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JAMB UTME - Chemistry - 2024

Swali 1 Ripoti

The number of geometrical isomers of butene are

2

3

5

7

Maelezo ya Majibu

To understand the geometrical isomers of butene, we need to explore its structure. Butene has four carbon atoms, and there are various structural forms that butene can take. These structural forms include linear or branched chains, with a double bond present between carbon atoms.

Geometric isomerism is a type of stereoisomerism. It occurs due to restricted rotation around the double bond, leading to different spatial arrangements of groups attached to the carbons forming the double bond. The geometric isomerism primarily occurs in alkenes like butene where the positions of substituents can vary.

Let's consider the different types of butene, focusing on the possibility of geometrical isomerism:

1-butene: This is a straight-chain isomer with the double bond starting at the first carbon. It doesn't show geometrical isomerism because the double bond is at the end of the chain, so there are no different spatial arrangements of groups possible.
2-butene: Here the double bond is between the second and third carbon. In this case, geometrical isomerism is possible because different groups (or atoms) attached to the carbon atoms can have different spatial arrangements. Specifically:
- cis-2-butene: Both methyl groups (CH₃) are on the same side of the double bond.
- trans-2-butene: The methyl groups are on opposite sides of the double bond.
Isobutene (or 2-methylpropene): This is a branched isomer, and it does not have geometric isomers because the substituents around the double bond do not allow for different spatial configurations on either side of the double bond.

In conclusion, for butene, only 2-butene has geometrical isomers (cis and trans). Therefore, the number of geometric isomers is 2.

Soma maelezo ya somo kwenye Organic Compounds (JAMB)

Organic Compounds

Tazama Jibu

Swali 3 Ripoti

H $_{2}$ SO $_{4}$

C $_{2}$ H $_{5}$ OH → C $_{2}$ H $_{4}$

170 $^{0}$ C

The reaction above illustrates

oxidation

esterification

dehydration

polymerization

Maelezo ya Majibu

This reaction illustrates dehydration. In chemistry, dehydration refers to the process of removing water (H₂O) from a compound. Let's break down the given reaction to understand this better.

The provided chemical equation is:

C₂H₅OH → C₂H₄ + H₂O

This equation indicates that ethanol (C₂H₅OH) is being transformed into ethylene (C₂H₄) with the production of water (H₂O).

The process involves the breaking of bonds in ethanol and the removal of a water molecule, as follows:

The ethanol molecule, C₂H₅OH, has the –OH group, which combines with a hydrogen atom from the rest of the molecule to form H₂O (water).
This results in the formation of C₂H₄ (ethylene), a smaller molecule, and the water molecule is removed.

This reaction is typically carried out under certain conditions, in this case at a high temperature of 1700°C, to facilitate the dehydration process.

Therefore, this is indeed a dehydration reaction as it involves converting ethanol into ethylene by removing water.

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Ghana Only) (WAEC)

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Soma maelezo ya somo kwenye Acids, Bases And Salts (JAMB)

Soma maelezo ya somo kwenye Air (JAMB)

Chemistry Of Carbon Compounds (Ghana Only)

Chemistry Of Carbon Compounds (Nigeria Only)

Acids, Bases And Salts

Tazama Jibu

Swali 6 Ripoti

The chemical formula for potassiumhexacyanoferrate(II) is

[Fe(CN)

_{6}

]

^{4 -}

K

_{3}

Fe(CN)

_{6}

Fe(CN)

_{6}

K

_{4}

Fe(CN)

_{6}

Metals And Their Compounds

Tazama Jibu

Swali 8 Ripoti

How many isomers has the organic compound represented by the formula C $_{3}$ H $_{8}$ O ?

2

3

4

5

Organic Compounds

Tazama Jibu

Swali 13 Ripoti

The volume occupied by 1 mole of an ideal gas at a temperature of 13 $^{0}$ C and a pressure of 1.58 atm is

[ R = 0.082 atm dm $^{3}$ K $^{- 1}$ mol $^{- 1}$ ]

14.84 dm

^{3}

14.84 cm

^{3}

0.64 dm

^{3}

0.64 cm

^{3}

Maelezo ya Majibu

According to the Ideal gas equation, PV = nRT

Given: P = 1.58 atm, V = ?, n = 1 mole, R = 0.082, T= 13 + 273K = 286K

Substituting all the given parameters,

V = $\frac{n R T}{P}$

V = $\frac{1 \times 0.082 \times 286}{1.58}$

V = 14.84 dm $^{3}$

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 17 Ripoti

C $_{x}$ H $_{y}$ O + 5O $_{2}$ → 4CO $_{2}$ + 4H $_{2}$ O

C $_{x}$ H $_{y}$ O in the equation is

CH

_{2}

O

C

_{2}

H

_{6}

O

CH

_{4}

O

C

_{2}

H

_{4}

O

Maelezo ya Majibu

C $_{x}$ H $_{y}$ O + 5O $_{2}$ → 4CO $_{2}$ + 4H $_{2}$ O

On balancing the equation, we should have

X = 4 , y = 8 and O = 2 ⇒ C $_{4}$ H $_{8}$ O $_{2}$

Since 2 is a common factor to the three atoms, we can divide through by 2, considering the fact that that formula is not in the option.

We finally have C $_{2}$ H $_{4}$ O

Soma maelezo ya somo kwenye Redox Reactions (WAEC)

Redox Reactions

Tazama Jibu

Swali 19 Ripoti

The general molecular formula C $_{n}$ H $_{2 n - 2}$ represents that of an

alkyne

alkene

alkane

alkanal

Organic Compounds

Tazama Jibu

Swali 22 Ripoti

The ions responsible for permanent hardness in water are sulphates of

Fe

^{3 +}

and Mg

^{2 +}

Ca

^{2 +}

and Mg

^{2 +}

Fe

^{2 +}

and Fe

^{3 +}

Na

^{+}

and Ca

^{2 +}

Tazama Jibu

Swali 23 Ripoti

An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [ C =12, H = 1, N = 14, O = 16].

C

_{5}

H

_{7}

NO

_{2}

C

_{5}

H

_{6}

NO

_{2}

C

_{3}

H

_{6}

NO

_{2}

C

_{3}

H

_{7}

NO

_{2}

Organic Compounds

Tazama Jibu

Swali 25 Ripoti

Which of the following is used in forming slag in the blast furnace for the extraction of iron?

Ca(CO

_{3}

)

_{2}

Ca(OH)

_{2}

CaCO

_{3}

CaO

Metals And Their Compounds

Tazama Jibu

Swali 26 Ripoti

In the extraction of Aluminium, the silica impurity is removed by

electrolysis

filtration

reaction with H

_{2}

SO

_{4}

precipitation

Maelezo ya Majibu

Aluminum is extracted from bauxite by electrolysis. The extraction proceeds in two stages;

1. Purification of the Bauxite: The impure bauxite is heated with sodium hydroxide solution to form soluble sodium tetrahydroxy aluminate (iii). The impurities in the ore which are iron (iii) oxide and trioxosilicate (iv) compounds are not soluble in the alkali. They are therefore filtered off as a sludge.

Aluminum hydroxide crystals is then added to filtrate, NaAl(OH) $_{4}$ solution to induce the precipitation of Aluminum hydroxide.

2. The electrolysis of the pure alumina

Soma maelezo ya somo kwenye Metals And Their Compounds (JAMB)

Metals And Their Compounds

Tazama Jibu

Swali 29 Ripoti

25.0g of potassium chloride were dissolved in 80g of distilled water at 30 $^{0}$ C. Calculate the solubility of the solute in mol dm $^{3}$ . [K =39, Cl = 35.5]

3.9 moldm

^{3}

4.6 moldm

^{3}

4.0 moldm

^{3}

4.2 moldm

^{3}

Tazama Jibu

Swali 35 Ripoti

Which of the following is an air pollutant?

O

_{2}

CO

H

_{2}

O

(_{g}

)

O

_{3}

Environmental Pollution

Tazama Jibu

Swali 36 Ripoti

In the conductance of aqueous CuSO $_{4}$ solution, the current carriers are the

mobile electrons

metallic ions

hydrated ions

hydrated electrons

Electrolysis

Tazama Jibu

Swali 37 Ripoti

The hybridization scheme in ethyne is

sp

sp

^{3}

sp

^{2}

d

sp

^{2}

s

Maelezo ya Majibu

Ethyne, also known as acetylene, is a simple alkyne with the chemical formula C₂H₂. In ethyne, each carbon atom is bonded to two other atoms: one hydrogen atom and the other carbon atom. The molecular structure of ethyne is linear, with a triple bond between the two carbon atoms.

To determine the hybridization scheme in ethyne, we need to examine the arrangement of the electron pairs around each carbon atom. In ethyne, each carbon atom is forming two sigma (σ) bonds and two pi (π) bonds. Let's explain:

The two carbon atoms are linked by a triple bond, consisting of one sigma bond and two pi bonds. The sigma bond is formed by the overlap of orbitals directly between the two carbon nuclei. The pi bonds are formed by the overlap of p orbitals above and below the plane of the molecule.
The carbon atoms in ethyne are bonded to only two atoms, which is indicative of a linear geometry.

When we consider the hybridization of the carbon atoms, we focus on the formation of sigma bonds and lone pairs. In ethyne, each carbon atom utilizes two orbitals to form sigma bonds: one with the hydrogen atom and one with the other carbon atom. This implies that each carbon atom in ethyne must use two hybrid orbitals.

The two hybrid orbitals formed by each carbon atom in ethyne are a result of mixing one s orbital with one p orbital. This hybridization is referred to as sp hybridization, characterized by a linear electron geometry. The remaining two unhybridized p orbitals on each carbon atom are responsible for forming the two pi bonds in the triple bond.

In conclusion, the hybridization scheme in ethyne is sp.

Soma maelezo ya somo kwenye Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

Tazama Jibu

Swali 38 Ripoti

Na $_{2}$ X ⇌ 2Na $^{+}$ + X $^{2 -}$

The bond between Na and X is likely to be

covalent

dative

ionic

metallic

Atomic Structure And Bonding

Tazama Jibu

Swali 39 Ripoti

The compound of Copper used as a fungicide is

CuCO

_{3}

Cu(NO

_{3}

)

_{2}

CuSO

_{4}

CuCl

_{2}

Metals And Their Compounds

Tazama Jibu

Swali 40 Ripoti

The molecular formular of a hydrocarbon with an empirical formula of CH $_{3}$ and a molar mass of 30 is

C

_{2}

H

_{4}

C

_{3}

H

_{3}

C

_{2}

H

_{6}

C

_{6}

H

_{6}

Organic Compounds

Tazama Jibu

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40