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Tambaya 1 Rahoto
Which of the following is an air pollutant?
Bayanin Amsa
An air pollutant is any substance in the air, introduced by natural or human activity, that causes harm or discomfort to living organisms, or damages the environment. Let's analyze the substances mentioned:
1. O2 (Oxygen)
Oxygen is the gas we need to breathe. It's not considered an air pollutant because it is essential for human and animal life, as well as many natural processes.
2. CO (Carbon Monoxide)
Carbon Monoxide is a colorless, odorless gas that is produced by burning fuel (like in cars and factories). This gas can be very dangerous if there is a lot of it, as it can prevent oxygen from entering the bloodstream. Because of its harmful effects, it is considered an air pollutant.
3. H2 (Hydrogen)
Hydrogen, while a flammable gas, is generally not harmful to the air or to organisms when it is released into the environment. Therefore, it is not considered an air pollutant.
4. O3 (Ozone)
Ozone is a bit tricky because it is both good and bad. Higher up in the atmosphere, it forms a layer that protects us from the sun’s UV radiation. However, at ground level, it is a harmful air pollutant. Ground-level ozone can cause health problems such as respiratory difficulties, so in this context, it is considered an air pollutant.
In conclusion, the substances that are considered air pollutants in this context are Carbon Monoxide (CO) and ground-level Ozone (O3).
Tambaya 2 Rahoto
In the treatment of water for municipal supply, chlorine is used to
Bayanin Amsa
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.
Tambaya 3 Rahoto
127g of sodium chloride was dissolved in 1.0dm3 of distilled water at 250 C . Determine the solubility in moldm−3 of sodium chloride at that temperature. [Na = 23, Cl = 35.5]
Bayanin Amsa
To determine the solubility of sodium chloride (NaCl) in mol/dm3 at the given temperature, you need to first calculate the number of moles of NaCl dissolved.
Step 1: Calculate the molar mass of NaCl.
The molar mass of a compound is found by adding the atomic masses of its constituent elements:
- Sodium (Na) has an atomic mass of 23.
- Chlorine (Cl) has an atomic mass of 35.5.
Thus, the molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.
Step 2: Calculate the number of moles of NaCl.
The formula to calculate moles is:
Number of moles = Mass (g) / Molar mass (g/mol)
Given mass of NaCl = 127 g,
Number of moles = 127 g / 58.5 g/mol ≈ 2.17 mol
Step 3: Calculate the solubility in mol/dm3.
Since the sodium chloride is dissolved in 1.0 dm3 of water, the solubility is the same as the number of moles, since the volume is already 1.0 dm3.
Therefore, the solubility of sodium chloride at that temperature is 2.17 mol/dm3.
Rounded to the options given, 2.17 mol/dm3 is approximately equal to 2.2 mol/dm3.
Tambaya 4 Rahoto
The Van der waals forces of attraction operates between
Bayanin Amsa
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.
Tambaya 5 Rahoto
Fats and oils are esters of fatty organic acids combined with a trihydric alkanol commonly referred to as
Bayanin Amsa
Fats and oils are types of lipids that belong to the category of esters of fatty acids. These are organic compounds formed when fatty acid molecules react with an alcohol. In the case of fats and oils, the alcohol involved is a trihydric alkanol, meaning it has three hydroxyl (-OH) groups.
The trihydric alkanol commonly found in fats and oils is glycerol. Glycerol, also known as glycerine, has the chemical formula C3H8O3 and has three carbon atoms, each of which is attached to a hydroxyl group, making it a perfect candidate to form esters with three fatty acid molecules.
When these fatty acids react with the hydroxyl groups of glycerol, they form compounds called triglycerides. These triglycerides are the primary constituents of both fats and oils. Therefore, the correct answer is that fats and oils are esters of fatty organic acids combined with glycerol as the trihydric alkanol.
Tambaya 6 Rahoto
CH3 -CH2 -OH and CH3 -O-CH3
The relationship between the two compounds above, is that they are
Bayanin Amsa
The relationship between the two compounds is that they are isomers.
To understand why these compounds are isomers, let's break down their structures and definitions:
1. Structures of the Compounds:
2. Definitions:
Both compounds have the same molecular formula: C2H6O. However, they have different arrangements of their atoms. Ethanol has a hydroxyl group (-OH) attached to an ethyl group (CH3-CH2-), while dimethyl ether involves two methyl groups (CH3-) bonded to an oxygen atom (O). This difference in structure leads to different chemical and physical properties, despite having the same molecular formula. Hence, these two compounds are classified as isomers.
Tambaya 7 Rahoto
Fog is a colloid in which
Bayanin Amsa
**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.
Tambaya 8 Rahoto
What accounts for the low melting and boiling points of covalent molecules?
Bayanin Amsa
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.
Tambaya 9 Rahoto
Biodegradable pollutants are not safe in water systems because they can cause
Bayanin Amsa
Biodegradable pollutants are substances that can be broken down by natural processes and microorganisms. However, when they are present in water systems, they can lead to several environmental and health issues. One of the main concerns is their potential to cause ill health. Here's why:
When biodegradable pollutants such as organic waste are introduced into water bodies, they are decomposed by bacteria and other microorganisms. This process consumes dissolved oxygen in the water. As the oxygen levels decrease, aquatic life such as fish and plants may suffer or die due to a lack of oxygen, disrupting the entire aquatic ecosystem.
This situation is known as eutrophication, which can lead to the excessive growth of algae, commonly referred to as algal blooms. These blooms often produce toxins that are harmful to both aquatic life and humans. Furthermore, when this polluted water is used for drinking, agriculture, or recreational purposes, it poses serious health risks to humans. These risks may include gastrointestinal infections, neurological disorders, and skin problems.
In addition, as the pollutants decompose, foul smells may be released, which can affect air quality in the vicinity, although the primary concern with biodegradable pollutants in water is related to how they affect water quality and health.
Therefore, it is crucial to properly manage and treat biodegradable pollutants before they enter water systems to prevent these health hazards. Failure to do so can result in significant environmental and health issues.
Tambaya 10 Rahoto
Cx Hy O + 5O2 → 4CO2 + 4H2 O
Cx Hy O in the equation is
Bayanin Amsa
Cx Hy O + 5O2 → 4CO2 + 4H2 O
On balancing the equation, we should have
X = 4 , y = 8 and O = 2 ⇒ C4 H8 O2
Since 2 is a common factor to the three atoms, we can divide through by 2, considering the fact that that formula is not in the option.
We finally have C2 H4 O
Tambaya 11 Rahoto
Heat of solution involves two steps that is accompanied by heat change. The energies involved in this steps are
Bayanin Amsa
The heat of solution refers to the overall energy change that occurs when a solute dissolves in a solvent. This process involves breaking and making of intermolecular forces, and it can be broken down into two main steps that are each accompanied by heat change. The energies involved in these steps are:
Lattice energy: This is the energy required to break the bonds between the ions in the solid crystal lattice of the solute. Breaking these bonds requires energy, and this step is usually endothermic, meaning it absorbs heat from the surroundings. The more energy needed to break the lattice, the higher the lattice energy.
Hydration energy: Once the lattice is broken, the ions are surrounded by solvent molecules, typically water, in a process known as hydration. The energy released when the solvent molecules interact with and stabilize the ions is called the hydration energy. This step is usually exothermic, meaning it releases heat into the surroundings.
In conclusion, the two energies involved in the heat of solution are lattice energy and hydration energy. The balance between these two energies determines whether the overall process of dissolving a solute in a solvent is endothermic or exothermic.
Tambaya 12 Rahoto
The percentage of carbon(IV) oxide in air is
Bayanin Amsa
The air we breathe is made up of a mixture of gases. The most abundant gases in the atmosphere are nitrogen and oxygen, but there are other gases present in smaller amounts, one of which is carbon dioxide, chemically known as carbon(IV) oxide.
Carbon dioxide makes up approximately 0.03% of the Earth's atmosphere by volume. This value can also be expressed in different terms, such as 300 parts per million (ppm). Even though it is a small percentage, carbon dioxide plays a significant role in maintaining the Earth's temperature through the greenhouse effect.
In summary, the percentage of carbon(IV) oxide in air is 0.03%.
Tambaya 13 Rahoto
If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are
Bayanin Amsa
To answer this question, let's break it down step by step:
Mass Number: The mass number is the total number of protons and neutrons in an atom's nucleus. In this case, the mass number is given as 31.
Stable Neutral Atom: A stable neutral atom has no overall electrical charge, meaning the number of protons (positively charged) must equal the number of electrons (negatively charged).
If we symbolize the number of protons by the atomic number (Z), we can say:
1. **Protons = Electrons** in a neutral atom.
2. **Mass Number (A) = Protons + Neutrons**.
Given that the mass number is 31, we have the equation:
A = Protons + Neutrons = 31.
Assuming a commonly known stable element like Phosphorus, which has an atomic number (Z) of 15, it means:
1. **Protons = 15**.
2. **Electrons = 15** (because it's a neutral atom).
3. To find Neutrons: Neutrons = Mass Number - Protons = 31 - 15 = 16.
So, in this scenario, the number of electrons is 15 and the number of neutrons is 16. This combination is found in the first option given.
Tambaya 14 Rahoto
The difference in molecular mass between an alkene and alkyne with six carbon per mole is
Bayanin Amsa
To determine the difference in molecular mass between an alkene and an alkyne, let's first take a look at their general formulas.
Alkene: An alkene is a hydrocarbon with at least one double bond between carbon atoms. For an alkene with six carbon atoms, the general formula is CnH2n. Therefore, for 6 carbon atoms, the molecular formula is C6H12.
Alkyne: An alkyne is a hydrocarbon with at least one triple bond between carbon atoms. For an alkyne with six carbon atoms, the general formula is CnH2n-2. Therefore, for 6 carbon atoms, the molecular formula is C6H10.
Now let's calculate the molecular masses:
Molecular mass of alkene (C6H12):
Molecular mass of alkyne (C6H10):
The **difference** in molecular mass between the alkene and alkyne is **84 g/mol - 82 g/mol** = 2 g/mol.
Tambaya 15 Rahoto
Alkanoates are naturally found in
Bayanin Amsa
Alkanoates, also known as fatty acid esters, are primarily found in lipids. Lipids are a broad group of naturally occurring molecules that include fats, waxes, sterols, fat-soluble vitamins (such as vitamins A, D, E, and K), and others. One of the main components of lipids is fatty acids and their derivatives, such as alkanoates.
To be more specific, alkanoates can be found in the form of triglycerides, which are the main constituents of body fat in humans and animals, as well as vegetable fat. Triglycerides are composed of glycerol bound to three fatty acids, and these fatty acids are usually present in the form of alkanoates.
Unlike proteins and rubber, which are made up of amino acids and polymers of isoprene respectively, lipids are the primary class of biomolecules where these alkanoate compounds can be found in significant amounts.
Tambaya 16 Rahoto
Hydrogen chloride gas and ammonia can be used to demonstrate the fountain experiment because they are
Bayanin Amsa
In the fountain experiment, hydrogen chloride gas (HCl) and ammonia (NH₃) are used to demonstrate the creation of a visible 'fountain' due to their high solubility in water. Here's a simple explanation:
When hydrogen chloride gas and ammonia gas come into contact with water, they dissolve very quickly and react vigorously. This is because both gases are very soluble in water. As they dissolve, a vacuum-like pressure is created inside the container where the gases are held, pulling water up into it, creating the 'fountain' effect.
Moreover, when HCl and NH₃ gases react with each other, they form a white, solid product known as ammonium chloride (NH₄Cl), which is a demonstration of how both gases can effectively dissolve and react with not just water, but also with each other.
Thus, the ability of these gases to create a fountain effect is primarily because they are very soluble in water, which allows them to dissolve rapidly and create the pressure differential necessary for the water to be pulled into the container dynamically.
Tambaya 17 Rahoto
What method is suitable for the separation of gases present in air?
Bayanin Amsa
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.
Tambaya 18 Rahoto
The chemical formula for potassiumhexacyanoferrate(II) is
Bayanin Amsa
The chemical formula for potassiumhexacyanoferrate(II) is K4Fe(CN)6.
Let's break down the name to understand why:
1. Potassium (K): The compound includes potassium ions. In this case, four potassium ions are present, indicated by the subscript 4 in K4.
2. Hexacyano: The prefix "hexa" means six, which signifies there are six cyanide ions (CN-) in the complex. This is represented as (CN)6.
3. Ferrate (II): The word "ferrate" suggests the presence of iron (Fe). The Roman numeral (II) indicates that the iron is in the +2 oxidation state.
Overall, the complex ion is [Fe(CN)6] with a charge of 4-, so to balance the charge, four potassium ions (each with a charge of +1) are needed, resulting in the formula K4Fe(CN)6.
Tambaya 19 Rahoto
In the conductance of aqueous CuSO4 solution, the current carriers are the
Bayanin Amsa
In the conductance of aqueous CuSO4 solution, the current carriers are the hydrated ions.
Here's why:
The other options can be understood as follows:
The correct answer is therefore hydrated ions because they enable the conduction of electricity through the aqueous solution.
Tambaya 20 Rahoto
Sulphur(IV)oxide can be used as a
Bayanin Amsa
Sulphur(IV) oxide has many uses including food preservation, refrigeration, laboratory reagent and solvent, sulphuric acid production, fumigant etc.Sulphur(IV) oxide is a good refrigerant because it has a high heat of evaporation and can be easily condensed.
Tambaya 21 Rahoto
How many isomers has the organic compound represented by the formula C3 H8 O ?
Bayanin Amsa
The molecular formula C3H8O represents organic compounds that contain 3 carbon atoms, 8 hydrogen atoms, and 1 oxygen atom. Let's elucidate the possible isomers, which are molecules with the same molecular formula but different structural arrangements.
1. Alcohols: One class of compounds that can form isomers for this formula are alcohols, which include a functional group -OH.
a. Propan-1-ol: This is a straight-chain alcohol where the -OH group is on the first carbon. The structure is as follows:
CH3-CH2-CH2-OH
b. Propan-2-ol: This is another alcohol where the -OH group is on the second carbon, giving it a different structure and properties:
CH3-CH(OH)-CH3
2. Ethers: This is another class of possible isomers, where the oxygen atom is bonded to two alkyl groups.
c. Methoxyethane: Also known as ethyl methyl ether, it has a structure where the oxygen is in a bridge position between a methyl group and an ethyl group:
CH3-O-CH2-CH3
These are the possible structural isomers for this molecular formula. Therefore, the compound C3H8O has three isomers overall:
Thus, the answer is three distinct isomers.
Tambaya 22 Rahoto
A type of isomerism that ClCH=CHCl can exhibit is
Bayanin Amsa
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.
Tambaya 23 Rahoto
Strong acids can be distinguished from weak acids by any of the following methods, EXCEPT
Bayanin Amsa
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.
Tambaya 24 Rahoto
The molecular formular of a hydrocarbon with an empirical formula of CH3 and a molar mass of 30 is
Bayanin Amsa
To find the molecular formula of a hydrocarbon given its empirical formula and molar mass, you need to compare the empirical formula mass with the given molar mass.
The empirical formula given is CH3. The molar mass of the empirical formula is calculated as follows:
Total empirical formula mass = 12 + 3 = 15 g/mol
The provided molar mass of the compound is 30 g/mol. To determine how many empirical units are in the molecular formula, divide the molecular mass (given) by the empirical formula mass:
Number of empirical units = 30 g/mol / 15 g/mol = 2
Therefore, the molecular formula is twice the empirical formula:
Empirical formula: CH3
Molecular formula: (CH3)2 = C2H6
The correct molecular formula is C2H6.
Tambaya 25 Rahoto
The combustion of candle under limited supply of air forms
Bayanin Amsa
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).
Tambaya 26 Rahoto
The product formed when ethyne is passed through a hot tube containing finely divided iron is
Bayanin Amsa
When **ethyne** (also known as acetylene) is passed through a hot tube containing finely divided iron, a process called decomposition occurs. The heat causes the ethyne molecules to break down, and under these conditions, they **re-combine** to form structures that result in more complex molecules.
The key transformation involves the conversion of these ethyne molecules into **aromatic compounds**. Aromatic compounds, such as **benzene**, have a distinct ring structure and are characterized by **stability** due to resonance (a phenomenon where electrons are delocalized over a certain structure, providing extra stability).
Thus, when ethyne is passed through a hot iron tube, it undergoes trimerization to form benzene, an **aromatic** compound. Therefore, the product formed is **aromatic**.
Tambaya 27 Rahoto
25.0g of potassium chloride were dissolved in 80g of distilled water at 300 C. Calculate the solubility of the solute in mol dm3 . [K =39, Cl = 35.5]
Bayanin Amsa
To calculate the solubility of potassium chloride (KCl) in mol dm3, we need to follow these steps:
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Moles of KCl = Mass of KCl / Molar mass of KCl = 25.0 g / 74.5 g/mol = 0.3356 mol
Convert ml to liters: 80 ml = 0.080 L
Concentration = Moles of solute / Volume of solvent in liters = 0.3356 mol / 0.080 L = 4.195 mol/dm3
The solubility of potassium chloride at 30°C in mol/dm3 is therefore approximately 4.2 mol/dm3.
Tambaya 28 Rahoto
The ions responsible for permanent hardness in water are sulphates of
Bayanin Amsa
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.
Tambaya 29 Rahoto
The law which states that a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is
Bayanin Amsa
The law that states a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is the law of definite proportion.
To explain this simply, let's consider water as an example. Water is made up of hydrogen and oxygen. According to the law of definite proportion, a sample of pure water taken from anywhere in the world will always contain the same ratio of hydrogen to oxygen by mass. Specifically, water will always have approximately 88.8% oxygen and 11.2% hydrogen by mass.
This is because a chemical compound has a fixed composition, regardless of the process used to create it or the source from which it is derived. The law of definite proportion, also known as the law of constant composition, is fundamental in chemistry because it supports the idea that chemical compounds are composed of elements in specific and fixed ratios. This does not change regardless of how the compound is prepared or where it is found.
Tambaya 30 Rahoto
Benzene formed nitrobenzene at temperature of 600 C when it reacts with mixture of concentrated trioxonitrate(V) acid and concentrated
Bayanin Amsa
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)**.
Tambaya 31 Rahoto
The term strong and weak acids is used to indicate the
Bayanin Amsa
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.
Tambaya 32 Rahoto
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?
Bayanin Amsa
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
Tambaya 33 Rahoto
Na2 X ⇌ 2Na+ + X2−
The bond between Na and X is likely to be
Bayanin Amsa
The bond between Na and X is most likely to be ionic. Let's break this down simply:
In the equation provided:
Na2X ⇌ 2Na+ + X2−
The sodium (Na) atoms become positively charged ions (Na+), while X becomes a negatively charged ion (X2−). This change in charge occurs because sodium atoms donate electrons to the X atom. The donation of electrons by sodium to X indicates a transfer of electrons, which is a hallmark of an ionic bond.
In an ionic bond, electrons are transferred from one atom to another, resulting in a positively charged ion and a negatively charged ion. These oppositely charged ions attract each other, forming a strong ionic bond.
In summary, since sodium (Na) donates electrons to X forming ions, the bond between Na and X is most likely to be ionic.
Tambaya 34 Rahoto
The hybridization scheme in ethyne is
Bayanin Amsa
Ethyne, also known as acetylene, is a simple alkyne with the chemical formula C2H2. In ethyne, each carbon atom is bonded to two other atoms: one hydrogen atom and the other carbon atom. The molecular structure of ethyne is linear, with a triple bond between the two carbon atoms.
To determine the hybridization scheme in ethyne, we need to examine the arrangement of the electron pairs around each carbon atom. In ethyne, each carbon atom is forming two sigma (σ) bonds and two pi (π) bonds. Let's explain:
When we consider the hybridization of the carbon atoms, we focus on the formation of sigma bonds and lone pairs. In ethyne, each carbon atom utilizes two orbitals to form sigma bonds: one with the hydrogen atom and one with the other carbon atom. This implies that each carbon atom in ethyne must use two hybrid orbitals.
The two hybrid orbitals formed by each carbon atom in ethyne are a result of mixing one s orbital with one p orbital. This hybridization is referred to as sp hybridization, characterized by a linear electron geometry. The remaining two unhybridized p orbitals on each carbon atom are responsible for forming the two pi bonds in the triple bond.
In conclusion, the hybridization scheme in ethyne is sp.
Tambaya 35 Rahoto
Scandium is not regarded as a transition metal because its ion has
Bayanin Amsa
Scandium is not regarded as a transition metal because its ion has no electron in the d-orbital.
To understand this, let's first define a transition metal. A transition metal is defined as an element that has an incomplete d-subshell in either its elemental form or in any of its common oxidation states.
When Scandium (Sc) loses electrons to form its most common ion (Sc3+), it loses three electrons. These electrons are removed from the 4s and 3d orbitals. The electron configuration for Scandium is [Ar] 3d1 4s2. Upon losing three electrons to form Sc3+, the resulting electron configuration is [Ar], which means there are:
As a result, there are no electrons in the d-orbital of the Scandium ion, which does not meet the criteria for a transition metal.
Tambaya 36 Rahoto
The IUPAC Nomenclature of CH3 CH2 C(CH3 )=C(CH3 )2 for the compound is
Bayanin Amsa
The compound in question is written as CH₃₃CH₂₂C(CH₃₃)=C(CH₃₃)₂₂, which seems to be intended as (CH3)3CH2CH=C(CH3)3. The IUPAC nomenclature of organic compounds follows specific rules to name the compound uniquely such that it is understood universally. Here is a comprehensive breakdown:
1. Select the longest carbon chain that includes the highest-order functional group, which, in this case, is the alkene group (double bond).
2. The longest chain consists of 5 carbons, which gives us the root name "pentene". We choose the carbon chain such that the double bond gets the lowest possible number, starting from the end of the chain closest to the double bond.
3. Number the carbon atoms in the chain from the end closest to the double bond. The numbering direction will determine the position of the double bond and substituents. The double bond starts on carbon 2.
4. Identify and name the substituents attached to the carbon chain. In this case, there are two methyl groups on carbon 3. This means it is dimethyl as there are two of them.
Thus, the complete name of the compound is 2,3-dimethylpent-2-ene. Here, "2,3-dimethyl" indicates the position and quantity of methyl groups, "pent" indicates the longest chain with 5 carbons, and "-2-ene" indicates a double bond starting at the second carbon.
Tambaya 37 Rahoto
Which of the following is used in forming slag in the blast furnace for the extraction of iron?
Bayanin Amsa
In the process of extracting iron in a blast furnace, CaCO3, or calcium carbonate, plays a crucial role in forming slag. Here is a simple and comprehensive explanation of how it works:
1. Role of Calcium Carbonate (CaCO3):
Calcium carbonate is commonly used as a flux in the blast furnace. When it is introduced into the furnace, it undergoes a decomposition reaction due to the high temperatures, breaking down into calcium oxide (CaO) and carbon dioxide (CO2).
2. Formation of Slag:
The calcium oxide (CaO) produced then reacts with silicon dioxide (SiO2) present in the iron ore. This reaction forms a liquid slag of calcium silicate. The slag serves two main functions:
Thus, calcium carbonate (CaCO3) is crucial for forming slag by providing the necessary calcium oxide (CaO) that reacts with impurities to form slag during the extraction of iron in a blast furnace.
Tambaya 38 Rahoto
During the fractional distillation of crude oil, the fraction that distills at 200 - 2500 C is
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The petroleum fractions that distill at 200–250°C are naphtha and kerosene,
Tambaya 39 Rahoto
Kerosene is used as solvent for
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Kerosene is commonly used as a solvent for paints. Let me explain why in a simple way:
Kerosene is a type of fuel that is composed of hydrocarbons, which are molecules made up of hydrogen and carbon atoms. These hydrocarbons give kerosene the ability to dissolve other similar substances.
Paints often contain oils and other hydrocarbon-based compounds. Since kerosene is also hydrocarbon-based, it can effectively dissolve and thin these compounds. This makes it suitable for use as a solvent in paints, allowing the paint to be thinned or cleaned up after use. This property makes kerosene a good choice for cleaning brushes and other painting tools or for dissolving dried paint.
On the other hand, sulphur, gums, and fats are typically not dissolved effectively by kerosene because of their different chemical properties. Therefore, kerosene as a solvent is primarily useful in the context of working with paints and similar hydrocarbon-based materials.
Tambaya 40 Rahoto
The shape of the molecule of Carbon(IV) oxide is
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The shape of the molecule of Carbon(IV) oxide, also known as carbon dioxide (CO2), is linear. This is because of the following reasons:
Due to this arrangement, carbon dioxide has a symmetric shape, making it non-polar despite having polar covalent bonds. The pulling forces of the two oxygen atoms on either side of the carbon atom cancel each other out, reinforcing its linear configuration.
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