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Tambaya 1 Rahoto
The oxidation number of Fe in [Fe(CN)\(_6\)]\(^{3-}\) is
Bayanin Amsa
In the compound [Fe(CN)\(_6\)]\(^{3-}\), the overall charge of the compound is negative 3 (\(^{3-}\)), which means the sum of the oxidation states of all the atoms in the compound must equal negative 3. We can start by looking at the cyanide ion (CN\(^-\)), which has a known oxidation state of negative 1 (\(-1\)). There are six of them in the compound, giving a total charge of negative 6 (\(-1 \times 6 = -6\)). To find the oxidation state of iron (Fe), we can use the fact that the sum of the oxidation states of all the atoms in a compound is equal to the overall charge of the compound. Let x be the oxidation state of iron (Fe). Therefore, x + (-6) = -3 Solving for x, we get: x = +3 Therefore, the oxidation state of iron (Fe) in [Fe(CN)\(_6\)]\(^{3-}\) is +3.
Tambaya 2 Rahoto
The formula of mercury (l) dioxonitrate (III) is
Bayanin Amsa
The formula of a compound gives the type and number of atoms that are present in a molecule of that compound. To determine the formula of mercury (l) dioxonitrate (III), we need to determine the oxidation states of the atoms present in the compound. The symbol Hg represents the element mercury, which has a valence of +1 in this compound. The symbol NO\(_2\) represents the nitrite ion, which has a valence of -1. The Roman numeral III indicates that the compound contains three nitrite ions. To balance the charges of the mercury ion and the nitrite ions, we need two nitrite ions for every mercury ion. This gives us the formula Hg\(_2\)(NO\(_2\))\(_2\), and the correct option is (C) Hg\(_2\)(NO\(_2\))\(_2\).
Tambaya 3 Rahoto
MgO does not readily dissolve in water because
Bayanin Amsa
MgO does not readily dissolve in water because its lattice energy is higher than its hydration energy. MgO is an ionic compound that has a strong ionic bond between magnesium and oxygen. When it is placed in water, the water molecules surround the ions and try to separate them from each other. This process is known as hydration, and it releases energy. However, the energy required to break the ionic bond in MgO is greater than the energy released during hydration. Therefore, MgO does not dissolve easily in water.
Tambaya 4 Rahoto
Consider the reaction represented by the following equation: xCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O. The values of x,y and z respectively, are
Bayanin Amsa
The chemical equation represents the combustion of methanol (CH\(_3\)OH) in the presence of oxygen (O\(_2\)). The reactants, methanol and oxygen, combine to produce carbon dioxide (CO\(_2\)) and water (H\(_2\)O). The balanced chemical equation for this reaction is: 2CH\(_3\)OH + 3O\(_2\) → 2CO\(_2\) + 4H\(_2\)O From this equation, we can see that the values of x, y, and z respectively are 2, 3, and 4. Therefore, the correct option is (A) 2, 3, and 4.
Tambaya 5 Rahoto
The complex compound formed by the combination of two elements with a large electronegativity difference is likely to be
Bayanin Amsa
When two elements with a large electronegativity difference combine, one element attracts electrons more strongly than the other, causing the electrons to be shared unequally. In such a case, the electron pair is pulled closer to the more electronegative element, resulting in partial charges on the atoms. This type of bonding is called polar covalent bonding, where there is a difference in electronegativity between the two atoms. However, if the electronegativity difference between the two elements is very large, one element will transfer its electrons completely to the other element to form ions. In this case, ionic bonding occurs, where there is a complete transfer of electrons from one element to another to form positively and negatively charged ions. Therefore, when two elements with a large electronegativity difference combine, the resulting compound is likely to be ionic in nature. Hence, the correct option is ionic.
Tambaya 6 Rahoto
At ordinary temperature H\(_2\)O is a liquid while H\(_2\)S is a gas. This is because H\(_2\)O has
Bayanin Amsa
H\(_2\)O is a liquid at ordinary temperature because it has strong hydrogen bonds holding its molecules together. Hydrogen bonds are the intermolecular forces of attraction between the hydrogen atoms of one molecule and the oxygen atom of another molecule. These hydrogen bonds are relatively strong compared to other intermolecular forces like induced dipole-induced dipole forces or ionic forces. On the other hand, H\(_2\)S is a gas at ordinary temperature because it has weak intermolecular forces holding its molecules together. Thus, the correct option is "strong hydrogen bonds holding its molecules together".
Tambaya 7 Rahoto
A substance L reacts with NH\(_4\)NO\(_{3(aq)}\) to generate ammonia gas. L is likely to be
Bayanin Amsa
The reaction of a substance with NH\(_4\)NO\(_{3(aq)}\) to generate ammonia gas is a test for identifying substances that are basic in nature. The reaction is known as the "Ammonia Test". NH\(_4\)NO\(_{3(aq)}\) is an acidic solution and it reacts with basic substances to produce ammonia gas. Among the given options, only NaOH is a strong base that can react with NH\(_4\)NO\(_{3(aq)}\) to generate ammonia gas. Therefore, the substance L is likely to be NaOH.
Tambaya 8 Rahoto
The volume occupied by 0.4 g of hydrogen gas at s.t.p. is [H = 1. 00; Molar volume at s.t.p. = 22.4 dm3]
Tambaya 9 Rahoto
The hydrolysis of NH\(_4\)CI gives
Tambaya 10 Rahoto
When a substance changes directly from the solid state to the gaseous state without forming a liquid, the substance is said to
Bayanin Amsa
When a substance changes directly from the solid state to the gaseous state without forming a liquid, the process is called sublimation. This means that the solid substance goes through a phase change and turns into gas without passing through the liquid phase. The term "sublime" is used to describe this process. For example, dry ice, which is solid carbon dioxide, undergoes sublimation when it is exposed to room temperature or higher temperatures, and turns directly into carbon dioxide gas without forming a liquid. Similarly, iodine crystals, which are solids, can sublime when heated, forming purple vapors of iodine gas.
Tambaya 11 Rahoto
The complex compound formed when aluminum dissolves in aqueous sodium hydroxide is
Bayanin Amsa
The complex compound formed when aluminum dissolves in aqueous sodium hydroxide is NaAl(OH)4. This is because when aluminum reacts with sodium hydroxide, it forms a complex ion [Al(OH)4]-. This complex ion has a negative charge and combines with a positive ion like Na+ to form the final compound NaAl(OH)4. The other options are not correct because they do not represent the correct combination of aluminum and sodium hydroxide.
Tambaya 12 Rahoto
Consider the following reaction equation: CaCO\(_{3(s)}\) + 2HCI\(_{(aq)}\) → CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\). What mass of CaCI\(_2\) would be obtained when 25.0g of CaCO\(_{3(s)}\) is reacted with excess HCI\(_{(aq)}\)? [CaCO\(_3 = 100; CaCI\(_2\) = 111]
Bayanin Amsa
To solve this problem, we need to use stoichiometry to relate the amount of CaCO\(_3\) reacted to the amount of CaCl\(_2\) produced. First, we need to balance the equation to make sure that the number of atoms of each element is equal on both sides: CaCO\(_3\) + 2HCl → CaCl\(_2\) + H\(_2\)O + CO\(_2\) Next, we need to determine the mole ratio between CaCO\(_3\) and CaCl\(_2\) from the balanced equation: 1 mole CaCO\(_3\) : 1 mole CaCl\(_2\) Now, we can use this mole ratio to calculate the number of moles of CaCl\(_2\) produced from 25.0g of CaCO\(_3\): 25.0 g CaCO\(_3\) x (1 mole CaCO\(_3\)/100 g CaCO\(_3\)) x (1 mole CaCl\(_2\)/1 mole CaCO\(_3\)) = 0.250 mole CaCl\(_2\) Finally, we can use the molar mass of CaCl\(_2\) to calculate the mass produced: 0.250 mole CaCl\(_2\) x 111 g CaCl\(_2\)/1 mole CaCl\(_2\) = 27.8 g CaCl\(_2\) Therefore, the mass of CaCl\(_2\) produced when 25.0g of CaCO\(_3\) is reacted with excess HCl\(_{(aq)}\) is 27.8 g. The correct answer is.
Tambaya 13 Rahoto
The energy evolved when magnesium burns in air is in the form of
Bayanin Amsa
When magnesium burns in air, it undergoes a chemical reaction with oxygen to form magnesium oxide. This reaction is highly exothermic, which means it releases a significant amount of energy in the form of heat and light. The heat energy is produced as a result of the exothermic reaction between magnesium and oxygen. This energy is released into the surroundings, which causes the temperature of the surroundings to increase. The amount of heat produced depends on the quantity of magnesium burned and the conditions under which the reaction occurs. The light energy is produced due to the high temperature of the reaction, which causes the magnesium and magnesium oxide to emit light in the form of a bright white flame. The intensity and color of the light depend on the temperature of the reaction and the composition of the materials involved. Therefore, the energy evolved when magnesium burns in air is in the form of light and heat, and the correct option is (C) light and heat.
Tambaya 14 Rahoto
The shape of a graphite crystal is
Bayanin Amsa
The shape of a graphite crystal is hexagonal. Graphite is made up of layers of carbon atoms arranged in a hexagonal lattice structure. Each carbon atom is bonded to three other carbon atoms in a flat plane, forming a hexagonal pattern. The layers are stacked on top of each other, held together by weak intermolecular forces, allowing the layers to slide past each other, making graphite a good lubricant.
Tambaya 15 Rahoto
The number of orbitals in a p-sub level of an atom is
Bayanin Amsa
An atom consists of electrons that occupy specific energy levels or shells around the nucleus. These shells are further divided into sub-levels. There are four types of sub-levels: s, p, d, and f. Each sub-level has a specific number of orbitals, which are regions of space where electrons are most likely to be found. The p-sub level has three orbitals. These orbitals are designated as px, py, and pz, each of which can hold a maximum of two electrons. Therefore, the p-sub level can hold a total of six electrons, which are distributed among the three orbitals. In summary, the number of orbitals in a p-sub level of an atom is three, and the correct option is (B) 3.
Tambaya 16 Rahoto
Which of the following statements about a radioactive substance is/are correct? l. lt emits radiation continuously and spontaneously. ll. The emitted radiations are affected by temperature and pressure III. The radiation can penetrate opaque matter
Bayanin Amsa
The correct statement(s) about a radioactive substance is/are l and lll only. Radioactive substances emit radiation continuously and spontaneously. The emitted radiations are not affected by temperature and pressure, hence statement ll is incorrect. Radiation can penetrate opaque matter, which is why it is important to handle radioactive substances with caution and protective gear. Therefore, statement lll is correct.
Tambaya 17 Rahoto
The d-block elements are paramagnetic because they
Bayanin Amsa
The correct answer is that the d-block elements are paramagnetic because they contain unpaired electrons in the partially filled 3d-orbital. Paramagnetism refers to the tendency of a substance to become magnetized when placed in an external magnetic field. This occurs because unpaired electrons in the d-orbitals of these elements are attracted to the magnetic field and align themselves with it, causing the substance as a whole to become magnetized. Therefore, the presence of unpaired electrons in the d-orbitals of these elements is responsible for their paramagnetic behavior.
Tambaya 20 Rahoto
Which of the following pollutants is not usually recycled?
Bayanin Amsa
Nuclear wastes are not usually recycled. Unlike other types of wastes such as aluminium cans, glass bottles, and paper wastes which can be recycled, nuclear wastes are highly radioactive and pose a significant threat to human health and the environment. Therefore, nuclear wastes are typically stored in special facilities designed to prevent their escape into the environment, rather than being recycled or reused.
Tambaya 21 Rahoto
The vapour pressure of a liquid depends on l. temperature. ll. rate of condensation. III. cohesive forces holding the particles together
Bayanin Amsa
The vapour pressure of a liquid is the pressure exerted by the molecules of the liquid when they evaporate and become a gas. The higher the temperature, the more energy the liquid molecules have, which means they move faster and are more likely to escape into the gas phase. So, option l is correct, as the vapour pressure of a liquid depends on the temperature. Rate of condensation refers to the process by which gas molecules collide and stick together to form a liquid. However, this process does not affect the vapour pressure of the liquid, as it is only the reverse of the process of evaporation. Therefore, option ll is not correct. Cohesive forces are the forces that hold the particles of a liquid together. The stronger these forces, the harder it is for the molecules to escape into the gas phase, and thus the lower the vapour pressure of the liquid. So, option lll is correct, as the vapour pressure of a liquid depends on the cohesive forces holding the particles together. In conclusion, the correct option is l and lll only. The vapour pressure of a liquid depends on the temperature and the cohesive forces holding the particles together.
Tambaya 22 Rahoto
The postulate that molecules are in constant random motion best explains why liquids
Bayanin Amsa
Tambaya 23 Rahoto
Which of the following electron configurations represents that of an atom in its ground state?
Bayanin Amsa
The electron configuration of an atom in its ground state is the arrangement of electrons in the lowest possible energy levels. The configuration that represents the ground state will have all the electrons in their respective orbitals according to the Aufbau principle, Pauli exclusion principle, and Hund's rule. Now, looking at the options: - has a partially filled p orbital, which is not allowed for an atom in its ground state. - also has a partially filled p orbital, which is again not allowed for an atom in its ground state. - has three electrons in the p sublevel, which violates the Pauli exclusion principle. - has all the electrons in their respective orbitals according to the Aufbau principle, Pauli exclusion principle, and Hund's rule. Thus, represents the electron configuration of an atom in its ground state. Therefore, the correct option is the one that shows the electron configuration with all the electrons in their respective orbitals according to the Aufbau principle, Pauli exclusion principle, and Hund's rule. In this case, it is: 1s22s22p63s0.
Tambaya 24 Rahoto
Consider the following reaction equation: C\(_{16}\)H\(_{34}\) → C\(_5\)H\(_{12}\) + C\(_{11}\)H\(_{22}\). The process represented by the equation is
Bayanin Amsa
The process represented by the given chemical equation is cracking. Cracking is a chemical process that involves breaking down larger hydrocarbon molecules into smaller ones. In this reaction, a long-chain hydrocarbon molecule (C16H34) is broken down into two smaller hydrocarbon molecules (C5H12 and C11H22) by the application of heat, which causes the bonds between the atoms in the hydrocarbon to break. Cracking is an important process in the petroleum industry because it allows for the production of smaller hydrocarbons that are more useful as fuels and other industrial applications.
Tambaya 25 Rahoto
Which of the following halogens is liquid at room temperature?
Bayanin Amsa
Bromine is the halogen that is liquid at room temperature. This is because it has the lowest boiling point of all the halogens, meaning it can easily evaporate to become a gas at higher temperatures. At room temperature, which is typically around 20-25 degrees Celsius (68-77 degrees Fahrenheit), bromine remains in its liquid state because its boiling point is 59 degrees Celsius (138 degrees Fahrenheit). Chlorine and fluorine are gases at room temperature because they have much lower boiling points than bromine. Iodine is a solid at room temperature, as it has a higher boiling point than bromine and therefore requires higher temperatures to turn into a liquid.
Tambaya 26 Rahoto
Which of the following substances would not produce ethanol when fermented?
Bayanin Amsa
Fermentation is a process by which microorganisms such as yeast convert sugars into alcohol, such as ethanol. The process occurs under anaerobic conditions, meaning in the absence of oxygen. The microorganisms break down the sugar molecules into ethanol and carbon dioxide. Out of the given options, vinegar would not produce ethanol when fermented because vinegar is a diluted solution of acetic acid, not a sugar. It is produced by the fermentation of ethanol by acetic acid bacteria, which convert the ethanol into acetic acid. On the other hand, cane sugar, glucose, and starch are all sources of fermentable sugars and can be used to produce ethanol when fermented. Therefore, the correct option is (D) Vinegar.
Tambaya 27 Rahoto
Two elements, X and Y are in the same group on the periodic table because they both have the same
Bayanin Amsa
The correct answer is "number of valence electrons". Elements in the same group on the periodic table have similar chemical and physical properties because they have the same number of valence electrons, which are the electrons in the outermost energy level of an atom. Valence electrons are important because they are involved in chemical bonding, which determines the reactivity and behavior of elements. Therefore, elements in the same group have similar chemical reactivity and tend to form similar types of chemical bonds.
Tambaya 29 Rahoto
A metal that is widely used in the manufacture of paints and overhead electric cables is
Bayanin Amsa
The metal that is commonly used in the production of both paints and overhead electric cables is aluminum. Aluminum is a lightweight and durable metal that is resistant to corrosion and can conduct electricity efficiently. It is also easy to shape and mold, which makes it a versatile material for various industrial applications. In the production of paints, aluminum is used as a pigment to provide a metallic or reflective finish. In the case of overhead electric cables, aluminum is preferred because it is lightweight, strong, and can conduct electricity effectively.
Tambaya 30 Rahoto
Brass is a mixture of
Bayanin Amsa
Brass is a metal alloy made by combining two main elements, copper (Cu) and zinc (Zn), in varying proportions. The exact proportions of copper and zinc can vary depending on the desired properties of the brass, such as its color, hardness, and durability. Other elements, such as lead (Pb), tin (Sn), or magnesium (Mg), can also be added to brass to enhance certain characteristics. However, they are not considered the main components of brass, which is always a mixture of copper and zinc. So, the correct answer is "Cu and Zn".
Tambaya 31 Rahoto
Which of the following gases has the lowest rate of diffusion under the same condition? [H = 1.00; He = 4. 00; O = 16.0; CI = 35.5]
Bayanin Amsa
The gas with the lowest rate of diffusion under the given conditions is the gas that is the heaviest, which means it has the highest molar mass. The molar masses of the gases listed in the question are: H2 (2.0 g/mol), He (4.0 g/mol), O2 (32.0 g/mol), and CI2 (71.0 g/mol). Therefore, the gas with the lowest rate of diffusion is CI2, because it has the highest molar mass. This means that CI2 molecules are larger and heavier than molecules of the other gases listed, and so they move more slowly and take longer to diffuse through a medium. In summary, the heavier the gas, the slower it diffuses, and the lighter the gas, the faster it diffuses.
Tambaya 32 Rahoto
Which of the following ions has an electron configuration different from the other?
Bayanin Amsa
The ion with an electron configuration different from the others is 17CI-. Chlorine (CI) has the electron configuration of 1s22s22p63s23p5 when neutral, but when it gains an electron to become an ion, it gains one more electron in the 3p subshell to have a full outer shell of 8 electrons, which gives it the electron configuration of 1s22s22p63s23p6. Oxygen (O), magnesium (Mg), and aluminum (AI) have electron configurations of 1s22s22p4, 1s22s22p63s2, and 1s22s22p63s23p1, respectively, when neutral. When they become ions, they either lose or gain electrons to achieve a full outer shell of 8 electrons for stability, which does not change their electron configuration in the same shell. Therefore, the correct answer is 17CI-.
Tambaya 33 Rahoto
On heating, the following trioxocarbonate (IV) salts decompose to give solid residue except
Bayanin Amsa
Tambaya 34 Rahoto
Which of the following devices function on redox reaction? l. Dry cell ll. Car battery III. Electric generator
Bayanin Amsa
Both the dry cell and car battery function on redox reactions, which involve the transfer of electrons between reactants. A dry cell contains an electrolyte paste that allows for the transfer of electrons between the zinc outer casing (anode) and the carbon rod (cathode) in the center of the cell. The zinc atoms lose electrons and become positively charged ions, while the carbon atoms gain those electrons and become negatively charged ions. This transfer of electrons creates an electric current that can power devices like flashlights. Similarly, a car battery uses a redox reaction to generate electricity. The battery contains two electrodes - a lead dioxide cathode and a lead anode - that are immersed in an electrolyte solution of sulfuric acid. When the battery is connected to a circuit, the lead dioxide electrode accepts electrons from the lead anode, creating a flow of electricity. An electric generator, on the other hand, does not rely on a redox reaction to function. Instead, it works by converting mechanical energy into electrical energy. The generator contains a wire coil that rotates within a magnetic field, creating an electric current in the wire. Therefore, the devices that function on redox reactions are l and ll only, and the answer is option C - l and ll only.
Tambaya 35 Rahoto
Which of the following elements is a metalloid?
Bayanin Amsa
A metalloid is an element that has properties of both metals and nonmetals. These elements are also called semimetals. Out of the options listed, only Silicon is a metalloid. Silicon is a chemical element with the symbol Si and atomic number 14. It is a hard, brittle crystalline solid with a blue-grey metallic lustre, and it is a member of the group 14 of the periodic table. Carbon, oxygen, and sodium are not metalloids. Carbon and oxygen are nonmetals, while sodium is a metal. In summary, Silicon is a metalloid because it has properties of both metals and nonmetals, while carbon, oxygen, and sodium are not metalloids because they do not have properties of both metals and nonmetals.
Tambaya 36 Rahoto
Which of the following substances is mainly responsible for the depletion of the ozone layer?
Bayanin Amsa
The substance mainly responsible for the depletion of the ozone layer is Chlorofluorocarbon or CFC. CFCs are a type of synthetic chemical compound that is commonly used in refrigerants, air conditioning systems, and aerosol sprays. When CFCs are released into the atmosphere, they rise up to the ozone layer and react with the ozone molecules. This chemical reaction breaks down the ozone molecules, resulting in the depletion of the ozone layer. The ozone layer is an essential layer in the Earth's atmosphere that protects us from harmful ultraviolet (UV) radiation from the sun. A decrease in the ozone layer could lead to an increase in skin cancer and other negative effects on human and animal health.
Tambaya 37 Rahoto
Consider the reaction represented by the following equation: CuO\(_{(s)}\) + H\(_2\)SO\(_{4(aq)}\) → CuSO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\). Which of the following factors will not affect the rate of the reaction?
Bayanin Amsa
The factor that will not affect the rate of the reaction is the pressure of the reaction system. The given chemical equation does not involve any gaseous reactants or products, so the pressure of the reaction system will not have any effect on the rate of the reaction. The other factors such as particle size of CuO, concentration of H\(_2\)SO\(_4(aq)\), and temperature of the reacting mixture can all affect the rate of the reaction. - The smaller the particle size of CuO, the larger the surface area, which leads to an increase in the frequency of collisions between reactant particles and hence an increase in the rate of the reaction. - The higher the concentration of H\(_2\)SO\(_4(aq)\), the more the number of H\(_2\)SO\(_4(aq)\) molecules available for reaction with CuO, which leads to an increase in the frequency of collisions and hence an increase in the rate of the reaction. - The higher the temperature of the reacting mixture, the higher the kinetic energy of the reactant molecules, which leads to an increase in the frequency of collisions and hence an increase in the rate of the reaction.
Tambaya 38 Rahoto
Which of the following industrial processes depends on the action of enzymes?
Bayanin Amsa
The industrial process that depends on the action of enzymes is the brewing of beer. Enzymes are proteins that act as catalysts, which means they speed up chemical reactions without being consumed in the process. In the brewing process, enzymes break down complex sugars in grains like barley into simpler sugars like glucose, which can then be fermented by yeast to produce alcohol and carbon dioxide. This process is called mashing and it relies on the action of enzymes naturally present in the grains, or added to the mash in the form of malted barley. So, enzymes play a crucial role in converting the complex carbohydrates in grains into the simple sugars needed to produce beer.
Tambaya 39 Rahoto
Which of the following metals would readily displace hydrogen from steam?
Bayanin Amsa
The metal that would readily displace hydrogen from steam is Magnesium. This is because Magnesium is a highly reactive metal that easily reacts with water vapor (steam) to produce magnesium oxide and hydrogen gas. The reaction is as follows: Mg (s) + H2O (g) → MgO (s) + H2 (g) The other metals listed, copper, lead, and silver, are less reactive and would not readily displace hydrogen from steam. Copper and silver are relatively stable and do not react with water or steam, while lead may react slowly but not as vigorously as magnesium.
Tambaya 40 Rahoto
A tertiary alkanol has a molecular formula C\(_4\)H\(_{10}\)O. What is the structural formula of the compound?
Bayanin Amsa
The molecular formula of the compound is C\(_4\)H\(_{10}\)O, which means it has four carbon atoms, ten hydrogen atoms, and one oxygen atom. Since it is a tertiary alkanol, it has a hydroxyl group (-OH) attached to a tertiary carbon atom, which means that the carbon atom is bonded to three other carbon atoms. Option (C) has the correct number of carbon atoms and a tertiary carbon atom attached to the hydroxyl group, which makes it a plausible option. The structural formula for (C) is (CH3)3COH, which represents a tertiary butanol. Therefore, option (C) is the correct answer.
Tambaya 41 Rahoto
Which of the following pH values indicates that a solution is a strong base?
Bayanin Amsa
The pH scale measures the acidity or basicity of a solution, with a pH of 7 being neutral, pH values below 7 being acidic, and pH values above 7 being basic (alkaline). Out of the options listed, pH value of 13 indicates that a solution is a strong base. A strong base is a substance that completely dissociates in water to produce hydroxide ions (OH\(^-\)) and has a pH greater than 7. A pH of 13 is highly alkaline, which indicates that the solution has a high concentration of hydroxide ions. In summary, a pH value of 13 indicates that a solution is a strong base because it is highly alkaline and has a high concentration of hydroxide ions.
Tambaya 42 Rahoto
The number of sulphur atoms in 3.20g of SO\(_{2(g)}\) is [O = 16.0;S = 32.0; Avogadro constant = 6.02 x 10\(^{23}\)]
Bayanin Amsa
To determine the number of sulfur atoms in 3.20g of SO\(_{2(g)}\), we first need to find the number of moles of SO\(_{2(g)}\) present in the given mass. This can be done using the molecular weight of SO\(_{2(g)}\), which is calculated by adding the atomic weights of sulfur and oxygen in the molecule: Molecular weight of SO\(_{2(g)}\) = (1 × atomic weight of S) + (2 × atomic weight of O) = (1 × 32.0) + (2 × 16.0) = 64.0 g/mol Next, we can calculate the number of moles of SO\(_{2(g)}\) present in 3.20g using the formula: Number of moles = Given mass ÷ Molecular weight Number of moles of SO\(_{2(g)}\) = 3.20 g ÷ 64.0 g/mol = 0.05 mol Finally, we can use Avogadro's constant to calculate the number of sulfur atoms present in 0.05 mol of SO\(_{2(g)}\): Number of sulfur atoms = Number of moles of SO\(_{2(g)}\) × Avogadro's constant × Number of sulfur atoms per molecule From the balanced chemical equation for the combustion of sulfur dioxide, we know that each molecule of SO\(_{2(g)}\) contains one atom of sulfur. Therefore, the number of sulfur atoms per molecule of SO\(_{2(g)}\) is 1. Number of sulfur atoms = 0.05 mol × 6.02 x 10\(^{23}\) mol\(^{-1}\) × 1 = 3.01 x 10\(^{22}\) Therefore, the number of sulfur atoms in 3.20g of SO\(_{2(g)}\) is 3.01 x 10\(^{22}\), and the correct option is (A) 3.01 x 10\(^{22}\).
Tambaya 43 Rahoto
The atomic radii of metals are usually
Bayanin Amsa
Metals usually have atomic radii greater than their ionic radii. This is because when a metal atom loses one or more electrons to form a cation, the effective nuclear charge (positive charge of the nucleus) increases, pulling the remaining electrons closer to the nucleus and reducing the size of the ion. However, in the metallic state, the outermost electrons are delocalized, meaning they are not associated with any one particular nucleus, but rather are free to move throughout the metal lattice. This results in the metallic radius being larger than the ionic radius.
Tambaya 44 Rahoto
A metal M forms two types of chlorine, MCl\(_2\) and MCl\(_3\) which of the following laws best explains the relationship between the chlorides? Law of
Bayanin Amsa
Tambaya 45 Rahoto
(a) (i) Define standard electrode potential.
(ii) State two factors that affect the value of standard electrode potential.
(iii) Give two uses of the values of standard electrode potential.
(iv) Draw and label a diagram for an electrochemic cell made up of Cu\(^{2+}\)/Cu; E° = +0.34 V Zn\(^{2+}\)/Zn; E° = -0.76 V
(v) Calculate the e.m.f. of the cell in (a)(iv) above.
(b) (i) In terms of electron transfer, define: I. oxidation; II. oxidizing agent.
(ii) Balance the following redox reaction: MnO\(^{-4}\) + l\(^{-}\) \(\to\) l\(_{2}\) + Mn\(^{2+}\)
(c) Classiify each of the following oxides as basic, amphoteric, acidic or neutral: (i) Carbon (II) oxide; (ii) Sulphu (IV) oxide; (iii) Aluminium oxide; (iv) Lithium oxide.
(d) What is hydrogen bonding?
Tambaya 46 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
Burette readings (initial and final) must be given to two decimal places. Volume of pipette used must also be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A solution containing 6.22 g of an acid H\(_2\)Y per dm\(^3\)
B contains 3.90 g of NaOH per dm\(^3\) of solution.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of acid A used. The equation for the reaction involved in the titration is:
H\(_2\)\(_{(aq)}\) + 2NaOH\(_({aq})\) \(\to\) NaY\(_{(aq)}\) + 2H\(_2\)O\(_{(l)}\)
[H = 1.00; O = 16.0; Na = 23.0]
(b) From your results and the information provided above, calculate the:
(i) The concentration of B in moldm\(^{-3}\)
(ii) concentration of A in moldm\(^3\)
(ii) molar mass of H\(_2\)Y.
(c) State whether the pH of each of the following solutions is lower than 7, greater than 7 or equal to 7. The
(I) solution A before titration
(ii) solution B before titration
Tambaya 47 Rahoto
(a) (i) What is diffusion?
(ii) State Charles' law:
(iii) Sketch a graph to illustrate Charles' law.
(iv) A given mass of a gas occupied 150 cm3 at 27 °C and a pressure of 1.013 x 105 Nm\(^{-2}\). Calculate the temperature at which its volume will be doubled at the same pressure.
(v) Arrange the three states of matter in order of increasing: (i) kinetic energy; (ii) forces of cohesion.
(b) (i) State Le Chatelier's principle. (ii) A metal M forms two oxides containing 11.1% and 20.0% of oxygen. Show that these figures agree with the law of multiple proportion.
(c) The table below shows the physical properties of substances A, B and C.
| Substance |
Melting point/°C | Boiling point/°C | Solubility in water at 25°C |
| A | 30 | 117 | Insolube |
| B | 31 | 160 | Insoluble |
| C | 1200 | 1200 | Insoluble |
(i) If A and B are miscible when melted and B and C react when heated, describe how a mixture of A, B, and C could be separated.
(ii) When 25.25g of the mixture A, B and C was separated, 7.52 g of A and 8.48 g of B were recovered. Assuming i there was no loss of components during separation, calculate the percentage by mass of C in the mixture
Tambaya 48 Rahoto
(a) (i) What is a structural isomer?
(ii) Write all the structural isomeric alkanols with the molecular formula C\(_4\)H\(_{10}\)O.
(iii) Which of the isomers from (a)(ii) above does not react easily on heating with acidified K\(_2\)Cr\(_2\)O\(_7\)?
(b) Chlorine reacted with excessentane in the presence of light. Chloropentane and a gas which fumes on cont with air were produced.
(i) Write an equation for the reaction.
(ii) Draw the structure of the major product.
(iii) What is the role of light in the reaction?
(iv) If a mixture of pentane and the major product is heated, which compounc would distil off first? Give a reason for your answer.
(v) Write the formula of the main product that would have be formed if but -1-ene has been used instead of pentane.
(c) Give the name and structural formula of the product which would be formed by hydration of each of the followinc compounds:
(i) CH\(_3\)CH(CH\(_3\))CH=CH\(_2\); (ii) CH\(_2\)=CHCOOH.
(d) (i) Write the structure of the amino acid, CH\(_3\)CH(NH\(_2\))COOH in: I. acidic medium; II. alkaline medium.
(ii) On analysis, an ammonium salt cf an alkanoic acid gave 60.5% carbon and 6.5% hydrogen. If 0.309 g of the salt yielded 0.0313 g of nitrogen, determine the empirical formula cf the salt. [H = 1.00; C =12.0; N =14.0; O = 16.0]
Bayanin Amsa
None
Tambaya 49 Rahoto
(a) Define each of the following terms: I. biotechnology; II. biogas.
(ii) State two applications of biotechnology
(b) (i) Describe briefly the production of ethanol from sugar cane juice.
(ii) State the by-product of the process in (b)(i)
(iii) Mention two uses of the by-product.
(iv) Ethanol can be produced from both cane sugar and petroleum. Explain briefly why the ethanol from cane sugar is renewable but that from petroleum is non-renewable.
(c) Distinguish between heavy chemicals and fine chemicals. Give one example of each chemical.
(d) Arrange the following gases in increasing order of deviation from ideal gas behaviour: HCl; O\(_2\); Cl\(_2\). Give reason(s) for your answer.
Bayanin Amsa
None
Tambaya 50 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
(a) Explain briefly the observations in each of the following processes:
(i) when carbon (IV) oxide is bubbled through lime water, It turns milky but the milkiness disappears when the gas is bubbled for a long time.
(ii) A precipitate of calcium hydroxide is insoluble in excess sodium hydroxide solution whereas that of lead (ii) hydroxide is soluble
(b)(i) What is a primary standard solution?
(ii) Calculate the mass of sodium trioxocarbonate (V) required to prepare 250 cm\(^3\) of 0.15gmoldm\(^3\) solution hydroxide is soluble. [Na = 23.0; O = 16.0; C = 12.0]
(c) Name one gas that can be collected by
(i) upward displacement of air
(ii) downward displacement of air
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