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Ibeere 2 Ìròyìn
The electronic configuration of element Z is 1s2 2s2 2p6 3s2 3p1 . What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.
Awọn alaye Idahun
Z = 1s2
2s2
2p6
3s2
3p1
?
We have Z3+
and SO2?4
The reaction : Z3+
+ SO2?4
?
Z2
(SO4
)3
.
Ibeere 3 Ìròyìn
Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?
2HCl + CaCO3 → CaCl2 + H2 O + CO2
Awọn alaye Idahun
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)
Ibeere 5 Ìròyìn
Which of the following does not support the fact that air is a mixture?
Awọn alaye Idahun
The option that does not support the fact that air is a mixture is "the constituents of air are in a fixed proportion by mass". Air is a mixture of different gases, primarily nitrogen (78%) and oxygen (21%), with small amounts of other gases such as carbon dioxide, argon, and neon. The proportion of each gas in air is not fixed and can vary depending on the location and other factors. For example, the amount of carbon dioxide in air can increase in areas with high levels of pollution, while the proportion of oxygen can decrease at high altitudes. Therefore, the composition of air is not in a fixed proportion by mass. On the other hand, the fact that air cannot be represented with a chemical formula and its constituents can be separated by physical means support the fact that air is a mixture. A chemical formula represents a pure substance, and since air is a mixture of gases, it cannot be represented by a single formula. Air can be separated into its individual components through physical means such as distillation or filtration, which is a characteristic of mixtures.
Ibeere 6 Ìròyìn
Hydrogen diffused through a porous plug
Awọn alaye Idahun
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.
Ibeere 7 Ìròyìn
Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?
Awọn alaye Idahun
The process involved in the turning of starch iodide paper blue-black by chlorine gas is option number 3: chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color. When chlorine gas comes in contact with iodide ions on the starch iodide paper, it oxidizes the iodide ion to form iodine. The iodine that is produced in this reaction is then able to react with the starch present on the paper to form a blue-black complex. This blue-black complex is formed due to the arrangement of the starch molecules and the iodine atoms in a way that causes them to absorb light at a specific wavelength, giving the blue-black color. Therefore, the blue-black color that is observed on the starch iodide paper is due to the reaction between iodine and starch, which is made possible by the oxidation of iodide ions by chlorine gas.
Ibeere 8 Ìròyìn
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
Awọn alaye Idahun
Ibeere 10 Ìròyìn
Which important nitrogen-containing compound is produced in Haber's process?
Awọn alaye Idahun
The important nitrogen-containing compound that is produced in Haber's process is NH3, which is also known as ammonia. Haber's process is a chemical process used to produce ammonia by reacting nitrogen gas (N2) and hydrogen gas (H2) under high pressure and temperature in the presence of an iron catalyst. The reaction between nitrogen and hydrogen produces ammonia as the main product, along with some nitrogen and hydrogen gases that do not react. NH3 is an important compound that is widely used in industry for the production of fertilizers, plastics, and other chemical products. It is also used as a cleaning agent, a refrigerant, and a fuel for engines. In addition, NH3 is an essential compound for life, as it is a key component of amino acids, which are the building blocks of proteins.
Ibeere 11 Ìròyìn
The emission of two successive beta particles from the nucleus 3215P will produce
Awọn alaye Idahun
Ibeere 12 Ìròyìn
The part of the total energy of a system that accounts for the useful work done in a system is known as
Awọn alaye Idahun
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.
Ibeere 13 Ìròyìn
Which quantum divides shells into orbitals?
Awọn alaye Idahun
The quantum that divides shells into orbitals is the "Azimuthal" quantum number, also known as the "angular momentum" quantum number. The azimuthal quantum number determines the shape of an electron's orbital, which is a region in space where there is a high probability of finding an electron. It describes the angular momentum of an electron in an atom and the number of subshells within a given shell. Each subshell is associated with a specific shape, and can hold a certain number of electrons. The azimuthal quantum number is represented by the letter "l" and can have integer values ranging from 0 to (n-1), where "n" is the principal quantum number. Each value of "l" corresponds to a different subshell shape: - l = 0 corresponds to an "s" subshell, which is spherical in shape. - l = 1 corresponds to a "p" subshell, which has a dumbbell shape with two lobes. - l = 2 corresponds to a "d" subshell, which has a more complex shape with four lobes and a doughnut-like ring. - l = 3 corresponds to an "f" subshell, which has an even more complex shape with eight lobes. The number of orbitals within a subshell is equal to 2l+1. For example, a "p" subshell (l = 1) has three orbitals (2l+1 = 3), which are labeled as "px", "py", and "pz". In summary, the azimuthal quantum number determines the shape of the electron's orbital and the number of subshells within a given shell, and it is represented by the letter "l".
Ibeere 14 Ìròyìn
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
Awọn alaye Idahun
Ionization energy and electron affinity increase across a period, and decrease down a group.
Ibeere 15 Ìròyìn
200cm3 of 0.50mol/dm3 solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is
Awọn alaye Idahun
To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.
Ibeere 16 Ìròyìn
The heat of formation of ethene, C2 H4 is 50 kJmol−1 , and that of ethane, C2 H6 is -82kJmol−1 . Calculate the heat evolved in the process:
C2 H4 + H2 → C2 H6
Awọn alaye Idahun
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.
Ibeere 17 Ìròyìn
Elements in the periodic table are arranged in the order of their
Awọn alaye Idahun
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
Ibeere 18 Ìròyìn
Consider the reaction
A(s) + 2B(g) → 2C(aq) + D(g)
What will be the effect of a decrease in pressure on the reaction?
Awọn alaye Idahun
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.
Ibeere 19 Ìròyìn
In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm3 of 1.25 molar tetraoxosulphate (vi) acid?
Awọn alaye Idahun
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
Ibeere 20 Ìròyìn
Burning magnesium ribbon in air removes which of the following
(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?
Awọn alaye Idahun
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.
Ibeere 21 Ìròyìn
The shapes of water, ammonia, carbon (iv) oxide and methane are respectively
Awọn alaye Idahun
Ibeere 22 Ìròyìn
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
Awọn alaye Idahun
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).
Ibeere 23 Ìròyìn
What technique is suitable for separating a binary solution of potassium chloride and potassium trioxochlorate (V)?
Awọn alaye Idahun
Fractional crystallization is the most suitable technique for separating a binary solution of potassium chloride and potassium trioxochlorate (V). This is because fractional crystallization is a process that separates a mixture of substances based on their solubility in a solvent at a particular temperature. In this case, potassium chloride and potassium trioxochlorate (V) have different solubilities in a solvent such as water at different temperatures. By carefully controlling the temperature, the solubility of each compound can be selectively increased or decreased, allowing them to be separated by crystallization. The less soluble compound will form crystals first and can be separated from the more soluble compound, which remains in the solution. Therefore, fractional crystallization can be used to separate potassium chloride and potassium trioxochlorate (V) in a binary solution.
Ibeere 24 Ìròyìn
Which of the following represents the kind of bonding present in ammonium chloride?
Awọn alaye Idahun
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.
Ibeere 25 Ìròyìn
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be
Awọn alaye Idahun
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
Ibeere 27 Ìròyìn
The IUPAC nomenclature of the compound
H3 C - CH(CH3 ) - CH(CH3 ) - CH2 - CH3
Awọn alaye Idahun
Ibeere 28 Ìròyìn
The IUPAC name for CH3 CH2 COOCH2 CH3 is
Awọn alaye Idahun
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.
Ibeere 29 Ìròyìn
In the reaction:
M + N → P
ΔH = +Q kJWhich of the following would increase the concentration of the product?
Awọn alaye Idahun
Increasing the temperature would increase the concentration of the product, P. The reaction rate, or the speed at which the reaction occurs, is influenced by temperature. An increase in temperature raises the kinetic energy of the reacting molecules, making it easier for them to collide and react. This leads to a higher rate of reaction and a higher concentration of the product, P. Adding a suitable catalyst can also increase the reaction rate, but it does not directly affect the concentration of the product. Increasing the concentration of P does not affect the reaction itself, but is a result of the reaction having taken place. Decreasing the temperature would slow down the reaction rate and reduce the concentration of the product.
Ibeere 30 Ìròyìn
A synthetic rubber is obtained from the polymerization of
Awọn alaye Idahun
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.
Ibeere 31 Ìròyìn
Hydrogen bond is a sort of
Awọn alaye Idahun
Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine.
Ibeere 32 Ìròyìn
A cell shorthand notation can be written as A / A+ // B2+ /B. The double slash in the notation represents the
Awọn alaye Idahun
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.
Ibeere 33 Ìròyìn
How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?
Awọn alaye Idahun
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.
Ibeere 34 Ìròyìn
For the general equation of the nature
XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is
Awọn alaye Idahun
The expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is: Kc = [R]m[S]n / [P]x[Q]y where Kc is the equilibrium constant, [R] and [S] are the concentrations of the products, and [P] and [Q] are the concentrations of the reactants, all raised to the stoichiometric coefficients (m, n, x, y) in the balanced equation. This equation is known as the equilibrium constant expression and it represents the ratio of the concentrations of the products and reactants at equilibrium for a particular chemical reaction. The equilibrium constant is a measure of how far a reaction proceeds towards completion, with a larger value indicating a greater extent of reaction. The equilibrium constant expression is derived from the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant expression represents the ratio of the rate constants for these two reactions. Therefore, the correct expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is Kc = [R]m[S]n / [P]x[Q]y.
Ibeere 36 Ìròyìn
The combustion of carbon(ii)oxide in oxygen can be represented by equation.
2CO + O2 ? 2CO2
Calculate the volume of the resulting mixture at the end of the reaction if 50cm3 of carbon(ii)oxide was exploded in 100cm3 of oxygen
Awọn alaye Idahun
Ibeere 37 Ìròyìn
By what amount must the temperature of 200cm3 of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm3 (Assume ideality)
Awọn alaye Idahun
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
Ibeere 38 Ìròyìn
A solution X, on mixing with AgNO3 solution gives a white precipitate soluble in aqueous NH3 , a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain
Awọn alaye Idahun
Ibeere 39 Ìròyìn
X is a substance which liberates CO2 on treatment with concentrated H2 SO4 . A warm solution of X can decolorize acidified KMnO4 . X is
Awọn alaye Idahun
It should be noted that for X to liberate CO2
, X must be a carbonate or an oxalate. Since X decolorizes KMnO4
, X must be an oxalate.
Therefore, X is H2
C2
O4
.
Ibeere 40 Ìròyìn
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
Awọn alaye Idahun
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.
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