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Ajụjụ 1 Ripọtì
Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?
Akọwa Nkọwa
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.
Ajụjụ 2 Ripọtì
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours, through molten magnesium chloride?
[1 faraday = 96500C, Mg = 24]
Akọwa Nkọwa
Current (I) = 2A; Time (t) = 2 hours = 7200 secs
Q = It
= 2 x 7200 = 14400C
1 F = 96500C
x = 14400C
x = 1440096500
= 0.15F
Mg2+
+ 2e−
→
Mg
2F →
24g
0.15F →
x
2x = 24 x 0.15
x = 24×0.152
= 1.8g
Ajụjụ 4 Ripọtì
How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?
Akọwa Nkọwa
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.
Ajụjụ 5 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 6 Ripọtì
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
Ajụjụ 7 Ripọtì
The IUPAC nomenclature of the compound
H3 C - CH(CH3 ) - CH(CH3 ) - CH2 - CH3
Ajụjụ 8 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 9 Ripọtì
Which of the following describes the chemical property of acids?
Ajụjụ 10 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 11 Ripọtì
Which of the following alkaline metals react more quickly spontaneously with water?
Akọwa Nkọwa
The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.
Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:
A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.
The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:
| Order of reactivity | Metal | Reactions with water or steam |
|---|---|---|
| most reactive | potassium (K) | very vigorous reaction with cold water |
| ↑ | sodium (Na) | vigorous reaction with cold water |
| ↓ | calcium (Ca) | less vigorous reaction with cold water |
| least reactive | magnesium (Mg) | slow reaction with cold water, vigorous with steam |
Ajụjụ 12 Ripọtì
When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called
Akọwa Nkọwa
When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.
Ajụjụ 13 Ripọtì
Which of the following could not be alkane?
Akọwa Nkọwa
An alkane is a type of hydrocarbon with only single bonds between the carbon atoms. It follows the general formula CnH2n+2, where "n" is the number of carbon atoms in the molecule. To determine whether a molecule is an alkane or not, we can calculate its molecular formula and check if it fits the general formula of alkane. Out of the given options, the third one (C7H14) cannot be an alkane. To see why, let's use the general formula of alkane, which is CnH2n+2. For C7H14 to be an alkane, it should have 2n+2 = 2(7) + 2 = 16 hydrogen atoms. However, C7H14 has only 14 hydrogen atoms, which means it does not follow the general formula of alkane. Therefore, C7H14 cannot be an alkane. The other options are as follows: - C4H10: This is butane, which is an alkane with four carbon atoms. - C5H12: This is pentane, which is an alkane with five carbon atoms. - C8H18: This is octane, which is an alkane with eight carbon atoms. In summary, the molecule C7H14 cannot be an alkane because it does not follow the general formula of alkane, while the other options are all examples of alkanes.
Ajụjụ 14 Ripọtì
The IUPAC name for CH3 CH2 COOCH2 CH3 is
Akọwa Nkọwa
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.
Ajụjụ 15 Ripọtì
The IUPAC name of the compound CF3 CHBrCl is
Ajụjụ 16 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 17 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 18 Ripọtì
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.
Akọwa Nkọwa
Z = 1s2
2s2
2p6
3s2
3p1
?
We have Z3+
and SO2?4
The reaction : Z3+
+ SO2?4
?
Z2
(SO4
)3
.
Ajụjụ 19 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 20 Ripọtì
What volume of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.
[Na = 23, N = 14, O = 16]
Akọwa Nkọwa
To calculate the volume of a solution, we need to use the formula: moles of solute = concentration x volume First, let's find the number of moles of sodium trioxonitrate (V) in 5g of the solute. The molar mass of NaNO3 is: Na = 23 N = 14 3 x O = 3 x 16 = 48 Molar mass = 23 + 14 + 48 = 85 g/mol The number of moles of NaNO3 in 5g is: moles = mass / molar mass = 5 / 85 = 0.0588 moles Now, we can use the formula above to find the volume of the solution: moles of solute = concentration x volume volume = moles of solute / concentration volume = 0.0588 moles / 0.100 M volume = 0.588 litres Therefore, the correct answer is 0.588 litres of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.
Ajụjụ 21 Ripọtì
Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (v) with
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
The shapes of water, ammonia, carbon (iv) oxide and methane are respectively
Ajụjụ 23 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 24 Ripọtì
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
Akọwa Nkọwa
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).
Ajụjụ 25 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 26 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 27 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 28 Ripọtì
Hydrocarbons which will react with Tollen's reagent conform to the general formula
Ajụjụ 29 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 30 Ripọtì
Consider the equation below:
Cr2 O2−7 + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2 O.
The oxidation number of chromium changes from
Akọwa Nkọwa
Cr2
O2−7
+ 6Fe2+
+ 14H+
→
2Cr3+
+ 6Fe3+
+ 7H2
O
The oxidation of Cr in Cr2
O2−7
:
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2 ⟹
2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3
Ajụjụ 31 Ripọtì
Consider the reaction
A(s) + 2B(g) → 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 32 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 33 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 34 Ripọtì
200cm3 of 0.50mol/dm3 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 35 Ripọtì
How many alkoxyalkanes can be obtained from the molecular formula C4 H10 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 36 Ripọtì
Which of the following statements about catalyst is false?
Akọwa Nkọwa
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.
Ajụjụ 37 Ripọtì
Which of the following pollutants will lead to the depletion of ozone layer?
Akọwa Nkọwa
The pollutant that leads to the depletion of the ozone layer is chlorofluorocarbon (CFCs). CFCs are man-made chemicals that were widely used in the past as refrigerants, solvents, and propellants. When CFCs are released into the atmosphere, they rise into the stratosphere, where they come into contact with ozone molecules. The chlorine atoms in CFCs react with ozone, breaking apart the ozone molecules and causing a reduction in the overall amount of ozone in the stratosphere. This process continues until all of the ozone-depleting chlorine atoms have been depleted. The resulting decrease in ozone in the stratosphere leads to an increase in the amount of harmful ultraviolet radiation that reaches the Earth's surface, which can have negative impacts on human health and the environment.
Ajụjụ 38 Ripọtì
If the cost of electricity required to discharge 10g of an ion X3+ is N20.00, how much would it cost to discharge 6g of ion Y2+ ?
[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 39 Ripọtì
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
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)
Ajụjụ 40 Ripọtì
Hydrogen bond is a sort of
Akọwa Nkọwa
Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine.
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