COMMON ERRORS
This topic is geared towards identifying those errors that most students are prone to in spoken and written English. In the same vein, efforts would be made to simultaneously provide the right versions of these errors. Examples of such errors abound.
🌹Divine upliftment/ Divine uplift
Divine upliftment (Wrong)
Divine uplift(Right)
Note: There is no word like 'upliftment' in the English language. It is one of the new coinagew that most public speakers use with impunity.
The word 'uplift' can also be used as a noun and as a verb. It means to raise, increase or promote. 'Uplift' is also a feeling of hope and happiness.
E.g
I. This is my year of divine upliftment (Wrong)
II. This is my year of divine uplift(Right)
III. The news that the Governor survived the plane crash gave us the mich needed uplift (Right)
🌹 godsent/godsend
godsent - Wrong
godsend - Right
Note: There is no such word like 'godsent' in the English language. Again, it does not mean "the person that God has sent" as it is weongky believed by most users of the word.
'Godsend' means something good that happens unexpectedly and helps somebody or something when they need help e.g.
I. This man is godsent to this family (Wrong)
II. This man is a godsend to this family (Right)
III. The new doctrine of giving to the needy ones has come as a godsend to the widows in our church (Right)
🌹Do you hear English?/Do you understand English
Do you hear English? (Wrong)
Do you understand the English language (Right)
Note: It is wrong to ask a person whether he or she hears the English language or any other language e.g. Yoruba, Hausa, Igbo, Efik etc. Every human being that does not have hearing problems hears all languages.
'Hearing' is one thing while 'understanding' is another. Therefore, if you are in doubt of your listener's ability to fathom what you are saying, the right question to ask is: 'Do you understand the English language?' not 'Do you hear English?. E.g
I. The Director advised all those who do not hear English to register for special English class (Wrong)
II. The Director advised all those who do not understand the English Language to register for special English class (Right)
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This topic is geared towards identifying those errors that most students are prone to in spoken and written English. In the same vein, efforts would be made to simultaneously provide the right versions of these errors. Examples of such errors abound.
🌹Divine upliftment/ Divine uplift
Divine upliftment (Wrong)
Divine uplift(Right)
Note: There is no word like 'upliftment' in the English language. It is one of the new coinagew that most public speakers use with impunity.
The word 'uplift' can also be used as a noun and as a verb. It means to raise, increase or promote. 'Uplift' is also a feeling of hope and happiness.
E.g
I. This is my year of divine upliftment (Wrong)
II. This is my year of divine uplift(Right)
III. The news that the Governor survived the plane crash gave us the mich needed uplift (Right)
🌹 godsent/godsend
godsent - Wrong
godsend - Right
Note: There is no such word like 'godsent' in the English language. Again, it does not mean "the person that God has sent" as it is weongky believed by most users of the word.
'Godsend' means something good that happens unexpectedly and helps somebody or something when they need help e.g.
I. This man is godsent to this family (Wrong)
II. This man is a godsend to this family (Right)
III. The new doctrine of giving to the needy ones has come as a godsend to the widows in our church (Right)
🌹Do you hear English?/Do you understand English
Do you hear English? (Wrong)
Do you understand the English language (Right)
Note: It is wrong to ask a person whether he or she hears the English language or any other language e.g. Yoruba, Hausa, Igbo, Efik etc. Every human being that does not have hearing problems hears all languages.
'Hearing' is one thing while 'understanding' is another. Therefore, if you are in doubt of your listener's ability to fathom what you are saying, the right question to ask is: 'Do you understand the English language?' not 'Do you hear English?. E.g
I. The Director advised all those who do not hear English to register for special English class (Wrong)
II. The Director advised all those who do not understand the English Language to register for special English class (Right)
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Projectiles
1) A ball is projected horizontally from the top of hill with a velocity of 20m/s it reaches the ground 4 seconds later, what is the height of the hill? (g=10m/s²).
A) 60m
B) 20m
C) 40m
D) 80m✅
Solution
The ball is projected horizontally, so it behaves like a projectile.
From the equation of motion, H = ut+½gt², when the ball is projected horizontally, the initial vertical velocity is zero, and therefore, the equation becomes, H = ½gt². By making t the subject, you have, t = √(2H/g)... Always remember this formula even when you're sleeping ✍️
So now, from the question, t = 4s, H = ?
Insert your values, 4 = √2H/10
16 = 2H/10
H = 80m✅
2. An object is projected with a velocity of 100m/s from the ground level at an angle to the vertical. If the total time of flight of the projectile is 10s, calculate the angle.
A) 30°
B) 45°
C) 60°
D) 105°
Solution
Time of flight= 2Usinθ/g when the angle is inclined to the horizontal and Time of flight= 2Ucosθ/g when the angle is inclined to the vertical.
From this now,
θ=?, U= 100m/s, T =10s
10 = 2×100cosθ/10
10 = 20cosθ
cosθ = 1/2
θ = ArcCos(1/2) = 60°✅
3. An object is projected with a velocity of 100m/s at an angle of 60° to the vertical. Calculate the time taken by the object to reach the highest point. (Take g=10m/s²).
A) 3.9s
B) 12.8s
C) 25s
D) 5s
Solution
Just like question 2,
Time of flight= 2Ucosθ/g when the angle is inclined to the vertical.
But we're asked time taken to reach the maximum height, which from the note I gave you is SMALL LETTER t.
t = Ucosθ/g
t = 100×cos60/10
t = 50/10
t = 5s✅
4. A stone is projected horizontally from the top of a tower with a speed of 5m/s. It lands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g-10m/s²).
A) 100m
B) 80m
C) 111m
D) 256m
Solution
We're given range. How did we know this?
The question says the distance from the foot of the tower, and the foot of a building is always on the horizontal ground ✍️
The horizontal distance covered, range (R) is the product of the initial horizontal velocity and the time taken.
R = ut
u = 5m/s, H=?, R= 20m
20 = 5t
t = 4s
From t = √(2H/g)...
4 = √2×H/10
Square both sides
16 = 2H/10
H = 80m✅
5. A stone thrown horizontally from the top of a vertical wall with a velocity of 15m/s, hits the horizontal ground at a point 45m from the base of the wall. Calculate the height of the wall.
A) 22m
B) 45m
C)9m
D) 25m
Solution
Use the same formula in question 4 for this.
Option B is correct ✅
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1) A ball is projected horizontally from the top of hill with a velocity of 20m/s it reaches the ground 4 seconds later, what is the height of the hill? (g=10m/s²).
A) 60m
B) 20m
C) 40m
D) 80m✅
Solution
The ball is projected horizontally, so it behaves like a projectile.
From the equation of motion, H = ut+½gt², when the ball is projected horizontally, the initial vertical velocity is zero, and therefore, the equation becomes, H = ½gt². By making t the subject, you have, t = √(2H/g)... Always remember this formula even when you're sleeping ✍️
So now, from the question, t = 4s, H = ?
Insert your values, 4 = √2H/10
16 = 2H/10
H = 80m✅
2. An object is projected with a velocity of 100m/s from the ground level at an angle to the vertical. If the total time of flight of the projectile is 10s, calculate the angle.
A) 30°
B) 45°
C) 60°
D) 105°
Solution
Time of flight= 2Usinθ/g when the angle is inclined to the horizontal and Time of flight= 2Ucosθ/g when the angle is inclined to the vertical.
From this now,
θ=?, U= 100m/s, T =10s
10 = 2×100cosθ/10
10 = 20cosθ
cosθ = 1/2
θ = ArcCos(1/2) = 60°✅
3. An object is projected with a velocity of 100m/s at an angle of 60° to the vertical. Calculate the time taken by the object to reach the highest point. (Take g=10m/s²).
A) 3.9s
B) 12.8s
C) 25s
D) 5s
Solution
Just like question 2,
Time of flight= 2Ucosθ/g when the angle is inclined to the vertical.
But we're asked time taken to reach the maximum height, which from the note I gave you is SMALL LETTER t.
t = Ucosθ/g
t = 100×cos60/10
t = 50/10
t = 5s✅
4. A stone is projected horizontally from the top of a tower with a speed of 5m/s. It lands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g-10m/s²).
A) 100m
B) 80m
C) 111m
D) 256m
Solution
We're given range. How did we know this?
The question says the distance from the foot of the tower, and the foot of a building is always on the horizontal ground ✍️
The horizontal distance covered, range (R) is the product of the initial horizontal velocity and the time taken.
R = ut
u = 5m/s, H=?, R= 20m
20 = 5t
t = 4s
From t = √(2H/g)...
4 = √2×H/10
Square both sides
16 = 2H/10
H = 80m✅
5. A stone thrown horizontally from the top of a vertical wall with a velocity of 15m/s, hits the horizontal ground at a point 45m from the base of the wall. Calculate the height of the wall.
A) 22m
B) 45m
C)9m
D) 25m
Solution
Use the same formula in question 4 for this.
Option B is correct ✅
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🔥JUST IN 🔥
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Projectiles
6. A stone, Q is thrown with velocity U at angle of 75° to the horizontal. Another stone R is thrown with the same velocity U but at an angle of 15° to the horizontal. The ranges covered by the stones will be
A. greater for Q
B. greater for R
C. greater for heavier of two stones.
D. the same for Q and R
Solution
θ₁=15° θ₂=75°
According to the note given to you, Range, R = U²sin2θ/g
meaning that, R is directly proportional to θ. At the same time, the value of the range depends on the value of sin2θ. Get your calculators, and punch sin(2×15)° and sin(2×75)°, you'll discover that they're the same. This is because in the second quadrant, the sine of any angle is positive. For this reason the ranges in the two different scenarios stated in the question is the same.
Option D is correct ✅
7. An object is projected with a velocity of 80m/s at an angle of 30° to the horizontal. The maximum height reached is
A. 20m
B. 80m
C. 160m
D. 320m
Solution
H = U²sin²θ/2g
H = 80²×sin²30/2×10
H = 6400×0.25/20
H = 80m
B✅
8. An object is projected from a height of 80m above the ground with a velocity of 40m/s at an angle of 30° to the horizontal. What is the time of flight?
A. 4s
B. 165
C. 10s
D. 8s
Solution
Time of flight= 2Usinθ/g when the angle is inclined to the horizontal.
T = 2×40×sin30°/10
= 80×0.5/10
T = 4s
A✅
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6. A stone, Q is thrown with velocity U at angle of 75° to the horizontal. Another stone R is thrown with the same velocity U but at an angle of 15° to the horizontal. The ranges covered by the stones will be
A. greater for Q
B. greater for R
C. greater for heavier of two stones.
D. the same for Q and R
Solution
θ₁=15° θ₂=75°
According to the note given to you, Range, R = U²sin2θ/g
meaning that, R is directly proportional to θ. At the same time, the value of the range depends on the value of sin2θ. Get your calculators, and punch sin(2×15)° and sin(2×75)°, you'll discover that they're the same. This is because in the second quadrant, the sine of any angle is positive. For this reason the ranges in the two different scenarios stated in the question is the same.
Option D is correct ✅
7. An object is projected with a velocity of 80m/s at an angle of 30° to the horizontal. The maximum height reached is
A. 20m
B. 80m
C. 160m
D. 320m
Solution
H = U²sin²θ/2g
H = 80²×sin²30/2×10
H = 6400×0.25/20
H = 80m
B✅
8. An object is projected from a height of 80m above the ground with a velocity of 40m/s at an angle of 30° to the horizontal. What is the time of flight?
A. 4s
B. 165
C. 10s
D. 8s
Solution
Time of flight= 2Usinθ/g when the angle is inclined to the horizontal.
T = 2×40×sin30°/10
= 80×0.5/10
T = 4s
A✅
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💎 COHESION
Cohesion is the force of attraction between molecules of thesame substance or kind e.g molecules of mercury
💎 ADHESION
Adhesion is the force of attraction between molecules of different substances or kinds e.g molecules of water and glass_
EFFECT OF COHESION AND ADHESION
Counting of papers with moist finger
water rises up in the capillary tube while mercury falls in a capillary tube
water forms a concave meniscus in a glass tube while mercury forms a convex meniscus
This phenomena is possible because The adhesion of water molecule to glass is greater than their cohesion while the cohesion of mercury molecules is greater than their adhesion to glass
💎ANGLE OF CONTACT
💫Angle of contact is the angle(measured through the liquid)between the wall of a capillary tube and the tangent to the meniscus of the liquid in the tube
NOTE : - The meniscus could be concave or convex depending on the nature of the liquid.Angle of contact depends on the neatness of the wall of the glass tube
When a liquid has concave meniscus, the angle of contact is acute. When it has a convex meniscus, the angle of contact is obtuse.
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Cohesion is the force of attraction between molecules of thesame substance or kind e.g molecules of mercury
💎 ADHESION
Adhesion is the force of attraction between molecules of different substances or kinds e.g molecules of water and glass_
EFFECT OF COHESION AND ADHESION
Counting of papers with moist finger
water rises up in the capillary tube while mercury falls in a capillary tube
water forms a concave meniscus in a glass tube while mercury forms a convex meniscus
This phenomena is possible because The adhesion of water molecule to glass is greater than their cohesion while the cohesion of mercury molecules is greater than their adhesion to glass
💎ANGLE OF CONTACT
💫Angle of contact is the angle(measured through the liquid)between the wall of a capillary tube and the tangent to the meniscus of the liquid in the tube
NOTE : - The meniscus could be concave or convex depending on the nature of the liquid.Angle of contact depends on the neatness of the wall of the glass tube
When a liquid has concave meniscus, the angle of contact is acute. When it has a convex meniscus, the angle of contact is obtuse.
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✅THE HOMOLOGOUS SERIES
A series of compounds in which successive members differ from each other by a specific structural unit (CH2) is called a homologous series e.g. alkanes, alkenes, alkynes, alkanols, etc. A member of a homologous series is called a homologue.
A homologous series has the following properties:
1. All members conform to a general formula.
e.g.
⚡Alkanes - CnH2n+2
⚡Alkenes - CnH2n
⚡Alkynes - CnH2n-2
⚡Alkanols - CnH2n+1OH; etc.
Where n is the number of carbon atoms.
2. Successive members differ in formula by -CH2 or 14grammes.
3. A general method of preparation exists for all members of a homologous series.
4. All members show similar chemical properties.
5. There is a gradual change in the physical properties of members of a homologous series with increasing number of carbon atoms.
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A series of compounds in which successive members differ from each other by a specific structural unit (CH2) is called a homologous series e.g. alkanes, alkenes, alkynes, alkanols, etc. A member of a homologous series is called a homologue.
A homologous series has the following properties:
1. All members conform to a general formula.
e.g.
⚡Alkanes - CnH2n+2
⚡Alkenes - CnH2n
⚡Alkynes - CnH2n-2
⚡Alkanols - CnH2n+1OH; etc.
Where n is the number of carbon atoms.
2. Successive members differ in formula by -CH2 or 14grammes.
3. A general method of preparation exists for all members of a homologous series.
4. All members show similar chemical properties.
5. There is a gradual change in the physical properties of members of a homologous series with increasing number of carbon atoms.
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🖨 OXIDES
Oxides are formed when elements combine with oxygen. They can be classified into several groups namely;
🔻 Basic Oxides: they are oxides of metals. Do you react with acids to form a salt and water only. E.g
Na₂O + 2HCl + 2NaCl + H₂O
🔻 Acidic Oxides: they are oxides of non-metals. They react with water to form acids. They react with alkalis to form a salt and water only. E g
CO₂ + 2NaOH ---> Na₂CO₃ + H₂O
🔻 Amphoteric Oxides: they are metallic oxides which can behave as basic oxides and as acidic oxides i.e they can react with both acids and alkalis to form salts and water only. They include;
📌 Zinc oxide
📌 Aluminum oxide
📌 Lead (II) oxide
📌 Tin (IV) oxide
Memory tip: ZALT oxides
🔻 Neutral Oxides: these oxides are neither acidic or basic. Hence, they are neutral to litmus. This include; water, carbon (II) oxide and nitrogen (I) oxide.
🔻 Other higher oxides: other higher oxides resemble the peroxide in that they contain a higher proportion of oxygen than the ordinary oxides. E.g
📌 Iron (II) diiron (III) oxide [Fe₃O₄]
📌 Dilead (II) lead (IV) oxide [Pb₃O₄]
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Oxides are formed when elements combine with oxygen. They can be classified into several groups namely;
🔻 Basic Oxides: they are oxides of metals. Do you react with acids to form a salt and water only. E.g
Na₂O + 2HCl + 2NaCl + H₂O
🔻 Acidic Oxides: they are oxides of non-metals. They react with water to form acids. They react with alkalis to form a salt and water only. E g
CO₂ + 2NaOH ---> Na₂CO₃ + H₂O
🔻 Amphoteric Oxides: they are metallic oxides which can behave as basic oxides and as acidic oxides i.e they can react with both acids and alkalis to form salts and water only. They include;
📌 Zinc oxide
📌 Aluminum oxide
📌 Lead (II) oxide
📌 Tin (IV) oxide
Memory tip: ZALT oxides
🔻 Neutral Oxides: these oxides are neither acidic or basic. Hence, they are neutral to litmus. This include; water, carbon (II) oxide and nitrogen (I) oxide.
🔻 Other higher oxides: other higher oxides resemble the peroxide in that they contain a higher proportion of oxygen than the ordinary oxides. E.g
📌 Iron (II) diiron (III) oxide [Fe₃O₄]
📌 Dilead (II) lead (IV) oxide [Pb₃O₄]
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Physics
Hooke’s law is an empirical law. True or False?
A) True✅
B) False
C) Cannot be determined
Solution
An empirical law is one which can be proven using an experiment, and Hooke's law happens to fit into this category. Option A is correct ✅
Hooke’s law states that?
A) stress is always proportional to strain✅
B) stress is proportional to strain before ultimate tensile strength
C) stress is proportional to strain under elastic limit
D) stress and strain are never directly proportional
Solution
A basic explanation of Hooke's law is that provided the elastic limit of an elastic material is not exceeded, the applied force is always proportional to the extension produced. However, it could be restated that *the stress experienced in an elastic material is proportional to the resulting strain*. For this reason, Stress α Strain, and the ratio of stress to strain is always CONSTANT. This constant is known as The Young's Modulus of the material. A is correct ✅
As the tension in an elastic string is increased from 100N to 180N, the string extends by 10cm. The work done in increasing the tension is ?
A) 8J
B) 10J
C) 14J✅
D) 18J
The load supported by a spring is increased from 50N to 90N . If the force constant for the spring is 2000N/m, the additional energy stored in the spring due to the increase in load is?
A)1.4J✅
B)0.9J
C)0.5J
D)0.4J
Solution
When the load is 50N
W = F²/2k
= 50²/2(2000)
= 0.625J
When the load is 90N
W = F²/2k
= 90²/2(2000)
= 2.025
∆W = 2.025-0.625
∆W = 1.4J✅
A 5m long wire is fixed to the ceiling. A weight of 10kg is hung at the lower end, and is 1m above the floor. The wire was elongated by 1mm. The energy stored in the wire due to stretching is?
A) zero B) 0.05J ✅C)100J D) 500J
Solution
elongation (extension) = 1mm = 1×10-³m
Force = 10kg×10m/s² = 100N
ENERGY stored = ½Fe
= ½×100×10-³
= 0.5×10-¹
= 0.05J✅
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Hooke’s law is an empirical law. True or False?
A) True✅
B) False
C) Cannot be determined
Solution
An empirical law is one which can be proven using an experiment, and Hooke's law happens to fit into this category. Option A is correct ✅
Hooke’s law states that?
A) stress is always proportional to strain✅
B) stress is proportional to strain before ultimate tensile strength
C) stress is proportional to strain under elastic limit
D) stress and strain are never directly proportional
Solution
A basic explanation of Hooke's law is that provided the elastic limit of an elastic material is not exceeded, the applied force is always proportional to the extension produced. However, it could be restated that *the stress experienced in an elastic material is proportional to the resulting strain*. For this reason, Stress α Strain, and the ratio of stress to strain is always CONSTANT. This constant is known as The Young's Modulus of the material. A is correct ✅
As the tension in an elastic string is increased from 100N to 180N, the string extends by 10cm. The work done in increasing the tension is ?
A) 8J
B) 10J
C) 14J✅
D) 18J
The load supported by a spring is increased from 50N to 90N . If the force constant for the spring is 2000N/m, the additional energy stored in the spring due to the increase in load is?
A)1.4J✅
B)0.9J
C)0.5J
D)0.4J
Solution
When the load is 50N
W = F²/2k
= 50²/2(2000)
= 0.625J
When the load is 90N
W = F²/2k
= 90²/2(2000)
= 2.025
∆W = 2.025-0.625
∆W = 1.4J✅
A 5m long wire is fixed to the ceiling. A weight of 10kg is hung at the lower end, and is 1m above the floor. The wire was elongated by 1mm. The energy stored in the wire due to stretching is?
A) zero B) 0.05J ✅C)100J D) 500J
Solution
elongation (extension) = 1mm = 1×10-³m
Force = 10kg×10m/s² = 100N
ENERGY stored = ½Fe
= ½×100×10-³
= 0.5×10-¹
= 0.05J✅
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📒Alloy
An alloy is a substance prepared by adding one or more element to a base or parent metal to obtain desirable products
🌟Brass -Cu and zn
Use for making moving parts of clocks and watches
🌟Bronze- Cu and Sn
For making coins and medal
🌟 Duralumin- Al, Cu ,Mg Mn
For construction of air craft, ships and machineries
🌟Steel- Fe and C
For construction of bridges ships and machineries
🌟 permallory- Fe and Ni
For making electromagnet
🌟Alnico -Fe,Al,Ni,Co
for making permanent magnet
🌟Soft solder - Pb and Sn
For welding and plumbing
🌟Type metal -Pb , Sb , Sn
For printing
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An alloy is a substance prepared by adding one or more element to a base or parent metal to obtain desirable products
🌟Brass -Cu and zn
Use for making moving parts of clocks and watches
🌟Bronze- Cu and Sn
For making coins and medal
🌟 Duralumin- Al, Cu ,Mg Mn
For construction of air craft, ships and machineries
🌟Steel- Fe and C
For construction of bridges ships and machineries
🌟 permallory- Fe and Ni
For making electromagnet
🌟Alnico -Fe,Al,Ni,Co
for making permanent magnet
🌟Soft solder - Pb and Sn
For welding and plumbing
🌟Type metal -Pb , Sb , Sn
For printing
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*Date- Jan. 11th- April 24th,2025*
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1.which of the following is use for the construction of bridges
A. Bronze
B. Steel✅
C. Soft solder
D. Alnico
2.coinage bronze is made up of cu, zn and
A. Al
B. Be
C. Sn✅
D. Na
3.an alloy is usually regarded as
A. A compound
B. A set of element
C. Uniform mixture ✅
D. None of the above
4.which of the following is use for printing
A. Type metal✅
B. Alnico
C. Duralumin
D. Brass
5.Galena is an ore of
A.copper
B.lead✅
C. Aluminum
D. Tin
6.potassium is very hard to extract with electrolysis because
A. It is soft
B. It is deliquescent
C. It evaporates fast
D. It is very reactive✅
7.Haber process is use to produce
A. Sulfur
B. Ammonia✅
C. Sodium bicarbonate
D. Nitrogen
8.calcium react directly with nitrogen to form
A. Ca3N3
B. Ca2N3
C. Ca3N2✅
D. Ca3N
9.most metal exist in nature as
A. Crust
B. Alloy
C. Ores✅
D. Felspar
10.Tin is purified by
A. Remelting✅
B. Concentrating
C. Roasting
C. All of the above
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A. Bronze
B. Steel✅
C. Soft solder
D. Alnico
2.coinage bronze is made up of cu, zn and
A. Al
B. Be
C. Sn✅
D. Na
3.an alloy is usually regarded as
A. A compound
B. A set of element
C. Uniform mixture ✅
D. None of the above
4.which of the following is use for printing
A. Type metal✅
B. Alnico
C. Duralumin
D. Brass
5.Galena is an ore of
A.copper
B.lead✅
C. Aluminum
D. Tin
6.potassium is very hard to extract with electrolysis because
A. It is soft
B. It is deliquescent
C. It evaporates fast
D. It is very reactive✅
7.Haber process is use to produce
A. Sulfur
B. Ammonia✅
C. Sodium bicarbonate
D. Nitrogen
8.calcium react directly with nitrogen to form
A. Ca3N3
B. Ca2N3
C. Ca3N2✅
D. Ca3N
9.most metal exist in nature as
A. Crust
B. Alloy
C. Ores✅
D. Felspar
10.Tin is purified by
A. Remelting✅
B. Concentrating
C. Roasting
C. All of the above
🎯Don't miss our ongoing intensive UTME tutorial. Enroll today and get access to all we have done.
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"❤Big Congratulations to Stephen!
Stephen, an undergraduate from another institution, joined our UTME tutorial last year and achieved an impressive 300+ score in JAMB. He's now admitted to study MB:BS at FUTA!
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Join us and make your academic dreams a reality!"
Stephen, an undergraduate from another institution, joined our UTME tutorial last year and achieved an impressive 300+ score in JAMB. He's now admitted to study MB:BS at FUTA!
This is what matters most to us - helping you achieve your academic goals. Read Stephen's inspiring testimony and discover how we can help.
Make this year's JAMB your last! Enroll in our tutorial and gain admission to your desired course. Our program has already delivered beyond expectations - check out our flyers to see the amazing benefits!
Don't miss this opportunity! *Note*
Our second month of tutorials begins today, February 13th, and will run until March 14th. We will complete the syllabus we started in January.
❤As a bonus, you will get free access to all our past sessions! When you enroll
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