Chemistry 2015 Jamb Examination

Chemistry 2015 Jamb Examination
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Question 1

Two equal bulbs, one containing ammonia and the other nitrogen are opened mouth-to-mouth to each other at room temperature. The entropy in the mixture of gases is likely to
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When two different gases such as ammonia and nitrogen are mixed, the system becomes more disordered because the molecules of each gas are now randomly distributed throughout the combined volume. Entropy is a measure of disorder or randomness, and mixing always increases randomness. Therefore, the total entropy of the system increases since the mixture allows a greater number of possible molecular arrangements.
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Question 2

A small quantity of solid ammonium chloride (NH4Cl) was heated gently in a test tube, the solid gradually disappears to produce two gases. Later, a white cloudy deposit was observed on the cooler part of the test tube. The ammonium chloride is said to have undergone
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Ammonium chloride is a substance that sublimes when heated, meaning it changes directly from solid to gas without passing through the liquid phase. When heated, it decomposes into ammonia and hydrogen chloride gases, which recombine on cooling to form solid ammonium chloride again. This process demonstrates sublimation as the solid disappears on heating and reforms as a solid deposit on cooling.
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Question 3

What volume of 0.5 M H2SO4 will exactly neutralize 20 cm3 of 0.1 M NaOH solution?
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The balanced equation for the reaction is:
H2SO4 + 2NaOH β†’ Na2SO4 + 2H2O
From the equation, 1 mole of H2SO4 reacts with 2 moles of NaOH.

Given:
\( M_1 = 0.5 \; M \), \( V_1 = ? \)
\( M_2 = 0.1 \; M \), \( V_2 = 20 \; cm^3 \)
\( n_1 = 1 \), \( n_2 = 2 \)

Using the equation:\[ \dfrac{M_1V_1}{n_1} = \dfrac{M_2V_2}{n_2} \]\[ V_1 = \dfrac{M_2V_2n_1}{M_1n_2} \]\[= \dfrac{0.1 \times 20 \times 1}{0.5 \times 2} = 2.0 \; cm^3 \]Hence, the required volume of acid is 2.0 cmΒ³.
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Question 4

Calculate the relative molecular mass of Limestone CaCO3 (Ca = 40, C = 12, O = 16)
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The molecular formula of limestone is CaCO3.
Its relative molecular mass is obtained by summing the atomic masses of all constituent atoms:\[ M_r = 40 + 12 + (3 \times 16) \]\[ M_r = 40 + 12 + 48 = 100 \]Thus, the relative molecular mass of limestone is 100.
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Question 5

One mole of a hydrocarbon contains 36 g of carbon and its density is 20. The structure of the hydrocarbon is (organic chemical)
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A hydrocarbon with 36 g of carbon means it contains 3 moles of carbon atoms (since the atomic mass of carbon is 12). Therefore, the empirical formula must have 3 carbon atoms, giving a molecular formula such as C3H4 or C3H6. Given its structure involves unsaturation, the correct formula corresponding to C3H4 is propyne (CH3C≑CH). This compound fits the data and molecular structure requirement.
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Question 6

S-block elements of the periodic table are made up of
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S-block elements are those in which the outermost electron enters an s-orbital. These include Group 1 elements (alkali metals) and Group 2 elements (alkaline earth metals). Therefore, the s-block of the periodic table comprises only Groups 1 and 2.
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Question 7

The principal constituent of natural gas is
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Natural gas is primarily composed of methane (CH4), which usually accounts for 80–95% of its total composition. Other hydrocarbons such as ethane, propane, and butane are present only in small quantities. Methane is a simple hydrocarbon and burns cleanly, making it the major component.
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Question 8

Rare gases are stable because they
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Noble gases are chemically inert because their outermost electron shells are completely filled, achieving a stable duplet (for helium) or octet (for others) configuration. This stable configuration makes them highly unreactive, explaining why they do not form ions or readily participate in chemical reactions.
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Question 9

Electronic configuration of an element X with atomic number 13 in the subsidiary energy level is
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The atomic number 13 corresponds to aluminium (Al). Its electrons fill the orbitals in the order of increasing energy:
1s2, 2s2, 2p6, 3s2, 3p1.
This gives a total of 13 electrons and matches option A.
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Question 10

Alloys are best prepared by
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Alloys are homogeneous mixtures of two or more metals (or a metal and a non-metal) prepared by melting the components together and then cooling the molten mixture. This allows the atoms of the constituent elements to mix uniformly, forming a solid solution with improved properties such as strength or resistance to corrosion.
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Chemistry 2015 Jamb Examination
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