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Tambaya 1 Rahoto
An example of a substance that does not change directly from solid to gas when heated is
Bayanin Amsa
When discussing the process of substances changing states, some substances can transition directly from a solid to a gas without passing through a liquid state. This process is called sublimation. However, not all substances exhibit this behavior. Let's examine the substances provided:
In conclusion, calcium carbonate (CaCO3) is the substance that does not change directly from a solid to a gas when heated, as it undergoes a decomposition process instead.
Tambaya 2 Rahoto
The pH of a 0.001 mol dm−3 of H2 SO4 is
[Log10 2 = 0.3]
Bayanin Amsa
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.
Tambaya 3 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 4 Rahoto
Alkylation of benzene is catalyzed by
Bayanin Amsa
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.
Tambaya 5 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 6 Rahoto
What would be the order of the electrolytic cell in an industry intending the production of silver plated spoons?
Bayanin Amsa
In the process of silver plating a spoon using an electrolytic cell, the correct configuration involves the following:
Cathode: The object to be plated, which in this case is the spoon. In an electrolytic cell, the cathode is where the reduction reaction occurs, and it is the surface on which the metal ions are deposited.
Anode: A rod made of silver. The anode is where oxidation occurs, meaning the silver rod will dissolve into the solution in the form of silver ions. These ions then move towards the cathode to be deposited as a thin layer on the spoon.
Electrolyte: A solution that contains a soluble silver salt (such as silver nitrate, AgNO3). The silver ions from this salt help in the process of transferring the silver from the anode to the cathode.
Thus, the proper order for silver plating a spoon in an electrolytic cell for industrial production is: "Cathode is the spoon; anode is a silver rod; electrolyte is a soluble silver salt."
Tambaya 7 Rahoto
Nitrogen obtained from air is not absolutely pure because it contains the following except
Bayanin Amsa
Nitrogen obtained from air is not absolutely pure because it contains other gases, including:
Tambaya 8 Rahoto
An organic compound with general formula RCOR' is an
Bayanin Amsa
The general formula RCOR' represents a class of organic compounds known as ketones. In this formula, R and R' are alkyl groups, which are chains of carbon and hydrogen atoms. The CO in the middle is a carbonyl group, which consists of a carbon atom double-bonded to an oxygen atom. Therefore, with the presence of two alkyl groups on either side of the carbonyl group, the compound is categorized as a ketone, scientifically referred to as an alkanone.
Here is a simple breakdown of the terms:
Hence, by looking at the general formula RCOR', the organic compound in question is undoubtedly an alkanone.
Tambaya 9 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 10 Rahoto
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].
Bayanin Amsa
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.
Tambaya 11 Rahoto
A gas that turns lime water milky is likely to be from
Bayanin Amsa
The gas that turns lime water milky is **Carbon Dioxide**. This is because carbon dioxide reacts with calcium hydroxide, which is the main component of lime water, to form calcium carbonate. This chemical reaction can be represented by the equation:
Ca(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)
In this equation, calcium hydroxide ({Ca(OH)2}) in the lime water reacts with carbon dioxide ({CO2}) to produce calcium carbonate ({CaCO3}) and water ({H2O}).
The result is a milky or cloudy appearance due to the formation of insoluble calcium carbonate precipitate in the lime water. This reaction is a common test for the presence of carbon dioxide gas.
Among the options given, **Trioxocarbonate(IV)** is another name for the Carbonate group involving the gas carbon dioxide ({CO2}). Hence, the gas related to Trioxocarbonate(IV) is the one that turns lime water milky.
Tambaya 12 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 13 Rahoto
Biuret test is a chemical test used for detecting the presence of
Bayanin Amsa
The Biuret test is a chemical test used for detecting the presence of proteins. When you perform a Biuret test, you are looking for peptide bonds, which are the connections between the amino acids in a protein. This is how it works:
The test is specifically tailored to proteins because carbohydrates, amines, and alkanoates do not exhibit the required peptide bonds necessary for this color change. Therefore, the Biuret test is not suitable for detecting these compounds.
Tambaya 14 Rahoto
The element which can combine with oxygen to form an acid anhydride of the form XO2 is
Bayanin Amsa
An Acid anhydride can be defined as a non-metal oxide which forms an acidic solution when reacted with water.
Sulphur is the element that can combine with oxygen to form an acid anhydride of the form XO2 .
An acid oxide is a compound that forms an acid when it reacts with water. Non-metals in groups 4–7 form acidic oxides.
Tambaya 15 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 16 Rahoto
If the solubility of KNO3 at 300 C is 3.10 mol/dm3 a solution containing 303g/dm3 KNO3 is likely to be
Bayanin Amsa
To determine the condition of the solution containing KNO3 at 300C, let's start by calculating the molarity of the given solution.
The molecular weight of KNO3 (Potassium Nitrate) is approximately:
Thus, KNO3 = 39 + 14 + (16 * 3) = 101 g/mol.
Now, to determine the molarity of the given solution:
Compare with the solubility at 300C:
If we compare the values:
Hence, the solution is unsaturated because it can still dissolve more KNO3 until it reaches the solubility limit of 3.10 mol/dm3.
Tambaya 17 Rahoto
In the extraction of Aluminium, the silica impurity is removed by
Bayanin Amsa
Aluminum is extracted from bauxite by electrolysis. The extraction proceeds in two stages;
1. Purification of the Bauxite: The impure bauxite is heated with sodium hydroxide solution to form soluble sodium tetrahydroxy aluminate (iii). The impurities in the ore which are iron (iii) oxide and trioxosilicate (iv) compounds are not soluble in the alkali. They are therefore filtered off as a sludge.
Aluminum hydroxide crystals is then added to filtrate, NaAl(OH)4 solution to induce the precipitation of Aluminum hydroxide.
2. The electrolysis of the pure alumina
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
Boyle's law can be expressed mathematically as
Bayanin Amsa
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.
Tambaya 20 Rahoto
Concentrated sodium chloride solution is electrolyzed using mercury cathode and graphite anode. The products at the anode and the cathode respectively are
Bayanin Amsa
When a concentrated sodium chloride solution is electrolyzed using a mercury cathode and graphite anode, the products are hydrogen gas at the cathode and chlorine gas at the anode
At the anode, 2Cl− → Cl2 + 2e−
At the cathode, 2H+ + 2e− → H2
During the electrolysis, hydrogen and chloride ions are removed from solution whereas sodium and hydroxide ions are left behind in solution. This means that sodium hydroxide is also formed during the electrolysis of sodium chloride solution.
Tambaya 21 Rahoto
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 ]
Bayanin Amsa
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.
Tambaya 22 Rahoto
Bayanin Amsa
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.
Tambaya 23 Rahoto
Bayanin Amsa
Silver and Gold are classified as noble metals. These metals are known for their resistance to corrosion and oxidation in moist air, unlike most other base metals. They can be found in the earth's crust as free, uncombined elements because they do not easily react with oxygen and other elements to form compounds. This property is what distinguishes noble metals from more reactive or corrosive ones. While the term "natural metals" seems applicable in that they occur naturally, the more precise and widely accepted term for metals like Silver and Gold is "noble metals".
Tambaya 24 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 25 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 26 Rahoto
At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that
Bayanin Amsa
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:
Tambaya 27 Rahoto
CuOs + H2 (g ) ⇌ Cus + H2 O(g )
In the equation above, the effect of increased pressure on the equilibrium position is that
Bayanin Amsa
To understand the effect of increased pressure on the equilibrium position of the given reaction:
CuO(s) + H2(g) ⇌ Cu(s) + H2O(g)
We need to consider Le Chatelier's Principle. According to this principle, if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust itself to counteract the effect of the change and re-establish equilibrium.
For the reaction in question, let's consider the number of gas molecules on each side of the equation:
Since both sides of the equation have the same number of gas molecules, an increase in pressure will not favor a shift to either the left or the right because the number of moles of gas on both sides of the equilibrium is the same.
Therefore, the effect of increased pressure on the equilibrium is that there is no effect. The position of the equilibrium remains unchanged, and pressure changes do not influence the production of more H2(g) or H2O(g) in this specific reaction.
Tambaya 28 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 29 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 30 Rahoto
An example of a physical change is
Bayanin Amsa
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.
Tambaya 31 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 32 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 33 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 34 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 35 Rahoto
Solubility curve is a plot of solubility against
Bayanin Amsa
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.
Tambaya 36 Rahoto
The term that is not associated with petroleum industry is ?
Bayanin Amsa
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.
Tambaya 37 Rahoto
When Sulphur(IV)oxide is passed into solution of acidified tetraoxomanganate(VII), the colour changes from
Bayanin Amsa
When Sulphur(IV) oxide (SO2) is passed into a solution of acidified tetraoxomanganate (VII) (KMnO4), it acts as a reducing agent. This reaction involves the reduction of potassium permanganate (KMnO4), which is characterized by a distinctive color change.
The tetraoxomanganate (VII) ion (MnO4-) is purple in color. During the reaction, SO2 gets oxidized while the MnO4- ion is reduced to Mn2+, which is almost colorless or pale pink, depending on the concentration.
Thus, the color of the solution changes from purple to almost colorless as the reaction progresses.
Tambaya 38 Rahoto
The composition of alloy permalloy is iron and
Bayanin Amsa
The alloy known as **permalloy** is composed primarily of **iron** and **nickel**. Permalloy is a well-known magnetic alloy that typically consists of about **80% nickel and 20% iron**. It is renowned for having high magnetic permeability, meaning it can become magnetized easily, which makes it extremely useful in a variety of electrical and magnetic applications, such as transformers, memory storage, and magnetic shielding. The nickel in permalloy enhances the magnetic properties of the iron, giving the alloy its unique characteristics.
Tambaya 39 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 40 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
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