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Tambaya 1 Rahoto
Tin is unaffected by air at ordinary temperature due to its
Bayanin Amsa
Tin is unaffected by air at ordinary temperature because of its high reactivity, which means it doesn't easily react or corrode with other elements such as oxygen, the main component of air. Tin has a high reactivity, which helps it resist corrosion and oxidation. This makes tin a great material for products that need to be resistant to corrosion, such as food containers and metal pipes.
Tambaya 2 Rahoto
For a general equation of the nature xP + yQ ? mR + nS, the expression for the equilibrium constant is
Bayanin Amsa
Expression for equilibrium constant
k = concentration of productconcentration of reactant
Tambaya 3 Rahoto
According to Charle's law, the volume of a gas becomes zero at
Bayanin Amsa
According to Charles's law, the volume of a gas would theoretically become zero at -273°C or -459.4°F or 0 Kelvin, which is known as absolute zero. Charles's law is a gas law that describes the relationship between the volume and temperature of a gas, while the pressure and amount of gas are kept constant. The law states that at a constant pressure, the volume of a gas is directly proportional to its absolute temperature. This means that as the temperature of a gas decreases, its volume will also decrease proportionally. However, according to the laws of thermodynamics, it is impossible to cool a gas down to absolute zero or to completely eliminate its volume. This is because at absolute zero, the gas molecules would have zero kinetic energy, and therefore they would not be moving. This is a state that cannot be reached due to the third law of thermodynamics. Therefore, Charles's law is only valid within a certain temperature range and it cannot be applied to gases at extremely low temperatures or high pressures.
Tambaya 4 Rahoto
Ethene is prepared industrially by
Bayanin Amsa
Ethene, also known as ethylene, is prepared industrially by a process called cracking. Cracking involves breaking down larger hydrocarbon molecules, typically obtained from crude oil or natural gas, into smaller ones. This is usually done by heating the hydrocarbons to very high temperatures (up to 850°C) in the presence of a catalyst. The cracking process breaks the long chains of hydrocarbons into shorter ones, including the production of ethene. This is because the high temperature and catalyst cause the chemical bonds in the hydrocarbon molecules to break apart, leading to the formation of smaller molecules. These smaller molecules, including ethene, can then be separated from the other products and used in various applications such as the production of plastics. Therefore, out of the given options, the industrial preparation of ethene is accomplished by the process of cracking.
Tambaya 5 Rahoto
The carbon atoms on ethane are
Bayanin Amsa
The carbon atoms in ethane are sp3 hybridized. Hybridization is a concept in chemistry that describes how the orbitals of an atom combine to form new hybrid orbitals. In the case of carbon, which has four valence electrons, the four orbitals (s, px, py, and pz) combine to form four hybrid orbitals that have equal energy and are oriented in a tetrahedral arrangement around the carbon atom. In ethane, each carbon atom is bonded to three other atoms: two other carbon atoms and three hydrogen atoms. Each of the carbon-hydrogen bonds is formed by overlapping the sp3 hybrid orbitals on the carbon atom with the 1s orbitals on the hydrogen atoms. The carbon-carbon bond is formed by overlapping two of the sp3 hybrid orbitals on each carbon atom. Overall, the sp3 hybridization of the carbon atoms in ethane allows for the formation of strong and stable covalent bonds between the carbon and hydrogen atoms, as well as between the two carbon atoms.
Tambaya 6 Rahoto
Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]
Bayanin Amsa
For an ideal gas PV = nRT
Amount in moles = n
Volume v = 10.5dm3
Pressure P = 6atm
Temperature T = 30°C + 273 = 303k
R, Gas constant = 0.082 atmdm3k-1 mol
Recall from ideal gas equation
pv = nRT
n = RVRT
n = 6×1050.082×303
n= 2.536mol
Tambaya 7 Rahoto
The reaction of halogens in the presence of sunlight is an example of
Bayanin Amsa
The reaction of halogens in the presence of sunlight is an example of a substitution reaction. In this type of chemical reaction, one atom or group of atoms in a molecule is replaced by another atom or group of atoms. In the case of halogens, such as chlorine or bromine, they can react with certain organic compounds by breaking the carbon-hydrogen bond and replacing the hydrogen atom with a halogen atom. This reaction is usually initiated by the energy from sunlight, which provides the necessary activation energy to break the bond. For example, in the presence of sunlight, chlorine gas can react with methane gas to form chloromethane and hydrogen chloride: CH4 + Cl2 -> CH3Cl + HCl In this reaction, one of the hydrogen atoms in methane is replaced by a chlorine atom, forming chloromethane, while hydrogen chloride is also produced as a byproduct. Overall, the reaction of halogens in the presence of sunlight is an example of a substitution reaction, where one atom or group of atoms is substituted for another.
Tambaya 8 Rahoto
The oxidation number of iodine in KIO3 is
Bayanin Amsa
The oxidation number of an atom is a measure of the number of electrons that it has gained or lost in a chemical reaction. To determine the oxidation number of iodine in KIO3, we need to use the fact that the sum of the oxidation numbers of all the atoms in a compound is equal to the overall charge of the compound. In KIO3, potassium (K) has an oxidation number of +1 because it belongs to the alkali metal group and it always has a +1 oxidation state. Oxygen (O) usually has an oxidation number of -2, unless it is combined with a more electronegative atom or a peroxide, which is not the case in KIO3. Let's assume that iodine (I) has an oxidation number of x. Since KIO3 is a neutral compound, the sum of the oxidation numbers of all its atoms must be zero. Therefore, we can write the following equation: (+1) + x + 3(-2) = 0 Simplifying the equation, we get: x = +5 Therefore, the oxidation number of iodine in KIO3 is +5.
Tambaya 9 Rahoto
The shape of ammonia molecules is
Bayanin Amsa
The shape of ammonia (NH3) molecules is tetrahedral. The ammonia molecule has one nitrogen atom and three hydrogen atoms. The nitrogen atom is at the center of the molecule, with the three hydrogen atoms arranged around it. The shape of the molecule is determined by the arrangement of the electrons in the outermost energy level of the nitrogen atom, which is called the valence shell. In the valence shell of the nitrogen atom, there are four pairs of electrons, one of which is a lone pair, while the other three are bonded to hydrogen atoms. These four pairs of electrons repel each other, and try to get as far away from each other as possible. This results in a tetrahedral arrangement of the atoms around the nitrogen atom. A tetrahedron is a three-dimensional shape with four triangular faces, and it looks like a pyramid with a triangular base. In the case of ammonia, the three hydrogen atoms form a triangular base, and the lone pair of electrons sits at the top of the pyramid, creating a tetrahedral shape. Therefore, the shape of ammonia molecules is tetrahedral.
Tambaya 10 Rahoto
The alkyl group is represented by the general formula
Bayanin Amsa
The alkyl group is represented by the general formula CnH2n+1, where "n" represents the number of carbon atoms in the group. An alkyl group is a type of organic molecule that consists of a chain of carbon atoms, each of which is bonded to two hydrogen atoms. The formula CnH2n+1 represents this structure, where "n" is the number of carbon atoms in the chain. The number of hydrogen atoms in the chain is always two more than the number of carbon atoms, hence the formula CnH2n+1.
Tambaya 11 Rahoto
An oxide XO2 has a vapour density of 32. What is the atomic mass of X?
Bayanin Amsa
Molecular mass = vapour density X2
Mm of XO2 = x + 16(2) = x + 32
vapour density = 32
∴ x + 32 = 32 × 2
x + 32 = 64
x = 64 - 32
x = 32
∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2
Tambaya 12 Rahoto
X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be
Bayanin Amsa
Enthalpy Change [?H]Entropy Change [?S]Gibbs free Energy[?G]PositivePositivedepends on T, may be + or -NegativePositivealways negativeNegativeNegativedepends on T, may be + or -PositiveNegativealways positive
?G= ?H ? T?S.
To determine whether ?G will be positive or negative, the value of ?H(change in enthalpy) and ?S (change in entropy) must be given. Likewise the temperature.
Tambaya 13 Rahoto
22688Ra → x86Rn + alpha particle
Bayanin Amsa
The given equation represents the alpha decay of ^226Ra into an unknown nuclide ^86Rn and an alpha particle. During alpha decay, an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons bound together, resulting in a decrease of two in the atomic number and a decrease of four in the mass number. To determine the atomic number and mass number of the unknown nuclide ^86Rn, we can subtract the atomic and mass numbers of the alpha particle from those of ^226Ra: Atomic number: 88 (Ra) - 2 (alpha particle) = 86 (Rn) Mass number: 226 (Ra) - 4 (alpha particle) = 222 (Rn) Therefore, the unknown nuclide produced in the alpha decay of ^226Ra is ^222Rn, and the correct option is (D) 222.
Tambaya 14 Rahoto
The gas that can be collected by downward displacement of air is
Bayanin Amsa
Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.
Tambaya 15 Rahoto
When few drops of concentrated trioxonitrate(V) acid is added to an unknown sample and wanned an intense yellow colouration is observed. The likely functional group present in the sample is
Bayanin Amsa
Xanthopreitic test for the presence of protein, when conc nitric acid is added to the drop, an intense yellow colouration is observed.
It contains all the functional group of protein which includes the amino, alkanol and the carboxylic group. Adding few drops of conc HNO3 to a protein, gives an intense yellow colouration.
It is called Xanthopreitic test.
Tambaya 16 Rahoto
The constituent common to duralumin and alnico is
Bayanin Amsa
The constituent common to duralumin and alnico is aluminum (Al). Duralumin is a type of aluminum alloy that typically contains around 4% copper and small amounts of other metals such as magnesium and manganese. It is known for its high strength-to-weight ratio and was commonly used in aircraft construction in the early 20th century. Alnico, on the other hand, is a type of permanent magnet alloy that contains aluminum, nickel, and cobalt (hence the name "alnico"). It also often contains small amounts of other metals such as copper and iron. Alnico magnets are known for their high magnetic strength and resistance to demagnetization. Although duralumin and alnico are very different in their properties and applications, they both contain aluminum as a common constituent. Aluminum is a versatile metal that is widely used in many different alloys due to its light weight, strength, and corrosion resistance.
Tambaya 17 Rahoto
Incomplete oxidation of ethanol yields
Bayanin Amsa
C2H5OH → CH3CHO → CH3COOH
Ethanol.....oxidation Ethanal........Ethanoic Acid
Primary alcohol oxidises to aldehyde and later to carboxylic acid.
Secondary alcohol oxidises to ketones.
Ethanol is an example of primary alcohol and primary alcohol can be oxidised to aldehyde and carboxylic acid. Wherein, incomplete oxidation of primary alcohol yields aldehyde also known as alkaline while complete oxidation of primary alcohol yields carboxylic acid.
Tambaya 18 Rahoto
The general formula of alkanones is
Bayanin Amsa
The general formula of alkanones is R2CO, which represents a functional group that consists of a carbon atom double-bonded to an oxygen atom (C=O) and attached to two alkyl (or aryl) groups represented by the "R" symbol. In simpler terms, alkanones are a type of organic compound that contain a carbon atom double-bonded to an oxygen atom and two additional carbon-containing groups. The "R" groups can be any combination of alkyl or aryl groups, which are made up of carbon and hydrogen atoms arranged in a specific way. The general formula R2CO applies to all alkanones, regardless of the specific alkyl or aryl groups attached to the central carbon atom. By knowing this formula, we can easily recognize an alkanone when we see one and predict its chemical properties based on the functional group it contains.
Tambaya 19 Rahoto
In the diagram above. X is
Bayanin Amsa
The setup represents the production of sulfur dioxide. And the cylinder marked X is SO2
Tambaya 20 Rahoto
Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]
Bayanin Amsa
M = Molar mass × Quantity of Electricity96500 × no of charge
= MmIT96500n
Copper II Chloride = CuCl2
CuCl2 → Cu2+ + 2Cl2
Mass of compound deposited = Molar mass × Quantity of Electricity96500 × no of charge
Q = IT
I = 0.5A
T = 45 × 60
T = 2700s
Q = 0.5 × 2700
= 1350c
Molarmass = 64gmol-1
no of charge = + 2
Mass = 64×135096500×2
Mass = 0.448g
Tambaya 21 Rahoto
Water for town supply is chlorinate to make it free from
Bayanin Amsa
Water for town supply is chlorinated to make it free from bacteria. Chlorine is added to water in small amounts to kill harmful bacteria and other microorganisms that may be present in the water. These microorganisms can cause illnesses such as typhoid, cholera, and dysentery if ingested. Chlorination is an effective and inexpensive method to disinfect water and protect public health. Chlorine works by breaking down the cell walls of microorganisms, which leads to their death. Chlorine also helps to control the growth of algae and other microorganisms in the water supply, which can cause taste and odor problems. Chlorination does not remove other impurities from the water such as bad color, temporary hardness, or permanent hardness. These impurities may require additional treatment methods such as filtration or water softening to be removed.
Tambaya 22 Rahoto
Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol-1]
Bayanin Amsa
The amount of silver deposited can be calculated using Faraday's law of electrolysis, which states that the amount of a substance deposited at an electrode during electrolysis is proportional to the number of moles of electric charge passed through the solution. We can start by using the formula: moles of substance = charge passed / Faraday's constant Where Faraday's constant is equal to 96500 Cmol^-1. Plugging in the given values: moles of silver = 9650 C / 96500 Cmol^-1 moles of silver = 0.1 So, the amount of silver deposited is 0.1 moles.
Tambaya 23 Rahoto
Due to the high reactivity of sodium, it is usually stored under
Bayanin Amsa
Na is kept under kerosene (paraffin) to avoid reactivity with air.
Paraffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.
Tambaya 24 Rahoto
If 100cm3 of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]
Bayanin Amsa
The rate at which a gas diffuses through a porous plug is inversely proportional to the square root of its molecular weight. The molecular weight of oxygen (O2) is 32 (16 x 2), while the molecular weight of hydrogen (H2) is 2 (1 x 2). Therefore, the square root of the ratio of the molecular weights of hydrogen and oxygen is: sqrt(2/32) = 0.25 This means that hydrogen diffuses through a porous plug 0.25 times as fast as oxygen. If it takes 50 seconds for 100cm3 of oxygen to pass through the porous plug, then it will take 0.25 times longer for the same volume of hydrogen to pass through the same porous plug. Thus, the time taken for the same volume of hydrogen to pass through the same porous plug is: 50 seconds x 0.25 = 12.5 seconds Therefore, the correct answer is 12.5s.
Tambaya 25 Rahoto
The reddish–brown rust on ion roofing sheets consists of
Bayanin Amsa
The reddish-brown rust on iron roofing sheets is mainly composed of Fe2O3.3H2O, which is a compound of iron and oxygen with water molecules. Iron roofing sheets, when exposed to air and moisture, undergo a chemical reaction. Oxygen in the air reacts with the iron to form a new compound called iron oxide (Fe2O3). As moisture (water) is also present, the iron oxide reacts with the water to form hydrated iron oxide (Fe2O3.3H2O), which is commonly known as rust. So, the reddish-brown rust on iron roofing sheets is mainly composed of Fe2O3.3H2O, which is a combination of iron, oxygen, and water molecules.
Tambaya 26 Rahoto
Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O2(l) + X. In the reaction above X is
Bayanin Amsa
In the given chemical equation, the reactants are solid calcium hydroxide (Ca(OH)2) and gaseous ammonium chloride (NH4Cl). When they react, they form solid calcium chloride (CaCl2), liquid water (H2O), and X, an unknown product. To determine the identity of X, we need to balance the chemical equation first. The balanced chemical equation is: Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O(l) + 2NH3(g) As per the balanced chemical equation, the product X is 2 molecules of ammonia (NH3). Therefore, the answer is (b) NH3.
Tambaya 27 Rahoto
CH2S(g) + O2(g) → 2Cn + SO2(g)
What is the change in the oxidation number of copper in the reaction?
Bayanin Amsa
In the reactant;
Cu2S
2 Cu - 2(1) = 0
2 Cu = 2
Cu = 22
Cu = +1
In the product, Cu
Cu = O
The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively
Tambaya 28 Rahoto
When ΔH is negative, a reaction is said to be
Bayanin Amsa
When ΔH is negative, a reaction is exothermic. In simpler terms, exothermic reactions release energy into their surroundings, which is often felt as heat. When ΔH is negative, it means that the enthalpy change of the reaction is negative, indicating that energy is being released. This can be seen as a decrease in temperature in the surrounding environment or a release of heat or light. Examples of exothermic reactions include combustion reactions, where a fuel combines with oxygen to release heat and light, or the reaction between an acid and a base to form a salt and water. In both cases, energy is released into the surrounding environment, and the enthalpy change is negative (ΔH < 0).
Tambaya 29 Rahoto
The tincture of iodine means iodine dissolved in
Bayanin Amsa
It is also called weak iodine solution. Tincture solutions are characterized by the presence of alcohol.
Tambaya 30 Rahoto
The process that occurs when two equivalent forms of a compound are in equilibrium is
Bayanin Amsa
Resonance involves two forms of a compound.
Isomerism involves two or more forms of an element.
Reforming involves the rearrangement of molecule.
This deals with a two forms of a molecule where the chemical connectivity is the same but the electrons are distributed differently around the structure.
Tambaya 31 Rahoto
Aluminium hydroxide is used in the dyeing industry as a
Bayanin Amsa
Aluminium hydroxide is used in the dyeing industry as a mordant. It combines with a dye and thereby fixes it in a material.
Tambaya 32 Rahoto
The acid anhydride that will produce weak acid in water is
Bayanin Amsa
H2CO3 is an example of a weak acid while H2SO4 and HNO3 are examples of a strong acid.
CO2 combines with water to give a weak trioxocarbonate (IV) acid.
CO2 + H2O ? H2CO3
Tambaya 33 Rahoto
A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?
Bayanin Amsa
The pH scale measures the acidity or basicity of a solution. The pH is defined as the negative logarithm (base 10) of the concentration of the hydronium ion (H3O+), which is given by the expression: pH = -log[H3O+] To calculate the concentration of the hydroxide ion (OH-) in the juice, we need to use the equation for the ion product of water: Kw = [H3O+][OH-] = 1.0 × 10-14 where Kw is the ion product constant of water. At 25°C, Kw has a value of 1.0 × 10-14. To solve for [OH-], we need to rearrange the equation: [OH-] = Kw / [H3O+] Now we can substitute the pH of the orange juice to calculate the concentration of hydronium ion [H3O+]. pH = -log[H3O+] 3.80 = -log[H3O+] [H3O+] = 10^(-3.80) [H3O+] = 1.58 × 10^(-4) mol/L Substituting this value into the ion product equation: [OH-] = Kw / [H3O+] [OH-] = 1.0 × 10^-14 / (1.58 × 10^-4) [OH-] = 6.33 × 10^-11 mol/L Therefore, the concentration of the hydroxide ion in the orange juice is 6.33 × 10^-11 mol/L.
Tambaya 34 Rahoto
In the laboratory preparation of ethyl ethanoate, the water present in the mixture is removed using a solution of
Bayanin Amsa
This is used in the preparation of ethyl ethanoate to remove the water present since it serves as a dehydrated agent.
Tambaya 35 Rahoto
The enzyme used in the hydrolysis of starch to dextrin and maltose is
Bayanin Amsa
This is any amylase or a mixture of amylases that converts starch to dextrin and maltose.
Tambaya 36 Rahoto
The constituent of air necessary in the rusting process are
Bayanin Amsa
The constituent of air includes O2, CO2, N2 and Noble gases. While for a rusting process to take place, the presence of O2, H2O and CO2 is important.
The constituent of air includes O2, CO2, N2 and Noble gases and for rusting process to take place.
The presence of O2 and CO2 as a constituent of air is indispensable
Tambaya 37 Rahoto
A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?
Bayanin Amsa
To solve this problem, we can use the combined gas law, which states that the product of pressure and volume divided by temperature is constant, as long as the amount of gas and the pressure remain constant. In this case, the pressure is constant and the initial conditions of the gas are: - Volume (V1) = 2 dm3 - Temperature (T1) = 300 K Let's call the final temperature T2, and the final volume V2 = 2V1 = 4 dm3, since we want the volume to be doubled. Using the combined gas law, we can set up the following equation: P * V1 / T1 = P * V2 / T2 Simplifying the equation, we get: V1 / T1 = V2 / T2 Substituting the values we know, we get: 2 / 300 = 4 / T2 Cross-multiplying and solving for T2, we get: T2 = 4 * 300 / 2 = 600 K Therefore, the temperature at which the gas will double its volume, keeping the pressure constant, is 600 K. Option (d) is the correct answer.
Tambaya 38 Rahoto
The enzyme used in the hydrolysis of starch to dextrin and maltose is
Bayanin Amsa
The enzyme used in the hydrolysis of starch to dextrin and maltose is called amylase. Starch is a complex carbohydrate found in many foods, including grains, potatoes, and vegetables. It consists of long chains of glucose molecules that are linked together. To digest starch, our bodies use the enzyme amylase, which breaks down the long chains into smaller molecules of dextrin and maltose. These smaller molecules are more easily absorbed into the bloodstream and used as energy by the body. Amylase is produced in the salivary glands and pancreas and is released into the digestive system when we eat. In the mouth, amylase begins breaking down starch as we chew our food, and then continues its work in the small intestine where it is released by the pancreas. In summary, amylase is the enzyme responsible for breaking down starch into smaller molecules of dextrin and maltose during digestion.
Tambaya 39 Rahoto
The furring of kettles is caused by the presence in water of
Bayanin Amsa
Furring of kettles is caused by the temporary hardness in water. Temporary hardness in water is caused by calcium and magnesium trioxocarbonate (IV)
CaCO3 causes the furring of kettles
Tambaya 40 Rahoto
In the diagram above, Y is
Bayanin Amsa
In the preparation of sulphur dioxide by the action of dilute acids on sulphates and bisulphites. conc H2SO4 helps to release SO2 from the mixture.
The setup represents the production of sulphur dioxide
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