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Wednesday, 24th May, 2023
Chemistry 2 (Essay)
– 09:30am – 11:30am
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Chemistry
1a) A transition element, also known as a transition metal, is an element that belongs to the d-block of the periodic table. The d-block is located between the s-block (groups 1 and 2) and the p-block (groups 13 to 18) in the periodic table. Transition elements have partially filled d orbitals in their atomic or ionic states, which gives them unique properties.
The transition elements include elements from group 3 to group 12.
1b)
i) Doubly charged cation:
To form a doubly charged cation, an element must lose two electrons. From the given electron configurations, we can see that element D (with the electron configuration 2:8:8:2) can lose its two outermost electrons to form a doubly charged cation. Therefore, element D forms a doubly charged cation.
bii) Soluble triosocarbonate (IV):
Looking at the electron configurations, we see that element B (with the electron configuration 2:8:2) can lose its two outermost electrons to achieve a +2 valence state. However, none of the given elements have a valence of +4, which is required to form a soluble triosocarbonate (IV). Therefore, none of the given elements can form a soluble triosocarbonate (IV).
1c)
1. Effective nuclear charge: The effective nuclear charge experienced by an electron increases as we move across a period. The effective nuclear charge is the net positive charge experienced by the valence electrons, taking into account shielding effects. With the addition of more protons in the nucleus, the positive charge increases. This stronger attraction between the positively charged nucleus and the negatively charged electrons makes it more difficult to remove an electron, thus increasing the ionization energy.
2. Atomic size: Across a period, the atomic size generally decreases. This reduction in atomic size is due to an increase in the number of protons and electrons, which leads to a stronger attractive force between the nucleus and the outermost electrons. As the atomic size decreases, the valence electrons are held more tightly, making it harder to remove them. Consequently, the ionization energy increases.
In summary, the general increase in the first ionization energies across a period can be attributed to the increasing effective nuclear charge and decreasing atomic size, both of which contribute to a stronger attraction between the nucleus and the valence electrons.
1d) Butanol (C4H10O)
. Ethane (C2H6)
1e) Alkanols are stronger bases than water because alkanols have a greater electron-donating capability due to the alkyl group. This increased electron density on the oxygen atom facilitates the acceptance of protons, making alkanols stronger bases than water. The alkyl group's electron-donating nature enhances the basicity of the alkanol, allowing it to react with acidic compounds more readily.
1f) Salt (NaCl)
Limestone (CaCO3)
Ammonia(NH3)
Carbon dioxide (CO2)
1g) Geometric isomerism is a form of stereoisomerism that arises due to the restricted rotation around a bond. In molecules with geometric isomerism, the atoms are arranged differently in space, leading to distinct isomeric forms with different physical and chemical properties. Geometric isomerism commonly occurs in compounds that have a double bond (C=C) or a ring structure.
The two primary types of geometric isomers are cis-isomers and trans-isomers:
1h)
Water gas is a better fuel than producer gas primarily due to its higher calorific value and a greater concentration of combustible gases
1i)
The term "heat of combustion" refers to the amount of heat energy released when a substance undergoes complete combustion with oxygen. It is also known as the enthalpy of combustion. The heat of combustion is typically expressed in units of energy per unit mass or energy per unit volume of the substance being burned. Common units include joules per gram (J/g), kilojoules per mole (kJ/mol)
1Ji)
Faraday's second law of electrolysis, also known as the law of equivalent proportions, states that during electrolysis, the amounts of different substances deposited or liberated at the electrodes are directly proportional to their chemical equivalent weights.
1jii)
To calculate the amount of silver deposited, we need to use Faraday's law
Given:
Amount of electric charge (Q) = 10920 coulombs
Faraday's constant (IF) = 96500 C/mol
using the equation:
Amount of substance (m) = Q / IF
Substituting the given values:
m = 10920 C / 96500 C/mol
m ≈ 0.113 mol
Since we're calculating the amount of silver deposited, and silver has an atomic weight of approximately 107.87 g/mol, we can further calculate the mass using the equation:
Mass (m) = Amount of substance (mol) × Atomic weight (g/mol)
Mass = 0.113 mol × 107.87 g/mol
Mass ≈ 12.17 g
Number 4
a)
Step 1) Prepare a solution of calcium chloride: Dissolve calcium chloride (CaCl2) in water to create a calcium chloride solution. This step involves measuring the appropriate amount of calcium chloride and adding it to a container of distilled water while stirring until the compound is completely dissolved.
Step 2) Add a source of carbonate ions: Introduce a source of carbonate ions to the calcium chloride solution. This can be achieved by adding sodium carbonate (Na2CO3) or sodium bicarbonate (NaHCO3) to the solution. The carbonate ions from these compounds will react with the calcium ions in the solution, forming calcium carbonate.
Step 3)Precipitation of calcium carbonate: A white precipitate of calcium carbonate will form as a result of the reaction between the calcium ions and carbonate ions. The reaction can be written as follows:
Ca2+ (aq) + CO3^2- (aq) → CaCO3 (s)
The solid calcium carbonate will precipitate out of the solution.
Step 4) Filtration and washing: Use filtration to separate the solid calcium carbonate from the liquid solution. Set up a filter paper in a funnel and pour the mixture through it. The calcium carbonate will be retained on the filter paper while the liquid, containing any remaining calcium chloride and byproducts, passes through.
Step 5) Drying: After filtration, transfer the wet calcium carbonate onto a watch glass or a suitable container. Allow the solid to air dry or use gentle heat to remove any remaining moisture. The resulting dry solid is calcium trioxocarbonate (IV), commonly known as calcium carbonate.
Chemistry 1 (Objective)
– 11:30pm – 12:30pm
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