0 rads-1
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #42853)
Physics
NECO SSCE 2001
Waves and Optics
Solution
A wave is represented by the equation y = 0.5 sin 0.4\pi(X - 60t), where the distance (X) is measured in centimetres and time (t) in seconds. What is the wavelength of the wave?
A
0.2cm
B
0.4cm
C
0.8cm
D
5.0cm
E
10.0cm
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The general equation for a wave is given by y = Asin(k(x−vt)), where A is the amplitude, k is the wave number, v is the wave speed, and x and t are the position and time, respectively.
The wave number k is related to the wavelength \lambda by the equation k = \(\frac{2\pi}{\lambda}\)
In the given wave equation, y=0.5sin(0.4\pi(x−60t)), the wave number k is 0.4\pi.
Setting k = \(\frac{2\pi}{\lambda}\) equal to 0.4\pi and solving for \lambda gives:
\lambda = \(\frac{2\pi}{0.4\pi}\) = 5cm
So, the wavelength of the wave is 5 cm.
Source: Myschool Past Questions Archive (Record #42854)
Physics
NECO SSCE 2001
Electricity and Magnetism
The magnetic effect of a current-carrying coil is NOT utilized in the making of a/an
A
carbon microphone
B
electric bells
C
magnetic relay
D
telephone earpiece
E
thermostat
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #42855)
Physics
NECO SSCE 2001
Modern Physics and Nuclear Physics
Solution
In 24 days a radioactive isotope decreases in mass from 128g to 2g. What is the half-life of the radioactive material?
A
2.54 days
B
4.00 days
C
4.80 days
D
6.00 days
E
8.00 days
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
N = N\(_o\)(\(\frac{1}{2}\))\(^{\frac{t}{T_{\frac{1}{2}}}}\)
N = 2g, N\(_o\) = 128g, t =24days and \(T_{\frac{1}{2}}\) = ?
2 = 128(\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
\(\frac{2}{128}\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
\(\frac{1}{64}\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
(\(\frac{1}{2}\))\(^6\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
6 = \(\frac{24}{T_{\frac{1}{2}}}\)
\(T_{\frac{1}{2}}\) = \(\frac{24}{6}\) = 4 days.
Source: Myschool Past Questions Archive (Record #42856)
Physics
NECO SSCE 2001
Heat Energy & Temperature
Solution
In 24 days a radioactive isotope decreases in mass from 128g to 2g. What is the half-life of the radioactive material?
A
2.54 days
B
4.00 days
C
4.80 days
D
6.00 days
E
8.00 days
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
N = N\(_o\)(\(\frac{1}{2}\))\(^{\frac{t}{T_{\frac{1}{2}}}}\)
N = 2g, N\(_o\) = 128g, t =24days and \(T_{\frac{1}{2}}\) = ?
2 = 128(\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
\(\frac{2}{128}\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
\(\frac{1}{64}\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
(\(\frac{1}{2}\))\(^6\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\)
6 = \(\frac{24}{T_{\frac{1}{2}}}\)
\(T_{\frac{1}{2}}\) = \(\frac{24}{6}\) = 4 days.
Source: Official Exam / Past Question Archive
Chemistry
NECO SSCE 2001
Stoichiometry & Mole Concept
Solution
20 cm\(^3\) of 0.09 mol dm\(^{-3}\) solution of tetraoxosuiphate (Vi) acid requires 30cm\(^3\) of sodium hydroxide solution for complete neutralization. The molar concentration of sodium hydroxide is [H = 1; S = 32; 0 = 16; Na = 23]
A
0.03 mol dm3
B
0.12 mol dm3
C
0.27 mol dm3
D
0.45 mol dm3
E
0.90 mol dm3
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
First, write a balanced chemical equation for the neutralization reaction.
H\(_2\)SO\(_4\) + 2 NaOH → Na\(_2\)SO\(_4\) + 2H\(_2\)O
From the balanced equation, N\(_A\) = 1, N\(_B\) = 2
V\(_A\) = 20 cm\(^3\) ; C\(_A\) = 0.09 moldm\(^{-3}\) ; V\(_B\) = 30 cm\(^3\) ; C\(_B\) = ?
Using the formula, \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\)
C\(_B\) = \(\frac{C_A V_A N_B}{V_B N_A}\)
= \(\frac{0.09 X 20 X 2}{30 X 1}\)
= 0.12 moldm\(^{-3}\)
Source: Myschool Past Questions Archive (Record #43317)
Chemistry
NECO SSCE 2001
Matter & Separation Techniques
The decreasing order of the oxidising power of the halogens is
A
Br>I>Cl>F
B
F>I>Cl>Br
C
F>Cl>Br>l
D
I>Br>Cl>F
E
Cl>Br>F>I
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 #43318)
Chemistry
NECO SSCE 2001
Atomic Structure and Periodic Table
Solution
Which statement BEST describes the process of oxidation?
A
addition of hydrogen
B
loss of oxygen
C
decrease in oxidation number
D
loss of electrons
E
gain of elections.
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
In terms of loss or gain of electron(s), the statement that best describes oxidation is the loss of electrons. Always use the acronym: OIL / RIG.
O- Oxidation
I - Is
L - Loss of electrons
R - Reduction
I - Is
G - Gain of electron
Source: Myschool Past Questions Archive (Record #43319)
Chemistry
NECO SSCE 2001
Matter & Separation Techniques
Solution
Animal charcoal differs from wood charcoal in that it
A
has a dark colour
B
has a high carbon content
C
is a good absorbent
D
is crystalline.
E
is rich in phosphate content
medium Difficulty
Verified Correct Answer: Option E
Chemical Principles & Mechanism
Animal charcoal or bone black is the carbonaceous residue obtained by the dry distillation of bones. It contains only about 10% carbon, the remaining being calcium and magnesium phosphates (80%) and other inorganic material originally present in the bones.
Source: Myschool Past Questions Archive (Record #43321)
Chemistry
NECO SSCE 2001
Atomic Structure and Periodic Table
Solution
The atomic number of iron is 56. The number of unpaired electrons in the Fe3+ ion is
A
1
B
2
C
3
D
4
E
5
medium Difficulty
Verified Correct Answer: Option E
Chemical Principles & Mechanism
-There are 4 unpaired electrons in Fe
-However; 5 unpaired electrons are present in Fe3+
Source: Myschool Past Questions Archive (Record #43322)
Chemistry
NECO SSCE 2001
Acids, Bases, and Salts
Solution
The use of sulphur to vulcanize rubber is made to make it
A
blacker.
B
cheaper. .
C
more elastic
D
tougher
E
softer.
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
The use of sulfur to vulcanize rubber is made to enhance its properties, making it more durable, elastic, and heat-resistant.
Source: Myschool Past Questions Archive (Record #43324)
Chemistry
NECO SSCE 2001
Stoichiometry & Mole Concept
Solution
Which element contains the greatest number of moles?
A
8 g of Helium
B
48 g of Carbon
C
78 g of Potassium
D
130g of Zinc
E
216g of Sliver
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
To find the number of moles, we use the formula n = \(\frac{mass}{Molar Mass}\)
N.B: [ He = 4, C = 12, K =39, Zn = 65, Ag = 108]
A. 8 g of Helium, n = \(\frac{8}{4}\) = 2 moles
B. 48 g of Carbon, n = \(\frac{48}{12}\) = 4 moles
C. 78 g of Potassium, n = \(\frac{78}{39}\) = 2 moles
D. 130g of Zinc, n = \(\frac{130}{65}\) = 2 moles
E. 216g of Sliver, n = \(\frac{216}{108}\) = 2moles
Comparing the results, the element with the greatest number of moles is B. 48 g of Carbon, containing 4 moles.
Source: Myschool Past Questions Archive (Record #43326)
Chemistry
NECO SSCE 2001
Matter & Separation Techniques
what dose the following equation illustrate? \( ^{238}U_{92} → ^{234}Th_{90} + ^4 He_2 \)
A
Artificial radioactivity
B
nuclear fission
C
nuclear fusion
D
radio-decomposition
E
spontaneous disintegration
medium Difficulty
Verified Correct Answer: Option E
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #43331)
Chemistry
NECO SSCE 2001
Matter & Separation Techniques
Which compound can be used as a reducing agent in a reaction?
A
Cl2
B
KMnO4
C
KCr2O7
D
MnO2
E
Na2SO3
medium Difficulty
Verified Correct Answer: Option E
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #43333)