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Ibeere 1 Ìròyìn
The principle which states that no two electrons in the same orbitals of an atom have same value for all four quantum numbers is the
Awọn alaye Idahun
The principle that states that no two electrons in the same orbitals of an atom can have the same value for all four quantum numbers is the Pauli Exclusion Principle.
To understand this principle, it's important to know a bit about the structure of an atom and what quantum numbers are:
Quantum Numbers:
1. **Principal Quantum Number (n):** This describes the energy level or shell of the electron.
2. **Angular Momentum Quantum Number (l):** This describes the subshell or shape of the orbital (s, p, d, f...).
3. **Magnetic Quantum Number (ml):** This describes the specific orbital within a subshell where the electron is located.
4. **Spin Quantum Number (ms):** This describes the spin direction of the electron, which can be either +1/2 or -1/2.
The Pauli Exclusion Principle asserts that each electron in an atom has a unique set of these four quantum numbers. While electrons can share the first three quantum numbers if they are in the same orbital (meaning they share the same energy level, the same subshell, and the same specific orbital within that subshell), they must have different Spin Quantum Numbers. This means that in any given orbital, one electron can have a spin of +1/2 and the other must have a spin of -1/2. This principle is fundamental in explaining the electronic structure of atoms and, consequently, the behavior and properties of elements.
Ibeere 2 Ìròyìn
If a salt weighs 2g and upon exposure to the atmosphere weighs 1.5g, this is as a result of
Awọn alaye Idahun
The observation that a salt initially weighs 2g, but reduces to 1.5g after exposure to the atmosphere is primarily due to the process called efflorescence.
Efflorescence occurs when a salt loses water molecules from its crystal structure when exposed to air, which is why the weight of the salt decreases over time. This loss of water is because some salts contain water of crystallization, and when such salts are exposed to the atmosphere, they can release this water, leading to a reduction in weight.
In this specific case, the salt has lost 0.5g of water, leading to the weight change from 2g to 1.5g. This process is different from hygroscopy, which involves absorbing moisture from the atmosphere, or deliquescence, where a substance absorbs moisture and eventually dissolves in it. It's also not related to effervescence, which is the escape of gas from an aqueous solution.
Ibeere 3 Ìròyìn
What method is suitable for the separation of gases present in air?
Awọn alaye Idahun
The suitable method for the separation of gases present in air is the fractional distillation of liquid air. This method is used due to the differing boiling points of the gases present in the air. Let me explain this in simple terms:
Air is a mixture of different gases, primarily nitrogen, oxygen, and argon, along with small amounts of other gases like carbon dioxide, neon, and krypton. Each of these gases turns into a liquid at different temperatures.
The process begins by cooling the air until it becomes a liquid. This is done at very low temperatures (around -200 degrees Celsius). Once the air is in liquid form, it is slowly warmed up in a distillation column. As it heats up, each gas boils off or evaporates at its respective boiling point and can be collected separately.
For example, nitrogen, which has a boiling point of about -196 degrees Celsius, will evaporate first and can be collected at the top of the distillation column. Following nitrogen, oxygen will evaporate at its boiling point of around -183 degrees Celsius. Finally, argon and other gases will do so at their respective temperatures.
In summary, fractional distillation of liquid air is effective because it takes advantage of the different boiling points to separate each gas from the air mixture.
Ibeere 4 Ìròyìn
In the treatment of water for municipal supply, chlorine is used to
Awọn alaye Idahun
In the treatment of water for municipal supply, chlorine is used to kill germs. This process is known as chlorination. Chlorine is a very effective disinfectant and is used to eliminate harmful microorganisms such as bacteria, viruses, and protozoans that may be present in the water. By doing so, chlorine helps to ensure that the water is safe for human consumption and protects public health by preventing waterborne diseases. It is important to note that **chlorine is not used to prevent tooth decay, prevent goitre, or to remove colour or odour** in water treatment for municipal supply.
Ibeere 5 Ìròyìn
An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [ C =12, H = 1, N = 14, O = 16].
Awọn alaye Idahun
To find the molecular formula of the compound, follow these steps:
1. Determine the Empirical Formula:
Start by assuming you have 100 grams of the compound. This means you have:
Now, convert these masses to moles using their atomic masses (C = 12, H = 1, N = 14, O = 16):
Next, divide each by the smallest number of moles to get the simplest ratio:
This gives us the empirical formula: C5H7NO2.
2. Determine the Molecular Formula:
The molecular formula is a multiple of the empirical formula. To determine this multiple, we need to find the empirical formula mass and compare it with the molar mass derived from the given vapor density.
Calculate the empirical formula mass:
The molar mass can be calculated from the vapor density:
Now, find the ratio of the molar mass to the empirical formula mass:
This ratio is approximately 1, indicating the molecular formula is the same as the empirical formula. Since empirical formulas typically should perfectly match the atomic proportions we derive from experiments, our calculations regarding the assumptions on the vapour and empirical formula mass remains our best match.
Therefore, the molecular formula is C5H7NO2.
Ibeere 6 Ìròyìn
H2 SO4
C2 H5 OH → C2 H4
1700 C
The reaction above illustrates
Awọn alaye Idahun
This reaction illustrates dehydration. In chemistry, dehydration refers to the process of removing water (H2O) from a compound. Let's break down the given reaction to understand this better.
The provided chemical equation is:
C2H5OH → C2H4 + H2O
This equation indicates that ethanol (C2H5OH) is being transformed into ethylene (C2H4) with the production of water (H2O).
The process involves the breaking of bonds in ethanol and the removal of a water molecule, as follows:
This reaction is typically carried out under certain conditions, in this case at a high temperature of 1700°C, to facilitate the dehydration process.
Therefore, this is indeed a dehydration reaction as it involves converting ethanol into ethylene by removing water.
Ibeere 7 Ìròyìn
The Van der waals forces of attraction operates between
Awọn alaye Idahun
The Van der Waals forces of attraction operate between molecules. These are weak forces of attraction that occur due to momentary changes in the electron distribution within molecules. Here's a simple explanation:
Therefore, the forces can affect the physical properties of molecular compounds, such as boiling and melting points, but do not generally involve charged particles like cations or anions.
Ibeere 8 Ìròyìn
The stability of atomic nucleus is determined by ratio of
Awọn alaye Idahun
The stability of an atomic nucleus is primarily determined by the neutron/proton ratio. This refers to the number of neutrons in relation to the number of protons within the nucleus. Let's break down why this ratio is crucial for nuclear stability:
The right balance between the number of neutrons and protons helps in achieving nuclear stability.
An imbalance in this ratio often results in an unstable nucleus, leading to radioactive decay as the nucleus attempts to reach a more stable form. This is why the neutron/proton ratio is a fundamental factor in the stability of the atomic nucleus.
Ibeere 9 Ìròyìn
The term strong and weak acids is used to indicate the
Awọn alaye Idahun
The terms strong and weak acids are used to indicate the extent of ionization of an acid. This means how completely an acid dissociates into its ions in water.
Strong acids completely dissociate in water. This means that nearly all the acid molecules break down into positive hydrogen ions (H+) and their respective anions. Examples include hydrochloric acid (HCl), sulfuric acid (H2SO4), and nitric acid (HNO3).
Weak acids, on the other hand, only partially dissociate in water. This means that only a small fraction of the acid molecules break down into ions. Most of the acid remains in its molecular form. An example of a weak acid is acetic acid (CH3COOH), which is found in vinegar.
Therefore, the strength of an acid in terms of its classification as strong or weak is about how fully it dissociates into ions in an aqueous solution, not about the number of H+ ions or the strength of its action on substances.
Ibeere 10 Ìròyìn
An example of a physical change is
Awọn alaye Idahun
A physical change involves a change in the physical properties of a substance, without a change in its chemical composition. This means that the substance remains the same at the molecular level, despite how it might appear differently.
An example of a physical change from the given options is the liquefaction of liquids. In this process, a substance transitions from a solid or gas to a liquid state. This change is purely physical because the molecular structure of the substance does not change; only its state or form does. Importantly, such a change is usually reversible, meaning the substance can return to its original state. For instance, water can change into ice (frozen) or steam (vapor), and can still revert back to liquid water.
On the other hand, the other options involve chemical changes, where the original substances undergo chemical reactions to form new substances with different properties, thus altering the molecular structure depending on the option.
Ibeere 11 Ìròyìn
The term that is not associated with petroleum industry is ?
Awọn alaye Idahun
Cracking, saponification and polymerization are all terminologies associated with the petroleum industry but fermentation is associated with the brewery industry.
Cracking is a chemical process that breaks down heavy hydrocarbon molecules into lighter, more useful ones.
Saponification is a chemical reaction that converts fats and oils into soap. It's not directly involved in petroleum, but it can be used to analyze petroleum products.
Polymerization is a process in the petroleum industry that converts light olefin gases into higher molecular weight hydrocarbons.
Fermentation is the process in which a substance breaks down into a simpler substance. Microorganisms like yeast and bacteria usually play a role in the fermentation process, creating beer, wine, bread,yogurt and other foods.
Ibeere 12 Ìròyìn
A type of isomerism that ClCH=CHCl can exhibit is
Awọn alaye Idahun
ClCH=CHCl can exhibit geometrical isomerism and positional isomerism. ClCH=CHCl can exhibit positional isomerism because the positions of the functional groups or substituent atoms are different. Positional isomerism occurs when compounds with the same molecular formula have different properties due to the difference in the position of a functional group, multiple bond, or branched chain.
Ibeere 13 Ìròyìn
A factor that does not affect the rate of a chemical reaction is
Awọn alaye Idahun
In evaluating the factors that affect the rate of a chemical reaction, we can look at each of the possible influences: surface area, temperature, volume, and catalyst.
Surface Area: When you increase the surface area of reactants, it allows more particles to collide with each other per unit of time, which in turn increases the rate of reaction. Imagine smaller particles like powders reacting faster than larger chunks because they have a greater surface exposed to the other reactants.
Temperature: Increasing the temperature usually increases the rate of reaction. Higher temperatures cause particles to move faster, increasing the energy of collisions, and therefore increasing the chance of successful reactions.
Catalyst: A catalyst is a substance that increases the rate of a chemical reaction without being consumed by it. It lowers the activation energy needed for the reaction to occur, thus allowing it to proceed faster.
Volume: The volume of the container or the amount of space in which a reaction occurs generally does not directly affect the rate of the reaction. While changing the volume can alter pressure or concentration in gaseous reactions, which in turn affects the rate, the volume itself is not a direct factor affecting reaction rate.
Therefore, the factor that does not directly affect the rate of a chemical reaction is volume. It indirectly affects reaction rates by altering concentration or pressure in certain reaction conditions, but it is not a direct influencing factor on its own.
Ibeere 14 Ìròyìn
H2 S(g) + Cl2 (g) → 2HCl(g) + S(s)
What is the change in oxidation state of sulphur from reactant to product?
Awọn alaye Idahun
To determine the change in oxidation state of sulfur, follow these steps:
In the given reaction:
H2S(g) + Cl2(g) → 2HCl(g) + S(s)
We observe:
Thus, the change in oxidation state of sulfur when moving from the reactants to the products is from **-2** to **0**. This indicates that sulfur is being oxidized.
The correct answer is that the oxidation state of sulfur changes from **-2 to 0**.
Ibeere 15 Ìròyìn
Awọn alaye Idahun
In the Contact Process, the catalyst used for the conversion of sulphur(IV) oxide (SO2) to sulphur(VI) oxide (SO3) is vanadium(V) oxide, also chemically represented as V2O5. This catalyst is preferred because it is more cost-effective and significantly more durable under reaction conditions than other catalysts such as platinum. Moreover, while platinum is also an effective catalyst, it is prone to poisoning by impurities that may be present in the reaction mixture. Vanadium(V) oxide, on the other hand, offers a better balance of efficiency, cost, and durability, making it the catalyst of choice in industrial applications of the Contact Process.
Ibeere 16 Ìròyìn
Strong acids can be distinguished from weak acids by any of the following methods, EXCEPT
Awọn alaye Idahun
To distinguish between strong acids and weak acids, we can employ several methods based on their chemical properties:
Conductivity Measurement: Strong acids dissociate completely in water, releasing more ions. Because ion concentration is directly related to electrical conductivity, strong acids exhibit higher conductivity than weak acids, which only partially dissociate.
Litmus Paper: This method helps determine if a solution is acidic or basic but does not provide detailed information about the strength (strong or weak) of an acid. Both strong and weak acids turn blue litmus red. Therefore, **litmus paper cannot effectively distinguish between a strong and a weak acid.**
Measurement of pH: Strong acids have a lower pH because they fully dissociate to release more hydrogen ions (H+), whereas weak acids have a relatively higher pH as they do not dissociate completely. Thus, pH measurement can distinguish the extent of acidity.
Measurement of Heat of Reaction: The heat of reaction can give insights into the strength of an acid because it involves the degree of ionization and the energetics associated with it. A strong acid will exhibit a different calorimetric response compared to a weak acid.
In summary, **litmus paper is not suitable for distinguishing between a strong and a weak acid**, as it only indicates acidity but does not reveal the strength of the acid.
Ibeere 17 Ìròyìn
Fog is a colloid in which
Awọn alaye Idahun
**Fog** is a type of colloid, which is a mixture where very small particles of one substance are evenly distributed throughout another substance. In the case of fog, it consists of tiny **liquid droplets** that are dispersed in a **gas**. Specifically, these are tiny droplets of water suspended in the air. When you walk through fog, you are essentially walking through air that contains these minute water droplets.
Thus, the correct description of fog as a colloid is that it consists of **liquid particles dispersed in a gas medium**. The liquid here is water, and the gas is air.
Ibeere 18 Ìròyìn
At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that
Awọn alaye Idahun
To solve the problem, we can use **Graham's law of effusion**. This law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is represented as:
Rate of diffusion of Gas X / Rate of diffusion of Gas Y = sqrt(Molar mass of Gas Y / Molar mass of Gas X)
According to the given information, gas X diffuses **twice as fast** as gas Y. This implies:
2 = sqrt(Molar mass of Gas Y / Molar mass of Gas X)
To eliminate the square root, square both sides of the equation:
(2)^2 = Molar mass of Gas Y / Molar mass of Gas X
This simplifies to:
4 = Molar mass of Gas Y / Molar mass of Gas X
Rearranging the equation, we find:
Molar mass of Gas Y = 4 * Molar mass of Gas X
This means that **Gas Y is four times as heavy as Gas X**. Therefore, the correct statement is:
Ibeere 19 Ìròyìn
The combustion of candle under limited supply of air forms
Awọn alaye Idahun
When a candle burns under a limited supply of air, it doesn't get enough oxygen to completely burn the hydrocarbons in the wax. In complete combustion (with enough air), the candle would ideally produce water (H2O) and carbon dioxide (CO2). However, under limited air supply, the process is incomplete and results in the formation of soot and carbon monoxide (CO).
Here's why:
In summary, under limited air conditions, the combustion of a candle primarily forms soot and carbon monoxide (CO).
Ibeere 20 Ìròyìn
The shape of ammonia molecule is
Awọn alaye Idahun
The shape of the ammonia molecule (NH3) is trigonal pyramidal. To understand why, let's explore the electron and molecular geometry using a simple explanation:
Ammonia consists of one nitrogen (N) atom bonded to three hydrogen (H) atoms. The nitrogen atom has five valence electrons requiring three more electrons to complete its octet. These are acquired by forming covalent bonds with three hydrogen atoms. In addition to the three bonding pairs, there is one lone pair of electrons on the nitrogen atom.
According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, electron pairs, including bonding pairs and lone pairs, repel each other and arrange themselves as far apart as possible to minimize repulsion. In ammonia:
The presence of the lone pair on nitrogen creates a slight distortion, causing the molecule's shape to be trigonal pyramidal rather than perfectly tetrahedral. The lone pair occupies more space and pushes the hydrogen atoms slightly closer together. This results in a pyramidal shape, with nitrogen at the apex, and the three hydrogen atoms forming the base of the pyramid.
The trigonal pyramidal shape of ammonia is a result of this molecular geometry, not to be confused with any of the other options like V-shaped, tetrahedral, or co-planar.
Ibeere 21 Ìròyìn
Benzene formed nitrobenzene at temperature of 600 C when it reacts with mixture of concentrated trioxonitrate(V) acid and concentrated
Awọn alaye Idahun
The reaction described is the nitration of benzene to form nitrobenzene. This is an example of an electrophilic aromatic substitution reaction. **Nitration** involves replacing a hydrogen atom on a benzene ring with a nitro group (NO2). This reaction requires a nitrating mixture composed of concentrated nitric acid (trioxonitrate(V) acid) and concentrated sulfuric acid (tetraoxosulphate(VI) acid). Let me explain why:
Nitration is typically carried out using a mixture of **concentrated nitric acid and concentrated sulfuric acid** at a temperature of around **60°C**. The role of sulfuric acid in this mixture is to act as a catalyst and a dehydrating agent. It helps generate the nitronium ion (NO2+), which is the active electrophile that attacks the benzene ring.
Here's a simplified mechanism for this reaction:
None of the other options listed (hydrochloric acid, phosphoric acid, and hydrogen iodide) contain the necessary combination of properties to generate the nitronium ion and facilitate the nitration of benzene.
Therefore, the correct mixture to carry out the nitration of benzene, forming nitrobenzene at a temperature of 60°C, is a combination of **concentrated nitric acid and concentrated sulfuric acid (tetraoxosulphate(VI) acid)**.
Ibeere 22 Ìròyìn
What is the vapour density of 560cm3 of a gas that weighs 0.4g at s.t.p?
[Molar Volume of gas at s.t.p = 22.4 dm3 ]
Awọn alaye Idahun
To find the vapour density of a gas, you can use the formula:
Vapour density = (Molar mass of gas) / 2
However, first, we need to determine the molar mass of the gas. One can find the molar mass using the given data:
We know that at standard temperature and pressure (s.t.p.), 1 mole of any gas occupies 22.4 dm3. We need to convert the volume from cm3 to dm3 because the molar volume is given in dm3:
560 cm3 = 0.560 dm3
Now, let's find the number of moles in 0.560 dm3:
The number of moles (n) = Volume of gas (dm3) / Molar volume at s.t.p. (dm3/mol)
n = 0.560 dm3 / 22.4 dm3/mol
n = 0.025 moles
Given that the mass of the gas is 0.4 grams, we can find the molar mass by using the relation:
Molar Mass = Mass / Number of Moles
Molar Mass = 0.4 g / 0.025 moles
Molar Mass = 16 g/mol
Now that we have the molar mass, we can find the vapour density:
Vapour density = Molar mass / 2
Vapour density = 16 g/mol / 2
Vapour density = 8.0
Hence, the vapour density of the gas is 8.0.
Ibeere 23 Ìròyìn
Alkylation of benzene is catalyzed by
Awọn alaye Idahun
Alkylation of benzene is a part of a reaction class called **Friedel-Crafts alkylation**. In this reaction, an alkyl group is transferred to the aromatic benzene ring, making it a more complex molecule. The catalyst used in this process is **aluminium chloride (AlCl3)**.
Here's how the reaction typically works:
In contrast, the other options wouldn't effectively catalyze alkylation of benzene for the following reasons:
Therefore, **aluminium chloride** is the catalyst used for the alkylation of benzene in Friedel-Crafts reactions.
Ibeere 24 Ìròyìn
The amount of water a substance chemically combined with is called water of
Awọn alaye Idahun
The amount of water that is chemically combined with a substance is referred to as water of crystallization. This is the water present in the crystalline form of a compound, necessary to maintain the structure of the crystals.
When certain substances crystallize from an aqueous solution, they incorporate a specific amount of water molecules into their crystal lattice structure. These water molecules are an integral part of the crystal and often affect its color, stability, and solubility. The water is combined in stoichiometric amounts, which means it is present in a fixed ratio relative to the rest of the molecule.
An example of this is copper(II) sulfate pentahydrate, which consists of copper(II) sulfate combined with five molecules of water per formula unit, represented as CuSO4·5H2O.
Ibeere 25 Ìròyìn
| COMPOUND | S | T | U | V | W |
| FORMULA | ROR' | RCOOH' | RCOR' | ROH' | RCOOR' |
From the table above, which of these two compounds can form functional group isomers?
Awọn alaye Idahun
ROH' and ROR' can form functional group isomers because they are the functional groups of alcohols and ethers, respectively.
Ethers have a pair of alkyl or aromatic groups attached to a linking oxygen atom. ROH is the functional group of alcohols, which are derivatives of water with one hydrogen atom replaced by an alkyl group.
Alcohols (ROH) and ethers (ROR') can form functional group isomers because they have the same chemical formula but different functional groups. E.g CH3 CH2 OH and CH3 OCH3
Ibeere 26 Ìròyìn
Determine the half-life of a first order reaction with constant 4.5 x 10−3 sec−1 .
Awọn alaye Idahun
To determine the half-life of a first-order reaction, you can use the formula:
Half-life (\(t_{1/2}\)) = \(\frac{0.693}{k}\)
where \(k\) is the rate constant of the reaction. For the given problem, the rate constant (\(k\)) is 4.5 x 10-3 s-1.
Substituting the value of \(k\) into the formula, we have:
\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}}\)
Perform the division:
\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}} \approx 154\) s
Therefore, the half-life of the reaction is 154 seconds.
Ibeere 27 Ìròyìn
The ions responsible for permanent hardness in water are sulphates of
Awọn alaye Idahun
Permanent hardness in water is mainly caused by the presence of certain metal ions, specifically the **sulfates (SO₄²⁻)** and **chlorides (Cl⁻)** of calcium (Ca) and magnesium (Mg). These compounds do not precipitate out when the water is boiled, which means they remain dissolved and continue to contribute to the hardness of the water.
Among the options you provided, the ions responsible for permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. The presence of calcium sulfate (CaSO₄) and magnesium sulfate (MgSO₄) in water keeps it hard.
When compared to temporary hardness, which can be removed by boiling the water to precipitate bicarbonates, **permanent hardness cannot be removed by boiling**. Instead, methods such as ion exchange or the use of water softeners are required to remove these ions from the water.
In summary, the ions causing permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. These ions remain dissolved and continue to make the water hard, despite boiling.
Ibeere 28 Ìròyìn
The constituent of petroleum fraction used in surfacing road is
Awọn alaye Idahun
Among the options listed, the constituent of petroleum used in surfacing roads is bitumen. Bitumen, also known as asphalt, is a sticky, black, and highly viscous liquid or semi-solid form of petroleum. It is the last fraction obtained when crude oil is distilled and is often left over after the lighter components are extracted.
Reasons why bitumen is used for road surfacing:
Due to these properties, bitumen is extensively used in road construction and surfacing, ensuring roads are durable, smooth, and safe for travel.
Ibeere 29 Ìròyìn
Boyle's law can be expressed mathematically as
Awọn alaye Idahun
Boyle's Law describes the relationship between the volume and pressure of a given amount of gas held at a constant temperature. It states that the pressure of a gas is inversely proportional to its volume. In simpler terms, if you decrease the volume of a gas, its pressure increases, provided the temperature remains constant, and vice versa.
The mathematical expression of Boyle's Law is PV = K, where:
This relationship implies that if you multiply the pressure by the volume, the result will always be the same constant as long as no other variables are changed. This is the classic formulation of Boyle's Law, illustrating the inverse relationship between pressure and volume for a gas at constant temperature.
Ibeere 30 Ìròyìn
C2 H4(g) + 3O2(g) → 2CO2(g) + 2H2 O(g)
The above equation represents the combustion of ethene.If 10cm3 of ethene is burnt in 50cm3 of oxygen, what would be the volume of oxygen that would remain at the end of the reaction?
Awọn alaye Idahun
Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.
C2 H4(g) + 3O2(g) → 2CO2(g) + 2H2 O(g)
1 mole : 3 moles
Total volume required: 10 cm3 50 cm3
Reacted Volume: 10 cm3 30 cm3
Residual volume: 0 (50 - 30) = 20 cm3
Ibeere 31 Ìròyìn
An example of highly unsaturated hydrocarbon is
Awọn alaye Idahun
To determine a highly unsaturated hydrocarbon, we must first understand the concept of saturation in hydrocarbons. **Saturated hydrocarbons** are compounds that contain the maximum possible number of hydrogen atoms, single-bonded to carbon atoms, and they are alkanes. **Unsaturated hydrocarbons** have one or more double or triple bonds between carbon atoms, which reduces the number of hydrogen atoms that can be bonded.
Examining the given options:
Based on this analysis, **C2H2** (ethyne) is a highly unsaturated hydrocarbon due to the presence of a **triple bond**. The triple bond signifies a greater level of unsaturation compared to double bonds in hydrocarbons like ethene (C2H4).
Ibeere 32 Ìròyìn
What accounts for the low melting and boiling points of covalent molecules?
Awọn alaye Idahun
The low melting and boiling points of covalent molecules are primarily due to the presence of weak intermolecular forces between the molecules. While covalent molecules consist of atoms bonded together by strong covalent bonds, the forces between separate molecules, known as van der Waals forces or London dispersion forces, are much weaker. These weak forces require significantly less energy to overcome, which explains why covalent molecules tend to have lower melting and boiling points.
Although covalent molecules have definite shapes and possess shared electron pairs, these characteristics have little influence on the melting and boiling points. The focus is instead on how much energy is needed to separate the molecules from one another.
Covalent molecules are not typically three-dimensional structures like ionic compounds or metals which form intricate lattices and require more energy to disrupt. Thus, the primary reason for their lower melting and boiling points is the presence of weak intermolecular forces that can be more easily overcome with minimal energy input.
Ibeere 33 Ìròyìn
When a specie undergoes oxidation, its
Awọn alaye Idahun
When a species undergoes oxidation, it experiences an increase in its oxidation number. Oxidation is a chemical process where a species loses electrons. In terms of oxidation number, electrons have a negative charge, so losing them results in an increase in charge. Thus, the oxidation number of the species becomes more positive or less negative.
To help understand, consider sodium (Na) reacting with chlorine (Cl2) to form sodium chloride (NaCl):
This change clearly shows that when sodium is oxidized, its oxidation number increases.
Therefore, the correct explanation is: a species undergoing oxidation will have its oxidation number increase.
Ibeere 34 Ìròyìn
The pH of a 0.001 mol dm−3 of H2 SO4 is
[Log10 2 = 0.3]
Awọn alaye Idahun
The question is asking about the pH of a 0.001 mol dm−3 solution of H2SO4 (sulfuric acid). To find the pH, we need to understand how sulfuric acid dissociates in water.
Step 1: Dissociation of H2SO4
Sulfuric acid, H2SO4, is a strong acid and dissociates completely in water in two steps:
1. The first dissociation: H2SO4 → H+ + HSO4-
2. The second dissociation: HSO4- → H+ + SO42-
For dilute solutions, particularly below 0.1 M, the first dissociation provides the major contribution to the H+ concentration. The second dissociation also contributes slightly to the acidity, but for simplicity and due to the dilute nature of this solution, the first step's contribution is primarily considered.
Step 2: Calculate the H+ Concentration
Since this is a strong acid and dissociates completely, for every 1 mole of H2SO4, we get 2 moles of H+. Therefore, for a 0.001 mol dm−3 solution of H2SO4, the concentration of H+ ions will be:
2 x 0.001 = 0.002 mol dm−3
Step 3: Calculate the pH
The pH is calculated using the formula: pH = -log[H+]
Substitute the H+ concentration:
pH = -log(0.002)
We know that log(10-2) = -2 and log(2) = 0.3 (as provided), so:
pH = -(log(2) + log(10-3))
pH = -(0.3 - 3)
pH = 3 - 0.3
pH = 2.7
Therefore, the pH of the 0.001 mol dm−3 H2SO4 solution is 2.7.
Ibeere 35 Ìròyìn
Awọn alaye Idahun
When a strong acid reacts with a strong base, the result is the formation of a neutral salt. This reaction is a part of a chemical process known as neutralization.
Let's break it down further:
During a neutralization reaction, the hydrogen ions (H⁺) from the acid combine with the hydroxide ions (OH⁻) from the base to form water (H₂O). Meanwhile, the remaining ions (for example, Na⁺ from NaOH and Cl⁻ from HCl) come together to form a compound known as a salt. This salt does not affect the acidity or basicity of the solution, hence it is considered neutral.
Therefore, the salt formed in such a reaction is a neutral salt, which is what is referred to as a normal salt in the options provided.
Ibeere 36 Ìròyìn
An oxide of nitrogen that can rekindle a glowing splint is
Awọn alaye Idahun
The ability to rekindle a glowing splint is an indicator of the presence of an oxidizing agent, typically oxygen or a substance that releases oxygen. Among oxides of nitrogen, only a few are capable of doing this.
Nitrogen(I) oxide, commonly known as nitrous oxide (N2O), is not a strong enough oxidizer to rekindle a glowing splint.
Nitrogen(II) oxide, known as nitric oxide (NO), is not stable in the presence of oxygen and does not have the ability to rekindle a glowing splint because it does not actively release oxygen.
Nitrogen(IV) oxide or nitrogen dioxide (NO2), can support combustion by releasing oxygen as it decomposes. It is a brown gas and an effective oxidizer.
Dinitrogen tetraoxide (N2O4) is in equilibrium with nitrogen dioxide (NO2). However, at standard conditions, it is not as effective an oxidizer for rekindling a glowing splint as pure NO2.
In conclusion, the oxide of nitrogen that can rekindle a glowing splint is nitrogen(IV) oxide or nitrogen dioxide (NO2) due to its ability to release oxygen and support combustion.
Ibeere 37 Ìròyìn
When the subsidiary quantum numbers (l) equals 1, the shape of the orbital is
Awọn alaye Idahun
The subsidiary quantum number, often referred to as the azimuthal quantum number or angular momentum quantum number, is denoted by l. This quantum number defines the shape of the atomic orbital. The value of l determines the type of orbital as follows:
For l = 1, the atomic orbital is a p orbital, which is characterized by its dumb-bell shape. This means that the electron density is concentrated in two lobes on opposite sides of the nucleus, resembling a dumb-bell.
Ibeere 38 Ìròyìn
Solubility curve is a plot of solubility against
Awọn alaye Idahun
A solubility curve is a plot of solubility against temperature. Let me explain in a simple way:
Solubility refers to the amount of a substance (solute) that can dissolve in a given quantity of solvent to form a homogeneous solution at a specified condition. The most common factor that affects solubility is the temperature.
Here's why a solubility curve typically involves temperature:
Therefore, plotting solubility against temperature in a solubility curve allows us to visualize and understand how solubility changes with variations in temperature.
Ibeere 39 Ìròyìn
The electronic configuration of an atom of Nitrogen is 1s2 2s2 2p1x 2p1y 2p1z because the atom is
Awọn alaye Idahun
The electronic configuration of nitrogen is given as: 1s2 2s2 2px1 2py1 2pz1.
This configuration suggests that nitrogen has 7 electrons, as follows:
This is the **ground state** electron configuration of nitrogen, meaning that the atoms have electrons in the **lowest possible energy levels**. It demonstrates nitrogen's **stable configuration**, where it has half-filled p orbitals, each with a single electron. This configuration obeys Hund's Rule, which states that every orbital in a subshell gets one electron before any one orbital gets two (due to electron repulsion). It also obeys the Aufbau principle which suggests electrons fill orbitals starting from the lowest energy level.
Therefore, this configuration indicates that the atom is simply obeying rules governing electron configuration. The electrons are in their lowest energy orbitals, consistent with the principles that direct electron arrangement in an atom, ensuring stability without being excited or unstable. There are no **energy changes** being depicted nor is the atom in an **excited state**—it is showing the normal ground state.
Ibeere 40 Ìròyìn
An example of a compound that is acidic in solution is
Awọn alaye Idahun
Phosphoric acid is a weak acid that can donate three hydrogen ions in water. Phosphoric acid partially ionizes when dissolved in an aqueous solution.
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