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Tambaya 1 Rahoto
An organic compound has the empirical formula \(CH_{2}\). If its molar mass is 42\(gmol^{-1}\), what is its molecular formula? (C = 12.0, H = 1.0)
Bayanin Amsa
The empirical formula of the organic compound is given as \(CH_{2}\), which means that the ratio of carbon to hydrogen atoms in the compound is 1:2. The molar mass of the compound is also given as 42\(gmol^{-1}\). To find the molecular formula, we need to determine the actual number of carbon and hydrogen atoms in the compound. Let's assume that the molecular formula of the compound is \(C_{x}H_{y}\). The molar mass of the compound can be calculated as follows: molar mass = (molar mass of carbon) x (number of carbon atoms) + (molar mass of hydrogen) x (number of hydrogen atoms) 42 = (12.0 g/mol) x (x) + (1.0 g/mol) x (y) Simplifying the equation, we get: 12x + y = 42 However, we also know that the ratio of carbon to hydrogen atoms in the compound is 1:2. This means that: x/y = 1/2 Multiplying both sides by y, we get: x = (1/2)y Substituting this value of x in the previous equation, we get: 12(1/2)y + y = 42 Simplifying the equation, we get: 18y = 42 y = 2.33 Since the number of atoms must be a whole number, we can round y to the nearest whole number, which is 2. This means that the molecular formula of the compound is \(C_{x}H_{2}\). Using the ratio of carbon to hydrogen atoms in the empirical formula, we know that x = 2. Therefore, the molecular formula of the compound is \(C_{2}H_{4}\). Hence, the answer is option A) \(C_{2}H_{4}\).
Tambaya 2 Rahoto
The preferential discharge of ions during electrolysis is influenced by the
Bayanin Amsa
The preferential discharge of ions during electrolysis is influenced by the nature of the electrode and electrolytic reactions. During electrolysis, the positive and negative ions in the electrolyte solution move towards the respective electrodes. The electrode's surface and the type of electrolytic reaction determine the preferential discharge of ions. When an electrode's surface has a high affinity for a particular ion, that ion will preferentially discharge at that electrode. For example, if the electrode is made of copper and the electrolyte solution contains copper ions and hydrogen ions, copper ions will preferentially discharge at the electrode's surface. Additionally, the type of electrolytic reaction can influence ion preference. For instance, in the electrolysis of water, hydrogen ions are preferentially discharged at the cathode, and oxygen ions are preferentially discharged at the anode. The mechanism of electrolysis and type of electrolytic cell can also play a role in the preferential discharge of ions.
Tambaya 3 Rahoto
\(Cu\) and \(HNO_{3}\) are not suitable for preparing hydrogen gas because of their
Bayanin Amsa
Copper (\(Cu\)) and nitric acid (\(HNO_{3}\)) are not suitable for preparing hydrogen gas because of their reactivity and oxidation properties, respectively. When copper is placed in nitric acid, a redox reaction occurs, where the copper is oxidized by the nitric acid. This reaction produces copper nitrate and nitrogen dioxide gas, but it does not produce hydrogen gas. Similarly, nitric acid is a strong oxidizing agent and can oxidize hydrogen gas to water, thereby preventing the formation of hydrogen gas. Therefore, neither copper nor nitric acid can be used to prepare hydrogen gas. Other metals, such as zinc or magnesium, can be used to react with acids to produce hydrogen gas because they are more reactive than copper and do not have the same oxidation properties as nitric acid.
Tambaya 4 Rahoto
What is the relative molecular mass of a compound which has empirical formula \(CH_{2}O\)? [C= 12, H= 1, O= 16]
Bayanin Amsa
Tambaya 6 Rahoto
Analysis of a hydrocarbon shows that it contains 0.93g of Carbon per gram of the compound. The mole ratio of carbon to hydrogen in the compound is [H = 1.0, C=12.0]
Bayanin Amsa
The ratio of carbon to hydrogen in the hydrocarbon can be determined using the given mass ratio of carbon and the fact that the molecular weight of the hydrocarbon is the sum of the atomic weights of its constituent elements. Let's assume a sample of 1 gram of the hydrocarbon, which contains 0.93 grams of carbon. This means that the mass of hydrogen in the sample is 1.0 - 0.93 = 0.07 grams. The number of moles of carbon in the sample can be calculated as: 0.93 g C x (1 mol C / 12.0 g C) = 0.0775 mol C Similarly, the number of moles of hydrogen in the sample can be calculated as: 0.07 g H x (1 mol H / 1.0 g H) = 0.07 mol H To find the mole ratio of carbon to hydrogen, we need to divide the number of moles of carbon by the number of moles of hydrogen: 0.0775 mol C / 0.07 mol H = 1.107 Rounding this to the nearest whole number, we get the mole ratio of carbon to hydrogen as 1:1. Therefore, the correct answer is 1:1.
Tambaya 7 Rahoto
Common Salt (NaCl) is used to preserve foods. Which of the following properties can be used to determine its purity before use?
Tambaya 8 Rahoto
The law of definite proportions states that
Bayanin Amsa
The law of definite proportions states that pure samples of the same compound contain the same elements combined in the same proportion by mass. This means that no matter where or how the compound is obtained, the ratio of the masses of the elements in the compound will always be the same. For example, if a water molecule contains 2 hydrogen atoms and 1 oxygen atom, the ratio of the masses of hydrogen to oxygen will always be 2:1 in any sample of water.
Tambaya 9 Rahoto
One of the criteria for confirming the purity of benzene is to determine its
Bayanin Amsa
One of the criteria for confirming the purity of benzene is to determine its boiling point. The boiling point of a substance is a physical property that is determined by the intermolecular forces between the particles in the substance. These forces are dependent on the molecular structure of the substance, and so the boiling point is unique for each substance. In the case of benzene, it has a known and well-defined boiling point of 80.1 °C. By determining the boiling point of a sample and comparing it to the known boiling point of pure benzene, we can determine the purity of the sample. If the boiling point of the sample is different from the known boiling point of pure benzene, it indicates that the sample is impure and contains other substances. So, determining the boiling point of a sample of benzene is a useful way to confirm its purity.
Tambaya 10 Rahoto
Pure water can be made to boil at a temperature lower than 100°C by
Tambaya 11 Rahoto
The most suitable substance for putting out petrol fire is
Bayanin Amsa
The most suitable substance for putting out a petrol fire is not water, carbon(IV) oxide, or a fire blanket, but sand. Sand can be used to smother a petrol fire by covering the flames and cutting off the oxygen supply, which will cause the fire to go out. However, it is important to note that sand should only be used on small fires and not on larger fires, as it can be difficult to spread a large amount of sand over the fire and it may not be effective. In such cases, it is best to evacuate the area and call the fire department.
Tambaya 12 Rahoto
Which of the following raw materials is used in the plastic industry?
Bayanin Amsa
In the plastic industry, ethene (also known as ethylene) is commonly used as a raw material. Ethene is a colorless and flammable gas that can be polymerized to produce a wide range of plastics, including polyethylene, which is one of the most widely used plastics in the world. Methane, sulfur, and hydrogen are not typically used as raw materials in the plastic industry.
Tambaya 13 Rahoto
Particles in a solid exibit
Bayanin Amsa
Particles in a solid exhibit vibrational motion. In a solid, the particles are closely packed together and held in a fixed position by strong intermolecular forces. However, the particles are not completely motionless. They can vibrate around their fixed positions due to thermal energy. The amount of energy available for this vibrational motion is determined by the temperature of the solid. As the temperature increases, the particles vibrate more vigorously. However, the particles do not move from their positions in the solid, and so they do not exhibit translational motion. Therefore, the correct answer is that particles in a solid exhibit vibrational motion. The other options (vibrational and translational motion, vibrational and random motion, and random and translational motion) are not correct because they describe the motion of particles in liquids or gases, where the particles are free to move and exhibit translational or random motion in addition to vibrational motion.
Tambaya 14 Rahoto
Ethene is produced from ethanol by
Bayanin Amsa
Ethene is produced from ethanol by: Dehydration. Dehydration is the process of removing water from a compound. In the case of ethanol, the process of dehydration involves removing a molecule of water (H2O) from two molecules of ethanol (C2H5OH) to form one molecule of ethene (C2H4). The reaction is typically carried out using a catalyst, such as concentrated sulfuric acid or aluminum oxide. The chemical equation for the reaction of ethanol to produce ethene by dehydration is: C2H5OH → C2H4 + H2O Therefore, the correct answer is "dehydration" - ethanol is converted to ethene by removing a molecule of water through the process of dehydration. The other options - decomposition, hydrolysis, and ozonolysis - do not accurately describe the process of producing ethene from ethanol.
Tambaya 15 Rahoto
The number of Hydrogen ions in 1.0\(dm^{3}\) of 0.02\(moldm^{-3}\) tetraoxosulphate(VI) acid is \([N_{A} = 6.02 \times 10^{23}]\)
Tambaya 16 Rahoto
Consider the following reaction equation: \(Br_{2} + 2KI \to 2KBr + I_{2}\). Bromine is acting as
Bayanin Amsa
In the given reaction, the bromine molecule (Br2) is reacting with two potassium iodide molecules (2KI) to form two potassium bromide molecules (2KBr) and one iodine molecule (I2). During this reaction, the bromine molecule is gaining electrons (receiving electrons) and being reduced, while the iodine molecule is losing electrons (donating electrons) and being oxidized. Therefore, we can conclude that the bromine molecule is acting as a oxidizing agent because it is causing the oxidation of the iodine molecule by accepting electrons, and therefore, itself being reduced.
Tambaya 17 Rahoto
How many atoms are contained in 0.2moles of nitrogen? \([N_{A} = 6.02 \times 10^{23}]\)
Bayanin Amsa
To determine the number of atoms in 0.2 moles of nitrogen, we can use Avogadro's number, which represents the number of particles in one mole of a substance. Avogadro's number is equal to 6.02 x 10^23 particles per mole. First, we need to calculate the number of nitrogen atoms in 0.2 moles of nitrogen. We can do this by multiplying the number of moles by Avogadro's number: 0.2 moles x 6.02 x 10^23 particles/mole = 1.204 x 10^23 particles Therefore, there are 1.204 x 10^23 atoms in 0.2 moles of nitrogen. The answer is \(1.20 \times 10^{23}\).
Tambaya 18 Rahoto
Which of the following CANNOT be an empirical formula?
Tambaya 19 Rahoto
The bonding pair of electrons in a Hydrogen Chloride molecule is pulled towards the chlorine atom because
Bayanin Amsa
The correct answer is that chlorine is more electronegative than hydrogen. Electronegativity is a measure of an atom's ability to attract electrons towards itself in a covalent bond. Chlorine has a higher electronegativity than hydrogen, which means that the bonding pair of electrons in a hydrogen chloride molecule is pulled more towards the chlorine atom than towards the hydrogen atom. This creates a partial positive charge on the hydrogen atom and a partial negative charge on the chlorine atom, forming a polar molecule.
Tambaya 20 Rahoto
Atoms are electrically neutral because they
Bayanin Amsa
The correct answer is that atoms are electrically neutral because they contain an equal number of protons and electrons. Protons have a positive charge, while electrons have a negative charge. In an atom, the number of protons in the nucleus is equal to the number of electrons orbiting around the nucleus. The positive charge of the protons is balanced by the negative charge of the electrons, resulting in an overall charge of zero for the atom. This balance of positive and negative charges is what makes atoms electrically neutral.
Tambaya 22 Rahoto
Which of the following statements is true about ionic radius? Ionic radius
Bayanin Amsa
Ionic radius is the measure of the size of an ion. It refers to the distance from the center of the nucleus to the outermost electron shell of an ion. When an atom loses or gains electrons to form an ion, its size changes. The trend in ionic radius is that it decreases as the nuclear charge increases. This is because as the nuclear charge increases, the number of protons in the nucleus increases, which leads to stronger attraction between the electrons and the nucleus. This stronger attraction pulls the electrons closer to the nucleus, reducing the size of the ion. Therefore, the correct answer is "decreases as nuclear charge increases".
Tambaya 23 Rahoto
When chlorine is passed through a sample of water, the pH of the water sample would be
Bayanin Amsa
When chlorine gas is passed through water, it reacts with water to form hydrochloric acid (HCl) and hypochlorous acid (HClO). Hypochlorous acid is a weak acid and can dissociate to form hydrogen ions (H+) and hypochlorite ions (ClO-). The formation of H+ ions in the water will increase the concentration of H+ ions, resulting in a decrease in pH. Therefore, the pH of the water sample will be less than 7 (i.e., acidic) when chlorine is passed through it. It is important to note that the pH of the water sample will depend on the amount of chlorine passed through it, the initial pH of the water, and other factors such as temperature and the presence of other chemicals.
Tambaya 24 Rahoto
In which of the following compounds does hydrogen form ionic compounds?
Bayanin Amsa
Out of the given compounds, only NaH (sodium hydride) contains ionic bonds involving hydrogen. In NaH, hydrogen has a negative charge (H\(^-\)) and sodium has a positive charge (Na\(^+\)). This results in the formation of an ionic compound. In the other compounds, hydrogen forms covalent bonds. In CH\(_4\) (methane), hydrogen shares electrons with carbon to form four covalent bonds. In HCl (hydrogen chloride), hydrogen shares electrons with chlorine to form a single covalent bond. In NH\(_3\) (ammonia), hydrogen shares electrons with nitrogen to form three covalent bonds. Therefore, only NaH contains ionic bonds involving hydrogen, while the others have covalent bonds.
Tambaya 25 Rahoto
A reaction is endothermic if the
Bayanin Amsa
An endothermic reaction is a reaction that requires energy to be absorbed from the surroundings. This means that the reaction absorbs heat and as a result, the surroundings feel cooler. Therefore, the correct option is "reaction vessel feels cool during the reaction". The other options do not necessarily define an endothermic reaction.
Tambaya 26 Rahoto
The change in the oxidation state of iron in the reaction represented by the equation: \(2FeCl_{3} + H_{2}S \to 2FeCl_{2} + 2HCl + S\) is
Bayanin Amsa
In the given chemical equation, the reactants are iron(III) chloride (\(FeCl_{3}\)) and hydrogen sulfide (\(H_{2}S\)), and the products are iron(II) chloride (\(FeCl_{2}\)), hydrogen chloride (\(HCl\)), and sulfur (\(S\)). Let's consider the oxidation state of iron in the reactants and products. Iron has a variable oxidation state, and it can exist in either +2 or +3 oxidation state in its compounds. In \(FeCl_{3}\), iron has an oxidation state of +3. Hydrogen sulfide (\(H_{2}S\)) has sulfur in -2 oxidation state. In the product \(FeCl_{2}\), iron has an oxidation state of +2. Sulfur in the product is in 0 oxidation state. Therefore, in the given chemical equation, the oxidation state of iron changes from +3 in the reactant (\(FeCl_{3}\)) to +2 in the product (\(FeCl_{2}\)). Hence the correct answer is "+3 to +2".
Tambaya 27 Rahoto
When zinc is added to AgNO\(_{3}\) solution, crystals of silver forms on the zinc surface. This indicates that zinc is
Bayanin Amsa
Tambaya 28 Rahoto
Consider the following equilibrium reaction: \(2AB_{{2}{(g)}} + B_{{2}{(g)}} \to 2AB_{{3}{(g)}}\). \(\Delta H= -X kJmol^{-1}\). The backward reaction will be favored by
Tambaya 29 Rahoto
The process of extraction of iron from its ore is
Bayanin Amsa
The process of extracting iron from its ore involves the reduction of the iron oxide (Fe2O3) in the ore to iron metal. This is achieved through a chemical reaction between the iron oxide and carbon (in the form of coke) in a blast furnace. First, the iron ore is crushed and mixed with coke and limestone. The coke, which is mostly carbon, serves as a reducing agent by combining with the oxygen in the iron oxide to form carbon dioxide (CO2). This reaction produces heat and carbon monoxide (CO). The carbon monoxide then reacts with the iron oxide to produce iron metal and more carbon dioxide. The limestone is added to the furnace to remove impurities such as silica (SiO2) that would otherwise interfere with the chemical reaction. The molten iron is then tapped off from the bottom of the furnace and can be further processed into steel.
Tambaya 30 Rahoto
Which of the following electron configurations represents the transition element Chromium \(_{24}Cr\)?
Bayanin Amsa
The electron configuration for Chromium \(_{24}Cr\) is \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} 3d^{5}\). Electron configuration is a description of the arrangement of electrons in an atom. It is represented by a series of numbers and letters that indicate the energy level, or shell, and the orbital, or subshell, that each electron resides in. In the case of Chromium, the electron configuration starts with the first energy level, which is filled with two electrons in the 1s orbital. The second energy level has two electrons in the 2s orbital and six electrons in the 2p orbital. The third energy level has two electrons in the 3s orbital and six electrons in the 3p orbital. Finally, the fourth energy level has one electron in the 4s orbital and five electrons in the 3d orbital. So, out of the given options, the electron configuration that represents Chromium \(_{24}Cr\) is \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} 3d^{5}\).
Tambaya 31 Rahoto
What is the mass of solute in 500\(cm^{3}\) of 0.005\(moldm^{-3}\) \(H_{2}SO_{4}\)? ( H = 1, S = 32.0, O=16.0)
Bayanin Amsa
n = cv where n = no of mole, c = molar concentration(mol/dm3) and v = volume( dm3)
volume = 500
cm3
= 0.5dm3, c = 0.005moldm−3
massmolarmass
= 0.0025 = mass98
(where 98g/mol is the molar mass of H2SO4)
number of mole(n) = c x v = 0.5 X 0.005 = 0.0025mol.
but n =
mass of H2SO4 = 0.0025 x 98 = 0.245g.
Tambaya 33 Rahoto
The empirical formula of a compound containing 0.067mol Cu and 0.066mol O is [Cu = 63.5, O = 16]
Bayanin Amsa
To determine the empirical formula of the compound, we need to find the simplest whole number ratio of the atoms present in the compound. Given that the compound contains 0.067 mol Cu and 0.066 mol O, we need to first convert the moles to whole numbers. Dividing both values by the smaller value (0.066 mol), we get: Cu: 0.067 mol ÷ 0.066 mol ≈ 1.02 ≈ 1 O: 0.066 mol ÷ 0.066 mol = 1 So, the ratio of Cu to O atoms is 1:1. Therefore, the empirical formula of the compound is CuO. We can check that the formula is correct by calculating the molar mass of CuO and comparing it to the experimental data. Molar mass of CuO = 63.5 g/mol + 16 g/mol = 79.5 g/mol The experimental data gives us: (0.067 mol)(63.5 g/mol) + (0.066 mol)(16 g/mol) = 4.2495 g The molar mass calculated from the experimental data is: (4.2495 g) / (0.133 mol) ≈ 31.97 g/mol This is close to the molar mass of CuO (79.5 g/mol), which confirms that the empirical formula is CuO.
Tambaya 34 Rahoto
What is the total number of shared pair of electrons in the compound above?
Bayanin Amsa
Tambaya 36 Rahoto
The diagram above illustrates a conical flask containing water and ice. Which of the following is correct about the diagram?
Bayanin Amsa
The correct statement about the diagram is: "Energy is absorbed when the ice changes to water." When ice changes to water, it undergoes a physical process known as "melting." During this process, energy is absorbed from the surroundings in order to overcome the attractive forces between the ice molecules and break them apart into individual water molecules. This energy is in the form of heat, and it is absorbed from the surroundings, which results in a decrease in temperature. Therefore, the water in the conical flask will be at a lower temperature than the ice after the ice starts melting. The water molecules do vibrate about a fixed point, but this statement is not relevant to the question being asked.
Tambaya 37 Rahoto
Stainless steel is an alloy comprising of
Tambaya 38 Rahoto
The position of equilibrium in a reversible reaction is affected by
Bayanin Amsa
The position of equilibrium in a reversible reaction is affected by the change in concentration of the reactants. When the concentration of reactants changes, the rate of the forward and reverse reactions will also change, which will shift the position of the equilibrium to counteract the change in concentration. For example, if the concentration of reactants increases, the rate of the forward reaction will increase, which will shift the position of the equilibrium to the right to counteract the increase in concentration. Conversely, if the concentration of reactants decreases, the rate of the reverse reaction will increase, which will shift the position of the equilibrium to the left to counteract the decrease in concentration.
Tambaya 39 Rahoto
Which of the following methods can be used to seperate blood cells from plasma?
Bayanin Amsa
The method that can be used to separate blood cells from plasma is: Centrifugation. Centrifugation is a process of separating components in a mixture based on their density differences. In the case of blood, when it is spun in a centrifuge, the heavier components, such as the blood cells, are forced to the bottom of the tube while the lighter components, such as plasma, stay at the top. The separation is based on the fact that blood cells are denser than plasma. Therefore, centrifugation is the method that can be used to separate blood cells from plasma. Filtration, chromatography, and distillation are other methods of separating mixtures, but they are not suitable for separating blood cells from plasma. Filtration is the process of separating solids from liquids using a filter. Chromatography is a laboratory technique for separating mixtures into their individual components. Distillation is the process of separating components of a mixture based on differences in their boiling points.
Tambaya 40 Rahoto
Consider the following reaction equation: \(2HCl + Ca(OH)_{2} \to CaCl_{2} + H_{2}O\). What is the volume of 0.1\(moldm^{-3}\) HCl that would completely neutralize 25\(cm^{3}\) of 0.3\(moldm^{-3}\) Ca(OH)\(_{2}\)?
Tambaya 41 Rahoto
When substance X was added to a solution of bromine water, the solution became colourless. X is likely to be
Tambaya 42 Rahoto
The following factors would contribute to environmental pollution except
Bayanin Amsa
Environmental pollution can be caused by a variety of factors, including the release of harmful chemicals and substances into the air, water, or soil. Of the options provided, photosynthesis is the factor that would not contribute to environmental pollution. Photosynthesis is the process by which plants use sunlight to convert carbon dioxide and water into oxygen and glucose. This natural process does not release any harmful chemicals or substances into the environment and, in fact, helps to improve air quality by producing oxygen. On the other hand, the production of ammonia, manufacture of cement, and combustion are all activities that can release harmful pollutants into the environment. Ammonia production can release nitrogen oxide and other harmful chemicals into the air and water, while cement manufacturing can release large amounts of carbon dioxide into the atmosphere. Combustion, whether in vehicles or power plants, can release harmful gases and particulate matter into the air. Therefore, photosynthesis is the only option among the given factors that would not contribute to environmental pollution.
Tambaya 43 Rahoto
What is the solubility of a salt if 0.4g of it is obtained on evaporating 200\(cm^{3}\) of its saturated solution to dryness?
Bayanin Amsa
Tambaya 44 Rahoto
At what temperature does the solubility of \(KNO_{3}\) equal that of \(NaNO_{3}\)?
Bayanin Amsa
Tambaya 45 Rahoto
Which of the following organic compounds can undergo both addition and substitution reactions?
Bayanin Amsa
Benzene is the organic compound that can undergo both addition and substitution reactions. Benzene is an aromatic compound with a unique structure that includes a ring of six carbon atoms, each of which is bonded to a hydrogen atom. The electrons in the carbon-carbon double bonds are delocalized, making the molecule more stable and less reactive than typical alkenes. However, the delocalized electrons can still participate in reactions, including addition and substitution. Addition reactions involve the breaking of a double or triple bond and the addition of new atoms or groups of atoms. Because benzene already has delocalized double bonds, it can undergo addition reactions such as hydrogenation and halogenation, in which hydrogen or a halogen is added to the ring. Substitution reactions involve the replacement of one atom or group of atoms with another atom or group of atoms. Benzene is particularly prone to substitution reactions because the delocalized electrons in the ring make it more susceptible to attacks by electrophiles, which are electron-deficient species that can accept a pair of electrons. Substitution reactions include nitration, sulfonation, and halogenation.
Tambaya 46 Rahoto
The valence electrons of \(_{12}Mg\) are in the
Bayanin Amsa
The valence electrons of magnesium, which has an atomic number of 12, are located in the 3s orbital. In an atom, electrons are arranged in shells surrounding the nucleus. The innermost shell, the 1s orbital, holds the electrons closest to the nucleus and is usually filled first. The next shell, the 2s orbital, is further away from the nucleus and is also filled before the third shell, the 3s orbital. In the case of magnesium, it has 12 electrons, and so the first two shells are completely filled. The 3rd shell, which is the valence shell, has only two electrons in the 3s orbital. These are the valence electrons, and they are the electrons that are involved in chemical reactions and bonding with other atoms.
Tambaya 47 Rahoto
The atomic number of an isotope of hydrogen is equal to its mass number because it
Tambaya 48 Rahoto
Which of the following properties would not influence electrovalent bond formation?
Bayanin Amsa
The property that would not influence electrovalent bond formation is catalytic ability. An electrovalent bond, also known as an ionic bond, is formed when atoms transfer electrons from one to another, resulting in the formation of ions with opposite charges. This bond is typically formed between atoms with significantly different electronegativities. The electronegativity of an atom is a measure of its ability to attract electrons towards itself, and it is a key factor in determining the type of bond that will form between two atoms. Electron affinity is the energy change that occurs when an electron is added to a neutral atom to form a negative ion. This property also influences the formation of electrovalent bonds. Ionization potential is the energy required to remove an electron from an atom. This property can also affect the formation of electrovalent bonds, as it can influence the ability of an atom to give up electrons. Catalytic ability, on the other hand, is the ability of a substance to speed up a chemical reaction without being consumed in the reaction. This property has no direct influence on the formation of electrovalent bonds. To summarize, the property that would not influence electrovalent bond formation is catalytic ability, while electronegativity, electron affinity, and ionization potential all have an impact on the formation of these types of bonds.
Tambaya 49 Rahoto
Which of the following statements best explains the difference between a gas and a vapour?
Bayanin Amsa
The best explanation for the difference between a gas and a vapor is: Unlike gases, vapors can easily be condensed into liquids. Gases and vapors are both in the gaseous state of matter, which means they have no fixed shape or volume and can expand to fill any container. However, vapors are substances that are normally in the liquid or solid state at room temperature and pressure, but are volatilized (turned into a gas) through evaporation or boiling. The main difference between gases and vapors is that gases are typically elements or compounds that are naturally in the gaseous state, whereas vapors are substances that have been volatilized from their liquid or solid state. Since vapors have been volatilized from their liquid or solid state, they can easily be condensed back into a liquid or solid form by cooling or compressing them. Therefore, the correct statement that explains the difference between a gas and a vapor is that "unlike gases, vapors can easily be condensed into liquids." The other options given are not true or do not accurately explain the difference between gases and vapors.
Tambaya 50 Rahoto
An acidic salt has
Bayanin Amsa
An acidic salt has hydrogen ions in its aqueous solution. When a salt is formed from a weak acid and a strong base, the resulting salt can still contain some of the acidic properties of the original acid. This means that when the salt is dissolved in water, it can release hydrogen ions (H+) into the solution, making it acidic. This is why the salt is called an acidic salt. In contrast, basic salts contain hydroxide ions (OH-) in their aqueous solution.
Tambaya 51 Rahoto
(a) Explain the statement, the standard electrode potential of zinc is -0.76 v.What is meant by the term periodic property of elements?
(b) Consider the following standard electrode potentials:
| \( \mathrm{Zn}^{2+}_{(aq)} + 2e^- \) | \( \mathrm{Zn(s)} \) | Eᶿ = - 0.76 V |
| \( \mathrm{Cu}^{2+}_{(aq)} + 2e^- \) | \( \mathrm{Cu(s)} \) | Eᶿ = + 0.34 V |
When the two half cells are connected:
(i) write the reaction equation at each electrode;
(ii) write the overall cell reaction equation;
(iii) state the type of reaction occurring at each electrode;
(iv) calculate the e.m.f. of the cell.
(c) (i) Name two chemical industries.
(ii) State two factors that should be considered when siting a chemical industry.
(iii) List two effects of a chemical industry on the community in which it is sited.
(d) Using chemical equations, explain briefly what would happen when hydrogen peroxide is added to:
(i) silver oxide;
(ii) chlorine gas.
(e) List three physical properties of nitrogen.
Tambaya 52 Rahoto
(a) (i) Define the term fermentation
(ii) Name the catalyst that can be used for this process
(b) Name two factors which determine the choice of an indicator for an acid-base titration
(c) Consider the following reaction equation: Fe + H\(_2\)SO\(_4\) \(\to\) FeSO\(_4\) + H\(_2\). Calculate the mass of unreacted iron when 5.0g of iron reacts with 10cm\(^3\) of 1.0 moldm\(^3\) H\(_2\)SO\(_4\), [Fe = 56.0]
(d) Name one:
(i) Heavy chemical used in electrolytic cells
(ii) Fine chemical used in textile industries
(e) Explain briefly how a catalyst increases the rate of a chemical reaction.
(f) (i) Write the chemical formula for the product formed when ethanoic acid reacts with ammonia
(ii) Give the name of the product formed in (f)(i)
(g) List three properties of aluminum that makes it suitable for the manufacture of drinks cans
(h) State two industrial uses of alkylalkanoates
(i) Name two steps involved in the crystallization of a salt from its solution
(j) List two effects of global warming
(a)(i) Fermentation is the slow breakdown of complex organic substances (especially carbohydrates/sugars) into simpler substances such as ethanol and carbon(IV) oxide by enzymes produced by micro-organisms.
(a)(ii) The enzyme zymase (from yeast).
(b) The choice of indicator depends on: (1) the strengths of the acid and base being titrated (i.e. the type of titration), and (2) the pH range at which the indicator changes colour, which must match the pH at the equivalence point.
(c) Mass of unreacted iron
\( n(\text{H}_2\text{SO}_4) = 1.0 \times \dfrac{10}{1000} = 0.01\ \text{mol} \)
Fe + H2SO4 \(\to\) FeSO4 + H2 (1 : 1), so Fe reacting = 0.01 mol.
Mass of Fe reacted = \(0.01 \times 56 = 0.56\ \text{g}\).
\[ \text{Unreacted Fe} = 5.0 - 0.56 = 4.44\ \text{g} \]
(d)(i) A heavy chemical: sodium chloride (brine) (or sulphuric acid / sodium hydroxide).
(d)(ii) A fine chemical: a dye.
(e) A catalyst provides an alternative reaction pathway with a lower activation energy, so a greater fraction of the colliding particles have enough energy to react, and the rate increases.
(f)(i) \( \text{CH}_3\text{COONH}_4 \)
(f)(ii) Ammonium ethanoate.
(g) Aluminium is light (low density), resistant to corrosion (protective oxide layer), and malleable/easily shaped and non-toxic.
(h) Alkylalkanoates (esters) are used as flavourings/perfumes and as solvents.
(i) (1) Evaporate the solution to the point of saturation; (2) cool the hot saturated solution to allow crystals to separate (then filter and dry).
(j) Melting of polar ice caps and rise in sea level, and climate change/flooding (or increased desertification).
Bayanin Amsa
(a)(i) Fermentation is the slow breakdown of complex organic substances (especially carbohydrates/sugars) into simpler substances such as ethanol and carbon(IV) oxide by enzymes produced by micro-organisms.
(a)(ii) The enzyme zymase (from yeast).
(b) The choice of indicator depends on: (1) the strengths of the acid and base being titrated (i.e. the type of titration), and (2) the pH range at which the indicator changes colour, which must match the pH at the equivalence point.
(c) Mass of unreacted iron
\( n(\text{H}_2\text{SO}_4) = 1.0 \times \dfrac{10}{1000} = 0.01\ \text{mol} \)
Fe + H2SO4 \(\to\) FeSO4 + H2 (1 : 1), so Fe reacting = 0.01 mol.
Mass of Fe reacted = \(0.01 \times 56 = 0.56\ \text{g}\).
\[ \text{Unreacted Fe} = 5.0 - 0.56 = 4.44\ \text{g} \]
(d)(i) A heavy chemical: sodium chloride (brine) (or sulphuric acid / sodium hydroxide).
(d)(ii) A fine chemical: a dye.
(e) A catalyst provides an alternative reaction pathway with a lower activation energy, so a greater fraction of the colliding particles have enough energy to react, and the rate increases.
(f)(i) \( \text{CH}_3\text{COONH}_4 \)
(f)(ii) Ammonium ethanoate.
(g) Aluminium is light (low density), resistant to corrosion (protective oxide layer), and malleable/easily shaped and non-toxic.
(h) Alkylalkanoates (esters) are used as flavourings/perfumes and as solvents.
(i) (1) Evaporate the solution to the point of saturation; (2) cool the hot saturated solution to allow crystals to separate (then filter and dry).
(j) Melting of polar ice caps and rise in sea level, and climate change/flooding (or increased desertification).
Tambaya 53 Rahoto
(a) (i) List the two gaseous fuels produced from coke.
(ii) Which of the two fuels listed in 5(a)(i) is a better fuel?
(iii) Give reasons for your answer in 5(a)(ii)
(iv) Write a balanced equation for the production of each of the fuels. [9 marks]
(b)(i) Differentiate between thermosets and thermoplastics.
(ii) Give one example of:
I. thermosets;
II. thermoplastics.
(iii) State three properties of plastics.
(c)(i) State the method of collecting gases which are denser than air.
(ii) Name two gases that could be used to perform the fountain experiment in the laboratory.
(iii) State the physical properties of the gases named in 5(c)(ii) which makes them suitable for the experiment. [4 marks]
(d) (i) State two compounds that could be used to test for water.
(ii) Give three disadvantages of hard water.
(a)(i) Water gas (CO + H2) and producer gas (CO + N2).
(a)(ii) Water gas is the better fuel.
(a)(iii) Water gas has a higher heating (calorific) value because it is a mixture of two combustible gases (CO and H2) and is not diluted by large amounts of non-combustible nitrogen, whereas producer gas contains much inert N2.
(a)(iv) Water gas: \[ \text{C} + \text{H}_2\text{O} \to \text{CO} + \text{H}_2 \] Producer gas: \[ 2\text{C} + \text{O}_2 \to 2\text{CO} \quad (\text{air over hot coke}) \]
(b)(i) Thermosets harden permanently once moulded and cannot be softened or re-shaped on further heating (they are cross-linked); thermoplastics soften on heating and can be re-moulded repeatedly.
(b)(ii) I. Thermoset: bakelite (or melamine). II. Thermoplastic: polythene (polyethene) (or PVC).
(b)(iii) Plastics are light, resistant to corrosion/chemicals, and poor conductors of heat and electricity (also easily moulded).
(c)(i) By downward delivery (upward displacement of air), collecting the gas in an upright jar.
(c)(ii) Ammonia and hydrogen chloride.
(c)(iii) They are extremely soluble in water, so the water rushes up to fill the partial vacuum created when the gas dissolves, producing the fountain.
(d)(i) Anhydrous copper(II) sulphate (white to blue) and anhydrous cobalt(II) chloride (blue to pink).
(d)(ii) Hard water wastes soap (forms scum), forms scale/fur in kettles, boilers and pipes, and the scale reduces heat efficiency and can block pipes.
Bayanin Amsa
(a)(i) Water gas (CO + H2) and producer gas (CO + N2).
(a)(ii) Water gas is the better fuel.
(a)(iii) Water gas has a higher heating (calorific) value because it is a mixture of two combustible gases (CO and H2) and is not diluted by large amounts of non-combustible nitrogen, whereas producer gas contains much inert N2.
(a)(iv) Water gas: \[ \text{C} + \text{H}_2\text{O} \to \text{CO} + \text{H}_2 \] Producer gas: \[ 2\text{C} + \text{O}_2 \to 2\text{CO} \quad (\text{air over hot coke}) \]
(b)(i) Thermosets harden permanently once moulded and cannot be softened or re-shaped on further heating (they are cross-linked); thermoplastics soften on heating and can be re-moulded repeatedly.
(b)(ii) I. Thermoset: bakelite (or melamine). II. Thermoplastic: polythene (polyethene) (or PVC).
(b)(iii) Plastics are light, resistant to corrosion/chemicals, and poor conductors of heat and electricity (also easily moulded).
(c)(i) By downward delivery (upward displacement of air), collecting the gas in an upright jar.
(c)(ii) Ammonia and hydrogen chloride.
(c)(iii) They are extremely soluble in water, so the water rushes up to fill the partial vacuum created when the gas dissolves, producing the fountain.
(d)(i) Anhydrous copper(II) sulphate (white to blue) and anhydrous cobalt(II) chloride (blue to pink).
(d)(ii) Hard water wastes soap (forms scum), forms scale/fur in kettles, boilers and pipes, and the scale reduces heat efficiency and can block pipes.
Tambaya 54 Rahoto
(a)(i) Draw and label a diagram for the laboratory preparation of a dry sample of sulphur(IV)oxide.
(ii) Write a balanced chemical equation for the reaction in (a)(i).
(iii) State the precaution that must be taken in the preparation of the gas stated in (a)(i).
(iv) Give a reason why the precaution stated in (a)(ii) must be taken.
(b)(i) State Dalton's law of partial pressures.
(ii) The volume of a sample of methane collected over-water at a temperature of 12°C and a pressure of 700 mmHg was 30cm\(^3\). Calculate the volume of the dry gas at s.t.p. [Saturated vapour pressure of water at 12°C is 10 mmHg] •
(c)(i) Write an equation for the reaction between chlorine and water.
(ii) Why does litmus paper turn red when put in the resulting solution in (c)(i)?
(d)(i) State the trend in the boiling points of chlorine, bromine and iodine.
(ii) Explain briefly why water has a higher boiling point than ammonia.
(a)(i) The apparatus for preparing and drying sulphur(IV) oxide is shown below.
(ii)
\[\mathrm{Na_2SO_3(aq)+2HCl(aq)\rightarrow 2NaCl(aq)+H_2O(l)+SO_2(g)}\]
(iii) The preparation should be carried out in a fume cupboard.
(iv) Sulphur(IV) oxide is poisonous and irritates the eyes and respiratory tract; the fume cupboard prevents inhalation of the gas.
(b)(i) Dalton's law of partial pressures states that the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.
(ii)
The methane was collected over water, so its pressure is:
\[P_1=700-10=690\ \mathrm{mmHg}\]
\[T_1=12+273=285\ \mathrm{K}\]
At s.t.p., \(P_2=760\ \mathrm{mmHg}\) and \(T_2=273\ \mathrm{K}\).
Using \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\):
\[V_2=\frac{P_1V_1T_2}{P_2T_1}=\frac{690\times30\times273}{760\times285}=26.1\ \mathrm{cm^3}\]
Therefore, the volume of dry methane at s.t.p. is \(26.1\ \mathrm{cm^3}\).
(c)(i)
\[\mathrm{Cl_2(g)+H_2O(l)\rightleftharpoons HCl(aq)+HClO(aq)}\]
(ii) Hydrogen chloride, \(\mathrm{HCl}\), ionises in water to produce \(\mathrm{H^+}\) ions. The resulting solution is acidic and turns blue litmus paper red.
(d)(i) The boiling points increase from chlorine to iodine:
\[\mathrm{Cl_2<Br_2<I_2}\]
(d)(ii) Water has a higher boiling point because its intermolecular hydrogen bonding is stronger and more extensive than that in ammonia. Oxygen is more electronegative than nitrogen, and each water molecule can form more hydrogen bonds; hence more energy is required to separate water molecules.
Bayanin Amsa
(a)(i) The apparatus for preparing and drying sulphur(IV) oxide is shown below.
(ii)
\[\mathrm{Na_2SO_3(aq)+2HCl(aq)\rightarrow 2NaCl(aq)+H_2O(l)+SO_2(g)}\]
(iii) The preparation should be carried out in a fume cupboard.
(iv) Sulphur(IV) oxide is poisonous and irritates the eyes and respiratory tract; the fume cupboard prevents inhalation of the gas.
(b)(i) Dalton's law of partial pressures states that the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.
(ii)
The methane was collected over water, so its pressure is:
\[P_1=700-10=690\ \mathrm{mmHg}\]
\[T_1=12+273=285\ \mathrm{K}\]
At s.t.p., \(P_2=760\ \mathrm{mmHg}\) and \(T_2=273\ \mathrm{K}\).
Using \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\):
\[V_2=\frac{P_1V_1T_2}{P_2T_1}=\frac{690\times30\times273}{760\times285}=26.1\ \mathrm{cm^3}\]
Therefore, the volume of dry methane at s.t.p. is \(26.1\ \mathrm{cm^3}\).
(c)(i)
\[\mathrm{Cl_2(g)+H_2O(l)\rightleftharpoons HCl(aq)+HClO(aq)}\]
(ii) Hydrogen chloride, \(\mathrm{HCl}\), ionises in water to produce \(\mathrm{H^+}\) ions. The resulting solution is acidic and turns blue litmus paper red.
(d)(i) The boiling points increase from chlorine to iodine:
\[\mathrm{Cl_2<Br_2<I_2}\]
(d)(ii) Water has a higher boiling point because its intermolecular hydrogen bonding is stronger and more extensive than that in ammonia. Oxygen is more electronegative than nitrogen, and each water molecule can form more hydrogen bonds; hence more energy is required to separate water molecules.
Tambaya 55 Rahoto
(a) Arrange the three states of matter in order of decreasing:
(i) kinetic energy;
(ii) force of cohesion.
(b) Consider the redox reaction equation:
(i) State the change in oxidation number of:
I. magnesium;
II. hydrogen.
(ii) Which of the species is being:
I. oxidized;
II. reduced?
(iii) Identify the oxidizing agent.
(c) (i) State two differences between boiling and evaporation.
(ii) What will be the effect of reduction of atmospheric pressure on the boiling point of water?
(d) For a given chemical equilibrium system, what is the significance of the equilibrium constant \(K\)?
(e) Consider the following organic compounds:
\(\mathrm{C_3H_7COOH}\); \(\mathrm{(CH_3)_3COH}\).
Give the IUPAC name of each compound.
(f) Why are organic compounds classified on the basis of functional groups?
(g) State three differences between the solubility of solids in liquids and gases in liquids.
(h) Write a balanced chemical equation for the reaction between fluorine and water.
(i) Define the term basicity of an acid.
(a)
(i) Decreasing kinetic energy: gas > liquid > solid.
(ii) Decreasing force of cohesion: solid > liquid > gas.
(b) For the reaction \( \text{Mg} + 2\text{HCl} \to \text{MgCl}_2 + \text{H}_2 \):
(c)(i) Two differences between boiling and evaporation:
(c)(ii) Reducing the atmospheric pressure lowers the boiling point of water, so it boils at a temperature below 100 °C.
(d) The equilibrium constant K indicates the extent to which a reaction proceeds and where the position of equilibrium lies: a large K means the products are favoured (reaction nearly complete), while a small K means the reactants are favoured.
(e) C3H7COOH is butanoic acid; (CH3)3COH is 2-methylpropan-2-ol.
(f) Because a functional group is the atom or group of atoms responsible for the characteristic chemical reactions of a compound. Compounds sharing the same functional group have similar chemical properties, so classifying by functional group makes their study systematic and their reactions predictable.
(g) Three differences between solubility of solids and of gases in liquids:
(h) \[ 2\text{F}_2 + 2\text{H}_2\text{O} \to 4\text{HF} + \text{O}_2 \]
(i) The basicity of an acid is the number of replaceable (ionizable) hydrogen ions produced by one molecule of the acid in aqueous solution.
Bayanin Amsa
(a)
(i) Decreasing kinetic energy: gas > liquid > solid.
(ii) Decreasing force of cohesion: solid > liquid > gas.
(b) For the reaction \( \text{Mg} + 2\text{HCl} \to \text{MgCl}_2 + \text{H}_2 \):
(c)(i) Two differences between boiling and evaporation:
(c)(ii) Reducing the atmospheric pressure lowers the boiling point of water, so it boils at a temperature below 100 °C.
(d) The equilibrium constant K indicates the extent to which a reaction proceeds and where the position of equilibrium lies: a large K means the products are favoured (reaction nearly complete), while a small K means the reactants are favoured.
(e) C3H7COOH is butanoic acid; (CH3)3COH is 2-methylpropan-2-ol.
(f) Because a functional group is the atom or group of atoms responsible for the characteristic chemical reactions of a compound. Compounds sharing the same functional group have similar chemical properties, so classifying by functional group makes their study systematic and their reactions predictable.
(g) Three differences between solubility of solids and of gases in liquids:
(h) \[ 2\text{F}_2 + 2\text{H}_2\text{O} \to 4\text{HF} + \text{O}_2 \]
(i) The basicity of an acid is the number of replaceable (ionizable) hydrogen ions produced by one molecule of the acid in aqueous solution.
Tambaya 56 Rahoto
(a)(i) Sketch a graphical representation of Charles' law.
(ii) Calculate the volume of oxygen that would be required for the complete combustion of 2.5 moles of ethanol at s.t.p. [molar volume at s.t.p. = 22.4 dm\(^3\)]
(b)(i) State the collision theory of reaction rates.
(ii) Using the collision theory, explain briefly how temperature can affect the rate of a chemical reaction.
(c)(i) Define esterification.
(ii) Give two uses of alkanoates.
(iii) Give the products of the alkaline hydrolysis of ethyl ethanoate.
(d) A tin coated plate and a galvanized plate were exposed for the same length of time.
(i) Which of the two plates corrodes faster
(ii) Explain briefly your answer in 2(d)(i)
(a)(i) Graphical representation of Charles’ law
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. A plot of volume against temperature in degrees Celsius is a straight line which, when extrapolated, cuts the temperature axis at 1730C.
Thus, \(V\propto T\) when \(T\) is measured in kelvin. A graph of \(V\) against temperature in kelvin passes through the origin.
(a)(ii)
The balanced equation for the combustion of ethanol is:
\[\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}\]
\(1\) mole of ethanol requires \(3\) moles of \(\mathrm{O_2}\).
\[n(\mathrm{O_2})=2.5\times3=7.5\text{ mol}\]
\[V(\mathrm{O_2})=7.5\times22.4=168.0\text{ dm}^3\]
Volume of oxygen required = \(168.0\text{ dm}^3\) at s.t.p.
(b)(i) Collision theory states that a reaction occurs only when reacting particles collide effectively, that is, with energy equal to or greater than the activation energy and in the appropriate orientation.
(b)(ii) On raising the temperature, reacting particles gain kinetic energy and move faster. Collisions become more frequent and a greater proportion of the collisions possess at least the activation energy. The number of effective collisions per second therefore increases, so the reaction rate increases.
(c)(i) Esterification is the reaction of an alkanol with an alkanoic acid, usually in the presence of a concentrated mineral acid catalyst, to form an alkanoate and water.
(c)(ii) Alkanoates are used:
(c)(iii) Alkaline hydrolysis of ethyl ethanoate produces ethanol and a salt of ethanoic acid. For example, with sodium hydroxide:
\[\mathrm{CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH}\]
The products are sodium ethanoate and ethanol.
(d)(i) The tin-coated plate corrodes faster.
(d)(ii) Tin is less reactive than iron, whereas zinc is more reactive than iron. If the coating is damaged, iron in contact with tin becomes the anode and corrodes rapidly. In a galvanized plate, zinc acts as a sacrificial metal and corrodes in preference to the iron, thereby protecting the iron plate.
Bayanin Amsa
(a)(i) Graphical representation of Charles’ law
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. A plot of volume against temperature in degrees Celsius is a straight line which, when extrapolated, cuts the temperature axis at 1730C.
Thus, \(V\propto T\) when \(T\) is measured in kelvin. A graph of \(V\) against temperature in kelvin passes through the origin.
(a)(ii)
The balanced equation for the combustion of ethanol is:
\[\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}\]
\(1\) mole of ethanol requires \(3\) moles of \(\mathrm{O_2}\).
\[n(\mathrm{O_2})=2.5\times3=7.5\text{ mol}\]
\[V(\mathrm{O_2})=7.5\times22.4=168.0\text{ dm}^3\]
Volume of oxygen required = \(168.0\text{ dm}^3\) at s.t.p.
(b)(i) Collision theory states that a reaction occurs only when reacting particles collide effectively, that is, with energy equal to or greater than the activation energy and in the appropriate orientation.
(b)(ii) On raising the temperature, reacting particles gain kinetic energy and move faster. Collisions become more frequent and a greater proportion of the collisions possess at least the activation energy. The number of effective collisions per second therefore increases, so the reaction rate increases.
(c)(i) Esterification is the reaction of an alkanol with an alkanoic acid, usually in the presence of a concentrated mineral acid catalyst, to form an alkanoate and water.
(c)(ii) Alkanoates are used:
(c)(iii) Alkaline hydrolysis of ethyl ethanoate produces ethanol and a salt of ethanoic acid. For example, with sodium hydroxide:
\[\mathrm{CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH}\]
The products are sodium ethanoate and ethanol.
(d)(i) The tin-coated plate corrodes faster.
(d)(ii) Tin is less reactive than iron, whereas zinc is more reactive than iron. If the coating is damaged, iron in contact with tin becomes the anode and corrodes rapidly. In a galvanized plate, zinc acts as a sacrificial metal and corrodes in preference to the iron, thereby protecting the iron plate.
Tambaya 57 Rahoto
(a)(i) Draw the structure of the sixth member of the alkenes.
(ii) Calculate the relative molecular mass of the sixth member of the alkene.
(iii) State one difference between cracking and reforming in the petroleum industry. [H = 1, C = 12]
(b)(i) Define the term enthalpy of neutralization.
(ii) Describe briefly how the enthalpy of neutralization of the reaction of dilute hydrochloric acid and aqueous potassium hydroxide could be determined.
(c) An electrochemical cell is constructed with copper and silver electrodes.
(i) State which of the electrodes will be the: 1. anode; II. cathode.
(ii) Give the reason for your answer in 3(c)(i).
(iii) State the type of reaction occurring at each electrode.
(iv) Write a balanced equation for the overall cell reaction.
(d)(i) Name the compound formed when iron is exposed to moist air for a long time.
(ii) Write a balanced chemical equation for the reaction in 3(d)(i).
(iii) Name one ore of iron.
(a)(i) Alkenes have the general formula \(\mathrm{C_nH_{2n}}\). Starting from ethene as the first member, the sixth member is heptene, \(\mathrm{C_7H_{14}}\). One correct structure is hept-1-ene:
(ii)
(iii) Cracking breaks large hydrocarbon molecules into smaller molecules, whereas reforming rearranges straight-chain hydrocarbons into branched-chain or cyclic hydrocarbons to improve petrol quality.
(b)(i) The enthalpy of neutralization is the enthalpy change when one mole of water is formed by the reaction of an acid with a base in dilute aqueous solution.
(ii) Prepare equimolar dilute hydrochloric acid and aqueous potassium hydroxide. Measure a known volume of the hydrochloric acid into a polystyrene cup calorimeter and record its initial temperature. Measure an equal volume of potassium hydroxide at the same initial temperature, add it quickly to the acid, stir, and record the maximum temperature reached.
Calculate the heat gained by the solution using \(q=mc\Delta T\), where \(m\) is the mass of the mixed solution and \(c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\). Since neutralization is exothermic, the heat of reaction is \(-q\). Divide this value by the number of moles of water formed to obtain the enthalpy of neutralization:
(c)(i)
(ii) Copper is more electropositive, or higher in the electrochemical series, than silver. Therefore copper loses electrons more readily, while \(\mathrm{Ag^+}\) ions gain electrons more readily.
(iii) Oxidation occurs at the copper anode and reduction occurs at the silver cathode.
(iv)
(d)(i) The compound formed is rust, hydrated iron(III) oxide, \(\mathrm{Fe_2O_3\cdot xH_2O}\).
(ii)
(iii) Haematite, \(\mathrm{Fe_2O_3}\), is an ore of iron.
Bayanin Amsa
(a)(i) Alkenes have the general formula \(\mathrm{C_nH_{2n}}\). Starting from ethene as the first member, the sixth member is heptene, \(\mathrm{C_7H_{14}}\). One correct structure is hept-1-ene:
(ii)
(iii) Cracking breaks large hydrocarbon molecules into smaller molecules, whereas reforming rearranges straight-chain hydrocarbons into branched-chain or cyclic hydrocarbons to improve petrol quality.
(b)(i) The enthalpy of neutralization is the enthalpy change when one mole of water is formed by the reaction of an acid with a base in dilute aqueous solution.
(ii) Prepare equimolar dilute hydrochloric acid and aqueous potassium hydroxide. Measure a known volume of the hydrochloric acid into a polystyrene cup calorimeter and record its initial temperature. Measure an equal volume of potassium hydroxide at the same initial temperature, add it quickly to the acid, stir, and record the maximum temperature reached.
Calculate the heat gained by the solution using \(q=mc\Delta T\), where \(m\) is the mass of the mixed solution and \(c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\). Since neutralization is exothermic, the heat of reaction is \(-q\). Divide this value by the number of moles of water formed to obtain the enthalpy of neutralization:
(c)(i)
(ii) Copper is more electropositive, or higher in the electrochemical series, than silver. Therefore copper loses electrons more readily, while \(\mathrm{Ag^+}\) ions gain electrons more readily.
(iii) Oxidation occurs at the copper anode and reduction occurs at the silver cathode.
(iv)
(d)(i) The compound formed is rust, hydrated iron(III) oxide, \(\mathrm{Fe_2O_3\cdot xH_2O}\).
(ii)
(iii) Haematite, \(\mathrm{Fe_2O_3}\), is an ore of iron.
Tambaya 58 Rahoto
(a) A compound of carbon, hydrogen and chlorine contains 0.48 g of carbon, 0.08 g of hydrogen and 1.42 g of chlorine.
(i) Determine the empirical formula of the compound.
(ii) If the molar mass of the compound is 99, calculate the molecular formula of the compound.
[H = 1.0, C = 12.0, Cl = 35.5]
(b) State three properties of NaCl (s) which shows that it is ionic.
(c) Consider the following reaction equation:
(i) On the same diagram, sketch and label a reaction profile for a catalysed and uncatalysed reaction between \(H_2\) and \(O_2\).
(ii) Indicate the possible positions of the activated complexes for the reaction profiles in (c)(i).
(d) The petrochemical industry produces addition polymers using one of the fractions obtained from crude oil.
(i) Name the fraction used as a raw material for the process.
(ii) What process is used to obtain the fraction from crude oil?
(iii) Name two gaseous hydrocarbons that can be used in making polymers.
(iv) Describe briefly how these hydrocarbons can be obtained.
(v) Name the polymer produced from one of the hydrocarbons named in (d)(iii).
(a) (i) Empirical formula
| Element | Mass/g | Moles | Simplest ratio |
|---|---|---|---|
| C | 0.48 | \(\frac{0.48}{12}=0.04\) | \(\frac{0.04}{0.04}=1\) |
| H | 0.08 | \(\frac{0.08}{1}=0.08\) | \(\frac{0.08}{0.04}=2\) |
| Cl | 1.42 | \(\frac{1.42}{35.5}=0.04\) | \(\frac{0.04}{0.04}=1\) |
Ratio of atoms \(=1:2:1\).
Empirical formula \(=\boxed{\mathrm{CH_2Cl}}\).
(a) (ii) Molecular formula
Empirical formula mass \(=12+2(1)+35.5=49.5\).
\[n=\frac{99}{49.5}=2\]
Molecular formula \(=(\mathrm{CH_2Cl})_2=\boxed{\mathrm{C_2H_4Cl_2}}\).
(b) Properties of sodium chloride showing that it is ionic
(c) (i) and (ii) Reaction profiles for the reaction between hydrogen and oxygen
The activated complexes are at the tops of the respective curves, as labelled. The catalysed route has a lower activation energy than the uncatalysed route.
(d)
(i) The fraction used as raw material is naphtha.
(ii) It is obtained from crude oil by fractional distillation.
(iii) Two gaseous hydrocarbons used in making polymers are ethene and propene.
(iv) They are obtained by cracking long-chain hydrocarbons in fractions such as naphtha. In catalytic cracking, the hydrocarbon vapour is heated to about \(600^\circ\mathrm{C}\) over a silica/alumina catalyst, producing smaller molecules including alkenes.
(v) Ethene produces poly(ethene) or polythene. Propene produces poly(propene).
Bayanin Amsa
(a) (i) Empirical formula
| Element | Mass/g | Moles | Simplest ratio |
|---|---|---|---|
| C | 0.48 | \(\frac{0.48}{12}=0.04\) | \(\frac{0.04}{0.04}=1\) |
| H | 0.08 | \(\frac{0.08}{1}=0.08\) | \(\frac{0.08}{0.04}=2\) |
| Cl | 1.42 | \(\frac{1.42}{35.5}=0.04\) | \(\frac{0.04}{0.04}=1\) |
Ratio of atoms \(=1:2:1\).
Empirical formula \(=\boxed{\mathrm{CH_2Cl}}\).
(a) (ii) Molecular formula
Empirical formula mass \(=12+2(1)+35.5=49.5\).
\[n=\frac{99}{49.5}=2\]
Molecular formula \(=(\mathrm{CH_2Cl})_2=\boxed{\mathrm{C_2H_4Cl_2}}\).
(b) Properties of sodium chloride showing that it is ionic
(c) (i) and (ii) Reaction profiles for the reaction between hydrogen and oxygen
The activated complexes are at the tops of the respective curves, as labelled. The catalysed route has a lower activation energy than the uncatalysed route.
(d)
(i) The fraction used as raw material is naphtha.
(ii) It is obtained from crude oil by fractional distillation.
(iii) Two gaseous hydrocarbons used in making polymers are ethene and propene.
(iv) They are obtained by cracking long-chain hydrocarbons in fractions such as naphtha. In catalytic cracking, the hydrocarbon vapour is heated to about \(600^\circ\mathrm{C}\) over a silica/alumina catalyst, producing smaller molecules including alkenes.
(v) Ethene produces poly(ethene) or polythene. Propene produces poly(propene).
Tambaya 59 Rahoto
a) (i) Describe, using the kinetic theory of matter, what happens when potassium chloride dissolves in water.
(ii) Give a reason why the process in (a) (i) is endothermic.
(b) (i) An underground iron pipe is less likely to corrode if it is bonded at intervals with magnesium rods. Give reasons for this observation.
(ii) State the stages involved in the rusting of iron.
(iii) State the condition for the rusting of iron in water.
(c) (i) What is a spontaneous reaction?
(ii) State two conditions that could make a reaction spontaneous.
(iii) Explain briefly why one gramme of sodium reacts more rapidly with water at 250C than one gramme of calcium at the same temperature.
(iv) Write equations for the reactions in (c)(iii).
(d) What mass of lead (II) trioxocarbonate (IV) would contain 35.0 g of lead?
[C=12.0, O = 16.0, Pb = 207.0]
(e) Name the type of intermolecular force present in:
(i) fluorine;
(ii) hydrogen fluoride.
(a)(i) The moving, polar water molecules collide with and surround the K+ and Cl- ions at the surface of the crystal, overcoming the electrostatic forces of the lattice. The ions are pulled away (hydrated) and, because of their kinetic motion, diffuse and become evenly spread throughout the water.
(a)(ii) The process is endothermic because the energy absorbed to break up the ionic lattice (lattice energy) is greater than the energy released when the ions are hydrated (hydration energy); the net heat is taken in from the surroundings.
(b)(i) Magnesium is more reactive (more electropositive) than iron, so it acts as a sacrificial anode: it corrodes in preference to the iron, protecting the pipe (cathodic protection).
(b)(ii) Stages of rusting: iron is oxidised at anodic areas, \( \text{Fe} \to \text{Fe}^{2+} + 2e^- \); oxygen and water are reduced at cathodic areas, \( \text{O}_2 + 2\text{H}_2\text{O} + 4e^- \to 4\text{OH}^- \); the Fe2+ is further oxidised by oxygen to hydrated iron(III) oxide (rust).
(b)(iii) Both water and oxygen (air) must be present.
(c)(i) A spontaneous reaction is one that, once started, proceeds on its own without a continuous supply of external energy.
(c)(ii) A negative enthalpy change (exothermic) and an increase in entropy/disorder (giving a negative free-energy change, \(\Delta G < 0\)).
(c)(iii) Sodium is more electropositive (more reactive) than calcium and loses its outer electron more readily; also 1 g of sodium contains more atoms than 1 g of calcium (Na = 23, Ca = 40), so it reacts faster.
(c)(iv) \[ 2\text{Na} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2 \] \[ \text{Ca} + 2\text{H}_2\text{O} \to \text{Ca(OH)}_2 + \text{H}_2 \]
(d) Mass of PbCO3
\( M(\text{PbCO}_3) = 207 + 12 + 48 = 267 \). In 267 g there is 207 g Pb, so:
\[ \text{mass} = \frac{267}{207} \times 35.0 = 45.1\ \text{g} \]
(e) (i) Fluorine: van der Waals (dispersion) forces. (ii) Hydrogen fluoride: hydrogen bonding.
Bayanin Amsa
(a)(i) The moving, polar water molecules collide with and surround the K+ and Cl- ions at the surface of the crystal, overcoming the electrostatic forces of the lattice. The ions are pulled away (hydrated) and, because of their kinetic motion, diffuse and become evenly spread throughout the water.
(a)(ii) The process is endothermic because the energy absorbed to break up the ionic lattice (lattice energy) is greater than the energy released when the ions are hydrated (hydration energy); the net heat is taken in from the surroundings.
(b)(i) Magnesium is more reactive (more electropositive) than iron, so it acts as a sacrificial anode: it corrodes in preference to the iron, protecting the pipe (cathodic protection).
(b)(ii) Stages of rusting: iron is oxidised at anodic areas, \( \text{Fe} \to \text{Fe}^{2+} + 2e^- \); oxygen and water are reduced at cathodic areas, \( \text{O}_2 + 2\text{H}_2\text{O} + 4e^- \to 4\text{OH}^- \); the Fe2+ is further oxidised by oxygen to hydrated iron(III) oxide (rust).
(b)(iii) Both water and oxygen (air) must be present.
(c)(i) A spontaneous reaction is one that, once started, proceeds on its own without a continuous supply of external energy.
(c)(ii) A negative enthalpy change (exothermic) and an increase in entropy/disorder (giving a negative free-energy change, \(\Delta G < 0\)).
(c)(iii) Sodium is more electropositive (more reactive) than calcium and loses its outer electron more readily; also 1 g of sodium contains more atoms than 1 g of calcium (Na = 23, Ca = 40), so it reacts faster.
(c)(iv) \[ 2\text{Na} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2 \] \[ \text{Ca} + 2\text{H}_2\text{O} \to \text{Ca(OH)}_2 + \text{H}_2 \]
(d) Mass of PbCO3
\( M(\text{PbCO}_3) = 207 + 12 + 48 = 267 \). In 267 g there is 207 g Pb, so:
\[ \text{mass} = \frac{267}{207} \times 35.0 = 45.1\ \text{g} \]
(e) (i) Fluorine: van der Waals (dispersion) forces. (ii) Hydrogen fluoride: hydrogen bonding.
Tambaya 60 Rahoto
1. (a ) Define the term compound.
(b) State two conditions necessary for the cracking of petroleum fractions.
(c) Name two transition elements that are used as catalyst.
(d) (i) Write an equation for the reaction between zinc dust and trioxonitrate (V) solution.
(ii) Which of the reactants in 1(d)(i) is:
I. reduced;
II. oxidized.
(e) Two isotopes of oxygen \(16O\) and \(18O\) have relative abundance of 90 % and 10 % respectively. Calculate the relative atomic mass of oxygen.
(f) List two ores of iron.
(g) (i) What is biotechnology?
(ii) Name one product that can be obtained using biotechnology.
(h) Define the term element.
(i) State two sources of methane in the atmosphere.
(j) Explain briefly why the trend of the boiling points for group VII elements is in the order \(I_2 > Brl_2 > Cl_2\).
(a) A compound is a pure substance formed when two or more elements combine chemically in a fixed proportion by mass.
(b) A high temperature and the presence of a catalyst (e.g. silica-alumina/zeolite). (High pressure is used in thermal cracking.)
(c) Iron and vanadium (also platinum or nickel).
(d)(i) (Taking the salt as silver trioxonitrate(V), AgNO3.)
\[ \text{Zn} + 2\text{AgNO}_3 \to \text{Zn(NO}_3)_2 + 2\text{Ag} \]
(d)(ii) I. Reduced: the silver ion, Ag+ (gains electrons). II. Oxidized: zinc (loses electrons).
(e) Relative atomic mass of oxygen
\[ A_r = \frac{(16 \times 90) + (18 \times 10)}{100} = \frac{1440 + 180}{100} = \frac{1620}{100} = 16.2 \]
(f) Haematite (Fe2O3) and magnetite (Fe3O4).
(g)(i) Biotechnology is the use of living organisms (or their enzymes/systems) to make or modify products useful to man.
(g)(ii) Ethanol (or antibiotics, insulin, yoghurt).
(h) An element is a substance made of only one kind of atom that cannot be split into simpler substances by ordinary chemical means.
(i) Decomposition of organic matter in swamps/marshes (biogas) and the digestion of food by ruminant animals (also natural gas and rice paddies).
(j) Boiling point increases in the order Cl2 < Br2 < I2 because, down the group, the molecular size and number of electrons increase, so the van der Waals (dispersion) forces between molecules become stronger and more energy is needed to separate them.
Bayanin Amsa
(a) A compound is a pure substance formed when two or more elements combine chemically in a fixed proportion by mass.
(b) A high temperature and the presence of a catalyst (e.g. silica-alumina/zeolite). (High pressure is used in thermal cracking.)
(c) Iron and vanadium (also platinum or nickel).
(d)(i) (Taking the salt as silver trioxonitrate(V), AgNO3.)
\[ \text{Zn} + 2\text{AgNO}_3 \to \text{Zn(NO}_3)_2 + 2\text{Ag} \]
(d)(ii) I. Reduced: the silver ion, Ag+ (gains electrons). II. Oxidized: zinc (loses electrons).
(e) Relative atomic mass of oxygen
\[ A_r = \frac{(16 \times 90) + (18 \times 10)}{100} = \frac{1440 + 180}{100} = \frac{1620}{100} = 16.2 \]
(f) Haematite (Fe2O3) and magnetite (Fe3O4).
(g)(i) Biotechnology is the use of living organisms (or their enzymes/systems) to make or modify products useful to man.
(g)(ii) Ethanol (or antibiotics, insulin, yoghurt).
(h) An element is a substance made of only one kind of atom that cannot be split into simpler substances by ordinary chemical means.
(i) Decomposition of organic matter in swamps/marshes (biogas) and the digestion of food by ruminant animals (also natural gas and rice paddies).
(j) Boiling point increases in the order Cl2 < Br2 < I2 because, down the group, the molecular size and number of electrons increase, so the van der Waals (dispersion) forces between molecules become stronger and more energy is needed to separate them.
Tambaya 61 Rahoto
(a)(i) Describe briefly the laboratory preparation of hydrogen gas from the action of steam on iron.
(ii) Write the equation for the reaction in (a)(i).
(iii) List three methods for the industrial preparation of hydrogen gas.
(b) Consider the following reaction equation:
\(\mathrm{CO(g)} + 2\mathrm{H}_{2(g)} \qquad \mathrm{CH}_3\mathrm{OH}_{(g)} + 201\) kJmol-J
Predict the effect of each of the following factors on the position of equilibrium:
(i) decrease in temperature;
(ii) increase in pressure;
(iii) increase in concentration of \(\mathrm{CO}_{(g)}\).
(c) (i) Define condensation polymerization.
(ii) Name two condensation polymers.
(iii) Write the formula of ethylethanoate.
(d) (i) Define the term allotropy.
(ii) Name two crystalline allotropes of carbon.
(iii) State one use of each of the allotropes named in (d)(ii).
(e) State two differences between nitrogen (I) oxide and oxygen.
Bayanin Amsa
None
Tambaya 62 Rahoto
(a) (i) State the collision theory of reaction rates.
(ii) Using the collision theory, explain briefly how temperature can affect the rate of a chemical reaction.
(b) (i) Sketch a graphical representation of Charles’ law.
(ii) Calculate the volume of oxygen that would be required for the complete combustion of 2.5 moles of ethanol at s.t.p.
[molar volume at s.t.p = \(22.4\ \text{dm}^3\)]
(c) (i) Define esterification.
(ii) Give two uses of alkanoates.
(iii) Give the products of the alkaline hydrolysis of ethyl ethanoate.
(d) A tin coated plate and a galvanized plate were exposed for the same length of time.
(i) Which of the two plates corrodes faster?
(ii) Explain briefly your answer in 2 (d) (i).
(a) (i) The collision theory states that reactant particles must collide before a reaction can occur. Only collisions with sufficient energy to overcome the activation energy, and with the correct orientation where necessary, result in reaction.
(a) (ii) When the temperature is increased, the reactant particles gain kinetic energy and move faster. Collisions occur more frequently, and a greater proportion of the collisions have energy equal to or greater than the activation energy. Therefore, the number of effective collisions per second increases and the reaction rate increases.
(b) (i) At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature in kelvin. Hence, a graph of volume against temperature in kelvin is a straight line passing through the origin.
(b) (ii)
The balanced equation for the complete combustion of ethanol is:
\[\mathrm{C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)}\]
From the equation, 1 mole of ethanol requires 3 moles of oxygen.
\[\text{Moles of }O_2 = 2.5 \times 3 = 7.5\text{ mol}\]
\[\text{Volume of }O_2 = 7.5 \times 22.4 = 168.0\text{ dm}^3\]
Therefore, the volume of oxygen required is 168.0 dm3 at s.t.p.
(c) (i) Esterification is the reaction between an alkanol and an alkanoic acid, usually in the presence of concentrated sulfuric acid, to form an ester and water.
(c) (ii) Uses of alkanoates include:
(c) (iii) Alkaline hydrolysis of ethyl ethanoate produces ethanol and an ethanoate salt.
\[\mathrm{CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH}\]
The products are ethanol and sodium ethanoate.
(d) (i) The tin-coated plate corrodes faster.
(d) (ii) Zinc is more reactive than iron and acts as a sacrificial metal in a galvanized plate. Zinc corrodes preferentially and protects the iron plate. Tin is less reactive than iron; therefore, when the tin coating is damaged or exposed, the iron corrodes readily in contact with tin.
Bayanin Amsa
(a) (i) The collision theory states that reactant particles must collide before a reaction can occur. Only collisions with sufficient energy to overcome the activation energy, and with the correct orientation where necessary, result in reaction.
(a) (ii) When the temperature is increased, the reactant particles gain kinetic energy and move faster. Collisions occur more frequently, and a greater proportion of the collisions have energy equal to or greater than the activation energy. Therefore, the number of effective collisions per second increases and the reaction rate increases.
(b) (i) At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature in kelvin. Hence, a graph of volume against temperature in kelvin is a straight line passing through the origin.
(b) (ii)
The balanced equation for the complete combustion of ethanol is:
\[\mathrm{C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)}\]
From the equation, 1 mole of ethanol requires 3 moles of oxygen.
\[\text{Moles of }O_2 = 2.5 \times 3 = 7.5\text{ mol}\]
\[\text{Volume of }O_2 = 7.5 \times 22.4 = 168.0\text{ dm}^3\]
Therefore, the volume of oxygen required is 168.0 dm3 at s.t.p.
(c) (i) Esterification is the reaction between an alkanol and an alkanoic acid, usually in the presence of concentrated sulfuric acid, to form an ester and water.
(c) (ii) Uses of alkanoates include:
(c) (iii) Alkaline hydrolysis of ethyl ethanoate produces ethanol and an ethanoate salt.
\[\mathrm{CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH}\]
The products are ethanol and sodium ethanoate.
(d) (i) The tin-coated plate corrodes faster.
(d) (ii) Zinc is more reactive than iron and acts as a sacrificial metal in a galvanized plate. Zinc corrodes preferentially and protects the iron plate. Tin is less reactive than iron; therefore, when the tin coating is damaged or exposed, the iron corrodes readily in contact with tin.
Tambaya 63 Rahoto
(a)(i) State two industrial uses of hdrogen.
(ii) Consider the equation below. Mg(HCO\(_3\))\(_{2(aq)}\) \(\to\) MgCO\(_{3(g)}\) + H\(_2\)0\(_{(l)}\) + CO\(_{2(g)}\)
1. State the type of hardness of water being removed as shown by the above equation.
2. Give two disadvantages of hardness of water.
(b)(i) In the extraction of aluminium by electrolysis, graphite electrodes are used. State the disadvantages of using this type of electrode.
(ii) Calcuim oxide reacts with water to form slaked line: I. Write a balanced equation for this reaction; II. State one use of slaked line.
(c)(i) What is meant by saponification?
(ii) List the raw materials needed for the manufacture of soap.
(iii) Name the main by-product obtained from the manufacture of soap.
(d) With the aid of chemical equations explain briefly how iron is extracted in the blast furnace using iron ore, coke and limestone as raw materials at the:
(i) bottom of the furnace; (ii) middle of the furnace (iii) top of the furnace.
(a)(i) Manufacture of ammonia (Haber process); hydrogenation of vegetable oils to make margarine. (Also manufacture of methanol, and use as a fuel.)
(a)(ii) 1. The hardness removed is temporary hardness. 2. Disadvantages of hard water: it wastes soap (forms an insoluble scum) and it forms scale/fur in kettles, boilers and pipes, reducing their efficiency.
(b)(i) The graphite (carbon) anodes react with the oxygen liberated during electrolysis, burning away as CO2, so they are used up and must be replaced frequently (adding to cost).
(b)(ii) I. \[ \text{CaO} + \text{H}_2\text{O} \to \text{Ca(OH)}_2 \] II. Slaked lime is used to neutralize soil acidity (also for making mortar and softening water).
(c)(i) Saponification is the alkaline hydrolysis of fats or oils (esters) by a hot alkali to produce soap and glycerol.
(c)(ii) Raw materials: a fat or oil (animal fat/vegetable oil) and a concentrated alkali (sodium hydroxide).
(c)(iii) The main by-product is glycerol (propane-1,2,3-triol).
(d) Extraction of iron in the blast furnace
Bayanin Amsa
(a)(i) Manufacture of ammonia (Haber process); hydrogenation of vegetable oils to make margarine. (Also manufacture of methanol, and use as a fuel.)
(a)(ii) 1. The hardness removed is temporary hardness. 2. Disadvantages of hard water: it wastes soap (forms an insoluble scum) and it forms scale/fur in kettles, boilers and pipes, reducing their efficiency.
(b)(i) The graphite (carbon) anodes react with the oxygen liberated during electrolysis, burning away as CO2, so they are used up and must be replaced frequently (adding to cost).
(b)(ii) I. \[ \text{CaO} + \text{H}_2\text{O} \to \text{Ca(OH)}_2 \] II. Slaked lime is used to neutralize soil acidity (also for making mortar and softening water).
(c)(i) Saponification is the alkaline hydrolysis of fats or oils (esters) by a hot alkali to produce soap and glycerol.
(c)(ii) Raw materials: a fat or oil (animal fat/vegetable oil) and a concentrated alkali (sodium hydroxide).
(c)(iii) The main by-product is glycerol (propane-1,2,3-triol).
(d) Extraction of iron in the blast furnace
Tambaya 64 Rahoto
(a) Outline the procedures for the treatment of water for town supply.
(b) (i) State two main impurities present in bauxite.
(ii) Give one reason why bauxite is usually preferred as the ore for the extraction of aluminium.
(iii) Outline the manufacture of aluminium form purified bauxite.
(c) (i) Explain briefly the term fine chemical industry.
(ii) State two differences between a fine chemical and a heavy chemical.
(d) (i) Write a balanced chemical equation for the combustion of coal.
(ii) If 5.4 g of coal is burnt completely, calculate the amount of oxygen measured at s.t.p. that would be required for the combustion.
[C = 12.0, O = 16.0, Molar volume of a gas at s.t.p. = \(22.4\ \text{dm}^3\)]
(e) Name two substances which can be used as electrodes during the electrolysis of acidified water.
(a) Treatment of water for town supply
(b)(i) Iron(III) oxide (Fe2O3) and silica (SiO2).
(b)(ii) Bauxite is preferred because it is rich in aluminium oxide and is the most abundant and economical source of the metal.
(b)(iii) The purified alumina (Al2O3) is dissolved in molten cryolite (to lower the melting point) and the mixture is electrolysed using carbon (graphite) anodes and a carbon-lined cathode. Aluminium is deposited at the cathode and oxygen is liberated at the anode:
Cathode: \( \text{Al}^{3+} + 3e^- \to \text{Al} \); \quad Anode: \( 2\text{O}^{2-} \to \text{O}_2 + 4e^- \).
(c)(i) A fine chemical industry manufactures chemicals in small quantities but of high purity and high value (e.g. drugs, dyes, cosmetics).
(c)(ii) (1) Fine chemicals are made in small quantities, heavy chemicals in large/bulk quantities. (2) Fine chemicals are of high purity and high cost per unit, whereas heavy chemicals are of lower purity and cheaper per unit.
(d)(i) \[ \text{C} + \text{O}_2 \to \text{CO}_2 \]
(d)(ii) \( n(\text{C}) = \dfrac{5.4}{12} = 0.45\ \text{mol} \); \( n(\text{O}_2) = 0.45\ \text{mol} \).
\[ V = 0.45 \times 22.4 = 10.08\ \text{dm}^3 \]
(e) Platinum and carbon (graphite) (inert electrodes).
Bayanin Amsa
(a) Treatment of water for town supply
(b)(i) Iron(III) oxide (Fe2O3) and silica (SiO2).
(b)(ii) Bauxite is preferred because it is rich in aluminium oxide and is the most abundant and economical source of the metal.
(b)(iii) The purified alumina (Al2O3) is dissolved in molten cryolite (to lower the melting point) and the mixture is electrolysed using carbon (graphite) anodes and a carbon-lined cathode. Aluminium is deposited at the cathode and oxygen is liberated at the anode:
Cathode: \( \text{Al}^{3+} + 3e^- \to \text{Al} \); \quad Anode: \( 2\text{O}^{2-} \to \text{O}_2 + 4e^- \).
(c)(i) A fine chemical industry manufactures chemicals in small quantities but of high purity and high value (e.g. drugs, dyes, cosmetics).
(c)(ii) (1) Fine chemicals are made in small quantities, heavy chemicals in large/bulk quantities. (2) Fine chemicals are of high purity and high cost per unit, whereas heavy chemicals are of lower purity and cheaper per unit.
(d)(i) \[ \text{C} + \text{O}_2 \to \text{CO}_2 \]
(d)(ii) \( n(\text{C}) = \dfrac{5.4}{12} = 0.45\ \text{mol} \); \( n(\text{O}_2) = 0.45\ \text{mol} \).
\[ V = 0.45 \times 22.4 = 10.08\ \text{dm}^3 \]
(e) Platinum and carbon (graphite) (inert electrodes).
Tambaya 65 Rahoto
(a)
(i) List two gaseous pollutants that can be generated by burning coal.
(ii) Explain briefly why coal burns more easily when it is broken into pieces than when it is in lumps.
(iii) What gas is responsible for most of the explosions in coal mines?
(iv) Name the non-volatile residue left behind after the destructive distillation of coal.
(b) State one oxide in each case which:
(i) is used in bleaching;
(ii) oxidizes hot concentrated HCl to chlorine;
(iii) dissolves in water to give a solution with pH greater than 7;
(iv) reacts with NaOH and also with HCl;
(v) is a reddish-brown gas.
(c)
(i) Write a balanced chemical equation for the reaction between chlorine gas and iron(II) chloride solution.
(ii) State the type of reaction in (c)(i).
(iii) Give a reason for your answer in (c)(ii).
(d) Consider the following set-up:
(i) Identify A and B.
(ii) Write a balanced chemical equation for the reaction.
(iii) Name the gas produced.
(iv) Why was the flask tilted downwards?
(v) What is the:
(I) function of B in the experiment;
(II) method of collection of the gas?
(e) Give one product obtained from refining petroleum that is solid.
(a) Coal
(i) Two gaseous pollutants produced by burning coal are sulphur(IV) oxide, \(SO_2\) and carbon(II) oxide (carbon monoxide), \(CO\). (Oxides of nitrogen, \(NO_x\), are also acceptable.)
(ii) Breaking coal into pieces greatly increases the total surface area of solid exposed to the oxygen of the air. Combustion is a surface reaction, so a larger area of contact means more collisions per second between the coal and oxygen molecules. The rate of burning therefore rises, and the broken coal ignites and burns more easily than the same mass held in one large lump.
(iii) Methane, \(CH_4\) (called firedamp) is responsible for most explosions in coal mines.
(iv) The non-volatile residue left after the destructive distillation of coal is coke.
(b) Oxides
| Property | Oxide |
|---|---|
| (i) used in bleaching | Sulphur(IV) oxide, \(SO_2\) |
| (ii) oxidises hot concentrated HCl to chlorine | Manganese(IV) oxide, \(MnO_2\) |
| (iii) dissolves in water to give a solution of pH > 7 | Sodium oxide, \(Na_2O\) (basic oxide) |
| (iv) reacts with both NaOH and HCl | Aluminium oxide, \(Al_2O_3\) (amphoteric) |
| (v) is a reddish-brown gas | Nitrogen(IV) oxide, \(NO_2\) |
The reaction in (ii) is:
\[MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O\]
(c) Chlorine and iron(II) chloride
(i) \[2FeCl_2 + Cl_2 \rightarrow 2FeCl_3\]
(ii) It is a redox (oxidation-reduction) reaction.
(iii) The iron is oxidised while the chlorine is reduced in the same reaction. \(Fe^{2+}\) loses one electron to become \(Fe^{3+}\) (oxidation), and each atom of the added chlorine gains an electron, changing from oxidation state \(0\) in \(Cl_2\) to \(-1\) in the chloride (reduction). Since oxidation and reduction occur together, the reaction is redox.
(d) The set-up shown
The diagram shows a round-bottom flask containing a mixture of calcium hydroxide, \(Ca(OH)_2\), and ammonium chloride, \(NH_4Cl\), mounted on a stand and heated with a burner. The flask is clamped so that its mouth points slightly downwards. A delivery tube A carries the gas evolved into a vertical tower B which is packed with granular solid, and the gas is finally collected in an inverted gas jar at the top.
(i) A is the delivery tube (which conveys the gas from the flask). B is a drying tower (drying column) packed with lumps of calcium oxide (quicklime), \(CaO\).
(ii) \[2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2NH_3 + 2H_2O\]
(iii) The gas produced is ammonia, \(NH_3\).
(iv) The flask was tilted with its mouth downwards so that the water (steam) formed in the reaction condenses near the cooler mouth and does not run back onto the hot base of the flask. If cold water ran onto the strongly heated glass, the sudden contraction would crack the flask.
(v)(I) The function of B is to dry the ammonia gas. The calcium oxide absorbs the water vapour carried over from the flask; ordinary drying agents such as concentrated \(H_2SO_4\) or anhydrous \(CaCl_2\) cannot be used because they react with the alkaline ammonia, so the basic drying agent \(CaO\) is used.
(v)(II) The gas is collected by downward displacement of air (upward delivery into an inverted gas jar). Ammonia is used this way because it is less dense than air and is very soluble in water, so it cannot be collected over water.
(e) Solid product from refining petroleum
A solid product obtained from refining petroleum is bitumen (asphalt). (Paraffin wax is also acceptable.)
Bayanin Amsa
(a) Coal
(i) Two gaseous pollutants produced by burning coal are sulphur(IV) oxide, \(SO_2\) and carbon(II) oxide (carbon monoxide), \(CO\). (Oxides of nitrogen, \(NO_x\), are also acceptable.)
(ii) Breaking coal into pieces greatly increases the total surface area of solid exposed to the oxygen of the air. Combustion is a surface reaction, so a larger area of contact means more collisions per second between the coal and oxygen molecules. The rate of burning therefore rises, and the broken coal ignites and burns more easily than the same mass held in one large lump.
(iii) Methane, \(CH_4\) (called firedamp) is responsible for most explosions in coal mines.
(iv) The non-volatile residue left after the destructive distillation of coal is coke.
(b) Oxides
| Property | Oxide |
|---|---|
| (i) used in bleaching | Sulphur(IV) oxide, \(SO_2\) |
| (ii) oxidises hot concentrated HCl to chlorine | Manganese(IV) oxide, \(MnO_2\) |
| (iii) dissolves in water to give a solution of pH > 7 | Sodium oxide, \(Na_2O\) (basic oxide) |
| (iv) reacts with both NaOH and HCl | Aluminium oxide, \(Al_2O_3\) (amphoteric) |
| (v) is a reddish-brown gas | Nitrogen(IV) oxide, \(NO_2\) |
The reaction in (ii) is:
\[MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O\]
(c) Chlorine and iron(II) chloride
(i) \[2FeCl_2 + Cl_2 \rightarrow 2FeCl_3\]
(ii) It is a redox (oxidation-reduction) reaction.
(iii) The iron is oxidised while the chlorine is reduced in the same reaction. \(Fe^{2+}\) loses one electron to become \(Fe^{3+}\) (oxidation), and each atom of the added chlorine gains an electron, changing from oxidation state \(0\) in \(Cl_2\) to \(-1\) in the chloride (reduction). Since oxidation and reduction occur together, the reaction is redox.
(d) The set-up shown
The diagram shows a round-bottom flask containing a mixture of calcium hydroxide, \(Ca(OH)_2\), and ammonium chloride, \(NH_4Cl\), mounted on a stand and heated with a burner. The flask is clamped so that its mouth points slightly downwards. A delivery tube A carries the gas evolved into a vertical tower B which is packed with granular solid, and the gas is finally collected in an inverted gas jar at the top.
(i) A is the delivery tube (which conveys the gas from the flask). B is a drying tower (drying column) packed with lumps of calcium oxide (quicklime), \(CaO\).
(ii) \[2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2NH_3 + 2H_2O\]
(iii) The gas produced is ammonia, \(NH_3\).
(iv) The flask was tilted with its mouth downwards so that the water (steam) formed in the reaction condenses near the cooler mouth and does not run back onto the hot base of the flask. If cold water ran onto the strongly heated glass, the sudden contraction would crack the flask.
(v)(I) The function of B is to dry the ammonia gas. The calcium oxide absorbs the water vapour carried over from the flask; ordinary drying agents such as concentrated \(H_2SO_4\) or anhydrous \(CaCl_2\) cannot be used because they react with the alkaline ammonia, so the basic drying agent \(CaO\) is used.
(v)(II) The gas is collected by downward displacement of air (upward delivery into an inverted gas jar). Ammonia is used this way because it is less dense than air and is very soluble in water, so it cannot be collected over water.
(e) Solid product from refining petroleum
A solid product obtained from refining petroleum is bitumen (asphalt). (Paraffin wax is also acceptable.)
Tambaya 66 Rahoto
(a) (i) Name the ore mostly used in the extraction of aluminium.
(ii) Name two major impurities in the ore named in (a)(i).
(iii) Name the material used in making the electrodes in the extraction of aluminium.
(iv) Give two reasons why aluminium is commonly recycled.
(v) Explain briefly why the anode has to be replaced at regular intervals during the extraction of aluminium.
(b) A current of 0.75 amperes was passed through an electrolysis containing chromium ions for one hour and four minutes. If the mass of chromium deposited was 0.52 g, calculate the:
(i) quantity of electricity passed;
(ii) moles of chromium deposited;
(iii) quantity of electricity required to deposit one mole of chromium;
(iv) charge on the chromium ion.
[Cr = 52.0, 1 F = 96500 C]
(c) In the contact process for the manufacture of tetraoxosulphate(VI) acid, the following reaction occurs:
\[2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \qquad 2\mathrm{SO}_{3(g)} \qquad H = -197\ \mathrm{kJ\ mol}^{-1}\]
(i) Name the catalyst used in the reaction;
(ii) State the optimum temperature for this reaction;
(iii) What would be the effect on the yield of \(\mathrm{SO}_3\) if a temperature higher than the optimum is used?
(d)(i) State two chemical methods by which temporary hardness of water can be removed.
(ii) Write a balanced chemical equation for each of the methods stated in (d)(i).
(a)(i) Bauxite (Al2O3·2H2O).
(a)(ii) Iron(III) oxide (Fe2O3) and silica (SiO2).
(a)(iii) Graphite (carbon).
(a)(iv) Recycling saves the large amount of electrical energy needed for extraction and conserves the limited bauxite reserves (also reduces waste/pollution).
(a)(v) The carbon anode reacts with the oxygen liberated at it, burning away as CO2, so it is gradually consumed and must be replaced.
(b) \(t = 1\,\text{h}\,4\,\text{min} = 3840\ \text{s}\).
(i) \( Q = It = 0.75 \times 3840 = 2880\ \text{C} \)
(ii) \( n(\text{Cr}) = \dfrac{0.52}{52} = 0.01\ \text{mol} \)
(iii) Quantity to deposit 1 mol = \( \dfrac{2880}{0.01} = 288000\ \text{C} \)
(iv) Number of Faradays per mole = \( \dfrac{288000}{96500} \approx 3 \), so the charge on the chromium ion is +3 (Cr3+).
(c)(i) Vanadium(V) oxide, V2O5.
(c)(ii) About 450 °C (400-450 °C).
(c)(iii) Since the forward reaction is exothermic, a temperature higher than the optimum decreases the yield of SO3 (equilibrium shifts backward).
(d)(i) & (ii) Removal of temporary hardness
Bayanin Amsa
(a)(i) Bauxite (Al2O3·2H2O).
(a)(ii) Iron(III) oxide (Fe2O3) and silica (SiO2).
(a)(iii) Graphite (carbon).
(a)(iv) Recycling saves the large amount of electrical energy needed for extraction and conserves the limited bauxite reserves (also reduces waste/pollution).
(a)(v) The carbon anode reacts with the oxygen liberated at it, burning away as CO2, so it is gradually consumed and must be replaced.
(b) \(t = 1\,\text{h}\,4\,\text{min} = 3840\ \text{s}\).
(i) \( Q = It = 0.75 \times 3840 = 2880\ \text{C} \)
(ii) \( n(\text{Cr}) = \dfrac{0.52}{52} = 0.01\ \text{mol} \)
(iii) Quantity to deposit 1 mol = \( \dfrac{2880}{0.01} = 288000\ \text{C} \)
(iv) Number of Faradays per mole = \( \dfrac{288000}{96500} \approx 3 \), so the charge on the chromium ion is +3 (Cr3+).
(c)(i) Vanadium(V) oxide, V2O5.
(c)(ii) About 450 °C (400-450 °C).
(c)(iii) Since the forward reaction is exothermic, a temperature higher than the optimum decreases the yield of SO3 (equilibrium shifts backward).
(d)(i) & (ii) Removal of temporary hardness
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