WAEC SSCE - Chemistry - 2009
Question 1 Report
Which of the following alloys does not contain copper? l. Brass II. Bronze III. Steel
Answer Details
Steel does not contain copper, so the correct answer is III only. Brass is an alloy of copper and zinc, and bronze is an alloy of copper and tin. Steel is an alloy of iron and carbon, and may also contain other elements such as manganese, nickel, and chromium.
Question 2 Report
Which of the following scientists discovered the neutron?
Answer Details
James Chadwick discovered the neutron. In the early 1930s, he conducted a series of experiments that led to the discovery of an uncharged subatomic particle within the nucleus of the atom. This particle was later identified as the neutron, which plays a critical role in nuclear reactions and the stability of the nucleus. Chadwick's discovery of the neutron earned him the Nobel Prize in Physics in 1935.
Question 3 Report
Consider the reaction represented by the equation: 2NaHCO3 → Na2CO 3 + H 2O + CO 2. What is the mass of sodium trioxocarbonate (lV) produced by the complete decomposition of 33.6 g of sodium hydrogen trioxocarbonate (IV)? [NaHCO3 = 84, Na2CO3 = 106]
Answer Details
The balanced equation for the reaction shows that 2 moles of NaHCO3 produce 1 mole of Na2CO3, and the molar mass of NaHCO3 is 84 g/mol. Therefore, 84 g of NaHCO3 will produce 106 g of Na2CO3. To determine the mass of Na2CO3 produced from 33.6 g of NaHCO3, we need to use stoichiometry. 33.6 g NaHCO3 x (1 mol NaHCO3 / 84 g NaHCO3) x (1 mol Na2CO3 / 2 mol NaHCO3) x (106 g Na2CO3 / 1 mol Na2CO3) = 21.2 g Na2CO3 Therefore, the mass of sodium trioxocarbonate (lV) produced by the complete decomposition of 33.6 g of sodium hydrogen trioxocarbonate (IV) is 21.2 g. The correct answer is 21.2 g.
Question 4 Report
The electron configuration of carbon atom in its excited state is [ 6C]
Answer Details
The electron configuration of an atom describes the arrangement of electrons in its energy levels or orbitals. The notation [ 6C ] means that the excited carbon atom has six electrons in its outermost energy level, which is the second energy level (n=2). The correct electron configuration for a carbon atom in its excited state is 1s22s12p2. The notation [ 6C ] can be understood as follows: the normal electron configuration for carbon is 1s22s22p2, meaning that it has four electrons in its outermost energy level. By exciting one of the 2s electrons to the empty 2p orbital, the atom now has a total of six electrons in the n=2 energy level, hence the notation [ 6C ]. Option D, 1s22s12px12py12pz1, represents the electron configuration of a carbon atom in its ground state.
Question 5 Report
Which of the following hydroxides will readily dissolved in water?
Answer Details
NaOH will readily dissolve in water. This is because it is a strong base and its ionic bonds with Na+ and OH- are easily broken in the presence of water, allowing the ions to separate and become hydrated. In contrast, Cu(OH)2, Pb(OH)2, and Zn(OH)2 are relatively insoluble in water, meaning that their ionic bonds are not easily broken and they do not readily dissociate into their constituent ions in water.
Question 6 Report
How many molecules are there in 14 g of nitrogen at s.t.p.? [N = 14, Avogadro's constant = 6.02 x 1023 mol-1]
Answer Details
Question 7 Report
Which of the following properties of covalent compounds is not correct? They
Answer Details
The correct option is "have high boiling points." Covalent compounds are formed by sharing of electrons between two non-metal atoms. These compounds generally have low melting and boiling points due to their weak intermolecular forces. They are mostly gaseous or volatile liquids. Covalent compounds also do not dissociate into ions in solution, and therefore they are non-electrolytes. However, some covalent compounds with strong intermolecular forces may have high boiling points.
Question 8 Report
Which of the following arrangements is in order of increasing ionization energy?
Answer Details
Question 9 Report
Consider the reaction represented by the equation: 2H2SO 4(aq0 + C(s) → 2H2O(l) + 2SO2(g) + CO 2(g) H2SO 4 is acting as
Question 10 Report
Which of the following noble gases has electronic structure similar to that of N in NH 3? [\(\frac{14}{7}\)N]
Answer Details
Question 11 Report
Which of the following oxides of nitrogen has oxidation number of + 1?
Answer Details
Question 12 Report
How many protons does 4020Ca contain?
Answer Details
The atomic number of an element is defined as the number of protons in the nucleus of its atom. In the symbol 4020Ca, 20 is written as a subscript, which represents the atomic number of calcium. Therefore, the number of protons in the nucleus of 4020Ca is 20. Answer: 20.
Question 13 Report
The separation of petroleum fractions depends on the differences in their
Answer Details
The separation of petroleum fractions is based on their boiling points. The different fractions of petroleum are separated by a process called fractional distillation, where crude oil is heated and vaporized. As the vapor rises through a column, it gradually cools down. As it cools, the vapor condenses into a liquid, and the different fractions with different boiling points condense at different levels in the column. The lighter fractions with lower boiling points, such as gases and liquids like gasoline, condense at the top of the column, while the heavier fractions with higher boiling points, such as liquids like diesel and lubricating oil, condense at lower levels. Thus, the separation of petroleum fractions is dependent on the differences in their boiling points.
Question 14 Report
Which of the following factors will affect the rate of the reaction represented by the equation? 2HCI(aq) + CaCO 3(s) → CaCI2(aq) + H2O (I) + CO2(g) l. Pressure ll. Concentration III. Nature of reactants IV. Temperature
Answer Details
The rate of the reaction represented by the equation 2HCI(aq) + CaCO3(s) → CaCI2(aq) + H2O(I) + CO2(g) can be affected by various factors: - Pressure: Pressure affects the rate of a reaction only if there are gases involved in the reaction. In this reaction, CO2 is produced as a gas, so increasing the pressure will increase the rate of the reaction. - Concentration: Increasing the concentration of the reactants will increase the number of collisions between the reactant particles and thus increase the rate of the reaction. - Nature of reactants: The nature of the reactants can affect the rate of the reaction. For example, a reaction involving reactive metals like magnesium will proceed faster than a reaction involving less reactive metals like copper. - Temperature: Increasing the temperature will increase the kinetic energy of the particles, resulting in more collisions and thus increasing the rate of the reaction. Therefore, the factors that will affect the rate of this particular reaction are ll, lll, and IV.
Question 15 Report
Which of the following molecules is not linear in shape?
Answer Details
NH3 is not linear in shape. It has a trigonal pyramidal shape because of the presence of one lone pair of electrons on the nitrogen atom. The other three bonded hydrogen atoms are at an angle of approximately 107 degrees to each other. CO2, O2, and HCl are all linear in shape.
Question 16 Report
What is the value of -14oC on the Kelvin temperature scale?
Answer Details
The Kelvin temperature scale is an absolute temperature scale where 0 Kelvin represents the lowest possible temperature (absolute zero). The Kelvin temperature scale is related to the Celsius scale by adding 273.15 to the Celsius value. To convert -14°C to Kelvin, we need to add 273.15 to the value. Thus, -14°C + 273.15 = 259.15 K. Therefore, the value of -14°C on the Kelvin temperature scale is 259 K (option a).
Question 18 Report
The number of particles in one mole of a chemical compound is the
Answer Details
The number of particles in one mole of a chemical compound is called Avogadro's number. Avogadro's number is a constant that represents the number of particles (such as atoms, molecules, or ions) in one mole of a substance, which is approximately 6.02 x 10^23 particles. This number is used to convert between the number of moles of a substance and the number of particles it contains. For example, one mole of water contains 6.02 x 10^23 water molecules. Therefore, Avogadro's number is the correct option.
Question 19 Report
Which of the following elements has the electronic configuration:1s22s22p63s23p1? [11Na, 13A1,
14N,16S]
Answer Details
Question 20 Report
Which of the following statement is correct? The solubility of
Question 21 Report
Which of the following elements exhibits the same chemical properties as the atom \(\frac{35}{17}\)X? An element with
Answer Details
The element with the symbol \(\frac{35}{17}\)X represents an atom of chlorine with 17 protons and 18 neutrons. Elements with the same number of protons have the same chemical properties. Therefore, the answer is (a) an element with atomic number 17, which is chlorine.
Question 22 Report
What is the molecular formula of a compound with empirical CH2O and vapour density of 90? [H = 1, C = 12, O = 16]
Answer Details
Question 23 Report
Which of the following electronic configurations correctly represents an element in period 3 of the periodic table?
Answer Details
Period 3 of the periodic table includes the elements sodium (Na), magnesium (Mg), aluminum (Al), silicon (Si), phosphorus (P), sulfur (S), and chlorine (Cl). The correct electronic configuration for an element in period 3 should have 3 energy levels and electrons distributed among those energy levels according to the Aufbau principle. Looking at the options given: - represents an element in period 2 as it has only 2 energy levels. - represents the noble gas configuration of the element neon (Ne) in period 2. - represents an element in period 3 with 3 energy levels, where the outermost energy level has 6 electrons (3s2 3p4). This configuration is consistent with the element sulfur (S). - represents an element in period 4 as it has 4 energy levels. Therefore, the correct option representing an element in period 3 is with the electronic configuration 1s2 2s2 2p6 3s2 3p4.
Question 24 Report
How many orbitals are associated with the p-sub energy level?
Answer Details
The p-sub energy level has a total of three orbitals. These orbitals are represented as Px, Py, and Pz. Each orbital can hold a maximum of two electrons, and together they can hold a total of six electrons. Therefore, the correct answer is 3.
Question 25 Report
A solution that contains as much solute as it can dissolve at a given temperature is said to be
Answer Details
A solution that contains as much solute as it can dissolve at a given temperature is said to be saturated. This means that no more solute can be dissolved in the solution without changing the temperature or pressure. Any additional solute added to a saturated solution will remain undissolved and settle at the bottom of the container. The saturation point is determined by the solubility of the solute in the solvent, which is the maximum amount of solute that can dissolve in a given amount of solvent at a given temperature and pressure.
Question 26 Report
Which of the following raw materials is used in the plastic industry ?
Answer Details
Ethene is the raw material used in the plastic industry. Ethene, also known as ethylene, is a hydrocarbon with the formula C2H4. It is produced from the cracking of petroleum fractions or natural gas. Ethene is a monomer that can be polymerized to form polyethylene, which is a versatile plastic used in a wide range of applications, from packaging materials to toys and medical devices. Calcium, hydrogen, and methane are not typically used as raw materials in the plastic industry.
Question 27 Report
Which of the following expressions correctly represents Charles' law?
Question 28 Report
Which of the following salts is insoluble in water?
Answer Details
The correct answer is AgCI. AgCI is insoluble in water because it is a sparingly soluble salt. When AgCI is added to water, it does not dissolve readily and forms a cloudy mixture. The cloudiness is due to the formation of small insoluble particles of AgCI in water. This is because the attractive forces between Ag+ ions and CI- ions in AgCI are stronger than the attractive forces between these ions and water molecules. In contrast, Pb(NO3)2, Na2CO3, and AgNO3 are soluble salts that dissolve readily in water to form clear solutions.
Question 29 Report
Consider the reaction represented by the following equation: H 2(g) + \(\frac{1}{2}\) O(g) → H 2O(g) ∆H =-285.8 KJmo -1 ∆H represents the enthalpy of
Answer Details
∆H represents the enthalpy change of a chemical reaction, which is defined as the difference between the enthalpy of the products and the enthalpy of the reactants. In this case, the equation given is the reaction of hydrogen gas and oxygen gas to form water, so ∆H represents the enthalpy change of the formation of water. Therefore, the correct answer is "formation".
Question 30 Report
Which of the following dilute acids does not react with metals to liberate hydrogen?
Answer Details
Question 31 Report
The major product in the solvay process is
Answer Details
The Solvay process is a method of producing sodium carbonate (Na₂CO₃) from brine and ammonia (NH₃). The process involves reacting brine with ammonia to form ammonium chloride (NH₄Cl), which is then reacted with limestone (CaCO₃) to produce carbon dioxide (CO₂), water (H₂O), calcium chloride (CaCl₂), and ammonia. The carbon dioxide is then reacted with the remaining ammonia to produce ammonium bicarbonate (NH₄HCO₃), which is then heated to produce sodium carbonate (Na₂CO₃), water, and carbon dioxide. Thus, the major product in the Solvay process is Na₂CO₃.
Question 32 Report
An important medical use of nuclear radiations is
Answer Details
One important medical use of nuclear radiations is radiotherapy. Radiotherapy uses high-energy radiation to destroy cancer cells and shrink tumors. This type of treatment is usually delivered by a machine called a linear accelerator, which produces high-energy X-rays or gamma rays that are directed towards the tumor. The radiation damages the DNA of the cancer cells, causing them to die. Radiotherapy is often used in combination with other treatments such as chemotherapy and surgery, and can be effective in treating a range of cancers including lung, breast, prostate, and brain cancer.
Question 33 Report
Which of the following substances is the anode in the dry Leclanche cell ?
Answer Details
In a dry Leclanche cell, the anode is the negative electrode that releases electrons into the external circuit during the cell's discharge. The anode of a dry Leclanche cell is the zinc container. It acts as the negative terminal of the cell and is where the oxidation of zinc metal to zinc ions takes place, producing electrons and zinc ions which move towards the cathode. Therefore, the correct option is zinc container.
Question 34 Report
In a mixture of gases which do not react chemically together, the pressure exerted by the individual gases is called
Answer Details
Question 35 Report
What is the product of the reaction between propene and 1 mole of hydrogen iodide?
Answer Details
The reaction between propene and 1 mole of hydrogen iodide results in the addition of the hydrogen iodide molecule to the carbon-carbon double bond in propene. The hydrogen atom adds to one carbon atom of the double bond, and the iodine atom adds to the other carbon atom. As a result, the product formed is CH 3CHICH3, which is 2-iodopropane. Therefore, the correct answer is: the product of the reaction between propene and 1 mole of hydrogen iodide is CH3CHICH3.
Question 36 Report
The electron configuration of carbon atom in its excited state is [ 6C]
Answer Details
The electron configuration of carbon in its ground state is 1s22s22p2. This means that there are two electrons in the 1s orbital, two in the 2s orbital, and two in the 2p orbital. However, the question is asking about the excited state of carbon, which means that at least one electron has been promoted from a lower energy level to a higher energy level. In the case of carbon, one of the electrons in the 2s orbital is promoted to the 2p orbital, resulting in the electron configuration [ 6C] 1s22s12p3. Therefore, the correct answer is 1s22s12px12py12pz1.
Question 37 Report
A catalyst increases the rate of chemical reaction by
Answer Details
A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with lower activation energy. Activation energy is the energy required to start a chemical reaction. By lowering the activation energy, a catalyst makes it easier for reactant molecules to react and form products. As a result, the reaction can occur more quickly, which increases the rate of the chemical reaction. Therefore, the correct option is "decreasing the activation energy of the reaction".
Question 39 Report
If 100 atoms of element X contains 70 atoms of 9X and 30 atoms of 11X, calculate the relative atomic mass of X
Answer Details
To calculate the relative atomic mass of X, we need to take into account the mass and abundance of each isotope of X. We know that element X has two isotopes: 9X and 11X. To calculate the relative atomic mass of X, we can use the formula: Relative atomic mass of X = (mass of 9X x % abundance of 9X) + (mass of 11X x % abundance of 11X) / 100 Substituting the values we have, we get: Relative atomic mass of X = (9 x 70) + (11 x 30) / 100 Relative atomic mass of X = 630 + 330 / 100 Relative atomic mass of X = 9.6 Therefore, the relative atomic mass of X is 9.6.
Question 40 Report
An element Q forms a compound QCI5. In which group of the periodic table is Q?
Answer Details
The Roman numeral in the compound QCI5 indicates the oxidation state of element Q in the compound, which is 5. From this information, we can determine the group number of element Q in the periodic table. Group number of an element is equal to its highest oxidation state. Since the oxidation state of element Q in the compound is 5, it belongs to group V of the periodic table. Therefore, the correct answer is V.
Question 41 Report
When protein is heated to a high temperature it undergoes
Answer Details
When proteins are heated to high temperatures, they undergo denaturation. Denaturation is the process by which a protein loses its three-dimensional structure, which is critical to its function. This change in structure is due to the breaking of the weak bonds holding the protein together, such as hydrogen bonds, ionic bonds, and hydrophobic interactions. The loss of these bonds causes the protein to unfold and lose its specific shape, which can lead to a loss of function. Denaturation can also be caused by other factors such as changes in pH, exposure to detergents or solvents, or high pressure.
Question 42 Report
A consequence of global warming is
Answer Details
A consequence of global warming is flooding. Global warming causes the temperature of the earth to rise, leading to the melting of glaciers and ice caps, which increases the volume of water in the oceans. This increase in water level leads to flooding in low-lying areas and coastal regions. Additionally, global warming can cause more intense and frequent weather events, such as hurricanes and heavy rainfall, which can also lead to flooding.
Question 43 Report
Which of the statements about the behaviour of the atom is correct?
Answer Details
The correct statement about the behavior of the atom is that cations are smaller than the parent atom. When an atom loses one or more electrons to form a cation, the number of electrons is less than the number of protons, resulting in a net positive charge. This positive charge attracts the remaining electrons more strongly towards the nucleus, causing the atomic radius to decrease. Therefore, cations are smaller than their parent atoms. Conversely, when an atom gains one or more electrons to form an anion, the negative charge causes the electron cloud to expand, resulting in a larger atomic radius than the parent atom. Hence, the other options are incorrect.
Question 44 Report
The following properties are characteristics of transition elements except
Answer Details
The question is asking which of the given properties is NOT a characteristic of transition elements. Transition elements are a group of elements in the periodic table that have partially filled d orbitals. They have several unique properties, including the ability to form complex ions, form coloured compounds, and exhibit catalytic abilities. However, one property that is not typically associated with transition elements is fixed oxidation states. Unlike many other elements, transition elements can have multiple oxidation states and are not limited to one fixed state. Therefore, the correct answer to the question is "fixed oxidation states."
Question 45 Report
Consider the reaction represented by the equation: 23892U + 10 → 139 56Ba + 94 36Kr + 31 0n. The reaction represents a
Answer Details
The reaction shown in the equation 23892U + 10 → 139 56Ba + 94 36Kr + 31 0n represents nuclear fission. Nuclear fission is a process in which a heavy nucleus is split into two or more smaller nuclei and typically releases a large amount of energy in the form of heat and radiation. In this particular reaction, uranium-238 is bombarded with a neutron, which causes it to split into two smaller nuclei (barium-139 and krypton-94) and three neutrons. This process also releases a significant amount of energy in the form of radiation.
Question 46 Report
The product of the reaction between C 2H5OH and concentrated H2SO 4 at 170oC is
Answer Details
Question 47 Report
Which of the following ions is isoelectronic with neon? [3Li, 3F, 10Ne, 17CI,19K]
Answer Details
The question is asking which of the given ions has the same electronic configuration as neon. Neon is a noble gas and has an electronic configuration of 1s22s22p6. Isoelectronic refers to ions or atoms that have the same number of electrons, thus the same electronic configuration. Among the given options, the ion that has the same number of electrons as neon is the one that has 10 electrons in total. Looking at the options, the only ion with 10 electrons is F-. Therefore, F- is isoelectronic with neon. Therefore, the correct option is F-.
Question 48 Report
Ethanol reacts with excess acidified K2Cr 2O7 to produce
Answer Details
Ethanol reacts with excess acidified K2Cr 2O7 to produce ethanoic acid. This reaction is an oxidation reaction, where ethanol is oxidized to ethanoic acid. The K2Cr 2O7 acts as an oxidizing agent and converts the ethanol to ethanoic acid by removing hydrogen atoms from ethanol and adding oxygen atoms to form a carboxylic acid. The other options listed in the question are not produced in this reaction.
Question 49 Report
Salts that contain water of crystallization are said to be
Answer Details
Salts that contain water of crystallization are said to be hydrated. Water of crystallization is the water molecules that are tightly bound to the salt molecules in a crystal structure. These water molecules can be removed by heating or by exposing the salt to dry air. When the water is removed, the salt becomes anhydrous. The number of water molecules present in a hydrated salt is indicated by a dot followed by a number in the chemical formula. For example, CuSO4.5H2O is copper sulfate pentahydrate, which means it contains five molecules of water of crystallization.
Question 50 Report
In bonded atoms, increase in electronegativity difference,
Answer Details
In bonded atoms, the electronegativity difference determines the distribution of electrons between the atoms. When there is an increase in the electronegativity difference between two atoms in a bond, one atom attracts electrons more strongly than the other, resulting in an uneven distribution of electrons. This uneven distribution of electrons leads to an increase in the polarity of the bond, as the atom that attracts electrons more strongly gains a partial negative charge, while the other atom gains a partial positive charge. Therefore, an increase in electronegativity difference increases the polarity of the bond.
Question 51 Report
What is the percentage by weight of calcium in Ca(OCI)2? [Ca = 40, Cl = 35.5, O = 16]
Answer Details
Question 52 Report
Which of the following compounds is not a double salt?
Answer Details
A double salt is a combination of two different salts which crystallize together to form a single substance with a unique formula. The formula of a double salt is a combination of the formulas of the two salts that make it up. From the given options, the compound that is not a double salt is Zn(OH)NO3. This is because it does not contain two different salts that have crystallized together to form a new substance. Rather, it is a single compound containing the elements zinc, oxygen, nitrogen, and hydrogen. Therefore, the answer is: Zn(OH)NO3
Question 53 Report
Dry hydrogen chloride when dissolved in methylbenzene
Answer Details
Dry hydrogen chloride gas (HCl) when dissolved in methylbenzene does not affect litmus paper. This is because methylbenzene is a non-polar solvent and does not contain water, which is required for litmus paper to change color. Therefore, litmus paper cannot be used to test the acidity or basicity of the solution.
Question 54 Report
(a)(1) Define each of the following terms: I. deliquescence; II. efflorescence.
(ii) Give one example I substance that undergoes each of the processes in (a)(i).
(b) Outline how three colourless gases suspected to be ethane, ethene and ethyne could be distinguished in the laboratory.
(c)(i) What is water crystallization?
(ii) Hydrated calcium chloride (CaCl\(_2\) = 111, xH\(_2\)O) contains 49.32% water of cystallization. Calm the value of x. [ (CaCl\(_2\) = 111, H\(_2\)O = 18 ]
(d) Name one calcium compound used (i) in the manufactur cement; (ii) as a dessicant; (iii) in the production of plaster of Paris (POP);
(iv) to neutralise acidic soils
(e) A concentrated solution of sodium chloride was electrolysed using graphite electrodes.
(i) State the ions present in the solution
(ii) Name the products at the I. anode, II. cathode.
(iii) Give the by-product of the electrolysis.
(a)(i) Definitions
- I. Deliquescence: the process by which a substance absorbs moisture (water vapour) from the atmosphere and dissolves in it to form a solution.
- II. Efflorescence: the process by which a hydrated crystalline substance loses part or all of its water of crystallization to the atmosphere and becomes powdery.
(a)(ii) Examples
- Deliquescence: sodium hydroxide (\(NaOH\)) or calcium chloride (\(CaCl_2\)).
- Efflorescence: washing soda, \(Na_2CO_3\cdot 10H_2O\) (or \(Na_2SO_4\cdot 10H_2O\)).
(b) Distinguishing ethane, ethene and ethyne
Pass each gas separately through bromine water: ethane does not decolourize it (no reaction), while ethene and ethyne both decolourize the red-brown bromine water. To separate ethene from ethyne, pass each of the two remaining gases through ammoniacal silver trioxonitrate(V) solution (Tollens' reagent): ethyne gives a white precipitate (silver dicarbide/acetylide), while ethene gives no precipitate. Thus ethane gives no change with bromine water, ethene decolourizes bromine water but gives no precipitate, and ethyne decolourizes bromine water and gives a white precipitate.
(c)(i) Water of crystallization
Water of crystallization is the definite/fixed number of water molecules chemically combined with each formula unit of a substance in its crystalline (hydrated) form.
(c)(ii) Value of x in \(CaCl_2\cdot xH_2O\)
Water is 49.32% of the hydrate, so anhydrous \(CaCl_2\) is \(100 - 49.32 = 50.68\%\).
\[x = \frac{49.32/18}{50.68/111} = \frac{2.740}{0.4566} = 6\]Therefore \(x = 6\); the hydrate is \(CaCl_2\cdot 6H_2O\).
(d) Calcium compounds
- (i) Manufacture of cement: calcium trioxocarbonate(IV), \(CaCO_3\) (limestone).
- (ii) As a desiccant: anhydrous calcium chloride, \(CaCl_2\).
- (iii) Production of plaster of Paris: calcium tetraoxosulphate(VI), \(CaSO_4\cdot\tfrac{1}{2}H_2O\) (from gypsum).
- (iv) To neutralise acidic soils: calcium oxide (quicklime, \(CaO\)) or calcium hydroxide (slaked lime).
(e) Electrolysis of concentrated sodium chloride solution (graphite electrodes)
- (i) Ions present: \(Na^+\), \(H^+\), \(Cl^-\) and \(OH^-\) (from the salt and from water).
- (ii) Products: I. Anode: chlorine gas (\(Cl_2\)); II. Cathode: hydrogen gas (\(H_2\)).
- (iii) By-product: sodium hydroxide (\(NaOH\)) solution.
Answer Details
(a)(i) Definitions
- I. Deliquescence: the process by which a substance absorbs moisture (water vapour) from the atmosphere and dissolves in it to form a solution.
- II. Efflorescence: the process by which a hydrated crystalline substance loses part or all of its water of crystallization to the atmosphere and becomes powdery.
(a)(ii) Examples
- Deliquescence: sodium hydroxide (\(NaOH\)) or calcium chloride (\(CaCl_2\)).
- Efflorescence: washing soda, \(Na_2CO_3\cdot 10H_2O\) (or \(Na_2SO_4\cdot 10H_2O\)).
(b) Distinguishing ethane, ethene and ethyne
Pass each gas separately through bromine water: ethane does not decolourize it (no reaction), while ethene and ethyne both decolourize the red-brown bromine water. To separate ethene from ethyne, pass each of the two remaining gases through ammoniacal silver trioxonitrate(V) solution (Tollens' reagent): ethyne gives a white precipitate (silver dicarbide/acetylide), while ethene gives no precipitate. Thus ethane gives no change with bromine water, ethene decolourizes bromine water but gives no precipitate, and ethyne decolourizes bromine water and gives a white precipitate.
(c)(i) Water of crystallization
Water of crystallization is the definite/fixed number of water molecules chemically combined with each formula unit of a substance in its crystalline (hydrated) form.
(c)(ii) Value of x in \(CaCl_2\cdot xH_2O\)
Water is 49.32% of the hydrate, so anhydrous \(CaCl_2\) is \(100 - 49.32 = 50.68\%\).
\[x = \frac{49.32/18}{50.68/111} = \frac{2.740}{0.4566} = 6\]Therefore \(x = 6\); the hydrate is \(CaCl_2\cdot 6H_2O\).
(d) Calcium compounds
- (i) Manufacture of cement: calcium trioxocarbonate(IV), \(CaCO_3\) (limestone).
- (ii) As a desiccant: anhydrous calcium chloride, \(CaCl_2\).
- (iii) Production of plaster of Paris: calcium tetraoxosulphate(VI), \(CaSO_4\cdot\tfrac{1}{2}H_2O\) (from gypsum).
- (iv) To neutralise acidic soils: calcium oxide (quicklime, \(CaO\)) or calcium hydroxide (slaked lime).
(e) Electrolysis of concentrated sodium chloride solution (graphite electrodes)
- (i) Ions present: \(Na^+\), \(H^+\), \(Cl^-\) and \(OH^-\) (from the salt and from water).
- (ii) Products: I. Anode: chlorine gas (\(Cl_2\)); II. Cathode: hydrogen gas (\(H_2\)).
- (iii) By-product: sodium hydroxide (\(NaOH\)) solution.
Question 55 Report
(a) State Gay Lussac's Law.
(b) Carbon (II) oxide reacted with oxygen to form carbon (IV) oxide in a see tube.
(i) Write a balanced equation for the reaction.
(ii) If 40 cm\(^3\) of the carbon (II) oxide were mixed with cm\(^3\) of oxygen,
I. calculate the volume of carbon (IV) oxide produced
II. which reactant is in excess and how much?
III. what was the total volume of the gaseous mixture at the end of the reaction?
(c) Consider the following oxides: CaO, SiO\(_2\), CO, NO\(_2\) and ZnO. Which of the oxide(s)
(i) is an acidic oxide that is insoluble in water?
(ii) reacts with water to give alkaline solution?
(iii) is amphoteric?
(iv) is neutral?
(v) is/are gaseous at room temperature?
(d) Explain why
(i) colourless concentrated trioxonitrate (V) acid turns yellow,
(ii) dilute trioxonitrate (V) acid does not liberate hydrogen when it reacts with magnesium.
(e) Write a chemical equation for the thermal decomposition of (i) Cu(NO\(_3\))\(_{2(g)}\)
(ii) NH\(_4\)NO\(_{3(g)}\)
Answer Details
None
Question 56 Report
(a)(i) Outline a suitable procedure for the preparation of ZnCl\(_2\) crystals stating from zinc granules.
(ii) Write a balanced equation for the reaction in (a)(i).
(b) Considei' the following solutions: Na\(_2\)SO\(_{4(aq)}\), CH3COOK\(_{(aq)}\), Pb(NO\(_3\))\(_{2(aq)}\) and MgCl\(_2\). Which of them has/have a pH
(i) greater than 7;
(ii) equal to 7;
(iii) less than 7?
(c)(i) Determine the oxidation number of Mn in I. MnO\(_2\); II. KMnO\(_4\).
(ii) State one laboratory use of each of the .compounds in (c)(i).
(d) Explain why oxidation and reduction processes are complementary.
(e) Consider the reaction represented by the equation: Fe\(_{(s)}\) + 2Ag\(^+_{(aq)}\) \(\to\) Fe\(^{2+}_{(aq)}\) + 2Ag\(_{(s)}\)
(i) Write a balanced ionic half equations for the reaction.
(ii) Which of the species is I. oxidized; II. reduced?
(iii) State the change in the oxidation number of silver during the reaction.
(a)(i) Preparation of ZnCl2 crystals from zinc granules
Add excess zinc granules to warm dilute hydrochloric acid in a beaker. The zinc reacts, giving off hydrogen gas (effervescence). When effervescence stops (all the acid is used up), filter off the excess unreacted zinc. Evaporate the filtrate over a water bath to the point of crystallization, allow it to cool so that crystals of zinc chloride form, then filter off and dry the crystals between filter papers.
(a)(ii) Equation
\[Zn_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_{2(g)}\](b) pH of the solutions
\(Na_2SO_4\) is the salt of a strong acid and strong base; \(CH_3COOK\) is the salt of a weak acid and strong base; \(Pb(NO_3)_2\) and \(MgCl_2\) are salts of strong acids and weak bases.
- (i) pH greater than 7: \(CH_3COOK\).
- (ii) pH equal to 7: \(Na_2SO_4\).
- (iii) pH less than 7: \(Pb(NO_3)_2\) and \(MgCl_2\).
(c)(i) Oxidation number of Mn
- I. In \(MnO_2\): \(Mn + 2(-2) = 0 \Rightarrow Mn = +4\).
- II. In \(KMnO_4\): \(+1 + Mn + 4(-2) = 0 \Rightarrow Mn = +7\).
(c)(ii) One laboratory use of each
- \(MnO_2\): used as a catalyst (e.g. in the decomposition of hydrogen peroxide) and as an oxidizing agent in the preparation of chlorine.
- \(KMnO_4\): used as an oxidizing agent in volumetric analysis (redox titrations), where it acts as its own indicator.
(d) Why oxidation and reduction are complementary
Oxidation is the loss of electrons and reduction is the gain of electrons. The electrons lost by the species being oxidized are exactly those gained by the species being reduced. Therefore the two processes must occur together and simultaneously (in equal number of electrons); one cannot take place without the other.
(e) The reaction \(Fe + 2Ag^+ \to Fe^{2+} + 2Ag\)
- (i) Half equations: \(Fe \to Fe^{2+} + 2e^-\) (oxidation); \(Ag^+ + e^- \to Ag\) (reduction).
- (ii) I. Oxidized: iron (\(Fe\)); II. Reduced: silver ion (\(Ag^+\)).
- (iii) The oxidation number of silver changes from \(+1\) to \(0\) (a decrease of 1).
Answer Details
(a)(i) Preparation of ZnCl2 crystals from zinc granules
Add excess zinc granules to warm dilute hydrochloric acid in a beaker. The zinc reacts, giving off hydrogen gas (effervescence). When effervescence stops (all the acid is used up), filter off the excess unreacted zinc. Evaporate the filtrate over a water bath to the point of crystallization, allow it to cool so that crystals of zinc chloride form, then filter off and dry the crystals between filter papers.
(a)(ii) Equation
\[Zn_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_{2(g)}\](b) pH of the solutions
\(Na_2SO_4\) is the salt of a strong acid and strong base; \(CH_3COOK\) is the salt of a weak acid and strong base; \(Pb(NO_3)_2\) and \(MgCl_2\) are salts of strong acids and weak bases.
- (i) pH greater than 7: \(CH_3COOK\).
- (ii) pH equal to 7: \(Na_2SO_4\).
- (iii) pH less than 7: \(Pb(NO_3)_2\) and \(MgCl_2\).
(c)(i) Oxidation number of Mn
- I. In \(MnO_2\): \(Mn + 2(-2) = 0 \Rightarrow Mn = +4\).
- II. In \(KMnO_4\): \(+1 + Mn + 4(-2) = 0 \Rightarrow Mn = +7\).
(c)(ii) One laboratory use of each
- \(MnO_2\): used as a catalyst (e.g. in the decomposition of hydrogen peroxide) and as an oxidizing agent in the preparation of chlorine.
- \(KMnO_4\): used as an oxidizing agent in volumetric analysis (redox titrations), where it acts as its own indicator.
(d) Why oxidation and reduction are complementary
Oxidation is the loss of electrons and reduction is the gain of electrons. The electrons lost by the species being oxidized are exactly those gained by the species being reduced. Therefore the two processes must occur together and simultaneously (in equal number of electrons); one cannot take place without the other.
(e) The reaction \(Fe + 2Ag^+ \to Fe^{2+} + 2Ag\)
- (i) Half equations: \(Fe \to Fe^{2+} + 2e^-\) (oxidation); \(Ag^+ + e^- \to Ag\) (reduction).
- (ii) I. Oxidized: iron (\(Fe\)); II. Reduced: silver ion (\(Ag^+\)).
- (iii) The oxidation number of silver changes from \(+1\) to \(0\) (a decrease of 1).
Question 57 Report
(a) Atoms of four non-metallic elements in the same group of the periodic table are arranged in order of increasing atomic radius as \(R < T < W < X\). Which of the elements
(i) would readily lose electron(s) from the outermost shell;
(ii) is most electronegative;
(iii) would T displace from aqueous solution;
(iv) is at the top of the group?
(b) The following table shows the electronic configuration of two elements Y and Z.
| Element | Electronic Configuration |
| Y | \(1s^2\ 2s^2\ 2p^5\) |
| Z | \(1s^2\ 2s^2\ 2p^5\ 3s^1\) |
(i) Name the I. group to which Y belongs II. period to which Z belongs.
(ii) What is the number of protons present in an atom of Z?
(iii) How many unpaired electrons are in an atom of Y?
(iv) Write the formula of the compound formed between Y and Z
(c) Name the type of bond(s) that exist(s) in each of the following compounds.
(i) \(\mathrm{CaCl}_2\); (ii) \(\mathrm{NH}_4\mathrm{Cl}\); \(\mathrm{CCl}_4\)
(d) Describe how the conductance of a molar solution of ammonia compares to that of sodium hydroxide solution.
(e) State the type of reaction illustrated by each of the following equations:
(i) \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}_{(/)} + \mathrm{CH}_3\mathrm{COOH}_{(aq)} \rightleftharpoons \mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5{}_{(/)} + \mathrm{H}_2\mathrm{O}_{(/)}\)
(ii) \(\mathrm{H}+_{Y^+_{(aq)}} + \mathrm{OH}^+_{(aq)} \to \mathrm{H}_2\mathrm{O}\)
(f) Determine the volume of the residual gas when \(20.0\text{cm}^3\) of hydrogen was sparked with \(15.0\text{cm}^3\) of oxygen and the resulting mixture cooled to room temperature.
(a) Four non-metals of the same group with increasing atomic radius \( R < T < W < X \), so R is at the top (smallest) and X at the bottom (largest).
- (i) Would most readily lose electrons from the outermost shell: X (largest atom, weakest hold on outer electrons, least electronegative).
- (ii) Most electronegative: R (smallest atom, at the top).
- (iii) For non-metals reactivity decreases down the group, so T can displace those below it, namely W and X, from aqueous solution.
- (iv) At the top of the group: R.
(b) Y: \( 1s^2\,2s^2\,2p^5 \) (fluorine). Z: \( 1s^2\,2s^2\,2p^6\,3s^1 \) (sodium), giving 11 electrons in the neutral atom.
| Element | Electronic configuration |
|---|---|
| Y | \( 1s^2\,2s^2\,2p^5 \) |
| Z | \( 1s^2\,2s^2\,2p^6\,3s^1 \) |
- (i) I. Y belongs to Group VII (the halogens). II. Z is in Period 3.
- (ii) Number of protons in an atom of Z \( = 11 \).
- (iii) In Y the \( 2p^5 \) sub-shell has one unpaired electron, so there is 1 unpaired electron.
- (iv) Y forms \( Y^- \) and Z forms \( Z^+ \), so the compound is ZY (that is \( NaF \)).
(c) Bond types
- (i) \( CaCl_2 \): ionic (electrovalent) bond.
- (ii) \( NH_4Cl \): ionic bond (between \( NH_4^+ \) and \( Cl^- \)) together with covalent and coordinate (dative) bonds within the ammonium ion.
- (iii) \( CCl_4 \): covalent bond.
(d) Conductance of ammonia compared with sodium hydroxide
Ammonia solution is a weak base that is only slightly ionized, so it produces few ions and conducts electricity poorly. Sodium hydroxide is a strong base that is fully ionized, producing many ions, so at the same molar concentration it conducts electricity much better than ammonia.
(e) Type of reaction
- (i) \( CH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \): esterification (a reversible condensation reaction).
- (ii) \( H^+ + OH^- \rightarrow H_2O \): neutralization.
(f) Volume of residual gas
\[ 2H_2 + O_2 \rightarrow 2H_2O \]\( 20.0\ cm^3 \) of \( H_2 \) reacts with \( \dfrac{20.0}{2} = 10.0\ cm^3 \) of \( O_2 \). Oxygen supplied is \( 15.0\ cm^3 \), so oxygen left over \( = 15.0 - 10.0 = 5.0\ cm^3 \). The water formed condenses to a liquid on cooling.
Residual gas = 5.0 cm\(^3\) of oxygen.
Answer Details
(a) Four non-metals of the same group with increasing atomic radius \( R < T < W < X \), so R is at the top (smallest) and X at the bottom (largest).
- (i) Would most readily lose electrons from the outermost shell: X (largest atom, weakest hold on outer electrons, least electronegative).
- (ii) Most electronegative: R (smallest atom, at the top).
- (iii) For non-metals reactivity decreases down the group, so T can displace those below it, namely W and X, from aqueous solution.
- (iv) At the top of the group: R.
(b) Y: \( 1s^2\,2s^2\,2p^5 \) (fluorine). Z: \( 1s^2\,2s^2\,2p^6\,3s^1 \) (sodium), giving 11 electrons in the neutral atom.
| Element | Electronic configuration |
|---|---|
| Y | \( 1s^2\,2s^2\,2p^5 \) |
| Z | \( 1s^2\,2s^2\,2p^6\,3s^1 \) |
- (i) I. Y belongs to Group VII (the halogens). II. Z is in Period 3.
- (ii) Number of protons in an atom of Z \( = 11 \).
- (iii) In Y the \( 2p^5 \) sub-shell has one unpaired electron, so there is 1 unpaired electron.
- (iv) Y forms \( Y^- \) and Z forms \( Z^+ \), so the compound is ZY (that is \( NaF \)).
(c) Bond types
- (i) \( CaCl_2 \): ionic (electrovalent) bond.
- (ii) \( NH_4Cl \): ionic bond (between \( NH_4^+ \) and \( Cl^- \)) together with covalent and coordinate (dative) bonds within the ammonium ion.
- (iii) \( CCl_4 \): covalent bond.
(d) Conductance of ammonia compared with sodium hydroxide
Ammonia solution is a weak base that is only slightly ionized, so it produces few ions and conducts electricity poorly. Sodium hydroxide is a strong base that is fully ionized, producing many ions, so at the same molar concentration it conducts electricity much better than ammonia.
(e) Type of reaction
- (i) \( CH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \): esterification (a reversible condensation reaction).
- (ii) \( H^+ + OH^- \rightarrow H_2O \): neutralization.
(f) Volume of residual gas
\[ 2H_2 + O_2 \rightarrow 2H_2O \]\( 20.0\ cm^3 \) of \( H_2 \) reacts with \( \dfrac{20.0}{2} = 10.0\ cm^3 \) of \( O_2 \). Oxygen supplied is \( 15.0\ cm^3 \), so oxygen left over \( = 15.0 - 10.0 = 5.0\ cm^3 \). The water formed condenses to a liquid on cooling.
Residual gas = 5.0 cm\(^3\) of oxygen.
Question 58 Report
(a) (i) State Boyle's law.
(ii) Give the mathematical expression of Boyle's law.
(iii) Sketch the graphical representation of Boyle's law.
(b) A given mass of gas occupied 500 cm3 at 30°C and 6.5 x 104 Nm\(^{-2}\). Calculate the volume of the gas at s.t.p. [ Standard pressure = 1.01 x 105 Nm\(^{-2}\)]
(c) Explain why an inflated balloon expands in warm water.
(d)(i) What is meant by the term vapour density? (ii) Determine the vapour density of SO\(_{2}\). [ S = 32, O = 16 ]
(e)(i) Which of the following gases, SO\(_{2}\) and NH\(_{3}\) can be collected by I. upward delivery; II. downward delivery?
(ii) What would be the nature of the solution formed by dissolving each of the following substances in water separately. I. SO\(_{2}\) II. NH\(_{3}\)
(iii) State the property exhibited by SO\(_{2}\) and NH\(_{3}\) in the following equations:
I. 2H\(_{2}\)S\(_{(aq)}\) + SO\(_{2(g)}\) \(\to\) 3S\(_{(s)}\) + 2H\(_{2}\)O\(_{(/)}\)
II. 3CuO\(_{(s)}\) + 2NH\(_{3(g)}\) \(\to\) 3Cu\(_{(s)}\) + 3H\(_{2}\)O\(_{(/)}\) + N\(_{2(s)}\).
(f) State two alloys of tin.
(ii) Give one use of each of the alloys in (f)(i).
(a)(i) Boyle's law states that, at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
(ii) \[V\propto\frac{1}{P}\] Hence, \[PV=k\] where \(k\) is a constant for a fixed mass of gas at constant temperature.
(iii) The graph of pressure against volume is a rectangular hyperbola.
(b) Using the combined gas law:
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\]
\[V_2=\frac{P_1V_1T_2}{P_2T_1}\]
\[V_2=\frac{(6.5\times10^4)(500)(273)}{(1.01\times10^5)(273+30)}\]
\[V_2=289.8\ \text{cm}^3\]
\[\boxed{V_2\approx290\ \text{cm}^3}\]
(c) In warm water, the air particles inside the balloon gain kinetic energy and move more rapidly. They collide more frequently and forcefully with the wall of the balloon, causing the balloon to expand until the internal and external pressures balance.
(d)(i) Vapour density is the ratio of the mass of a given volume of a gas or vapour to the mass of an equal volume of hydrogen at the same temperature and pressure.
(d)(ii)
\[M_r(\mathrm{SO_2})=32+(2\times16)=64\]
\[\text{Vapour density}=\frac{M_r}{2}=\frac{64}{2}=\boxed{32}\]
(e)(i)
- Upward delivery: \(\mathrm{NH_3}\).
- Downward delivery: \(\mathrm{SO_2}\).
(e)(ii)
- \(\mathrm{SO_2}\) dissolves in water to form an acidic solution.
- \(\mathrm{NH_3}\) dissolves in water to form a basic or alkaline solution.
(e)(iii)
- \(\mathrm{SO_2}\) acts as an oxidising agent.
- \(\mathrm{NH_3}\) acts as a reducing agent.
(f)(i) Two alloys of tin are:
- Bronze, an alloy of copper and tin.
- Soft solder, an alloy of tin and lead.
(f)(ii)
- Bronze is used for making coins, medals and ornaments.
- Soft solder is used for joining metals, for example in electrical connections and plumbing.
Answer Details
(a)(i) Boyle's law states that, at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
(ii) \[V\propto\frac{1}{P}\] Hence, \[PV=k\] where \(k\) is a constant for a fixed mass of gas at constant temperature.
(iii) The graph of pressure against volume is a rectangular hyperbola.
(b) Using the combined gas law:
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\]
\[V_2=\frac{P_1V_1T_2}{P_2T_1}\]
\[V_2=\frac{(6.5\times10^4)(500)(273)}{(1.01\times10^5)(273+30)}\]
\[V_2=289.8\ \text{cm}^3\]
\[\boxed{V_2\approx290\ \text{cm}^3}\]
(c) In warm water, the air particles inside the balloon gain kinetic energy and move more rapidly. They collide more frequently and forcefully with the wall of the balloon, causing the balloon to expand until the internal and external pressures balance.
(d)(i) Vapour density is the ratio of the mass of a given volume of a gas or vapour to the mass of an equal volume of hydrogen at the same temperature and pressure.
(d)(ii)
\[M_r(\mathrm{SO_2})=32+(2\times16)=64\]
\[\text{Vapour density}=\frac{M_r}{2}=\frac{64}{2}=\boxed{32}\]
(e)(i)
- Upward delivery: \(\mathrm{NH_3}\).
- Downward delivery: \(\mathrm{SO_2}\).
(e)(ii)
- \(\mathrm{SO_2}\) dissolves in water to form an acidic solution.
- \(\mathrm{NH_3}\) dissolves in water to form a basic or alkaline solution.
(e)(iii)
- \(\mathrm{SO_2}\) acts as an oxidising agent.
- \(\mathrm{NH_3}\) acts as a reducing agent.
(f)(i) Two alloys of tin are:
- Bronze, an alloy of copper and tin.
- Soft solder, an alloy of tin and lead.
(f)(ii)
- Bronze is used for making coins, medals and ornaments.
- Soft solder is used for joining metals, for example in electrical connections and plumbing.
Question 59 Report
(a)(i) Define polymerization.
(ii) Mention two types of polymerization.
(iii) Give one example of I. naturally occurring polymer, II. synthetic polymer.
(b)(i) Describe how ethyne could be prepared in the laboratory.
(ii) Write an equation for the I. polymerization of ethyne II. complete reaction of ethyne with hydrogen bromide.
(iii) Name each of the products formed in (b)(ii) II.
(c)(i) Define each of the following terms: I. pollution; Il. pollutant.
(ii) Mention three types of pollution.
(iii) Give two ways by which pollution can be prevented.
(a)(i) Polymerization
Polymerization is the chemical process by which many small molecules (monomers) combine together to form one very large molecule (a polymer).
(a)(ii) Two types of polymerization
Addition polymerization and condensation polymerization.
(a)(iii) Examples
- I. Naturally occurring polymer: rubber (also starch, cellulose or protein).
- II. Synthetic polymer: polythene (also polyvinyl chloride, PVC, or nylon).
(b)(i) Laboratory preparation of ethyne
Place a few lumps of calcium dicarbide (calcium carbide) in a flask and add water drop by drop. Ethyne gas is evolved rapidly and is collected over water.
\[CaC_{2(s)} + 2H_2O_{(l)} \to Ca(OH)_{2(aq)} + C_2H_{2(g)}\](b)(ii) Equations
- I. Polymerization of ethyne (to benzene):
- II. Complete reaction of ethyne with hydrogen bromide:
(b)(iii) Name of the product
The product of the complete reaction with hydrogen bromide is 1,1-dibromoethane (\(CH_3CHBr_2\)).
(c)(i) Definitions
- I. Pollution: the introduction of harmful substances (pollutants) into the environment, causing damage to living things and their surroundings.
- II. Pollutant: the harmful substance that causes pollution.
(c)(ii) Three types of pollution
Air pollution, water pollution and land (soil) pollution. (Noise pollution is also acceptable.)
(c)(iii) Two ways of preventing pollution
- Proper treatment and safe disposal of industrial and domestic wastes (and recycling).
- Fitting factories and vehicles with filters/catalytic converters to reduce harmful emissions, and afforestation.
Answer Details
(a)(i) Polymerization
Polymerization is the chemical process by which many small molecules (monomers) combine together to form one very large molecule (a polymer).
(a)(ii) Two types of polymerization
Addition polymerization and condensation polymerization.
(a)(iii) Examples
- I. Naturally occurring polymer: rubber (also starch, cellulose or protein).
- II. Synthetic polymer: polythene (also polyvinyl chloride, PVC, or nylon).
(b)(i) Laboratory preparation of ethyne
Place a few lumps of calcium dicarbide (calcium carbide) in a flask and add water drop by drop. Ethyne gas is evolved rapidly and is collected over water.
\[CaC_{2(s)} + 2H_2O_{(l)} \to Ca(OH)_{2(aq)} + C_2H_{2(g)}\](b)(ii) Equations
- I. Polymerization of ethyne (to benzene):
- II. Complete reaction of ethyne with hydrogen bromide:
(b)(iii) Name of the product
The product of the complete reaction with hydrogen bromide is 1,1-dibromoethane (\(CH_3CHBr_2\)).
(c)(i) Definitions
- I. Pollution: the introduction of harmful substances (pollutants) into the environment, causing damage to living things and their surroundings.
- II. Pollutant: the harmful substance that causes pollution.
(c)(ii) Three types of pollution
Air pollution, water pollution and land (soil) pollution. (Noise pollution is also acceptable.)
(c)(iii) Two ways of preventing pollution
- Proper treatment and safe disposal of industrial and domestic wastes (and recycling).
- Fitting factories and vehicles with filters/catalytic converters to reduce harmful emissions, and afforestation.
