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Question 1 Report
Elements X and Y have electronic configurations \(1s^2 2s^2 2p^4\) and \(1s^2 2s^2 2p^6 3s^2 3p^1\) respectively. When they combine, the formula of the compound formed is
Answer Details
Question 2 Report
Which of the following represents the kind of bonding present in ammonium chloride?
Answer Details
Ammonium chloride contains both ionic and covalent bonds. In ammonium chloride, the ammonium ion (NH4+) is positively charged and the chloride ion (Cl-) is negatively charged. These ions are held together by ionic bonds, which are formed between positively and negatively charged ions. However, the bond between the hydrogen atom in the ammonium ion and the nitrogen atom in the ammonium ion is also a covalent bond. This type of covalent bond is known as a dative covalent bond, or a coordinate covalent bond, because the electron pair being shared is supplied by one atom only (the nitrogen atom in this case). So, the kind of bonding present in ammonium chloride is both ionic and dative covalent. In simple terms, ammonium chloride contains both ionic bonds between its positive and negative ions, and a dative covalent bond between the hydrogen atom and nitrogen atom within the ammonium ion.
Question 3 Report
Consider the reaction: \( \mathrm{A + 2B(g) \rightleftharpoons 2C + D(g)} \) (\( \Delta H = +\mathrm{ve} \))
What will be the effect of decrease in temperature on the reaction?
Answer Details
The effect of a decrease in temperature on the reaction will be that the rate of the backward reaction will increase. In a chemical reaction, the rate of the forward and backward reactions are determined by the activation energy required for each step and the temperature of the system. When the temperature is decreased, the rate of the reaction decreases, and the rate of the backward reaction increases. This shift in the rate of the backward reaction means that there will be a shift in the position of the equilibrium of the reaction. As the rate of the backward reaction increases, the concentration of the reactants will increase and the concentration of the products will decrease, leading to a decrease in the overall yield of the products. In this reaction, as ΔH (the change in enthalpy) is positive, which means that the reaction is endothermic. Endothermic reactions absorb heat from the surroundings to proceed, so a decrease in temperature will lead to a decrease in the rate of the forward reaction and an increase in the rate of the backward reaction. This shift in the rate of the backward reaction will shift the position of the equilibrium of the reaction to the left, leading to an increase in the concentration of the reactants and a decrease in the concentration of the products.
Question 4 Report
The heat of formation of ethene, \( \mathrm{C_2H_4} \) is 50 \( \mathrm{kJmol^{-1}} \), and that of ethane, \( \mathrm{C_2H_6} \) is -82\( \mathrm{kJmol^{-1}} \). Calculate the heat evolved in the process:
\[ \mathrm{C_2H_4 + H_2 \rightarrow C_2H_6} \]
Answer Details
The heat evolved in a chemical reaction can be calculated by subtracting the heat of formation of the reactants from the heat of formation of the products. In this case, the reactants are ethene (C2H4) and hydrogen (H2), and the product is ethane (C2H6). The heat of formation of ethene is 50 kJ/mol and that of hydrogen is 0 kJ/mol (because hydrogen is a reference element). The heat of formation of ethane is -82 kJ/mol. So, the heat evolved in the reaction is given by: Heat evolved = (Heat of formation of products) - (Heat of formation of reactants) = (-82 kJ/mol) - (50 kJ/mol + 0 kJ/mol) = -82 kJ/mol - 50 kJ/mol = -132 kJ/mol. Therefore, the heat evolved in the process is -132 kJ.
Question 5 Report
Consider the reaction
\( \mathrm{A(s) + 2B(g) \rightarrow 2C(aq) + D(g)} \)
What will be the effect of a decrease in pressure on the reaction?
Answer Details
Given: The equation below
A(s) + 2B(g) → 2C(aq) + D(g)
Since we have a higher number of moles of gaseous species on the LHS, i.e 2, a decrease in pressure will favor the forward reaction.
Question 6 Report
If the cost of electricity required to discharge 10g of an ion \(X^{3+}\) is N20.00, how much would it cost to discharge 6g of ion \(Y^{2+}\)?
[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]
Answer Details
X3+
+ 3e−
→
X
3F = 27g
xF = 10g
x3=1027⟹x=109F
109
F ≡
N20.00
1F is equivalent to x
1109=x20
910=x20⟹x=N18.00
1F is equivalent to N18.00.
Y2+
+ 2e−
→
Y
2F = 24g
xF = 6g
x = 6×224=12F
1F = N18.00
12
F = 12×N18.00
= N9.00
Question 7 Report
Which of the following statements about catalyst is false?
Answer Details
The false statement about catalysts is: "catalysts do not alter the mechanism of the reaction and never appear in the rate law." Catalysts are substances that speed up chemical reactions without being consumed in the process. They achieve this by reducing the activation energy needed for the reaction to occur. Enzymes are a type of biological catalysts. In a chemical reaction, a catalyst is not consumed and does not appear in the overall balanced equation. However, catalysts can alter the mechanism of a reaction by providing an alternative pathway with a lower activation energy. This alternative pathway can have a different rate-determining step, which means that the presence of the catalyst can change the rate law of the reaction. Therefore, the statement that catalysts do not alter the mechanism of the reaction and never appear in the rate law is false.
Question 9 Report
\( \mathrm{SO_3} \) is not directly dissolved in water in the industrial preparation of \( \mathrm{H_2SO_4} \) by the contact process because
Answer Details
Question 10 Report
How many alkoxyalkanes can be obtained from the molecular formula \( \mathrm{C_4H_{10}O} \)?
Answer Details
Alkoxyalkanes have a general formula of R-O-R', where R and R' are alkyl groups. From the given molecular formula C4H10O, we can see that there are four carbon atoms, so the longest possible alkyl group is butyl (C4H9-). To form alkoxyalkanes, we need to attach an oxygen atom to the alkyl group. This can be done in three ways - by attaching the oxygen to one of the terminal carbon atoms (forming a primary alcohol), by attaching it to one of the central carbon atoms (forming a secondary alcohol), or by attaching it to the carbonyl carbon atom (forming an ester). So, we can obtain a maximum of three alkoxyalkanes from the given molecular formula. However, we need to take into account that there are different isomers possible for each type of alcohol or ester, depending on which carbon atom the oxygen is attached to. Therefore, the correct answer is (at least) 3.
Question 11 Report
The emission of two successive beta particles from the nucleus \( {}^{32}_{15}\mathrm{P} \) will produce
Answer Details
Question 12 Report
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be
Answer Details
In the question above an Hydrocarbon combust to give CO2 and H20
Let Hydrocarbon be
CxHy + x+Y/4O2= xCO2 + Y/2H2O
Mass of C0=44g and H2O=18g
at STP vol= 22.4
Therefore, 1mole of CO2 contains 44g and 22.4dm³ at STP
1mole = 22.4dm³
xmole = 89.6dm³
Cross multiplying x=89.6/22.4 =4mole of CO2 produce
1mole of H2O = 18g
Xmole = 56g
Cross multiplying
X = 56/18 = 3mole of H20
Then....
CxHy + X + y/4O2 = 4CO2+ 3H2O
Balancing
C4H6 + 6O2 = 4CO2 + 3H2O
Question 13 Report
Burning magnesium ribbon in air removes which of the following
(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?
Answer Details
Burning magnesium ribbon in air will remove oxygen (option i) from the air, but not nitrogen (option ii), argon (option iii), or carbon dioxide (option iv). When magnesium burns, it reacts with oxygen in the air to form magnesium oxide. The reaction can be represented by the following equation: 2Mg(s) + O2(g) → 2MgO(s) The magnesium in the ribbon combines with oxygen in the air to form solid magnesium oxide. This reaction is exothermic, which means that it releases heat and light energy. So, when magnesium ribbon is burned in air, it consumes the oxygen in the air to form magnesium oxide. However, nitrogen, argon, and carbon dioxide are not chemically reactive with magnesium, and therefore are not removed from the air by the burning of magnesium ribbon. In summary, the correct option is (i) only - burning magnesium ribbon in air removes oxygen only.
Question 14 Report
The hybridization in the compound \( \mathrm{CH_3-CH_2-C\equiv H} \) is
Answer Details
The hybridization in a and b is sp3 hybridization while in c and d is sp hybridization.
Question 15 Report
When chlorine water is exposed to bright sunlight, the following products are formed
Question 16 Report
Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?
\(2HCl + CaCO_3 \rightarrow CaCl_2 + H_2O + CO_2\)
Answer Details
The following factors increase a reaction rate
- Increase in concentration of reactants
- Increase in temperature
- Addition of catalyst
- Increase in the surface area of reactant(s)
Question 17 Report
Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?
I. Atomic Number ii. Ionization energy iii. Metallic character iv. Electron affinity
Answer Details
Ionization energy and electron affinity increase across a period, and decrease down a group.
Question 18 Report
The part of the total energy of a system that accounts for the useful work done in a system is known as
Answer Details
The part of the total energy of a system that accounts for the useful work done in a system is known as "Gibbs free energy". Gibbs free energy is a thermodynamic property that represents the amount of energy that can be converted into useful work in a system. It takes into account both the energy of the system and the entropy, or disorder, of the system. In other words, Gibbs free energy is a measure of the energy available to do work, taking into account the energy that is unavailable due to entropy. In simple terms, if a system has a high Gibbs free energy, it has a lot of energy available to do work, and if a system has a low Gibbs free energy, it has little energy available to do work.
Question 19 Report
A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years
Answer Details
The half-life of a radioactive nucleus is the time it takes for half of its particles to decay. This means that after 20 years, 100,000 particles will become 50,000 particles. After 40 years, we can find the number of particles remaining by counting the number of half-lives that have passed. Since 40 years is double the half-life of 20 years, this means that two half-lives have passed, so the number of particles will be halved twice. Starting with 100,000 particles: - After 1 half-life (20 years), there will be 50,000 particles remaining. - After 2 half-lives (40 years), there will be 25,000 particles remaining. So, exactly at the end of 40 years, there will be 25,000 particles remaining.
Question 20 Report
Which two gases can be used for the demonstration of the fountain experiment?
Answer Details
Two gases that can be used in the study of fountain experiment is ammonia gas and hydrogen chloride gas. The experiment introduces concepts like solubility and the gas laws at the entry level.
Question 21 Report
A synthetic rubber is obtained from the polymerization of
Answer Details
A synthetic rubber is obtained from the polymerization of isoprene. Isoprene is a type of hydrocarbon that can be polymerized, or chemically joined together, to form long chains. This process is called polymerization, and the resulting material is called a polymer. When isoprene is polymerized, it forms a synthetic rubber, which is a type of polymer that is used in a wide range of products, including tires, hoses, and adhesives. Synthetic rubber offers several advantages over natural rubber, including improved durability and resistance to heat, ozone, and chemicals.
Question 22 Report
In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm\(^{3}\) of 1.25 molar tetraoxosulphate (vi) acid?
Answer Details
Equation of reaction : 2NaOH + H2 SO4 → Na2 SO4 + 2H2 O
Concentration of a base, CB = 0.5M
Volume of acid, VA = 10cm3
Concentration of an acid, CA = 1.25M
Volume of base, VB = ?
Recall:
CAVACBVB=nAnB
... (1)
N.B: From the equation,
nAnB=12
From (1)
1.25×100.5×VB=12
12.50.5VB=12
25 = 0.5VB
VB = 50.0 cm3
Question 23 Report
How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?
Answer Details
Electrons are not found in the nucleus of an atom. The nucleus of an atom only contains protons and neutrons, while electrons are located outside the nucleus in the electron cloud. The mass number of an atom is equal to the sum of the number of protons and the number of neutrons in the nucleus. Therefore, if an atom has a mass number of 23 and 17 neutrons, then the number of protons in the nucleus can be calculated as: Protons = Mass number - Neutrons Protons = 23 - 17 Protons = 6 This means that the nucleus of the atom contains 6 protons. The number of electrons in a neutral atom is equal to the number of protons, so the atom also contains 6 electrons in the electron cloud surrounding the nucleus. In summary, the answer is that there are 6 protons and 6 electrons in the atom.
Question 25 Report
The oxidation state(s) of nitrogen in ammonium nitrite is/are
Answer Details
Ammonium nitrite = NH4
NO2
NH+4
: Let the oxidation number of Nitrogen = x
x + 4 = 1 ⟹
x = 1 - 4
x = -3
NO−2
: x - 4 = -1
x = -1 + 4 ⟹
x = +3.
The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.
Question 26 Report
The electronic configuration of element Z is \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1\). What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.
Answer Details
Z = 1s2
2s2
2p6
3s2
3p1
?
We have Z3+
and SO2?4
The reaction : Z3+
+ SO2?4
?
Z2
(SO4
)3
.
Question 27 Report
The velocity, V of a gas is related to its mass, M by (k = proportionality constant)
Answer Details
Recall:
V = √3RTM
∴V∝1√M
V=k√M
V = kM12
Question 28 Report
Which of the following gases contains the least number of atoms at s.t.p?
Answer Details
At standard temperature and pressure (s.t.p), all gases have the same number of atoms or molecules. What changes between them is the volume they occupy, and this is dependent on their molecular mass and the number of moles. Comparing the number of moles between the gases listed above, 7 moles of argon will contain the most number of atoms, followed by 4 moles of chlorine, then 3 moles of ozone, and finally 1 mole of butane would contain the least number of atoms. In summary, the number of atoms in a gas sample depends on the number of moles, but at s.t.p, the volume occupied by each gas depends on its molecular mass and the number of moles.
Question 29 Report
The IUPAC name for \( \mathrm{CH_3CH_2COOCH_2CH_3} \) is
Answer Details
The IUPAC name for the given molecule is ethyl propanoate. To arrive at the IUPAC name, we first identify the longest continuous chain of carbon atoms, which in this case is a 4-carbon chain (propane). We then identify and name the substituent groups attached to this chain, which are a methyl group (CH3) attached to the second carbon atom and an ethoxy group (OC2H5) attached to the third carbon atom. The ethoxy group is named as an ethyl group, and the entire molecule is named as ethyl propanoate, following the standard IUPAC naming conventions for esters.
Question 30 Report
The IUPAC nomenclature of the compound
\(\mathrm{H_3C - CH(CH_3) - CH(CH_3) - CH_2 - CH_3}\)
Question 31 Report
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
Answer Details
Question 32 Report
A cell shorthand notation can be written as \(A / A^{+} // B^{2+} / B\). The double slash in the notation represents the
Answer Details
The double slash in the cell shorthand notation represents the salt bridge. A salt bridge is a component of an electrochemical cell that connects the two half-cells and allows the flow of ions between them. It consists of an inert electrolyte solution (usually a salt) that is placed between the two half-cells. The purpose of the salt bridge is to maintain electrical neutrality in each half-cell by allowing the flow of ions to balance the charge buildup in the half-cells. In the cell shorthand notation, the double slash "//" represents the salt bridge that connects the two half-cells of the electrochemical cell. The first half-cell is represented on the left-hand side of the slash and the second half-cell is represented on the right-hand side of the slash. The anode (where oxidation occurs) is represented on the left side, and the cathode (where reduction occurs) is represented on the right side. Therefore, the correct answer is option number 3: salt bridge.
Question 33 Report
Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?
Answer Details
The metal that is most essential in the regulation of blood volume, blood pressure, and osmotic equilibrium is sodium. Sodium is a key electrolyte that helps maintain the balance of fluids in the body, including blood volume and blood pressure. Sodium ions are positively charged and are attracted to negatively charged ions, such as chloride (Cl-) and bicarbonate (HCO3-), which together help regulate the pH of the blood. Sodium is also essential for maintaining osmotic equilibrium, which refers to the balance of solutes between cells and the extracellular fluid. Osmotic equilibrium is critical for proper cellular function and is regulated by the movement of water and electrolytes, including sodium, in and out of cells. While the other metals listed (zinc, manganese, and iron) are important for various functions in the body, such as enzyme activity and oxygen transport, they are not directly involved in regulating blood volume, blood pressure, and osmotic equilibrium in the same way that sodium is. Therefore, the answer is not options 1, 2, or 4, and the correct answer is: sodium.
Question 34 Report
2-methylprop-1-ene is an isomer of
Answer Details
2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene. An isomer is a molecule that has the same molecular formula as another molecule, but a different arrangement of atoms. In this case, 2-methylprop-1-ene has the molecular formula C4H8, and so do 3-methyl but-1-ene and 2-methyl but-1-ene. The difference between these three molecules is in the arrangement of the carbon and hydrogen atoms. 2-methylprop-1-ene has a branched structure with a double bond between the first and second carbon atoms. 3-methyl but-1-ene is also a branched molecule, but the double bond is between the second and third carbon atoms. Similarly, 2-methyl but-1-ene has a double bond between the first and second carbon atoms, but it has a different branching pattern. On the other hand, pent-2-ene has five carbon atoms, so it has a different molecular formula than 2-methylprop-1-ene. Therefore, 2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene, but not of pent-2-ene, because it has the same molecular formula and a different arrangement of atoms compared to the other two isomers.
Question 35 Report
A solution X, on mixing with \( \mathrm{AgNO_3} \) solution gives a white precipitate soluble in aqueous \( \mathrm{NH_3} \), a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain
Question 36 Report
Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are
Answer Details
The chemical constituent of soda lime used to prepare methane in the laboratory is Ca(OH)2 (calcium hydroxide) and NaOH (sodium hydroxide). Soda lime is a mixture of these two compounds. When sodium ethanoate (NaC2H3O2) is heated with soda lime, it undergoes a reaction known as the Kolbe's reaction, which produces methane gas (CH4) as one of the products. The reaction can be represented as follows: 2NaC2H3O2 + 2Ca(OH)2 → 2CH4 + 2NaOH + 2CaCO3 In this reaction, the sodium ethanoate reacts with the calcium hydroxide to form calcium acetate (Ca(C2H3O2)2) and sodium hydroxide. The calcium acetate then decomposes to produce methane gas and calcium carbonate (CaCO3), which is a solid precipitate. Therefore, the chemical constituents of soda lime used to prepare methane in the laboratory are calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH).
Question 38 Report
Elements in the periodic table are arranged in the order of their
Answer Details
Elements in the periodic table are arranged in the order of their atomic numbers. The atomic number of an element is the number of protons in the nucleus of an atom of that element. The elements are arranged in order of increasing atomic number from left to right and from top to bottom in the periodic table. The elements in each row, also known as a period, have the same number of electron shells, while the elements in each column, also known as a group or family, have the same number of valence electrons. This arrangement makes it possible to predict the chemical and physical properties of an element based on its position in the periodic table. Therefore, the correct answer is: - atomic numbers
Question 39 Report
Hydrogen diffused through a porous plug
Answer Details
Hydrogen gas (H2) diffuses faster than oxygen gas (O2) through a porous plug. This is because the rate of diffusion of a gas through a porous plug is inversely proportional to the square root of its molar mass. Since the molar mass of hydrogen (2 g/mol) is much smaller than that of oxygen (32 g/mol), the rate of diffusion of hydrogen through a porous plug is much faster than that of oxygen. To be more specific, the ratio of the diffusion rates of two gases through a porous plug is given by the equation: Rate of diffusion of gas A / Rate of diffusion of gas B = √(Molar mass of gas B / Molar mass of gas A) Using the molar masses of hydrogen and oxygen, we get: Rate of diffusion of hydrogen / Rate of diffusion of oxygen = √(32 g/mol / 2 g/mol) = √16 = 4 Therefore, hydrogen diffuses through a porous plug four times as fast as oxygen. Thus, the correct answer is: four times as fast as oxygen.
Question 40 Report
By what amount must the temperature of 200cm\(^{3}\) of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm\(^{3}\) (Assume ideality)
Answer Details
Using the ideal gas law and equation:
P1V1T1=P2V2T2
P1×200cm3300K=2P×150cm3T2
Cross multiply:
T2=300×150×2P200×P
=450K
or 177∘C
Don't forget to convert to ∘C
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