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Ajụjụ 1 Ripọtì
What technique is suitable for separating a binary solution of potassium chloride and potassium trioxochlorate (V)?
Akọwa Nkọwa
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.
Ajụjụ 2 Ripọtì
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?
Akọwa Nkọwa
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.
Ajụjụ 3 Ripọtì
Elements in the periodic table are arranged in the order of their
Akọwa Nkọwa
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
Ajụjụ 4 Ripọtì
Which of the following pairs cannot be represented with a chemical formula?
Akọwa Nkọwa
The pair that cannot be represented with a chemical formula is air and bronze. Air is a mixture of several gases, primarily nitrogen (N₂) and oxygen (O₂), with small amounts of other gases such as argon (Ar), carbon dioxide (CO₂), and neon (Ne). Since air is a mixture and not a pure substance, it cannot be represented by a chemical formula. Bronze, on the other hand, is an alloy composed mainly of copper (Cu) and tin (Sn) with small amounts of other metals. The composition of bronze can vary depending on the specific alloy, but it can be represented by a chemical formula such as CuSn. Sodium chloride (NaCl) is a compound composed of sodium (Na) and chlorine (Cl) in a fixed ratio of 1:1, and it can be represented by a chemical formula. Similarly, copper (Cu) and sodium chloride (NaCl) can each be represented by a chemical formula. Cu is an element, so its chemical formula is simply its symbol, while NaCl is a compound with a fixed ratio of sodium and chlorine atoms. Caustic soda (sodium hydroxide, NaOH) and washing soda (sodium carbonate, Na₂CO₃) are both compounds that can be represented by chemical formulas. NaOH consists of one sodium atom, one oxygen atom, and one hydrogen atom, while Na₂CO₃ consists of two sodium atoms, one carbon atom, and three oxygen atoms.
Ajụjụ 5 Ripọtì
The velocity, V of a gas is related to its mass, M by (k = proportionality constant)
Akọwa Nkọwa
Recall:
V = √3RTM
∴V∝1√M
V=k√M
V = kM12
Ajụjụ 6 Ripọtì
The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?
[X = 63, Y = 27, 1F = 96,500C]
Akọwa Nkọwa
For X: X2+
+ 2e−
→
X
2F = 63g
xF = 6g
x = 6×263=421F
421
F = N12.00
1F = 12421
= N63.00
1F is equivalent to N63.00.
For Y: Y3+
+ 3e−
→
Y
3F = 27g
xF = 9g
x = 3×927
= 1F
1F = N63.00
Ajụjụ 7 Ripọtì
The IUPAC nomenclature of the compound
\(\mathrm{H_3C - CH(CH_3) - CH(CH_3) - CH_2 - CH_3}\)
Ajụjụ 8 Ripọtì
2-methylprop-1-ene is an isomer of
Akọwa Nkọwa
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.
Ajụjụ 9 Ripọtì
A secondary alkanol can be oxidized to give an
Akọwa Nkọwa
A secondary alkanol is an alcohol with two carbon atoms attached to the carbon bearing the hydroxyl group (-OH). Secondary alkanols can be oxidized by a strong oxidizing agent, such as potassium dichromate (K2Cr2O7), to give an alkanone. During the oxidation process, the oxygen atom from the oxidizing agent replaces the hydroxyl group of the secondary alkanol to form a carbonyl group (C=O) in the alkanone. Since alkanones contain a carbonyl group, they are also known as ketones. Therefore, the answer to the question is alkanone, as secondary alkanols can be oxidized to form ketones.
Ajụjụ 10 Ripọtì
The shapes of water, ammonia, carbon (iv) oxide and methane are respectively
Ajụjụ 11 Ripọtì
Which important nitrogen-containing compound is produced in Haber's process?
Akọwa Nkọwa
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.
Ajụjụ 12 Ripọtì
When chlorine water is exposed to bright sunlight, the following products are formed
Ajụjụ 13 Ripọtì
The following are isoelectronic ions except
Akọwa Nkọwa
Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na+
= 10 electrons = 2, 8
Mg2+
= 10 electrons = 2,8
O2−
= 10 electrons = 2,8
Si2+
= 12 electrons = 2,8,2
⟹
Si2+
is not isoelectronic with the rest.
Ajụjụ 14 Ripọtì
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.
[H = 1, C = 12, O = 16]
Akọwa Nkọwa
The molecular formula of a compound is determined by the number of atoms of each element present in the molecule. To find the molecular formula, we need to determine the number of atoms of each element in the compound. First, we convert the percent composition to grams. For example, 40.0% carbon means 40.0 g of carbon per 100 g of compound. Then we divide the number of grams of each element by the molar mass of each element. For example, 40.0 g of carbon divided by the molar mass of carbon (12 g/mol) gives us 3.33 mol of carbon. Next, we convert the number of moles of each element to the number of atoms by multiplying the number of moles by Avogadro's number (6.022 x 10^23 atoms/mol). Finally, we balance the numbers of atoms of each element by dividing them by the smallest number of atoms of all the elements and rounding to the nearest whole number. In this case, the smallest number of atoms is 2, which is the number of hydrogen atoms. So, we divide the number of atoms of carbon and oxygen by 2 to balance the numbers of atoms of all the elements. Therefore, the molecular formula of the compound is C6H12O6.
Ajụjụ 15 Ripọtì
How many alkoxyalkanes can be obtained from the molecular formula \( \mathrm{C_4H_{10}O} \)?
Akọwa Nkọwa
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.
Ajụjụ 16 Ripọtì
A synthetic rubber is obtained from the polymerization of
Akọwa Nkọwa
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.
Ajụjụ 19 Ripọtì
Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (v) with
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
At 27°C, 58.5g of sodium chloride is present in 250cm\(^3\) of a solution. The solubility of sodium chloride at this temperature is?
(molar mass of sodium chloride = 111.0gmol\(^{-1}\))
Akọwa Nkọwa
Given the Mass of the salt = 58.5g
Volume = 250 cm3
= 0.25 dm3
Mass concentration = MassVolume
= 58.50.25
= 234 gdm−3
Solubility (in moldm−3
= 234111
= 2.11 moldm−3
≊
2.0 moldm−3
Ajụjụ 21 Ripọtì
Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?
Akọwa Nkọwa
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.
Ajụjụ 22 Ripọtì
The part of the total energy of a system that accounts for the useful work done in a system is known as
Akọwa Nkọwa
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.
Ajụjụ 23 Ripọtì
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
Ajụjụ 24 Ripọtì
Which two gases can be used for the demonstration of the fountain experiment?
Akọwa Nkọwa
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.
Ajụjụ 25 Ripọtì
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
Akọwa Nkọwa
Ionization energy and electron affinity increase across a period, and decrease down a group.
Ajụjụ 26 Ripọtì
Which of the following is a set of neutral oxides?
Ajụjụ 27 Ripọtì
The emission of two successive beta particles from the nucleus \( {}^{32}_{15}\mathrm{P} \) will produce
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
Which of the following represents the kind of bonding present in ammonium chloride?
Akọwa Nkọwa
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.
Ajụjụ 29 Ripọtì
The molecular shape and bond angle of water are respectively
Akọwa Nkọwa
The shape of water molecule = Bent/ V- shaped
The bond angle of water = 104.5°/ 105°
Ajụjụ 30 Ripọtì
The IUPAC name of the compound \(\mathrm{CF_3CHBrCl}\) is
Ajụjụ 31 Ripọtì
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?
Akọwa Nkọwa
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.
Ajụjụ 32 Ripọtì
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be
Akọwa Nkọwa
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
Ajụjụ 33 Ripọtì
When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is
Akọwa Nkọwa
The compound formed when the end alkyl groups of ethyl ethanoate are interchanged is ethyl propanoate. This is because ethyl ethanoate consists of two parts: the "ethyl" group and the "ethanoate" group. The ethyl group is a two-carbon chain, and the ethanoate group is a combination of a one-carbon chain and a carbonyl group (C=O) that is also attached to an oxygen atom. When the end alkyl groups are interchanged, the "ethyl" group is moved from the second carbon to the first carbon of the ethanoate group, and the "propanoate" group is formed. The "propanoate" group consists of a three-carbon chain and the carbonyl group. Therefore, the resulting compound is ethyl propanoate, which has a chemical formula of CH3CH2COOCH2CH3. This compound is commonly used as a flavoring agent and has a fruity odor reminiscent of pears.
Ajụjụ 34 Ripọtì
The combustion of carbon(ii)oxide in oxygen can be represented by equation.
\(2\mathrm{CO} + \mathrm{O}_2\ ?\ 2\mathrm{CO}_2\)
Calculate the volume of the resulting mixture at the end of the reaction if \(50\mathrm{cm}^3\) of carbon(ii)oxide was exploded in \(100\mathrm{cm}^3\) of oxygen
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
\(200\text{cm}^3\) of \(0.50\text{mol}/\text{dm}^3\) solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is
Akọwa Nkọwa
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.
Ajụjụ 36 Ripọtì
Which quantum divides shells into orbitals?
Akọwa Nkọwa
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".
Ajụjụ 37 Ripọtì
The oxidation state(s) of nitrogen in ammonium nitrite is/are
Akọwa Nkọwa
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.
Ajụjụ 38 Ripọtì
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:
\[ \mathrm{C_2H_4 + H_2 \rightarrow C_2H_6} \]
Akọwa Nkọwa
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.
Ajụjụ 39 Ripọtì
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]
Akọwa Nkọwa
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
Ajụjụ 40 Ripọtì
Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?
\(2HCl + CaCO_3 \rightarrow CaCl_2 + H_2O + CO_2\)
Akọwa Nkọwa
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)
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