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Swali 1 Ripoti
If 6.5g of magnesium ribbon reacts completely with Hydrochloric acid, how many grams of hydrogen gas would be liberated?
Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)
[ H = 1.0, Mg = 24.0 ]
Maelezo ya Majibu
Swali 2 Ripoti
which of the following organic compounds has the lowest boiling point?
Maelezo ya Majibu
The organic compound with the lowest boiling point is (CH\(_3\))\(_3\). This is because it is a gas at room temperature and has a very low molecular weight. As the molecular weight of the organic compound increases, the boiling point also increases. Therefore, C\(_4\)H\(_10\) has a higher boiling point than (CH\(_3\))\(_3\), C\(_5\)H\(_12\) has a higher boiling point than C\(_4\)H\(_10\), and C\(_6\)H\(_14\) has the highest boiling point of all.
Swali 3 Ripoti
The properties of a good primary standard solution include the following except?
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Swali 4 Ripoti
consider the following reaction equation: 2Al(OH)\(_3\) + 3H\(_2\)SO\(_4\) →Al\(_2\)(SO4)\(_3\)(aq)+6H\(_2\)O. How many moles of tetraoxosulphate(vi)acid is needed to produce 0.5mol of aluminium tetraoxosulphate(vi)?
Maelezo ya Majibu
Swali 6 Ripoti
If the oxidation potential of Mg and Al are +2.37 volts and + 1.66 volts respectively, then the Mg would?
Maelezo ya Majibu
Oxidation potential is a measure of the tendency of a substance to lose electrons and get oxidized (or become positively charged). A higher oxidation potential means that the substance is more likely to lose electrons, while a lower oxidation potential means that it is less likely to lose electrons. In the case of Mg and Al, Mg has a higher oxidation potential of +2.37 volts compared to Al's +1.66 volts. This means that Mg is more likely to lose electrons and get oxidized, while Al is less likely to do so. Therefore, Mg will replace Al in a reaction. In other words, Mg will take the place of Al, meaning that Mg will be oxidized while Al will remain in the solution.
Swali 7 Ripoti
The process by which alkanoic acid reacts reversibly with alkanols is known as?
Maelezo ya Majibu
The process by which alkanoic acid reacts reversibly with alkanols is known as esterification. During this chemical reaction, a molecule of alkanoic acid and a molecule of an alcohol combine to form an ester and a molecule of water. This reaction is reversible, which means that the products can react with each other to produce the starting materials. Esters are a group of organic compounds that are commonly found in fragrances, flavors, and plastics. They have a characteristic fruity or flowery smell, and they are responsible for the flavor and aroma of many fruits and flowers. Esterification is an important industrial process used to produce a wide variety of products such as perfumes, flavors, and plastics.
Swali 8 Ripoti
When excess chlorine is mixed with ethane at room temperature, the product formed is?
Maelezo ya Majibu
When excess chlorine is mixed with ethane at room temperature, the product formed is 1,2-dichloroethane. This is because ethane reacts with chlorine in the presence of ultraviolet light or heat to form a mixture of 1,2-dichloroethane and 1,1,2,2-tetrachloroethane. However, in the absence of light or heat, only 1,2-dichloroethane is produced as the major product.
Swali 9 Ripoti
The atoms of \(^{64}_{29}\)Cu and \(^{65}_{30}\)Zn have the same number of?
Maelezo ya Majibu
Swali 10 Ripoti
A solid with higher melting and boiling points is likely to be?
Maelezo ya Majibu
A solid with higher melting and boiling points is likely to be an electrovalent compound, also known as an ionic compound. Electrovalent compounds are made up of positively charged ions (cations) and negatively charged ions (anions) held together by electrostatic attractions. These attractions are strong and require a significant amount of energy to break, which is why ionic compounds have higher melting and boiling points compared to covalent compounds, which are held together by weaker intermolecular forces. Covalent compounds consist of molecules with covalent bonds between atoms, where atoms share electrons to form a bond. Covalent compounds typically have low melting and boiling points, as the intermolecular forces are weak. A dative compound, also known as a coordinate compound, contains a coordinate bond where one atom donates both electrons to the bond. These types of compounds can have varying melting and boiling points depending on the strength of the bond. Non-metals can form both covalent and ionic compounds, so their identity alone does not determine the melting and boiling points of a solid. Therefore, the correct option is "An Electrovalent compound".
Swali 11 Ripoti
Stoichiometry is based on the law?
Maelezo ya Majibu
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, the total mass of the reactants must equal the total mass of the products. This means that in a chemical reaction, the number of atoms of each element involved must be conserved. This can be used to calculate the amounts of reactants and products needed to produce a given reaction, and to predict the products that will result from a given reaction. The law of conservation of mass is the fundamental principle upon which stoichiometry is based. Therefore, stoichiometry is based on the law of conservation of mass.
Swali 12 Ripoti
The most suitable indicator for the reaction between NaOH and HCOOH is?
Maelezo ya Majibu
The best indicator for the reaction between NaOH (sodium hydroxide) and HCOOH (formic acid) is phenolphthalein. Phenolphthalein is a pH indicator that changes color based on the acidity of the solution. In an acidic solution, it is colorless, while in a basic (alkaline) solution, it turns pink or red. This makes it ideal for monitoring the progress of a reaction between an acid and a base. As the reaction between NaOH and HCOOH proceeds and the concentration of hydroxide ions increases, the solution will become more basic, causing the phenolphthalein to change color, indicating that the reaction is complete.
Swali 13 Ripoti
One of the deductions from Rutherford's alpha particle scattering experiment is that?
Maelezo ya Majibu
Rutherford's alpha particle scattering experiment was conducted by bombarding a thin gold foil with alpha particles. The experiment led to the conclusion that most of the alpha particles passed through the gold foil with little or no deflection, but some were scattered at various angles. Rutherford deduced that the positive charge of an atom is concentrated in a tiny nucleus at the center of the atom, which is why most of the alpha particles passed through the gold foil with little or no deflection. Therefore, one of the deductions from Rutherford's alpha particle scattering experiment is that the nuclei of atoms are positively charged.
Swali 14 Ripoti
Which of the following substances can be classified as heavy chemicals?
Maelezo ya Majibu
Among the options given, H\(_2\)SO\(_4\) (sulfuric acid) can be classified as a heavy chemical. Heavy chemicals refer to chemicals that are produced in large quantities and are used as raw materials in various industries, such as fertilizers, pharmaceuticals, plastics, and dyes. Sulfuric acid is one of the most widely produced chemicals in the world, and it is used in the production of fertilizers, detergents, and other industrial chemicals. Therefore, H\(_2\)SO\(_4\) can be considered a heavy chemical. CaOCl\(_2\) (calcium hypochlorite), AgNO\(_3\) (silver nitrate), and CaO (calcium oxide) are also chemicals but they are not typically classified as heavy chemicals because they are not produced in the same large quantities as sulfuric acid and are not as widely used as raw materials in various industries.
Swali 15 Ripoti
The group to which elements belong in the periodic table is determined by the number?
Maelezo ya Majibu
The group to which elements belong in the periodic table is determined by the number of valence electrons. Valence electrons are the electrons in the outermost shell (valence shell) of an atom that are involved in chemical bonding. The group number of an element in the periodic table corresponds to the number of valence electrons in its atom. For example, elements in Group 1 (also known as the alkali metals) have one valence electron, elements in Group 2 (also known as the alkaline earth metals) have two valence electrons, and so on. Elements in the same group have similar chemical and physical properties due to their similar valence electron configurations. Therefore, the correct option is "valence electrons".
Swali 16 Ripoti
Kipps apparatus can be used to prepare?
Maelezo ya Majibu
Kipps apparatus can be used to prepare H2S. Kipps apparatus is a type of apparatus used in chemistry laboratories to prepare and store gases. It consists of a glass bulb that is connected to a tap and a gas generator. The bulb can be filled with a solution of a substance that will react with a base to produce a gas, such as H2S. To prepare H2S, the bulb is filled with a solution of sulfur and an acid, which will react to produce H2S gas. The gas can then be collected and stored in the bulb, and the tap can be used to release the gas as needed. Overall, Kipps apparatus is a useful tool for preparing and storing gases, and is commonly used in chemistry and other scientific fields to produce and store gases like H2S for research and experimentation.
Swali 17 Ripoti
If a catalyst is added to a system, the temperature and the pressure remain constant, there would be no effect on the?
Maelezo ya Majibu
A catalyst is a substance that speeds up the rate of a chemical reaction without being consumed in the process. When a catalyst is added to a system, the activation energy of the reaction remains unchanged, but the rate of the reaction increases. The activation energy of a reaction is the minimum amount of energy that must be supplied to the reactants for the reaction to occur. The addition of a catalyst does not change this energy requirement, but it can provide an alternative reaction pathway with a lower activation energy, allowing the reaction to proceed more easily and quickly. The rate of the reverse reaction, which is the rate at which the products of the reaction revert back to the reactants, may also be affected by the addition of a catalyst, but this depends on the specific reaction and the type of catalyst being used. The heat of reaction, which is the change in enthalpy (heat content) that occurs during a chemical reaction, is also not affected by the addition of a catalyst. This is because a catalyst only speeds up the reaction rate, but does not change the energy required for the reaction to occur. Therefore, the addition of a catalyst to a system with constant temperature and pressure would affect the rate of the reaction, but would not affect the activation energy, rate of the reverse reaction, or heat of reaction.
Swali 18 Ripoti
Consider the following reaction represented by the following equation: C\(_2\)H\(_2\) \(^{H\(_2\)}_{→}\) X \(^{H\(_2\)}_{→}\) Y. X and Y respectively are?
Maelezo ya Majibu
The reactant in the equation is C\(_2\)H\(_2\), which is the chemical formula for ethyne (also known as acetylene). The products are formed by adding hydrogen gas (H\(_2\)) to the ethyne molecule. When ethyne reacts with hydrogen, it undergoes a reduction reaction and forms ethene (also known as ethylene) as the first product, which still has a carbon-carbon double bond, and then ethane, which is a saturated hydrocarbon and has only single bonds between carbon atoms, as the second product. Therefore, the products X and Y, respectively, are ethene and ethane. So, the correct option is "ethene and ethane".
Swali 19 Ripoti
The oxidation of chromium in Cr\(_2\)O\(_7)\(-2\) ?
Maelezo ya Majibu
The oxidation state of chromium in the Cr\(_2\)O\(_7)\(-2\) ion is +6. The oxidation state of an element refers to the number of electrons it has lost or gained in a chemical reaction. In the Cr\(_2\)O\(_7)\(-2\) ion, there are two chromium atoms, each with a different oxidation state. To determine the oxidation state of chromium in the Cr\(_2\)O\(_7)\(-2\) ion, we need to consider the overall charge of the ion, which is -2. Oxygen has an oxidation state of -2, so the seven oxygen atoms in the ion contribute a total oxidation state of -14. We also know that the oxidation state of the entire Cr\(_2\)O\(_7)\(-2\) ion is -2, so the two chromium atoms must contribute a total oxidation state of +12 to balance out the negative charge. Since there are two chromium atoms, each chromium atom must have an oxidation state of +6. This is because +6 + +6 = +12, which balances out the -2 charge on the ion. Therefore, the oxidation state of chromium in the Cr\(_2\)O\(_7)\(-2\) ion is +6.
Swali 20 Ripoti
Which of the following pairs of components have the same molecular mass?
[ H = 1.0, C = 12.0, O = 16.0 ]
Maelezo ya Majibu
Molecular mass is the sum of the atomic masses of all the atoms in a molecule. To determine which pair of components have the same molecular mass, we need to calculate the molecular mass of each component and compare them. The molecular mass of each component is calculated by adding up the atomic masses of all the atoms in the molecule. - Ethene: C\(_2\)H\(_4\) = (2 x 12.0) + (4 x 1.0) = 28.0 - Ethyne: C\(_2\)H\(_2\) = (2 x 12.0) + (2 x 1.0) = 26.0 - Ethane: C\(_2\)H\(_6\) = (2 x 12.0) + (6 x 1.0) = 30.0 - Carbon (ii) Oxide: CO = (1 x 12.0) + (1 x 16.0) = 28.0 - Carbon (iv) Oxide: CO\(_2\) = (1 x 12.0) + (2 x 16.0) = 44.0 From the calculations above, we can see that ethene and carbon (ii) oxide have the same molecular mass of 28.0.
Swali 21 Ripoti
Which of the substances is non-polar?
Maelezo ya Majibu
Non-polar substances are those that have an equal distribution of electrons and therefore have no positive or negative poles. An example of a non-polar substance is Hydrogen Bromide. On the other hand, polar substances have a positive and negative pole, meaning that the electrons are not evenly distributed. Examples of polar substances are Water and Ammonia. Bromine, like Hydrogen Bromide, is a non-polar substance.
Swali 22 Ripoti
consider the reaction represented by the following equation: 2SO\(_2\) + O\(_2\) ⇌ 2SO\(_3\)
A decrease in the pressure of the reaction would
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Swali 23 Ripoti
Which of the following metals reacts slowly with cold water?
Maelezo ya Majibu
Of the given options, calcium is the metal that reacts slowly with cold water. Calcium reacts with water, producing hydrogen gas and calcium hydroxide. The reaction is exothermic but the heat produced is not sufficient to ignite the hydrogen gas. The other metals listed in the options - iron, silver, and potassium - are more reactive with water than calcium. Iron reacts with steam, while silver and potassium react vigorously with cold water. Therefore, the correct option is "Calcium".
Swali 24 Ripoti
What is the name of the gas liberated when diluted hydrochloric acid is added to iron(ii)sulphide?
Maelezo ya Majibu
Swali 25 Ripoti
What is the solubility of a salt X at 25ºC if the saturated solution of the salt contains 0.28g in 100cm\(^3\) of the solution in mol\dm\(^3\)?
[ X = 56.0g ]
Maelezo ya Majibu
The solubility of a salt in a solution is the maximum amount of the salt that can dissolve in the solution at a given temperature to form a homogeneous mixture. It is typically expressed as the concentration of the salt in the solution, in units of moles per liter (mol/L) or grams per liter (g/L). To determine the solubility of a salt, you need to know the mass of the salt that dissolves in a given volume of solution at a given temperature. In this case, the saturated solution of the salt contains 0.28 g of the salt in 100 cm\(^3\) of the solution, which means that 0.28 g of the salt dissolves in 100 cm\(^3\) of the solution. To convert the mass of the salt to moles, you need to know the molecular weight of the salt. Given that the molecular weight of the salt is 56.0 g, 0.28 g of the salt is equal to 0.28/56.0 = 0.005 moles of the salt. The concentration of the salt in the solution is then given by the number of moles of the salt divided by the volume of the solution, which is 0.005 moles/100 cm\(^3\) = 0.05 mol/L. Therefore, the solubility of the salt X at 25ºC is 0.05 mol/L or 0.05 M.
Swali 26 Ripoti
Biotechnology is applied in the?
Maelezo ya Majibu
Biotechnology is applied in the treatment of sewage, treatment of water, and purification of chemicals. Biotechnology is the application of living organisms or their products to improve human health and the environment. In the case of sewage treatment, biotechnology is used to degrade organic matter and remove pathogens from wastewater, making it safe for discharge into the environment. This can be done using microbes such as bacteria and fungi that break down organic matter through processes such as aerobic and anaerobic digestion. Similarly, in water treatment, biotechnology can be used to remove pollutants and pathogens from water sources to make them safe for drinking or other uses. This can involve the use of living organisms such as bacteria or algae to remove contaminants or the use of biological processes such as bioremediation to degrade pollutants. In addition, biotechnology can be applied in the purification of chemicals. This can involve the use of enzymes or other biological molecules to catalyze chemical reactions or the use of microbes to produce chemicals such as antibiotics or enzymes. Purification of metals, on the other hand, typically involves physical and chemical methods such as smelting, leaching, and electrolysis, and does not typically involve biotechnology.
Swali 27 Ripoti
The method used to collect gas in the laboratory depends on its?
I. boiling point II. density III. smell IV. solubility in water
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Swali 28 Ripoti
Which of the following methods is NOT used for the separation of mixtures?
Maelezo ya Majibu
Electrolysis is NOT used for the separation of mixtures. Electrolysis is a process in which an electric current is passed through a solution or molten compound to decompose it into its constituent elements or ions. It is used for the extraction of metals from their ores and for the purification of some chemicals. On the other hand, the other three methods mentioned - chromatography, crystallization, and distillation - are commonly used for the separation of mixtures. - Chromatography is a technique used to separate different components of a mixture based on their properties such as size, polarity, and solubility. It is used in fields such as chemistry, biochemistry, and environmental science. - Crystallization is the process of forming solid crystals from a solution, melt, or gas. It is used to purify solids and to separate a mixture of solids into individual components. - Distillation is a process that separates the components of a mixture based on their boiling points. It is used to purify liquids, to separate a mixture of liquids into individual components, and to remove impurities from a liquid. Therefore, electrolysis is NOT used for the separation of mixtures, while the other three methods - chromatography, crystallization, and distillation - are commonly used.
Swali 29 Ripoti
The number of electrons in the 3d orbital of \(_24\)Cr is
Maelezo ya Majibu
The number of electrons in the 3d orbital of \(_{24}\)Cr is 5. In the electronic configuration of chromium (Cr), the atomic number 24 indicates that there are 24 electrons present in the atom. The electronic configuration of Cr is: 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^1\) 3d\(^5\). The electronic configuration tells us that there are five electrons present in the 3d orbital of Cr. This is because the 3d orbital can hold up to 10 electrons, and in the case of Cr, only five of those slots are occupied by electrons. Therefore, the number of electrons in the 3d orbital of \(_{24}\)Cr is 5.
Swali 30 Ripoti
which of the following pairs of substances has the highest heat of neutralization?
Maelezo ya Majibu
The pair of substances that has the highest heat of neutralization is the one composed of a strong acid and a strong base. When an acid reacts with a base, they undergo a neutralization reaction, in which an acid donates a proton (H+) to a base to form water and a salt. The heat released during this reaction is called the heat of neutralization. Strong acids and bases are highly reactive, meaning that they will release a large amount of heat when they neutralize each other. This is because they completely dissociate in water to form ions, making their reaction highly exothermic. On the other hand, weak acids and bases have a lower degree of dissociation and are less reactive, meaning that they will release less heat when they neutralize each other. This is because only a small fraction of the weak acid or base molecules will react, and the reaction will not be as exothermic as that of strong acids and bases. Therefore, the highest heat of neutralization will be observed in the reaction between a strong acid and a strong base.
Swali 31 Ripoti
which of the following reagents could be used to distinguish between propan-1-ol and propan-2-ol
Maelezo ya Majibu
Swali 32 Ripoti
which of the following elements has the highest ionization energy?
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Swali 33 Ripoti
The name of the compound CH\(_3\)C(CH\(_3\))CH\(_2\)CH\(_2\)CH\(_3\)
Maelezo ya Majibu
The name of a organic compound is based on its molecular structure, which includes the arrangement of its carbon and hydrogen atoms. The compound CH\(_3\)C(CH\(_3\))CH\(_2\)CH\(_2\)CH\(_3\) has a total of seven carbon atoms, which form a straight chain. The first carbon atom has three methyl groups (CH\(_3\)) attached to it, while the second carbon atom has only one methyl group attached to it. Based on these characteristics, the name of the compound is 2,2-dimethylpentane. The prefix "2,2-dimethyl" indicates that there are two methyl groups attached to the second carbon atom, and the suffix "pentane" indicates that there are five carbon atoms in the chain.
Swali 34 Ripoti
Vinegar is an aqueous solution of?
Maelezo ya Majibu
Vinegar is an aqueous solution of ethanoic acid.
Swali 35 Ripoti
The bleaching action of chlorine is through the process?
Maelezo ya Majibu
The bleaching action of chlorine is through the process of oxidation. Chlorine is a strong oxidizing agent, which means that it is able to remove electrons from other molecules. In the presence of water, chlorine reacts with water molecules to form hypochlorous acid, which then breaks down into hypochlorite ions and hydrogen ions. The hypochlorite ions are responsible for the bleaching action. When hypochlorite ions come into contact with a colored molecule, such as a dye or a pigment, they oxidize the molecule by removing electrons. This changes the chemical structure of the molecule, making it colorless. This is why chlorine is such an effective bleaching agent. Oxidation is a chemical process in which a molecule loses electrons. This process is opposite to reduction, in which a molecule gains electrons. Chlorine is a strong oxidizing agent because it has a high tendency to attract electrons from other molecules. This makes it useful in a variety of applications, including water treatment and bleaching.
Swali 36 Ripoti
consider the following reaction equation: 2NH\(_3 \) .+ 3Cl\(_2\) → N2 + 6HCl.
NH\(_3\) is acting as?
Maelezo ya Majibu
NH\(_3\) is acting as a reducing agent in the given reaction. A reducing agent is a substance that donates electrons, and as a result, it causes the reduction of another substance in a chemical reaction. In the given reaction equation, NH\(_3\) donates electrons to Cl\(_2\), which gets reduced to HCl. At the same time, NH\(_3\) gets oxidized to N\(_2\). Therefore, NH\(_3\) acts as a reducing agent in the given reaction. It is important to note that a reducing agent always gets oxidized in a chemical reaction as it donates electrons to another substance. Similarly, an oxidizing agent always gets reduced as it accepts electrons from another substance.
Swali 37 Ripoti
A sample of air was bubbled into water. The PH of the water slowly changed from 7 to 6. Which of the following gases in the samples caused the change?
Maelezo ya Majibu
The pH of water is a measure of its acidity, with a lower pH indicating a more acidic solution and a higher pH indicating a more basic solution. When the pH of water changes from 7 to 6, it means that the water has become more acidic. Carbon dioxide (CO\(_2\)) is known to dissolve in water and form carbonic acid, which is a weak acid. This weak acid will increase the acidity of the water, causing its pH to decrease. Oxygen and nitrogen are not acidic gases and would not cause the pH of the water to change. Carbon(II) oxide, also known as carbon monoxide, is a toxic gas that can interfere with the ability of the blood to transport oxygen, but it is not an acidic gas and would not cause the pH of the water to change. Therefore, the gas that caused the change in the pH of the water from 7 to 6 is carbon dioxide (CO\(_2\)).
Swali 38 Ripoti
When an equilibrium is established between dissolved and undissolved solutes, the solution is said to be?
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Swali 39 Ripoti
In a mixture of gases which do not react chemically together, the pressure of the individual gas is?
Maelezo ya Majibu
In a mixture of gases that do not react chemically together, the pressure of each individual gas is called its partial pressure. When different gases are mixed together in a container, each gas behaves independently of the others. The total pressure exerted by the mixture of gases is equal to the sum of the partial pressures of each gas present in the mixture. This is known as Dalton's law of partial pressures. Mathematically, this can be represented as: Total pressure = Partial pressure of gas A + Partial pressure of gas B + ... + Partial pressure of gas N The partial pressure of each gas in the mixture depends on its mole fraction, which is the ratio of the number of moles of that gas to the total number of moles of all gases present in the mixture. The partial pressure of a gas can be calculated using the following formula: Partial pressure of gas A = Mole fraction of gas A x Total pressure Therefore, in a mixture of gases that do not react chemically together, the pressure of the individual gas is known as its partial pressure, and it can be calculated using the above formula.
Swali 40 Ripoti
Which of the following compounds is formed by the oxidation of ethanol?
Maelezo ya Majibu
The compound that is formed by the oxidation of ethanol is acetic acid, which has the chemical formula CH\(_3\)CO\(_2\)H. Ethanol, which has the chemical formula C\(_2\)H\(_5\)OH, can be oxidized by various oxidizing agents, such as potassium dichromate or acidified potassium permanganate, to form acetic acid. During the oxidation process, the two carbon atoms in ethanol are oxidized to form a carbonyl group (C=O) in acetic acid, and two hydrogen atoms are removed to form a carboxyl group (-COOH). The other options listed do not correspond to the oxidation product of ethanol. C\(_2\)H\(_4\)CO\(_2\)H is oxalic acid, which is not derived from ethanol. C\(_2\)H\(_5\)CO\(_2\)H is ethanoic acid, which is another name for acetic acid and is the correct answer. CH\(_3\)OH is methanol, which is not an oxidation product of ethanol. Finally, CH\(_3\)CO\(_2\)H is also acetic acid, but it is not an oxidation product of ethanol since it is already in its fully oxidized form.
Swali 41 Ripoti
which of the following solutions containing one mole per dm\(^3\) of the compound would have the lowest PH?
Maelezo ya Majibu
The pH of a solution is a measure of its acidity, with a lower pH indicating a more acidic solution and a higher pH indicating a more basic solution. The pH scale ranges from 0 to 14, with 7 being neutral. Of the options given, hydrochloric acid has the lowest pH because it is a strong acid. Ethanoic acid (also known as acetic acid) is a weak acid, so its pH is higher than that of hydrochloric acid. Sodium chloride is a salt and does not contribute to the acidity of a solution, so its pH would be neutral (7). Sodium hydrogen trioxocarbonate (IV), also known as sodium carbonate, is a basic compound, so its pH would be higher than 7, meaning it is not as acidic as the other options. Therefore, hydrochloric acid would have the lowest pH of the solutions listed.
Swali 42 Ripoti
When quicklime dissolves in water?
Maelezo ya Majibu
When quicklime (calcium oxide, CaO) dissolves in water, heat is evolved. This process is known as slaking of lime. When CaO is added to water, it undergoes a vigorous reaction in which it combines with water (H2O) to form calcium hydroxide [Ca(OH)2]. The balanced chemical equation for this reaction is: CaO + H2O -> Ca(OH)2 + heat The heat is released because the reaction is exothermic. This means that the products (Ca(OH)2) have a lower energy than the reactants (CaO and H2O). The excess energy is released in the form of heat. The heat evolved during the slaking of lime makes it useful in a variety of applications, such as in construction and in the treatment of water and wastewater. It can also be used as a source of heat in some industrial processes. Therefore, when quicklime dissolves in water, heat is evolved.
Swali 43 Ripoti
If the value of ΔH is positive for a reaction, it means that the reaction is?
Maelezo ya Majibu
If the value of \(\Delta H\) is positive for a reaction, it means that the reaction is endothermic. \(\Delta H\) stands for the change in enthalpy, which is a measure of the amount of heat absorbed or released during a chemical reaction. A positive value of \(\Delta H\) indicates that the reaction absorbs heat from the surroundings. This means that the reaction requires an input of energy in order to proceed. Endothermic reactions are those that absorb heat from the surroundings. In an endothermic reaction, the products have a higher energy than the reactants. The energy is absorbed from the surroundings, so the surroundings become cooler. For example, melting ice is an endothermic process because it absorbs heat from the surroundings. On the other hand, exothermic reactions release heat to the surroundings, and have a negative \(\Delta H\) value. These reactions are often spontaneous and occur without the need for an external energy input. For example, burning wood is an exothermic process because it releases heat to the surroundings. Therefore, if the value of \(\Delta H\) is positive for a reaction, it means that the reaction is endothermic.
Swali 44 Ripoti
How many moles of carbon(iv)oxide contains 16.0g of oxygen?
[ C = 12.0, O = 16.0 ]
Maelezo ya Majibu
Swali 45 Ripoti
Which of the following compounds has coordinate bonds in its structure?
Maelezo ya Majibu
The compound that has coordinate bonds in its structure is Al\(_2\)Cl\(_6\). A coordinate bond (also known as a dative bond) is a type of covalent bond in which both electrons in the bond are donated by one atom. In Al\(_2\)Cl\(_6\), each aluminum atom shares its three valence electrons with three chlorine atoms, and each chlorine atom shares one of its electrons with the aluminum atom. This results in a compound with two aluminum atoms that are both surrounded by six chlorine atoms, and each aluminum atom has six shared electrons, three from its own valence electrons and three from the chlorine atoms' lone pairs of electrons. MgCl\(_2\), AgCl, and NaCl do not have coordinate bonds in their structures. MgCl\(_2\) and NaCl have ionic bonds, where the electrons are transferred from one atom to another. AgCl has a polar covalent bond, where the electrons are shared unequally between the two atoms.
Swali 46 Ripoti
Calcium and magnesium belong to the group of the periodic table because both?
Maelezo ya Majibu
Calcium and magnesium belong to the same group in the periodic table because they have the same number of valence electrons in their outermost shell. They both have two valence electrons, which makes them very similar in terms of their chemical behavior. This similarity is also reflected in their properties, such as their ability to form cations and their tendency to form colorless salts. Therefore, the correct option is "have the same number of valence electrons".
Swali 47 Ripoti
An organic compound contains 72% carbon, 12% hydrogen and 16% oxygen by mass. What is the empirical formula of the compound?
[ H = 1.0. C = 12.0, O = 16.0 ]
Maelezo ya Majibu
The empirical formula of a compound is the simplest whole-number ratio of atoms in the compound. To find the empirical formula, we need to convert the percent composition by mass of each element to its corresponding number of moles, and then divide each number by the smallest number of moles to find the smallest whole-number ratio. Given the following percent composition by mass: - Carbon: 72% = 72 g / (12.0 g/mol) = 6.0 mol - Hydrogen: 12% = 12 g / (1.0 g/mol) = 12.0 mol - Oxygen: 16% = 16 g / (16.0 g/mol) = 1.0 mol Dividing each number of moles by the smallest number of moles (1.0 mol for oxygen), we get: - Carbon: 6.0 mol / 1.0 mol = 6.0 - Hydrogen: 12.0 mol / 1.0 mol = 12.0 - Oxygen: 1.0 mol / 1.0 mol = 1.0 Rounding each number to the nearest whole number, we get: - Carbon: 6.0 - Hydrogen: 12.0 - Oxygen: 1.0 The smallest whole-number ratio of atoms in the compound is C\(_6\)H\(_12\)O, which is the empirical formula of the compound.
Swali 48 Ripoti
The periodic table is an arrangement of elements according to their?
Maelezo ya Majibu
The periodic table is an arrangement of elements according to their atomic number, which is the number of protons in the nucleus of an atom. The elements are arranged in order of increasing atomic number in rows and columns, with elements in the same row (period) having the same number of electron shells and elements in the same column (group/family) having similar chemical and physical properties due to their similar electronic configurations. The periodic table provides a systematic way to organize and study the properties of elements, allowing chemists and scientists to predict the behavior of elements and their compounds. The atomic number is a fundamental property of an element that determines its identity and properties, so it is the most important factor in arranging the elements in the periodic table. Therefore, the correct option is "Atomic number".
Swali 49 Ripoti
Which of the following apparatus is NOT used in volumetric analysis?
Maelezo ya Majibu
A desiccator is not used in volumetric analysis. A desiccator is a container used for preserving moisture-sensitive items by creating a dry and airtight environment. It is commonly used in laboratory settings to store chemicals and other materials that are sensitive to moisture. On the other hand, a pipette and a burette are essential apparatus in volumetric analysis. A pipette is a laboratory tool used to measure and dispense precise volumes of liquid, while a burette is a long, graduated glass tube with a precisely adjustable valve used for adding a solution of known volume to a reaction mixture. A conical flask is also commonly used in volumetric analysis. It is a type of laboratory flask with a cone-shaped body and a narrow neck. It is used to mix and store solutions, as well as to hold reactions while they are being observed.
Swali 51 Ripoti
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
(a) State what would be observed when BaCl, the solution is a portion of a saturated Na\(_2\)CO\(_2\) followed by dilute HCI in excess.
(ii) A gas Q decolorized acidified kM\(_n\)O\(_4\) Solution. Suggest what Q could be
(b) Name one substance used in the laboratory for drying each of the following substances:
(i) ammonia gas:
(ii) carbon (IV) oxide.
(c) Give a reason why a given mass of sodium hydroxide pellets cannot be used to prepare a standard solution.
(a)
BaCl2(aq) + Na2 CO3(aq) followed saturated by dilute HCl in excess
- A white precipitate is formed.
- Effervescence occurs and a gas evolved and the white precipitate dissolves in excess dil HCl
(ii)
Gas Q that decolourized acidified KMnO2 solution could be any of these.
SO2(g) , H2 S(g) , unsaturated hydrocarbon (C2 H2 , C2 H4 ....)
(b)
One substance used in the laboratory for drying each of the following:
(i) ammonia gas - Calcium oxide
(ii) carbon (IV) Oxide - Calcium chloride
- Concentrated tetraoxosulphate( VI) acid.
(c)
Sodium hydroxide pellets cannot be used to prepare a standard solution because it is deliquescent, it absorbs water vapour (moisture) and carbon (IV) oxide from the air, and this would add to its mass.
Maelezo ya Majibu
(a)
BaCl2(aq) + Na2 CO3(aq) followed saturated by dilute HCl in excess
- A white precipitate is formed.
- Effervescence occurs and a gas evolved and the white precipitate dissolves in excess dil HCl
(ii)
Gas Q that decolourized acidified KMnO2 solution could be any of these.
SO2(g) , H2 S(g) , unsaturated hydrocarbon (C2 H2 , C2 H4 ....)
(b)
One substance used in the laboratory for drying each of the following:
(i) ammonia gas - Calcium oxide
(ii) carbon (IV) Oxide - Calcium chloride
- Concentrated tetraoxosulphate( VI) acid.
(c)
Sodium hydroxide pellets cannot be used to prepare a standard solution because it is deliquescent, it absorbs water vapour (moisture) and carbon (IV) oxide from the air, and this would add to its mass.
Swali 52 Ripoti
(a) (i) What is meant by hardness of water?
(ii) Describe briefly how you would determine what proportion of hardness in a given sample of water is due to permanent hardness.
(iii) Give two reasons why hardness of water is an undesirable property.
(b) State the:
(i) reagents;
(ii) conditions for the laboratory preparation of trioxonitrate (V) acid.
(iii) How does concentrated trioxonitrate (V) acid reacts with:
(I) sulphur;
(II) aluminium.
(c) Name one amphoteric oxide.
(a)
(i) Hardness of water refers to the presence of dissolved metal ions, mainly calcium and magnesium, in water. These ions can form insoluble salts with soap, reducing its effectiveness, and can also cause scale buildup in pipes and appliances.
(ii) To determine the proportion of hardness due to permanent hardness in a sample of water, we can first determine the total hardness by titrating the water sample with a solution of ethylenediaminetetraacetic acid (EDTA) using an indicator such as Eriochrome Black T. The EDTA will react with all the metal ions present in the water. Then, we can remove the temporary hardness by boiling the water sample to precipitate out the calcium and magnesium bicarbonates. The difference between the total hardness and the hardness after boiling will give us the proportion of permanent hardness.
(iii) Hardness of water is undesirable because it can cause scale buildup in pipes and appliances, leading to decreased efficiency and increased maintenance costs. It can also reduce the effectiveness of soaps and detergents, requiring more of these products to achieve the desired cleaning results.
(b)
(i) Reagents for the laboratory preparation of trioxonitrate (V) acid include concentrated sulfuric acid and a nitrate salt, such as potassium nitrate or sodium nitrate.
(ii) Trioxonitrate (V) acid can be prepared in the laboratory by adding a nitrate salt to concentrated sulfuric acid while cooling the mixture in an ice bath. The mixture is then gently warmed until trioxonitrate (V) acid fumes appear.
(iii)
(I) When concentrated trioxonitrate (V) acid reacts with sulfur, it produces nitrogen dioxide gas, sulfur dioxide gas, and oxygen gas.
(II) When concentrated trioxonitrate (V) acid reacts with aluminum, it produces aluminum nitrate, water, and nitrogen dioxide gas.
(c)
One example of an amphoteric oxide is aluminum oxide (Al2O3). An amphoteric oxide is a type of oxide that can react with both acids and bases to form salts and water. Aluminum oxide can react with acids to form salts such as aluminum nitrate, and can also react with bases to form salts such as sodium aluminate.
Maelezo ya Majibu
(a)
(i) Hardness of water refers to the presence of dissolved metal ions, mainly calcium and magnesium, in water. These ions can form insoluble salts with soap, reducing its effectiveness, and can also cause scale buildup in pipes and appliances.
(ii) To determine the proportion of hardness due to permanent hardness in a sample of water, we can first determine the total hardness by titrating the water sample with a solution of ethylenediaminetetraacetic acid (EDTA) using an indicator such as Eriochrome Black T. The EDTA will react with all the metal ions present in the water. Then, we can remove the temporary hardness by boiling the water sample to precipitate out the calcium and magnesium bicarbonates. The difference between the total hardness and the hardness after boiling will give us the proportion of permanent hardness.
(iii) Hardness of water is undesirable because it can cause scale buildup in pipes and appliances, leading to decreased efficiency and increased maintenance costs. It can also reduce the effectiveness of soaps and detergents, requiring more of these products to achieve the desired cleaning results.
(b)
(i) Reagents for the laboratory preparation of trioxonitrate (V) acid include concentrated sulfuric acid and a nitrate salt, such as potassium nitrate or sodium nitrate.
(ii) Trioxonitrate (V) acid can be prepared in the laboratory by adding a nitrate salt to concentrated sulfuric acid while cooling the mixture in an ice bath. The mixture is then gently warmed until trioxonitrate (V) acid fumes appear.
(iii)
(I) When concentrated trioxonitrate (V) acid reacts with sulfur, it produces nitrogen dioxide gas, sulfur dioxide gas, and oxygen gas.
(II) When concentrated trioxonitrate (V) acid reacts with aluminum, it produces aluminum nitrate, water, and nitrogen dioxide gas.
(c)
One example of an amphoteric oxide is aluminum oxide (Al2O3). An amphoteric oxide is a type of oxide that can react with both acids and bases to form salts and water. Aluminum oxide can react with acids to form salts such as aluminum nitrate, and can also react with bases to form salts such as sodium aluminate.
Swali 53 Ripoti
(a) A hydrocarbon having the formula C\(_{10}\)H\(_{22}\) was cracked to produce C\(_6\)H\(_{14}\) and another hydrocarbon P.
(i) Give the molecular formula of P.
(ii) Draw the structures of two isomers of P.
(ii) Give a reason why P could be polymerized.
(b) State the guiding principles which are used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals.
(c)Consider each of the following substances: NaH, H\(_2\), H\(_2\)S, NH\(_4\)Cl.
(i) Describe the nature of the intermolecular forces holding the units or molecules together in the condensed (liquid or solid) state.
(ii) Explain briefly what happens when a sample of each of the substances is added to water.
(iii) Write the chemical equations of any reactions occurring or of any equilibria established.
(d) Element J has the following electron configuration: Is\(^2\)2s\(^2\)2p\(^6\)3s\(^2\).
(i) How many unpaired electrons can be found in J?
(ii) State whether J would be a good oxidizing or reducing agent.
(iii) Give a reason for the answer in (d)(ii).
(a)
(i) The molecular formula of hydrocarbon P can be determined by balancing the carbon and hydrogen atoms in the reactant and products of the cracking reaction. The balanced equation is:
C10H22 → C6H14 + C4H8
Therefore, the molecular formula of hydrocarbon P is C4H8.
(ii) Two isomers of hydrocarbon P are:
1. But-1-ene:
H H | | H-C=C-C-C-H | | H H
2. But-2-ene:
H H | | H-C-C=C-C-H | | H H
(iii) Hydrocarbon P could be polymerized because it contains a carbon-carbon double bond, which can undergo addition polymerization to form a long chain polymer.
(b)
The guiding principles used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals are the Aufbau principle, the Pauli exclusion principle, and Hund's rule.
The Aufbau principle states that electrons fill atomic orbitals in order of increasing energy level, starting with the lowest energy level.
The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.
Hund's rule states that for a set of orbitals of equal energy (degenerate), the lowest energy is attained when the number of electrons with the same spin is maximized.
(c)
(i) NaH has ionic bonds holding the units together, while H2, H2S, and NH4Cl have covalent bonds holding the molecules together.
(ii) When NaH is added to water, it reacts to form hydrogen gas and sodium hydroxide:
NaH + H2O → NaOH + H2
When H2 is added to water, it dissolves and forms a homogeneous mixture.
When H2S is added to water, it reacts to form hydrosulfuric acid:
H2S + H2O ⇌ H3O+ + HS-
When NH4Cl is added to water, it dissolves and dissociates into ammonium cations and chloride anions:
NH4Cl → NH4+ + Cl-
(iii) The chemical equations for the reactions are:
NaH + H2O → NaOH + H2
H2S +
Maelezo ya Majibu
(a)
(i) The molecular formula of hydrocarbon P can be determined by balancing the carbon and hydrogen atoms in the reactant and products of the cracking reaction. The balanced equation is:
C10H22 → C6H14 + C4H8
Therefore, the molecular formula of hydrocarbon P is C4H8.
(ii) Two isomers of hydrocarbon P are:
1. But-1-ene:
H H | | H-C=C-C-C-H | | H H
2. But-2-ene:
H H | | H-C-C=C-C-H | | H H
(iii) Hydrocarbon P could be polymerized because it contains a carbon-carbon double bond, which can undergo addition polymerization to form a long chain polymer.
(b)
The guiding principles used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals are the Aufbau principle, the Pauli exclusion principle, and Hund's rule.
The Aufbau principle states that electrons fill atomic orbitals in order of increasing energy level, starting with the lowest energy level.
The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.
Hund's rule states that for a set of orbitals of equal energy (degenerate), the lowest energy is attained when the number of electrons with the same spin is maximized.
(c)
(i) NaH has ionic bonds holding the units together, while H2, H2S, and NH4Cl have covalent bonds holding the molecules together.
(ii) When NaH is added to water, it reacts to form hydrogen gas and sodium hydroxide:
NaH + H2O → NaOH + H2
When H2 is added to water, it dissolves and forms a homogeneous mixture.
When H2S is added to water, it reacts to form hydrosulfuric acid:
H2S + H2O ⇌ H3O+ + HS-
When NH4Cl is added to water, it dissolves and dissociates into ammonium cations and chloride anions:
NH4Cl → NH4+ + Cl-
(iii) The chemical equations for the reactions are:
NaH + H2O → NaOH + H2
H2S +
Swali 54 Ripoti
(a) Explain briefly why \(_2^4He\) has a stable electron configuration compared to \(_4^9Be\)
(b) Consider the following elements: 1H and 3Li.
(i) State the number of electrons that an atom of each element would have after forming an ionic bond.
(ii) Give a reason for each of your answers stated in (b)(i).
(c) State two factors that should be considered when siting a chemical industry.
(d) State two advantages of using a catalyst instead of high temperatures in chemical reactions.
(e) Turpentine burns in chlorine according to the following equation:
C\(_{10}\)H\(_{16}\)\(_{(I)}\) + 8Cl\(_{2(g)}\) \(\to\) 10C\(_{(s)}\) + 16HCl\(_{(g)}\)
Calculate the mass of turpentine that would completely burn in 21.3 g of chlorine.
[Molar mass of chlorine = 71 gmol\(^{-1}\); Molar mass of Turpentine = 136 gmol\(^{-1}\)]
(f) What is cracking?
(g) State two factors that may influence the value of electron affinity.
(h) What are carbohydrates?
(i) State two differences between a simple sugar and starch.
(j) Write an equation to show the dissociation of each of the following acids:
(i) H\(_2\)CO\(_3\);
(ii) CH\(_3\)COOH.
(a) The stability of an atom's electron configuration is determined by the number of electrons in the outermost energy level. Helium has two electrons in its outermost energy level, which is the maximum number it can hold, so its electron configuration is stable. Beryllium, on the other hand, has four electrons in its outermost energy level, which is not stable, so it tends to lose or gain electrons to achieve stability.
(b)(i) An atom of hydrogen would have one electron, while an atom of lithium would have no electrons after forming an ionic bond.
(b)(ii) When hydrogen forms an ionic bond, it loses its single electron to become a positively charged ion (H+). Lithium, with three electrons in its outermost energy level, readily donates one electron to become a positively charged ion (Li+). This electron donation allows both elements to achieve a stable electron configuration.
(c) Two factors to consider when siting a chemical industry are access to raw materials and transportation infrastructure.
(d) Two advantages of using a catalyst instead of high temperatures in chemical reactions are that it reduces the energy required for the reaction and it can increase the selectivity of the reaction, meaning that it can produce more of the desired product and less of unwanted byproducts.
(e) To calculate the mass of turpentine that would completely burn in 21.3 g of chlorine, we need to use stoichiometry. First, we need to calculate the number of moles of chlorine present:
21.3 g Cl2 x (1 mol Cl2/71 g Cl2) = 0.3 mol Cl2
Next, we need to use the balanced chemical equation to determine the number of moles of turpentine needed to react with 0.3 mol Cl2:
8 mol Cl2 / 1 mol turpentine
0.3 mol Cl2 x (1 mol turpentine / 8 mol Cl2) = 0.0375 mol turpentine
Finally, we can use the molar mass of turpentine to convert the number of moles to mass:
0.0375 mol turpentine x 136 g/mol = 5.1 g turpentine
Therefore, 5.1 g of turpentine would completely burn in 21.3 g of chlorine.
(f) Cracking is the process of breaking down large hydrocarbons into smaller ones by breaking the carbon-carbon bonds in the molecules. This is typically done by heating the hydrocarbons to high temperatures in the presence of a catalyst.
(g) Two factors that may influence the value of electron affinity are the atomic radius and the effective nuclear charge. A smaller atomic radius and a higher effective nuclear charge will increase the attraction between the nucleus and electrons, leading to a higher electron affinity.
(h) Carbohydrates are a group of biomolecules that include sugars, starches, and cellulose. They are made up of carbon, hydrogen, and oxygen atoms and serve as a source of energy for living organisms.
(i) Simple sugars are monosaccharides, meaning they consist of a single sugar molecule, while starch is a polysaccharide, meaning it is made up of multiple
Maelezo ya Majibu
(a) The stability of an atom's electron configuration is determined by the number of electrons in the outermost energy level. Helium has two electrons in its outermost energy level, which is the maximum number it can hold, so its electron configuration is stable. Beryllium, on the other hand, has four electrons in its outermost energy level, which is not stable, so it tends to lose or gain electrons to achieve stability.
(b)(i) An atom of hydrogen would have one electron, while an atom of lithium would have no electrons after forming an ionic bond.
(b)(ii) When hydrogen forms an ionic bond, it loses its single electron to become a positively charged ion (H+). Lithium, with three electrons in its outermost energy level, readily donates one electron to become a positively charged ion (Li+). This electron donation allows both elements to achieve a stable electron configuration.
(c) Two factors to consider when siting a chemical industry are access to raw materials and transportation infrastructure.
(d) Two advantages of using a catalyst instead of high temperatures in chemical reactions are that it reduces the energy required for the reaction and it can increase the selectivity of the reaction, meaning that it can produce more of the desired product and less of unwanted byproducts.
(e) To calculate the mass of turpentine that would completely burn in 21.3 g of chlorine, we need to use stoichiometry. First, we need to calculate the number of moles of chlorine present:
21.3 g Cl2 x (1 mol Cl2/71 g Cl2) = 0.3 mol Cl2
Next, we need to use the balanced chemical equation to determine the number of moles of turpentine needed to react with 0.3 mol Cl2:
8 mol Cl2 / 1 mol turpentine
0.3 mol Cl2 x (1 mol turpentine / 8 mol Cl2) = 0.0375 mol turpentine
Finally, we can use the molar mass of turpentine to convert the number of moles to mass:
0.0375 mol turpentine x 136 g/mol = 5.1 g turpentine
Therefore, 5.1 g of turpentine would completely burn in 21.3 g of chlorine.
(f) Cracking is the process of breaking down large hydrocarbons into smaller ones by breaking the carbon-carbon bonds in the molecules. This is typically done by heating the hydrocarbons to high temperatures in the presence of a catalyst.
(g) Two factors that may influence the value of electron affinity are the atomic radius and the effective nuclear charge. A smaller atomic radius and a higher effective nuclear charge will increase the attraction between the nucleus and electrons, leading to a higher electron affinity.
(h) Carbohydrates are a group of biomolecules that include sugars, starches, and cellulose. They are made up of carbon, hydrogen, and oxygen atoms and serve as a source of energy for living organisms.
(i) Simple sugars are monosaccharides, meaning they consist of a single sugar molecule, while starch is a polysaccharide, meaning it is made up of multiple
Swali 55 Ripoti
(a) (i) Name three different methods for preparing salts.
(ii) Give one example of a balanced equation for each of the methods named in (a)(i).
(iii) State two uses of sodium trioxocarbonate (IV).
(b) If you were given some impure copper, describe how you would
obtain a specimen of the pure metal by electrolysis.
(c) Given that sodium chloride has a solubility of 36.3 at 30 and 39.0 at
100 and that of silver nitrate is 297.0 at 30 and 952.0 at 100.
(i) Calculate the percentage of each substance in the saturated solution at 100 that is deposited on cooling to 30
(ii) Deduce which of the two salts can be purified more efficiently by crystallization.
a)
b)
c)
To calculate the percentage of each substance in the saturated solution at 100 that is deposited on cooling to 30:
Based on the percentage of each substance that is deposited, sodium chloride can be purified more efficiently by crystallization.
Maelezo ya Majibu
a)
b)
c)
To calculate the percentage of each substance in the saturated solution at 100 that is deposited on cooling to 30:
Based on the percentage of each substance that is deposited, sodium chloride can be purified more efficiently by crystallization.
Swali 56 Ripoti
(a) In the Solvay process, explain briefly with equations the functions of the following substances:
(b) (i) Write a chemical equation for the fermentation of glucose.
(ii) Explain briefly why a tightly-corked glass bottle filled to the brim with fresh palm-wine shatters on standing for some time.
(c) Consider the following metals: Na, Fe, K and Cu.
(i) Arrange the metals in order of increasing reactivity.
(ii) Which of the metals will react with cold water?
(ii) Which of the metals could form coloured salts?
(d)(i) What is a redox reaction?
(ii) Identify which of the following reaction equations are redox.
(i) 2Na + Cl\(_2\) ? 2NaCl
(ii) AgCl + 2NH\(_3\) ? [Ag(NH\(_3\))\(_2\)]Cl
(iii) C\(_2\)H\(_2\) + H\(_2\) ? C\(_2\)H\(_4\)
(IV) HCl + KOH ? KCl + H\(_2\)O
(V) 2FeCl\(_3\) + 2KI ? 2FeCl\(_2\) + 2KCl + I\(_2\)
(iii) Give a reason for each of the answers in (d)(ii).
(iv) Write balanced equations of the half reactions for any two of the redox reactions in (d)(ii).
(a) Functions in the Solvay Process:
(b) Fermentation and Shattered Bottle:
(i) Glucose Fermentation:
C6H12O6 (aq) -> 2C2H5OH (aq) + 2CO2 (g) (Glucose ferments to ethanol and carbon dioxide)
(ii) Shattered Bottle:
Fresh palm wine undergoes fermentation by yeast. During fermentation, yeast consumes sugar and produces carbon dioxide gas. In a tightly sealed bottle, the CO2 gas gets trapped and builds up pressure. As pressure increases, the glass bottle can no longer withstand the force and shatters.
(c) Metal Reactivity:
(i) Increasing Reactivity: K > Na > Fe > Cu
(ii) Reacting with Cold Water: K, Na (These alkali metals react vigorously with water)
(iii) Colored Salts: Cu (Copper forms characteristic colored salts, like blue copper sulfate)
(d) Redox Reactions and Half Reactions:
(i) Redox Reaction:
A redox reaction is a type of chemical reaction where there is a transfer of electrons between the participating atoms or ions. Oxidation is the loss of electrons, and reduction is the gain of electrons.
(ii) Identifying Redox Reactions:
(iii) Explanation for Redox Reactions:
(iv) Balanced Half Reactions (Examples):
Ag -> Ag+ + e-
I- -> I2 + e- (Note: I- needs to be balanced with Cl- from the reactant side)
Maelezo ya Majibu
(a) Functions in the Solvay Process:
(b) Fermentation and Shattered Bottle:
(i) Glucose Fermentation:
C6H12O6 (aq) -> 2C2H5OH (aq) + 2CO2 (g) (Glucose ferments to ethanol and carbon dioxide)
(ii) Shattered Bottle:
Fresh palm wine undergoes fermentation by yeast. During fermentation, yeast consumes sugar and produces carbon dioxide gas. In a tightly sealed bottle, the CO2 gas gets trapped and builds up pressure. As pressure increases, the glass bottle can no longer withstand the force and shatters.
(c) Metal Reactivity:
(i) Increasing Reactivity: K > Na > Fe > Cu
(ii) Reacting with Cold Water: K, Na (These alkali metals react vigorously with water)
(iii) Colored Salts: Cu (Copper forms characteristic colored salts, like blue copper sulfate)
(d) Redox Reactions and Half Reactions:
(i) Redox Reaction:
A redox reaction is a type of chemical reaction where there is a transfer of electrons between the participating atoms or ions. Oxidation is the loss of electrons, and reduction is the gain of electrons.
(ii) Identifying Redox Reactions:
(iii) Explanation for Redox Reactions:
(iv) Balanced Half Reactions (Examples):
Ag -> Ag+ + e-
I- -> I2 + e- (Note: I- needs to be balanced with Cl- from the reactant side)
Swali 57 Ripoti
(a) The following reaction scheme is an illustration of the contact process. Study the scheme and answer the questions that follow.
(i) Name X and Y
(ii) Write a balanced chemical equation for each of the processes I, II, III and IV
(iii) Name the catalyst used in process II
(iv) Using Le Chatelier's principle, explain briefly why increasing the temperature would not favour the reaction in II
(v) State two uses of SO\(_2\)
(b) Consider the following equation: 2H\(_{2(g)}\) + O\(_{2(g)}\) \(\to\) 2H\(_2\)O\(_{(g)}\)
Calculate the volume of unused oxygen gas when 40 cm\(^3\) of hydrogen gas is sparked with 30cm\(^3\) of oxygen gas
(c) Calcium carbonate of mass 1.0 g was heated until there was no further change.
a)
(i) X is sulfur dioxide (SO2) and Y is sulfur trioxide (SO3).
(ii)
Process I: Sulfur (S) is burned to form sulfur dioxide (SO2).
S (s) + O2 (g) → SO2 (g)
Process II: Sulfur dioxide (SO2) is oxidized to form sulfur trioxide (SO3) using oxygen (O2) in the presence of a catalyst, vanadium(V) oxide (V2O5).
2SO2 (g) + O2 (g) → 2SO3 (g)
Process III: Sulfur trioxide (SO3) is dissolved in concentrated sulfuric acid (H2SO4) to form oleum (H2S2O7).
SO3 (g) + H2SO4 (l) → H2S2O7 (l)
Process IV: Oleum (H2S2O7) is diluted with water (H2O) to form concentrated sulfuric acid (H2SO4).
H2S2O7 (l) + H2O (l) → 2H2SO4 (l)
(iii) The catalyst used in process II is vanadium(V) oxide (V2O5).
(iv) In process II, the reaction is exothermic (it releases heat). According to Le Chatelier's principle, increasing the temperature would shift the equilibrium in the opposite direction, favoring the reactants (SO2 and O2). Therefore, increasing the temperature would not favor the forward reaction in process II.
(v) Two uses of SO2 are:
b)
The equation given is:
2H2 (g) + O2 (g) → 2H2O (g)
To solve the problem, we can use the mole ratio
Maelezo ya Majibu
a)
(i) X is sulfur dioxide (SO2) and Y is sulfur trioxide (SO3).
(ii)
Process I: Sulfur (S) is burned to form sulfur dioxide (SO2).
S (s) + O2 (g) → SO2 (g)
Process II: Sulfur dioxide (SO2) is oxidized to form sulfur trioxide (SO3) using oxygen (O2) in the presence of a catalyst, vanadium(V) oxide (V2O5).
2SO2 (g) + O2 (g) → 2SO3 (g)
Process III: Sulfur trioxide (SO3) is dissolved in concentrated sulfuric acid (H2SO4) to form oleum (H2S2O7).
SO3 (g) + H2SO4 (l) → H2S2O7 (l)
Process IV: Oleum (H2S2O7) is diluted with water (H2O) to form concentrated sulfuric acid (H2SO4).
H2S2O7 (l) + H2O (l) → 2H2SO4 (l)
(iii) The catalyst used in process II is vanadium(V) oxide (V2O5).
(iv) In process II, the reaction is exothermic (it releases heat). According to Le Chatelier's principle, increasing the temperature would shift the equilibrium in the opposite direction, favoring the reactants (SO2 and O2). Therefore, increasing the temperature would not favor the forward reaction in process II.
(v) Two uses of SO2 are:
b)
The equation given is:
2H2 (g) + O2 (g) → 2H2O (g)
To solve the problem, we can use the mole ratio
Swali 58 Ripoti
(a) State the conditions necessary for the cracking of long-chain hydrocarbons to produce more gasoline.
(b) State two reasons why metallic objects are electroplated
(c) (i) Explain briefly why calcium oxide cannot be used to dry hydrogen chloride gas.
(ii) State one drying agent for hydrogen chloride gas.
(d) Concentrated trioxonitrate (V) acid was added to a solution of iron (II) tetraoxosulphate (VI) and the mixture heated. The mixture turned from pale green to yellow with the evolution of a brown gas. Explain briefly these observations.
(e) (i) Write the equation for the reaction between zinc oxide and
(ii) State which property of zinc oxide is shown by the reaction in (e)(i).
(f) Two isotopes of chlorine are \(^{35}_{17}\)Cl and \(^{37}_{17}\)Cl State one:
(g) State the two products formed when chlorine water is exposed to sunlight.
(h) Consider the reaction represented by the following equation:
State the:
(i) What is meant by carbon-12 scale?
(j) State two properties of a chemical system in equilibrium.
(a) The cracking of long-chain hydrocarbons to produce more gasoline requires the following conditions:
(b) Metallic objects are electroplated for two main reasons:
(c)
(d) The pale green color of the iron (II) tetraoxosulphate (VI) solution indicates the presence of Fe2+ ions. Upon adding concentrated trioxonitrate (V) acid and heating, the Fe2+ ions are oxidized to Fe3+ ions, which give the yellow color to the solution. The brown gas evolved is likely to be nitrogen dioxide (NO2), which is a product of the reaction between the trioxonitrate (V) acid and the Fe2+ ions.
(e)
(f) Similarity: Both isotopes of chlorine have the same number of electrons and protons, and hence have the same chemical properties.
Difference: The isotopes have different atomic masses due to the presence of different numbers of neutrons in their nuclei.
(g) When chlorine water is exposed to sunlight, the two products formed are:
(h) The reaction can be represented as follows:
Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
Species undergoing oxidation: Fe2+ ions
Oxidizing agent: Cr2O72- ions
(i) The carbon-12 scale is a standard used to define the relative atomic masses of elements. It is based on the isotope carbon-12
Maelezo ya Majibu
(a) The cracking of long-chain hydrocarbons to produce more gasoline requires the following conditions:
(b) Metallic objects are electroplated for two main reasons:
(c)
(d) The pale green color of the iron (II) tetraoxosulphate (VI) solution indicates the presence of Fe2+ ions. Upon adding concentrated trioxonitrate (V) acid and heating, the Fe2+ ions are oxidized to Fe3+ ions, which give the yellow color to the solution. The brown gas evolved is likely to be nitrogen dioxide (NO2), which is a product of the reaction between the trioxonitrate (V) acid and the Fe2+ ions.
(e)
(f) Similarity: Both isotopes of chlorine have the same number of electrons and protons, and hence have the same chemical properties.
Difference: The isotopes have different atomic masses due to the presence of different numbers of neutrons in their nuclei.
(g) When chlorine water is exposed to sunlight, the two products formed are:
(h) The reaction can be represented as follows:
Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
Species undergoing oxidation: Fe2+ ions
Oxidizing agent: Cr2O72- ions
(i) The carbon-12 scale is a standard used to define the relative atomic masses of elements. It is based on the isotope carbon-12
Swali 59 Ripoti
(a)(i) Draw and label a diagram to illustrate the preparation and collection of dry chlorine gas in the laboratory.
(ii) State two uses of chlorine.
(b) Describe the preparation of hydrogen from water gas.
(i) Name the chief ore of aluminium.
(ii) Why is the ore purified?
(iii) Name the electrode used in the electrolysis.
(iv) Give one reason why cryolite, NaAlF\(_6\), is added to the electrolyte.
(c) Name three products obtained directly from the destructive distillation of coal.
(a) (i) To prepare and collect dry chlorine gas in the laboratory, a setup as shown in the diagram below can be used:
In the diagram, a container of hydrochloric acid (HCl) is connected to a gas delivery tube which is placed over a container of calcium hydroxide (Ca(OH)2) through a delivery tube. The hydrochloric acid and calcium hydroxide react to produce chlorine gas, which is collected in the container over the calcium hydroxide. The container is sealed so that the chlorine gas does not escape.
(ii) Two uses of chlorine are:
(b) The preparation of hydrogen from water gas can be carried out by a process called water gas shift reaction. This reaction involves the reaction of carbon monoxide (CO) and steam (H2O) over a catalyst to produce hydrogen (H2) and carbon dioxide (CO2). The reaction can be represented by the equation:
CO + H2O ? H2 + CO2
(i) The chief ore of aluminum is bauxite.
(ii) The ore is purified to remove impurities, such as silica and iron, which would interfere with the electrolysis process used to extract aluminum from the ore.
(iii) The electrode used in the electrolysis is a carbon electrode.
(iv) Cryolite, NaAlF6, is added to the electrolyte to lower the melting point of the electrolyte, allowing the aluminum to be extracted at a lower temperature.
(c) Three products obtained directly from the destructive distillation of coal are:
Maelezo ya Majibu
(a) (i) To prepare and collect dry chlorine gas in the laboratory, a setup as shown in the diagram below can be used:
In the diagram, a container of hydrochloric acid (HCl) is connected to a gas delivery tube which is placed over a container of calcium hydroxide (Ca(OH)2) through a delivery tube. The hydrochloric acid and calcium hydroxide react to produce chlorine gas, which is collected in the container over the calcium hydroxide. The container is sealed so that the chlorine gas does not escape.
(ii) Two uses of chlorine are:
(b) The preparation of hydrogen from water gas can be carried out by a process called water gas shift reaction. This reaction involves the reaction of carbon monoxide (CO) and steam (H2O) over a catalyst to produce hydrogen (H2) and carbon dioxide (CO2). The reaction can be represented by the equation:
CO + H2O ? H2 + CO2
(i) The chief ore of aluminum is bauxite.
(ii) The ore is purified to remove impurities, such as silica and iron, which would interfere with the electrolysis process used to extract aluminum from the ore.
(iii) The electrode used in the electrolysis is a carbon electrode.
(iv) Cryolite, NaAlF6, is added to the electrolyte to lower the melting point of the electrolyte, allowing the aluminum to be extracted at a lower temperature.
(c) Three products obtained directly from the destructive distillation of coal are:
Swali 60 Ripoti
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
C is a mixture of two inorganic compounds. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusions you draw from the result of each test.
(a) Put all of C in a boiling tube and add about 10 cm\(^3\) of distilled water. Shake thoroughly and filter. Keep both the residue and the filtrate.
(b)(i) To about 2 cm\(^3\) of the filtrate add a few drops of Silver trioxoitrate (V) followed by dilute HNO\(_{3}\)
(ii) Add excess NH\(_3\) solution to the resulting mixture in (b)(i).
(C)(i) Put the residue in a test tube, add about 2 cm\(^3\) of dilute HCl and shake.
(ii) Add NH\(_3\) Solution in drops to the mixture from (c)(i) and then in excess.
Experiment:
This experiment involves analyzing a mixture (C) containing two inorganic compounds. You will perform tests on the filtrate (soluble part) and residue (insoluble part) after filtration with water.
Observations and Conclusions:
(a) Filtration:
(b) Tests on Filtrate (Solution):
(i) Silver Nitrate (AgNO3) and Nitric Acid (HNO3):
(ii) Excess Ammonia (NH3) Solution:
(c) Tests on Residue:
(i) Dilute Hydrochloric Acid (HCl):
(ii) Ammonia Solution (NH3):
Important Note:
These are general observations and conclusions based on common inorganic compounds. The actual behavior depends on the specific components present in mixture C.
Maelezo ya Majibu
Experiment:
This experiment involves analyzing a mixture (C) containing two inorganic compounds. You will perform tests on the filtrate (soluble part) and residue (insoluble part) after filtration with water.
Observations and Conclusions:
(a) Filtration:
(b) Tests on Filtrate (Solution):
(i) Silver Nitrate (AgNO3) and Nitric Acid (HNO3):
(ii) Excess Ammonia (NH3) Solution:
(c) Tests on Residue:
(i) Dilute Hydrochloric Acid (HCl):
(ii) Ammonia Solution (NH3):
Important Note:
These are general observations and conclusions based on common inorganic compounds. The actual behavior depends on the specific components present in mixture C.
Swali 61 Ripoti
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
A is a solution containing 15.8 g dm\(^3\) of Na\(_2\)S\(_2\)O\(_3\). B was obtained by dissolving 9.0 g of an impure sample of I\(_2\) in aqueous Kl and the solution made up to 1 dm\(^3\).
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B. Use starch solution as indicator. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of A used. The equation for the reaction involved in the titration is I\(_2\) + 2S\(_2\)O\(_3\) \(\to\) 2I\(^-\) + S\(_4\)O\(_6^{2-}\).
(b) From your results and the information provided, calculate the:
(i) concentration of A in mol dm\(^{-3}\)
(ii) concentration of I\(_2\) in B in mol dm\(^{-3}\);
(iii) percentage by mass of I\(_2\) in the sample
(c) Give reasons why the starch indicator was not added to the titration mixture at the beginning of the titration. [O = 16.0, Na = 23.0, S = 32.0, 1 = 127.0] Credit will be given for strict adherence to the instructions for observations precisely recorded and for accurate inferences. AIl tests, observations and inferences must be clearly entered in the booklet in ink at the time they are made.
Experiment:
You are titrating a solution of sodium thiosulfate (Na2S2O3, solution A) against a solution containing iodine (I2, solution B) obtained from an impure sample.
Requirements:
(a) Average Volume of Solution A:
(b) Concentration Calculations:
i) Concentration of Solution A (Na2S2O3):
ii) Concentration of Iodine (I2) in Solution B:
iii) Percentage by Mass of I2 in the Sample:
(c) Why Not Add Starch Indicator at the Beginning?
Starch forms a blue-black complex with free iodine molecules. If added at the beginning, the solution would appear blue even before the endpoint. By adding starch near the endpoint (when free I2 concentration is low), a faint blue color appears, indicating the complete reaction of I2 with Na2S2O3. This allows for a more accurate determination of the titration endpoint.
Maelezo ya Majibu
Experiment:
You are titrating a solution of sodium thiosulfate (Na2S2O3, solution A) against a solution containing iodine (I2, solution B) obtained from an impure sample.
Requirements:
(a) Average Volume of Solution A:
(b) Concentration Calculations:
i) Concentration of Solution A (Na2S2O3):
ii) Concentration of Iodine (I2) in Solution B:
iii) Percentage by Mass of I2 in the Sample:
(c) Why Not Add Starch Indicator at the Beginning?
Starch forms a blue-black complex with free iodine molecules. If added at the beginning, the solution would appear blue even before the endpoint. By adding starch near the endpoint (when free I2 concentration is low), a faint blue color appears, indicating the complete reaction of I2 with Na2S2O3. This allows for a more accurate determination of the titration endpoint.
Swali 62 Ripoti
(a) (i) State the important points put forward in Dalton’s atomic theory.
(ii) How does the theory explain the law of multiple proportion?
(b) (i) Name three commercially useful products that may be obtained by chemical transformation of vegetable oils.
(ii) With the aid of equations, describe the chemical reaction involved in the transformations stated in (b)(i).
(c) Certain properties of Beryllium (Be) and its compounds differ from those of Magnesium (Mg) and its compounds, but rather resembles those of Aluminium and its compounds. Explain briefly why this is so.
(d) Write balanced equation for the reaction between iodine and aqueous sodium trioxothiosulphate(VI).
a)
(i) The important points put forward in Dalton's atomic theory are:
(ii) The theory explains the law of multiple proportion by stating that when two elements combine to form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other element are ratios of small whole numbers.
b)
(i) Three commercially useful products that may be obtained by chemical transformation of vegetable oils are soap, biodiesel, and margarine.
(ii) The chemical reactions involved in the transformations stated in (b)(i) are:
c) Beryllium (Be) and its compounds differ from those of Magnesium (Mg) and its compounds, but rather resemble those of Aluminium and its compounds because:
d) The balanced equation for the reaction between iodine and aqueous sodium trioxothiosulphate(VI) is:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
This reaction is a redox reaction, in which iodine is reduced to iodide ions and trioxothiosulphate(VI) is oxidized to tetrathionate ions. Sodium ions are spectator ions and do not participate in the reaction. The reaction is commonly used as a titration to determine the concentration of iodine in a sample.
Maelezo ya Majibu
a)
(i) The important points put forward in Dalton's atomic theory are:
(ii) The theory explains the law of multiple proportion by stating that when two elements combine to form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other element are ratios of small whole numbers.
b)
(i) Three commercially useful products that may be obtained by chemical transformation of vegetable oils are soap, biodiesel, and margarine.
(ii) The chemical reactions involved in the transformations stated in (b)(i) are:
c) Beryllium (Be) and its compounds differ from those of Magnesium (Mg) and its compounds, but rather resemble those of Aluminium and its compounds because:
d) The balanced equation for the reaction between iodine and aqueous sodium trioxothiosulphate(VI) is:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
This reaction is a redox reaction, in which iodine is reduced to iodide ions and trioxothiosulphate(VI) is oxidized to tetrathionate ions. Sodium ions are spectator ions and do not participate in the reaction. The reaction is commonly used as a titration to determine the concentration of iodine in a sample.
Swali 63 Ripoti
(a) (i) Write an equation for the reaction by which sulphur dioxide in solution could be converted to tetraoxosulphate (VI) acid.
(ii) State one test to confirm the conversion of sulphur dioxide to tetraoxosulphate (VI) acid.
(iii) State the reaction of concentrated tetraoxosulphate (VI) acid with:
(I) oxalic acid;
(II) copper.
(iv) What property of concentrated tetraoxosulphate (VI) acid does each of the reactions stated in (a)(iii) illustrate?
(b) (i) Explain briefly why water is referred to as a universal solvent.
(ii) Give one chemical test for water.
(c) (i) What is the major component of synthetic gas?
(ii) Give one reason why synthetic gas is not a major source of air pollution.
(d) Name one product of destructive distillation of coal that is:
(i) solid;
(ii) liquid;
(iii) gas.
(e) (i) Write a balanced chemical equation for the complete combustion of carbon.
(ii) State one:
(I) physical;
(II) chemical property of the products in (e)(i).
(a)
(i) 2SO2 + O2 + 2H2 O → 2H2 SO4
(ii)
Use BaCl2 solution it gives white insoluble solid / precipitate to form bariumtetraoxosulphate (VI)
(iii) I. C2 H2 O4 −H2OH2SO4 CO + CO2
(iv)
I. H2 SO4 is acting as a dehydrating agent
II. Cu + 2H2 2
SO4 → SO2 + CuSO4 + 2H2 O
III. H2
SO4
is acting as an oxidizing agent
(b)
(i) Water is referred to as a universal solvent because it is able to dissolve more substances than any other liquid. This has to do with the polarity of each water molecule. The hydrogen side of each water molecule carries a slight positive electric charge, while the oxygen side carries a slight negative electric charge and this helps water dissociate ionic compounds into their positive and negative ions.
(ii) Water will turn cobalt (II) chloride paper from blue to pink / anhydrous copper (II) tetraoxosulphate (VI) crystals from white to blue
(c)
(i) CO / Carbon (II) Oxide
(ii) Because it is combustible / can be used as fuel
(d)
(i) coke
(ii) coal tar / ammoniacal liquor / pitch
(iii) coal gas
(e)
(i) C(s) + O2(g) →CO2(g )
(ii)
i)
- It is odourless
- It is colourless
- It is denser than air
- It is soluble in water
- It is a gas at room temperature
- It is easily compressible
(ii)
It does not support combustion
It reacts with red hot carbon (to form CO)
it is weakly acidic in water
it reacts with alkali (to form trioxocarbonate (IV) ) / turns lime water milky
Reacts with burning magnesium
Maelezo ya Majibu
(a)
(i) 2SO2 + O2 + 2H2 O → 2H2 SO4
(ii)
Use BaCl2 solution it gives white insoluble solid / precipitate to form bariumtetraoxosulphate (VI)
(iii) I. C2 H2 O4 −H2OH2SO4 CO + CO2
(iv)
I. H2 SO4 is acting as a dehydrating agent
II. Cu + 2H2 2
SO4 → SO2 + CuSO4 + 2H2 O
III. H2
SO4
is acting as an oxidizing agent
(b)
(i) Water is referred to as a universal solvent because it is able to dissolve more substances than any other liquid. This has to do with the polarity of each water molecule. The hydrogen side of each water molecule carries a slight positive electric charge, while the oxygen side carries a slight negative electric charge and this helps water dissociate ionic compounds into their positive and negative ions.
(ii) Water will turn cobalt (II) chloride paper from blue to pink / anhydrous copper (II) tetraoxosulphate (VI) crystals from white to blue
(c)
(i) CO / Carbon (II) Oxide
(ii) Because it is combustible / can be used as fuel
(d)
(i) coke
(ii) coal tar / ammoniacal liquor / pitch
(iii) coal gas
(e)
(i) C(s) + O2(g) →CO2(g )
(ii)
i)
- It is odourless
- It is colourless
- It is denser than air
- It is soluble in water
- It is a gas at room temperature
- It is easily compressible
(ii)
It does not support combustion
It reacts with red hot carbon (to form CO)
it is weakly acidic in water
it reacts with alkali (to form trioxocarbonate (IV) ) / turns lime water milky
Reacts with burning magnesium
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