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Physics
WAEC / WASSCE 2014
Current Electricity & Electric Circuits
Diagram
Solution
You are provided with a potentiometer AB, a 102\(\Omega\) standard resistor R, a battery of emf 4.5V, a jockey J, and other necessary materials.
Connect a circuit as shown in the diagram above.
Close key K. Without J making contact with AB, read and record the ammeter reading I. Open the key.
Use the jockey to make contact with AB at the 20cm mark such that AJ = x = 20cm. Close the key, read and record the ammeter reading.
Evaluate x\(^{-1}\).
Repeat the procedure for values of x= 35cm, 45cm, 60cm, and 80cm respectively.
Tabulate your readings.
Plot a graph with x\(^{-1}\) on the vertical axis and \(l_i\) on the horizontal axis, starting both axes from the origin (0, 0).
Determine the slope, s, of the graph.
From your graph, determine the value \(l_o\) of \(I_1\) for which x\(^{-1}\)= 0.
Evaluate \(\frac{I_{o}}{I}\).
State two precautions taken to obtain accurate results.
(b)i. Define the emf of a battery.
ii. A cell X of emf 1.018V is balanced by a length of 50.0cm on a potentiometer wire. Another cell Y is balanced by a length of 75.0cm on the same wire. Calculate the emf of Y.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Electricity
I = 0.36A
\(\frac{x}{cm}\)
\(\frac{I}{A}\)
\(\frac{X^{-1}}{cm^{-1}}\)
20.0
1.14
0.050
35.0
0.90
0.029
45.0
0.84
0.022
60.0
0.78
0.017
80.0
0.66
0.013
Slope (s) = \(\frac{\bigtriangleup {X}^{-1}}{\bigtriangleup {I}}\) = \(\frac{0.08-0.01}{1.68-0.644} = \frac{0.07}{1.036}\)
From the graph,
When X\(^{-1}\) = 0
\(I_0\) of \(I_1\) = 0.448A
x. Evaluate \(\frac{I_{0}}{I}\)
= \(\frac{0.448}{0.36}\) = 1.244
Precaution;
The key was opened when readings were not taken.
I ensured tight connections of the connecting wire
I avoided parallax errors when reading the potentiometer.
(bi) Define emf of a battery: The e.m.f of a cell is the potential difference across its terminals when it is in an open circuit. i.e. not Supplying current to an external circuit.
ii. A cell x emf is 1.018V.
Lengt\(h_1\) on the potentiometer x = 50.0cm
Lengt\(h_2\) on the potentiometer, y,= 75.0cm
A cell Y emf is?
Solution
By formula;
\(\frac{E_{1}}{E_{2}}\) = \(\frac{L_{2}}{L_{1}}\)
\(\frac{1.018}{E_{2}} = \frac{75.0}{50.0}\)
75.\(E_2\) = 1.018 x 50
\(E_2\) = \(\frac{1.018\times 50}{75}\)
\(E_2\) = 0.612V.
see graph above
Source: Myschool Past Questions Archive (Record #59978)
Chemistry
UTME / JAMB 2014
Stoichiometry, Mole Concept & Volumetric Analysis
A mixture is different from a compound because
A
the properties of a compound are those of its individual constituents while those of a mixture differ from its constituents
B
a mixture is always homogeneous while a compound is not
C
the constituent of a compound are chemically bound together while those of a mixture are not
D
a mixture can be represented by a chemical formula while a compound cannot
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 #30675)
Chemistry
UTME / JAMB 2014
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
What is the percentage of sulphur in sulphur (IV) oxide?
A
66%
B
25%
C
40%
D
50%
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
molar mass of SO 2=64 :.%S=32/64 x 100 = 50%
Source: Myschool Past Questions Archive (Record #30676)
Chemistry
UTME / JAMB 2014
Particulate Nature of Matter, Laws & Formulae
Solution
A gas X diffuses twice as fast as gas Y. if the relative molecular mass of X is 32, calculate the relative molecular mass of Y.
A
128
B
8
C
16
D
64
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
\(\frac{RY}{RX}\) = \(\frac{\sqrt{My}}{\sqrt{My}}\) 1/2 = \(\frac{\sqrt{32}}{\sqrt{y}}\) \(\frac{32}{y}\) = 1/4
Y = \(\frac{32x4}{1}\) = 128
Source: Myschool Past Questions Archive (Record #30677)
Chemistry
UTME / JAMB 2014
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
200 cm3 of a gas at 25ºC exerts a pressure of 700 mmHg. Calculate its pressure if its volume increase 350 cm3 at 75ºC.
A
342.53 mmHg
B
1430.54 mmHg
C
467.11 mmHg
D
400.00 mmHg
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
\(\frac{p1V 1}{T1}\) = \(\frac{P2V2}{T2}\) \(\frac{700x200}{298}\) = \(\frac{P2X 350}{348}\) P2 = \(\frac{700x200x348}{298x350}\) = 467.11mmHg
Source: Myschool Past Questions Archive (Record #30678)
Chemistry
UTME / JAMB 2014
Atomic Structure, Orbitals & Nuclear Chemistry
An element X has electron configuration 1s2 2s2 2p6 3s2 3p5. Which of the following statements is correct about the element?
A
lt has a completely filled p-orbital
B
lt has 5 electrons in its outermost shell.
C
lt belongs to group ll on the periodic table
D
lt is a halogen
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 #30680)
Chemistry
UTME / JAMB 2014
Periodic Table & Periodicity
If the difference in electronegativity of elements P and Q is 3.0. The bond that will be formed between them is
A
metallic
B
covalent
C
co-ordinate
D
ionic
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 #30683)
Chemistry
UTME / JAMB 2014
Atomic Structure, Orbitals & Nuclear Chemistry
How many protons, neutrons and electrons respectively are present in the element 60/27 Co?
A
27, 33 and 33
B
33, 27 and 27
C
27, 33, and 27
D
60, 33 and 60
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 #30688)
Chemistry
UTME / JAMB 2014
Stoichiometry, Mole Concept & Volumetric Analysis
The radio active radiation used in studying the arrangement of particles in giant organic molecules is
A
\(\gamma\)- rays
B
\(\alpha\)- particles
C
X-rays
D
\(\beta\) - particles
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 #30689)
Chemistry
UTME / JAMB 2014
Particulate Nature of Matter, Laws & Formulae
Solution
A silicon -containing ore has 92% 28Si, 5% 29Si and 3% 30Si. Calculate the relative atomic mass of the silicon.
A
14.00
B
29.00
C
28.11
D
28.00
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
(\(\frac{92}{100}\) x 28) + (\(\frac{5}{100}\) x 29) + (\(\frac{3}{100}\) x 30) = 28.11
Source: Myschool Past Questions Archive (Record #30690)
Chemistry
UTME / JAMB 2014
Water, Solutions & Solubility
Solution
Water is said to be temporarily hard when it contains
A
Ca (HCO3)2 and Mg (HCO3)2 salts
B
Ca (HCO3)2 and CaCO3 SALTS
C
Mg(HCO3)2 and CaSO4 salts
D
CaSO4 and Ca(HCO3)2 salts
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Temporary hardness is caused by the presence of dissolved calcium hydrogentrioxocarbonate(IV) and magnesium hydrogentrioxocarbonate(IV) which decomposes on heating to form calcium trioxocarbonate(IV) and magnesium trioxocarbonate(IV) respectively.
Source: Myschool Past Questions Archive (Record #30693)
Chemistry
UTME / JAMB 2014
Water, Solutions & Solubility
On exposure to the atmosphere, a hydrated salt loses its water of crystallization to become anhydrous. This phenomenon is referred to as
A
efflorescence
B
deliquescence
C
hygroscopy
D
hydrolysis
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 #30694)
Chemistry
UTME / JAMB 2014
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
16.55g of lead (ll) trioxonitrate (V) was dissolved in 100g of distilled water at 20oC, calculate the solubility of the solute in moldm-3
[Pb = 207, N = 14, O = 16]
A
0.05 g
B
2.00 g
C
1.00 g
D
0.50g
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
molar mass of pb (N0 3) 2 = 331
Mole = \(\frac{16.55}{331}\)= 0.05 mole
∴0.05mole pb (No3)2 dissolved in 1000g H2O x = \(\frac{0.05\times1000}{100}\)= 0.5 mole /dm 3
Source: Myschool Past Questions Archive (Record #30695)
Chemistry
UTME / JAMB 2014
Acids, Bases, pH & Salts
Which of the following salts will turn blue litmus red?
A
Sodium tetrahydroxozincate (ll)
B
Sodium hydrogentetraoxosulphate (VI)
C
Sodium trioxocarbonate (lV)
D
Zinc chloride hydroxide
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 #30706)
Chemistry
UTME / JAMB 2014
Oxidation, Reduction & Redox Reactions
Zn (s) + CuSO 4(aq) → ZnSO 4(aq) + Cu (s)
In the reaction above, the oxidation number of the reducing agent changes from
A
0 to + 4
B
0 to + 2
C
+ 1 to + 2
D
+ 1 to + 3
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 #30707)