WAEC SSCE - Chemistry - 2016
Tambaya 1 Rahoto
Which of the following substances is a non-electrolyte?
Bayanin Amsa
A non-electrolyte is a substance that does not conduct electricity when dissolved in water. It is usually a covalent compound that does not dissociate into ions when dissolved in water. Among the given substances, only C6H12O6 (glucose) is a non-electrolyte because it is a covalent compound made up of molecules that do not dissociate into ions in water. The other substances are electrolytes because they either ionize or dissociate into ions in water and conduct electricity. H2SO4 and NH4Cl are strong electrolytes because they dissociate completely into ions in water, while CH3COOH is a weak electrolyte because it only partially ionizes in water.
Tambaya 2 Rahoto
Consider the following ionic equation:
Cr2O72- + 14H+ + ne- \(\to\) 2Cr3+ + 7H2O.
The value of n in the equation is_______?
Bayanin Amsa
Tambaya 3 Rahoto
The mass of potassium hydroxide required to make 300.ocm3 of 0.4 moldm-3 solution is {KOH = 56.0}
Bayanin Amsa
To calculate the mass of potassium hydroxide required, we can use the formula: mass = moles × molar mass First, let's calculate the number of moles of potassium hydroxide required to make 300 cm3 of 0.4 mol dm-3 solution: moles = concentration × volume moles = 0.4 mol dm-3 × 0.3 dm3 = 0.12 moles The molar mass of potassium hydroxide (KOH) is 56.0 g mol-1, so we can now calculate the mass of KOH required: mass = moles × molar mass mass = 0.12 moles × 56.0 g mol-1 = 6.72 g Therefore, the mass of potassium hydroxide required to make 300 cm3 of 0.4 mol dm-3 solution is 6.72 g. So, the correct option is (C) 6.72g.
Tambaya 4 Rahoto
When a compound X is heated with concentrated tetraoxosulphate(VI)acid, it produces an alkene. X ia an
Tambaya 5 Rahoto
Which of the following salts when dissolved in water will form a solution that will change blue litmus to red?
Bayanin Amsa
The salt that will change blue litmus to red when dissolved in water is NH4Cl. This is because NH4Cl undergoes hydrolysis in water to form NH4OH and HCl. NH4OH is a weak base and will react with water to produce NH4+ and OH- ions. The OH- ions will increase the concentration of hydroxide ions in the solution, making it slightly basic. Litmus paper is an indicator that changes color in response to the pH of a solution. Blue litmus paper turns red in acidic solutions, and since the NH4Cl solution is slightly basic, it will change the blue litmus paper to red.
Tambaya 6 Rahoto
When element \(_{20}\)Y combines with element \(_8\)Z, it forms
Bayanin Amsa
The answer to the question depends on the electronic configuration and reactivity of the elements Y and Z. Element Y has an atomic number of 20, indicating it has 20 protons in its nucleus, while element Z has an atomic number of 8, indicating it has 8 protons in its nucleus. To determine the type of compound formed, we need to consider the electronegativity of each element. Electronegativity is the measure of an atom's ability to attract electrons towards itself when it forms a chemical bond. Generally, when elements with a large difference in electronegativity combine, they form an ionic compound, while elements with similar electronegativity combine to form covalent compounds. Since Y has a larger atomic number than Z, it has more electrons in its valence shell, making it less electronegative than Z. Therefore, Y is more likely to donate its electrons to Z, forming an ionic compound with the formula YZ. So, the correct option is "An ionic compound, YZ is formed."
Tambaya 7 Rahoto
The bond formed between H2O and H+ to form the hydroxonium H3O+ is
Tambaya 8 Rahoto
Which of the following compounds is/are secondary alkanols?
i) CH3CH(CH3)CH3OH
ii) CH3CH2CH(OH)CH3
iii) CH3C(OH)(CH3)CH3
Tambaya 9 Rahoto
An aqueous solution of (NH4)2 SO4 is
Bayanin Amsa
An aqueous solution of (NH4)2SO4 is acidic. When ammonium sulfate ((NH4)2SO4) is dissolved in water, it dissociates into its ions: 2 NH4+ and SO4 2-. The ammonium ion (NH4+) is a weak acid and tends to donate a hydrogen ion (H+) to the water molecules, resulting in the formation of hydronium ions (H3O+). This increases the concentration of H+ ions in the solution and makes it acidic. The chemical equation for the dissociation of ammonium sulfate in water is as follows: (NH4)2SO4 + H2O → 2 NH4+ + SO4 2- + H2O Therefore, among the options provided, the correct answer is acidic.
Tambaya 10 Rahoto
Chemical equilibrium is attained when
Tambaya 12 Rahoto
A compound with molecular formula CH2O2 is
Tambaya 13 Rahoto
Which of the following bonds are broken when ethanol is boiled?
I. Covalent bonds
II. Ionic bonds
III. Hydrogen bonds
Bayanin Amsa
The correct answer is III only - hydrogen bonds are broken when ethanol is boiled. Ethanol is a molecule that contains both polar and nonpolar regions due to the presence of the hydroxyl (-OH) group. The polar region of ethanol allows it to form hydrogen bonds with neighboring ethanol molecules. These hydrogen bonds are relatively weak and require less energy to break than covalent or ionic bonds. When ethanol is boiled, the increased thermal energy breaks the hydrogen bonds between the ethanol molecules, allowing the molecules to escape into the gas phase. The covalent bonds within the ethanol molecules and the ionic bonds between any ions present in the solution are not affected by boiling and remain intact.
Tambaya 14 Rahoto
Which of the following factors will affect the rate of the reaction represented by the following equation?
2HCl(aq) + CaCO3(s) \(\rightarrow\) CaCl2(aq) + H2O(1) + CO2(g)
I. Pressure
II.Concentration
III. Nature of reactants
IV. Temperature
Bayanin Amsa
Tambaya 15 Rahoto
On evaporation to dryness, 3503 of saturated solution of salt Z gave 55.5g of salt. What is the solubility of the salt? [Z = 101]
Bayanin Amsa
The solubility of a salt is defined as the maximum amount of solute that can be dissolved in a given amount of solvent at a specified temperature and pressure. Given that 3503 of saturated solution of salt Z gave 55.5g of salt. We can find the solubility of the salt using the formula: Solubility = Mass of solute / Volume of solution First, we need to convert the given mass of salt to moles using its molar mass: Molar mass of Z = 101g/mol Number of moles of Z = 55.5g / 101g/mol = 0.5495 mol Now, we can calculate the volume of the solution using the given volume of 3503: Volume of solution = 3503 = 0.35 dm3 Substituting the values into the formula, we get: Solubility = 0.5495 mol / 0.35 dm3 = 1.57 moldm-3 Therefore, the solubility of salt Z is 1.57 moldm-3.
Tambaya 16 Rahoto
The general gas equation \(\frac{PV}{T}\) = K is a combination of
Bayanin Amsa
The general gas equation \(\frac{PV}{T}\) = K is a combination of Boyle's law and Charles' law. Boyle's law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. Charles' law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its temperature. The general gas equation combines these two laws and adds in the ideal gas law constant, K, to relate the pressure, volume, and temperature of a gas. It is important to note that the ideal gas law assumes that the gas behaves ideally, which means that its particles are infinitely small and there are no attractive or repulsive forces between them.
Tambaya 17 Rahoto
Which of the following arrangements shows increasing order of reactivity of the halogens?
Tambaya 18 Rahoto
The alkanol obtained from the production of soap is
Tambaya 19 Rahoto
Metals can be stretched into wires because they are
Bayanin Amsa
Metals can be stretched into wires because they are ductile. Ductility is the ability of a metal to be stretched into thin wires without breaking. This property is due to the ability of the metallic bonds between atoms to stretch without breaking. As a result, metals can be drawn into thin wires, making them useful in electrical wiring and other applications.
Tambaya 20 Rahoto
Amino acids are obtained from proteins by
Bayanin Amsa
Amino acids are the building blocks of proteins. They are linked together by peptide bonds to form a polypeptide chain, which then folds into a three-dimensional protein structure. Proteins can be broken down into their constituent amino acids by hydrolysis, which is the addition of water to break the peptide bonds. Therefore, the correct answer is hydrolysis.
Tambaya 21 Rahoto
The refreshing characteristic taste of fizzy drinks is due ot the presence of
Bayanin Amsa
Tambaya 22 Rahoto
Propane carbon(IV)oxide diffuse at the same rate because [H = 1.00, C = 12.0, O = 16.0]
Bayanin Amsa
The diffusion rate of a gas is directly proportional to its molecular mass. Propane (C3H8) has a higher molecular mass than carbon dioxide (CO2). However, since the question asks about carbon(IV)oxide (CO2), which has a molecular mass of 44, the answer is that propane and carbon(IV)oxide diffuse at the same rate because they have the same relative molecular mass. Therefore, "They have the same relative molecular mass" is the correct answer.
Tambaya 23 Rahoto
The oxidation number of sulphur is +4 in
Bayanin Amsa
The oxidation number of an atom is a measure of the number of electrons it has gained or lost, assuming that the electrons in a chemical bond are assigned to the more electronegative atom. In the compound H2SO3, there are two hydrogen atoms, each with an oxidation number of +1, and one oxygen atom, which has an oxidation number of -2. The sum of the oxidation numbers in H2SO3 is zero, since the compound is neutral. Therefore, we can calculate the oxidation number of sulfur as follows: 2(+1) + x + 3(-2) = 0 Solving for x, we get: x = +4 Therefore, the oxidation number of sulfur in H2SO3 is +4. In summary, the correct option in the question is H2SO3, where the oxidation number of sulfur is +4.
Tambaya 24 Rahoto
How many moles of oxygen would contain 1.204 x 1024 molecules?
[Avogadro's constant (NA) = 6.02 x 1023]
Bayanin Amsa
To determine the number of moles of oxygen, we need to use Avogadro's constant, which is the number of particles in one mole of a substance. Given that the number of oxygen molecules is 1.204 x 10^24, we can use this information to calculate the number of moles of oxygen using the following formula: Number of moles = Number of particles / Avogadro's constant Plugging in the values, we get: Number of moles = 1.204 x 10^24 / 6.02 x 10^23 Number of moles = 2 Therefore, the answer is 2. Hence, is the correct answer.
Tambaya 25 Rahoto
The following atoms of carbon 126C, 136C and 146C can be described as?
Bayanin Amsa
The atoms of carbon 126C, 136C and 146C can be described as isotopes. Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei. These isotopes of carbon all have six protons (which define them as carbon), but they have different numbers of neutrons. 126C has 6 neutrons, 136C has 7 neutrons and 146C has 8 neutrons. Isotopes have identical chemical properties but have different atomic masses due to their different numbers of neutrons.
Tambaya 26 Rahoto
The boiling points of HF, H2O and NH3 increase in the order of
Bayanin Amsa
The boiling point of a substance is the temperature at which it changes its state from liquid to gas. The boiling point of a substance is dependent on its intermolecular forces, which include van der Waals forces, hydrogen bonding, and dipole-dipole interactions. The stronger the intermolecular forces, the higher the boiling point. In the given options, NH3, H2O, and HF are all polar molecules, which exhibit hydrogen bonding, the strongest of intermolecular forces. Among the given options, HF molecules have the strongest hydrogen bonding as compared to H2O and NH3 molecules. Therefore, HF has the highest boiling point among the given options. Similarly, NH3 molecules have the weakest hydrogen bonding as compared to HF and H2O molecules, so NH3 has the lowest boiling point among the given options. Therefore, the order of increasing boiling points is NH3 < HF < H2O. Hence, the correct option is: "NH3 < HF < H2O".
Tambaya 27 Rahoto
Consider the following reaction equation
C2H4(g) + 3O2(g) \(\to\) 2CO2(g) + 2H2O(l).
The volume of oxygen at s.t.p that will be required to burn 14g of ethene is
[C2H4 = 28; Molar volume of gas at s.t.p = 22.4dm3
Bayanin Amsa
To solve this problem, we need to first balance the chemical equation: C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)
From the balanced equation, we can see that 1 mole of ethene reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide gas and 2 moles of liquid water.
We can calculate the number of moles of ethene using the given mass and molar mass: n(C2H4) = m/M = 14/28 = 0.5 moles
Using the mole ratio from the balanced equation, we can calculate the number of moles of oxygen required: n(O2) = 3n(C2H4) = 1.5 moles
Finally, we can use the ideal gas law to calculate the volume of oxygen at s.t.p: PV = nRT V = nRT/P = (1.5 mol)(8.31 J/mol K)(273 K)/(101,325 Pa) = 33.6 dm3
Therefore, the volume of oxygen at s.t.p that will be required to burn 14g of ethene is 33.6 dm3.
Answer: 33.6dm3.
Tambaya 28 Rahoto
Chemicals that are produced in small quantities and with very degree of purity are
Bayanin Amsa
Chemicals that are produced in small quantities and with a very high degree of purity are called "fine chemicals". Fine chemicals are also known as "specialty chemicals" or "performance chemicals". They are typically produced in relatively small quantities (usually less than 1000 tons per year) and are used for specific purposes in industries such as pharmaceuticals, biotechnology, agrochemicals, and electronics. Fine chemicals are produced through complex and highly specialized processes that require a high degree of technical expertise and advanced manufacturing techniques. The production of fine chemicals requires strict control of the purity and quality of the raw materials, as well as the process conditions, in order to ensure that the final product meets the required specifications and standards. Overall, the production of fine chemicals requires a high level of skill and expertise, and is often carried out by specialized companies that focus on producing these highly specialized chemicals.
Tambaya 29 Rahoto
The minimum amount of energy required for effective collisions between reacting particles is known as
Bayanin Amsa
Activation energy is the minimum amount of energy required for effective collisions between reacting particles. In a chemical reaction, reactant molecules must collide with a certain minimum amount of kinetic energy in order for the reaction to occur. The reactant molecules must also collide with the correct orientation in order for the reaction to take place. The minimum amount of energy required for these effective collisions is known as the activation energy. Activation energy is a barrier that must be overcome for a reaction to occur. This barrier can be overcome by increasing the temperature of the reaction, since this increases the kinetic energy of the reactant molecules, making them more likely to collide with sufficient energy to react. Alternatively, a catalyst can be used to lower the activation energy barrier and increase the rate of the reaction. In summary, the correct option in the question is activation energy, which is the minimum amount of energy required for effective collisions between reacting particles.
Tambaya 30 Rahoto
The pH of four solutions M,N,Q and R are 2,6,8 and 11 respectively.
Which of the following deductions about the solutions is correct?
Tambaya 31 Rahoto
Chlorine water is used as a bleaching agent because it is
Bayanin Amsa
Chlorine water is used as a bleaching agent because it is an oxidizing agent. Bleaching is a chemical process that involves the breaking down of chromophores, which are responsible for the coloration of substances. Chlorine water is able to oxidize these chromophores and break them down, which results in the removal of the color from the substance being bleached. The oxidizing power of chlorine water comes from the fact that chlorine is a highly electronegative element, which means that it has a strong tendency to attract electrons. This makes it an effective oxidizing agent, as it is able to strip electrons from other substances and cause them to undergo oxidation.
Tambaya 32 Rahoto
The shape of the water molecule is_______?
Bayanin Amsa
The shape of the water molecule is V-shaped. The water molecule has two hydrogen atoms and one oxygen atom, and the three atoms are not in a straight line, but form a V-shape. This is because of the arrangement of the electron pairs in the molecule. The two lone pairs of electrons on the oxygen atom repel the two hydrogen atoms, pushing them down and creating a V-shape. This shape gives water many of its unique properties, such as its ability to form hydrogen bonds and its high surface tension.
Tambaya 33 Rahoto
How many electron are present in 94Be2+?
Bayanin Amsa
Beryllium has 4 electrons and a 2+ ion means it has lost 2 electrons. Therefore, the number of electrons present in 94Be2+ ion can be calculated by subtracting 2 from the original number of electrons in beryllium. Thus, the number of electrons in 94Be2+ ion is 2. Therefore, the answer is (a) 2.
Tambaya 34 Rahoto
The quality of electricity required to discharge 1 mole of univalent ion is
Bayanin Amsa
The quality of electricity required to discharge 1 mole of univalent ion is 96500C. This quantity of electricity is known as Faraday's constant, which is equivalent to the charge on one mole of electrons (6.022 x 10^23 electrons). When one mole of an univalent ion is discharged, it means that one mole of electrons is transferred, and therefore, the charge required for this transfer is equal to the charge on one mole of electrons, which is 96500C.
Tambaya 35 Rahoto
An atom X consist of 6 protons, 6 electrons and 7 neutrons. Which of the following representations of the atom is correct?
Tambaya 36 Rahoto
Which of the following metals is common to both brass and bronze?
Bayanin Amsa
Copper is the metal that is common to both brass and bronze. Brass is an alloy made of copper and zinc, while bronze is an alloy made of copper and tin. Since copper is a primary component of both alloys, it is the metal that is common to both brass and bronze.
Tambaya 37 Rahoto
Which of the following sources of energy contributes to green-house effect?
Bayanin Amsa
Natural gas contributes to the greenhouse effect. When natural gas is burned, it produces carbon dioxide (CO2), which is a greenhouse gas that contributes to global warming. The combustion of natural gas also releases small amounts of other greenhouse gases, such as methane. Methane is a potent greenhouse gas that is about 25 times more effective at trapping heat than carbon dioxide over a 100-year time span. Therefore, the use of natural gas as a source of energy contributes to the greenhouse effect.
Tambaya 38 Rahoto
The spreading of the scent of a flower in a garden is an example of_____?
Bayanin Amsa
The spreading of the scent of a flower in a garden is an example of diffusion. Diffusion is the movement of particles (molecules, ions or atoms) from an area of higher concentration to an area of lower concentration. In this case, the scent molecules of the flower move from the area of high concentration (the flower) to an area of lower concentration (the surrounding air) until they are evenly distributed throughout the space. This spreading of the scent is an example of the random movement of particles due to diffusion.
Tambaya 39 Rahoto
The energy which accompanies the addition of an electron to an isolated gaseous atom is
Bayanin Amsa
The energy which accompanies the addition of an electron to an isolated gaseous atom is called electron affinity. When an atom gains an electron, it becomes a negatively charged ion. The energy released or absorbed when this process occurs is the electron affinity. If energy is released when an electron is added to the atom, then the electron affinity is negative, and if energy is absorbed, then the electron affinity is positive. Therefore, the correct option is "electron affinity."
Tambaya 40 Rahoto
Aluminium is used in the manufacture or aircraft because it
Bayanin Amsa
Aluminium is used in the manufacture of aircraft because it is light and resists corrosion. Aircraft need to be as light as possible in order to fly efficiently, and aluminium is a lightweight metal. Additionally, aluminium naturally forms a layer of oxide on its surface which protects it from further corrosion, making it an ideal material for use in aircraft which are exposed to harsh environmental conditions.
Tambaya 41 Rahoto
When NH4Cl(s) is dissolved in water, the container feels cold to touch. This implies
Bayanin Amsa
When NH4Cl(s) is dissolved in water, the container feels cold to touch, indicating that the process is endothermic. Endothermic processes absorb energy from the surroundings to proceed. In this case, the dissolution of NH4Cl in water requires energy to break the ionic bonds between the NH4+ and Cl- ions in the solid and to separate the water molecules from each other. This energy is absorbed from the surroundings, causing the container to feel cold to the touch. The formation of a saturated solution or the solubility of NH4Cl in water is not indicated by the cooling effect.
Tambaya 42 Rahoto
An unsaturated solution differs from a saturated solution at a given temperature because it
Bayanin Amsa
An unsaturated solution differs from a saturated solution at a given temperature because it can still dissolve more solute at that temperature, while a saturated solution has already dissolved the maximum amount of solute that it can hold at that temperature. Therefore, an unsaturated solution has a lower concentration of solute compared to a saturated solution. The ability of a solution to dissolve more solute depends on the nature of the solute, the solvent, and the temperature of the solution.
Tambaya 43 Rahoto
Which of the following elements would produce coloured ion in agueous solution?
Bayanin Amsa
Copper is the element that would produce a colored ion in aqueous solution. Copper ions have a tendency to form complex ions with water or other molecules, which can result in a range of colors depending on the specific ligands involved. For example, copper(II) sulfate forms a blue-colored solution when dissolved in water, while copper(II) chloride forms a green-colored solution. In general, transition metals such as copper are more likely to produce colored ions in solution due to the presence of unpaired electrons in their d orbitals.
Tambaya 44 Rahoto
Which of the following statements about solids is correct?
Bayanin Amsa
The correct statement about solids is that solid particles cannot be easily compressed. This is because the particles in a solid are tightly packed and have fixed positions due to strong intermolecular forces holding them together. Any attempt to compress a solid would require a significant amount of force to overcome these forces and move the particles closer together. In contrast, liquids and gases can be compressed more easily because their particles are not as closely packed and have more room to move around. Therefore, option D is the correct statement.
Tambaya 45 Rahoto
An element X froms the following oxides X2O, XO and XO2. This phenomenon illustrates the law of________?
Bayanin Amsa
The phenomenon illustrated in this question is the law of multiple proportion. According to this law, when two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers. In the given example, element X combines with oxygen to form three different oxides, and the ratios of the masses of X to oxygen in these oxides are in the ratio of small whole numbers (2:1, 1:1, and 1:2). This is consistent with the law of multiple proportion, which states that when two elements (in this case, X and oxygen) form more than one compound, the ratios of the masses of one element (X) that combine with a fixed mass of the other element (oxygen) are in the ratio of small whole numbers.
Tambaya 46 Rahoto
Consider the following reaction equation
H3O+(aq) + OH(aq) \(\rightarrow\) 2H2O(1)
The reaction represents
Bayanin Amsa
The reaction equation H3O+ + OH- \(\rightarrow\) 2H2O represents a neutralization reaction, where a hydrogen ion (H+) from an acid (H3O+) reacts with a hydroxide ion (OH-) from a base to produce water (H2O) and form a neutral solution. Therefore, the answer is neutralization.
Tambaya 47 Rahoto
Ripening of fruits is hastened by using
Bayanin Amsa
The ripening of fruits is hastened by using ethene. Ethene is a plant hormone that regulates the process of ripening in fruits. It is produced naturally by plants, including fruits, and can also be produced synthetically. When ethene gas is introduced to a fruit, it triggers a series of biochemical reactions that lead to the ripening of the fruit. The gas promotes the breakdown of chlorophyll, which gives fruits their green color, and the conversion of starches into sugars. This process makes the fruit softer and sweeter, and also stimulates the production of more ethene gas, which in turn hastens the ripening of the fruit.
Tambaya 48 Rahoto
Which of the following apparatus can be used to measure a specific volume of a liquid accurately?
Bayanin Amsa
Out of the given options, the apparatus that can be used to measure a specific volume of a liquid accurately is a pipette. A pipette is a glass or plastic tube that is used to transfer a measured volume of liquid from one container to another. It is designed to dispense a very precise volume of liquid, typically ranging from 1 to 100 milliliters, depending on the type of pipette. Pipettes are often used in chemistry and biology experiments where precise measurements of liquid volumes are critical to the success of the experiment. Beakers and conical flasks can hold liquids but are not designed for accurate measurement of volumes. A measuring cylinder can measure volumes, but it may not be as precise as a pipette.
Tambaya 49 Rahoto
The compound that makes palm wine taste sour after exposure to the air for few days is
Tambaya 50 Rahoto
Consider the following half-cell reactions.
Al(s) \(\to\) Al3+(aq) + 3e-
Cu2+(aq) + 2e- \(\to\) Cu(s)
The overall equation for the reaction is
Tambaya 51 Rahoto
(a) Consider the following atoms: \(^R_T\)X; \(^S_T\)X.
(i) State the phenomenon exhibited by the two atoms.
(ii) What is the difference between the atoms?
(iii) Give two examples of elements that exhibit the phenomenon stated in (ai)
(iv) lf T is 17, write the electron configuration of the element
(b)(i) State two differences between metals and-non-metals with respect to their:
I. physical properties;
2. chemical properties.
(ii) Give one example of each for the following compounds:
I. an amphoteric oxide;
II. a hydride which evolves hydrogen when reacted with water;
Ill. a trioxocarbonate (IV) salt which is readily decomposed on heating;
IV. a chloride salt which is readily hydrolyzed in water.
(c)(i) State three characteristic properties of transition metals.
(ii) Write the electron configuration of \(_{30}Zn\)
(iii.) Explain briefly why zinc is not considered as a typical transition element.
(d) Consider the reaction represented by the following equation:- Na\(_2\)CO\(_{3(aq)}\) + MgCl\(_{2(aq)}\) ----> 2NaCl\(_{(aq)}\) + MgCO\(_{3(aq)}\). Calculate the mass of sodium trioxocarbonate (IV) needed to produce 3.36 of magnesium trioxocarbonate (IV). [C = 12.0, O = 16.0, Na = 23.0, Mg = 24:0 ]
(a) The atoms \(^R_TX\) and \(^S_TX\) have the same atomic number T but different mass numbers R and S.
- (i) The phenomenon exhibited is isotopy (they are isotopes).
- (ii) The difference is that they have different numbers of neutrons (hence different mass numbers), while having the same number of protons.
- (iii) Two examples of elements exhibiting isotopy: chlorine (\(^{35}Cl\) and \(^{37}Cl\)) and carbon (\(^{12}C\) and \(^{14}C\)). (Hydrogen and oxygen are also acceptable.)
- (iv) If T = 17, the electron configuration is 2, 8, 7.
(b)(i) Metals compared with non-metals:
| Metals | Non-metals | |
|---|---|---|
| Physical | Shiny/lustrous, malleable and ductile, good conductors of heat and electricity, mostly solids with high melting points. | Dull, brittle (if solid), poor conductors (except graphite), many are gases or low-melting solids. |
| Chemical | Form basic oxides; are electropositive, lose electrons to form positive ions (reducing agents). | Form acidic oxides; are electronegative, gain electrons to form negative ions (oxidizing agents). |
(ii) Examples:
- I. Amphoteric oxide: zinc oxide, \(ZnO\) (or \(Al_2O_3\), PbO).
- II. Hydride evolving hydrogen with water: sodium hydride, \(NaH\) (\(NaH + H_2O \rightarrow NaOH + H_2\)).
- III. Trioxocarbonate(IV) salt readily decomposed on heating: copper(II) trioxocarbonate(IV), \(CuCO_3\) (or \(ZnCO_3\)).
- IV. Chloride salt readily hydrolysed in water: aluminium chloride, \(AlCl_3\) (or \(FeCl_3\)).
(c)(i) Three characteristic properties of transition metals: they show variable oxidation states, form coloured ions/compounds, and act as catalysts (they also form complex ions).
(ii) Electron configuration of \(_{30}Zn\): 2, 8, 18, 2 (\(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\)).
(iii) Zinc is not regarded as a typical transition element because its 3d subshell is completely filled (3d^{10}); it shows only one oxidation state (+2) and forms mainly white/colourless compounds, so it lacks the characteristic transition-metal properties.
(d) \(Na_2CO_3 + MgCl_2 \rightarrow 2NaCl + MgCO_3\). To produce 3.36 g of \(MgCO_3\):
Molar mass of \(MgCO_3 = 24 + 12 + (3 \times 16) = 84\,g\,mol^{-1}\)
\[\text{moles of } MgCO_3 = \frac{3.36}{84} = 0.04\,mol\]
From the equation, 1 mole \(Na_2CO_3\) gives 1 mole \(MgCO_3\), so moles of \(Na_2CO_3 = 0.04\,mol\).
Molar mass of \(Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106\,g\,mol^{-1}\)
\[\text{mass of } Na_2CO_3 = 0.04 \times 106 = 4.24\,g\]
Bayanin Amsa
(a) The atoms \(^R_TX\) and \(^S_TX\) have the same atomic number T but different mass numbers R and S.
- (i) The phenomenon exhibited is isotopy (they are isotopes).
- (ii) The difference is that they have different numbers of neutrons (hence different mass numbers), while having the same number of protons.
- (iii) Two examples of elements exhibiting isotopy: chlorine (\(^{35}Cl\) and \(^{37}Cl\)) and carbon (\(^{12}C\) and \(^{14}C\)). (Hydrogen and oxygen are also acceptable.)
- (iv) If T = 17, the electron configuration is 2, 8, 7.
(b)(i) Metals compared with non-metals:
| Metals | Non-metals | |
|---|---|---|
| Physical | Shiny/lustrous, malleable and ductile, good conductors of heat and electricity, mostly solids with high melting points. | Dull, brittle (if solid), poor conductors (except graphite), many are gases or low-melting solids. |
| Chemical | Form basic oxides; are electropositive, lose electrons to form positive ions (reducing agents). | Form acidic oxides; are electronegative, gain electrons to form negative ions (oxidizing agents). |
(ii) Examples:
- I. Amphoteric oxide: zinc oxide, \(ZnO\) (or \(Al_2O_3\), PbO).
- II. Hydride evolving hydrogen with water: sodium hydride, \(NaH\) (\(NaH + H_2O \rightarrow NaOH + H_2\)).
- III. Trioxocarbonate(IV) salt readily decomposed on heating: copper(II) trioxocarbonate(IV), \(CuCO_3\) (or \(ZnCO_3\)).
- IV. Chloride salt readily hydrolysed in water: aluminium chloride, \(AlCl_3\) (or \(FeCl_3\)).
(c)(i) Three characteristic properties of transition metals: they show variable oxidation states, form coloured ions/compounds, and act as catalysts (they also form complex ions).
(ii) Electron configuration of \(_{30}Zn\): 2, 8, 18, 2 (\(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\)).
(iii) Zinc is not regarded as a typical transition element because its 3d subshell is completely filled (3d^{10}); it shows only one oxidation state (+2) and forms mainly white/colourless compounds, so it lacks the characteristic transition-metal properties.
(d) \(Na_2CO_3 + MgCl_2 \rightarrow 2NaCl + MgCO_3\). To produce 3.36 g of \(MgCO_3\):
Molar mass of \(MgCO_3 = 24 + 12 + (3 \times 16) = 84\,g\,mol^{-1}\)
\[\text{moles of } MgCO_3 = \frac{3.36}{84} = 0.04\,mol\]
From the equation, 1 mole \(Na_2CO_3\) gives 1 mole \(MgCO_3\), so moles of \(Na_2CO_3 = 0.04\,mol\).
Molar mass of \(Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106\,g\,mol^{-1}\)
\[\text{mass of } Na_2CO_3 = 0.04 \times 106 = 4.24\,g\]
Tambaya 52 Rahoto
(a)(i) Name a suitable drying agent for the preparation of carbon (IV) oxide in the laboratory.
(ii) Using one chemical test, distinguish between carbon (II) oxide and carbon (IV) oxide.
(b)(i) Describe briefly how oxygen and nitrogen could be obtained separately from air on an industrial scale
(ii) State how a lighted splint can be used to distinguish between samples of oxygen and nitrogen.
(c)(i) Give one reason why bauxite is usually preferred as the ore for the extraction of aluminium.
(ii). List two main impurities. usually present in bauxite.
(iii) State the function of sodium hydroxide solution in the extraction of aluminium from its ore.
(iv) Explain briefly why it is difficult to extract aluminium by chemical reduction of aluminium oxide
(v) Write an equation for the reaction of aluminium oxide with aqueous sodium hydroxide.
(d) (i) The melting and boiling points of sodium chloride are 801 °C and 141.3 °C respectively. Explain briefly why sodium chloride does not conduct electricity at 25°C but does so between 801 °C and 1413 °C.
(ii) State the reason why sodium metal is stored under paraffin oil in the laboratory.
(e)(i) State what would be observed when aqueous sodium trioxocarbonate(IV) is added to a solution containing iron (III) ions
(ii) Write a balanced equation for the reaction in (e)(i).
Bayanin Amsa
None
Tambaya 53 Rahoto
(a) What are nucleons?
(b) State Graham's law of diffusion.
(c) Explain briefly why aluminium does not corrode easily.
(d) State three examples of periodic properties.
(e) State two reasons why real gases deviate from ideal gas behaviour.
(f) List three uses of fractional distillation in industry.
(g) What factors determine the selective discharge of ions at the electrodes during electrolysis?
(h) State the type of reaction represented by each of the following equations:
(i) C\(_2\)H\(_6\) + Br\(_2\) ---> C\(_2\)H\(_5\)Br + HBr;
(ii) C\(_2\)H\(_4\) + .Br\(_2\) ---> CH\(_4\)Br\(_2\)
(i) Name the products formed when butane burns in limited supply of air.
(j) List three methods of separating a solid from a liquid.
(a) Nucleons are the particles found in the nucleus of an atom, that is, the protons and neutrons.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass).
\[\text{rate} \propto \frac{1}{\sqrt{\rho}}\]
(c) Aluminium does not corrode easily because it forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide layer sticks firmly to the metal and prevents air and moisture from reaching the metal underneath, protecting it from further attack.
(d) Three periodic properties: atomic radius, ionization energy and electronegativity (electron affinity and metallic/non-metallic character are also acceptable).
(e) Two reasons real gases deviate from ideal behaviour:
- The molecules of a real gas have an appreciable (finite) volume, which is not negligible compared with the volume of the container, especially at high pressure.
- There are intermolecular forces of attraction between real gas molecules, especially at high pressure and low temperature.
(f) Three industrial uses of fractional distillation: refining of crude petroleum into its various fractions; separation of liquid air into oxygen and nitrogen; separation of ethanol from a water-ethanol mixture (rectification of spirits).
(g) Factors that determine selective discharge of ions during electrolysis: the position of the ion in the electrochemical (discharge) series, the concentration of the ion in the electrolyte, and the nature of the electrode used.
(h) Types of reaction:
- (i) \(C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr\) is a substitution reaction.
- (ii) \(C_2H_4 + Br_2 \rightarrow C_2H_4Br_2\) is an addition reaction.
(i) When butane burns in a limited supply of air, incomplete combustion occurs, giving carbon(II) oxide (and carbon/soot) and water vapour.
(j) Three methods of separating a solid from a liquid: filtration, evaporation and decantation (centrifugation and crystallization are also acceptable).
Bayanin Amsa
(a) Nucleons are the particles found in the nucleus of an atom, that is, the protons and neutrons.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass).
\[\text{rate} \propto \frac{1}{\sqrt{\rho}}\]
(c) Aluminium does not corrode easily because it forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide layer sticks firmly to the metal and prevents air and moisture from reaching the metal underneath, protecting it from further attack.
(d) Three periodic properties: atomic radius, ionization energy and electronegativity (electron affinity and metallic/non-metallic character are also acceptable).
(e) Two reasons real gases deviate from ideal behaviour:
- The molecules of a real gas have an appreciable (finite) volume, which is not negligible compared with the volume of the container, especially at high pressure.
- There are intermolecular forces of attraction between real gas molecules, especially at high pressure and low temperature.
(f) Three industrial uses of fractional distillation: refining of crude petroleum into its various fractions; separation of liquid air into oxygen and nitrogen; separation of ethanol from a water-ethanol mixture (rectification of spirits).
(g) Factors that determine selective discharge of ions during electrolysis: the position of the ion in the electrochemical (discharge) series, the concentration of the ion in the electrolyte, and the nature of the electrode used.
(h) Types of reaction:
- (i) \(C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr\) is a substitution reaction.
- (ii) \(C_2H_4 + Br_2 \rightarrow C_2H_4Br_2\) is an addition reaction.
(i) When butane burns in a limited supply of air, incomplete combustion occurs, giving carbon(II) oxide (and carbon/soot) and water vapour.
(j) Three methods of separating a solid from a liquid: filtration, evaporation and decantation (centrifugation and crystallization are also acceptable).
Tambaya 54 Rahoto
(a) (i) Draw the structures of the isomers of the alkene with molecular formurat C\(_4\)H\(_8\)
(ii) State the class of alkanols to which each of the following compounds belongs:
I. CH\(_3\)C(CH\(_3\))\(_2\)OH;
II. CH\(_3\)CH(CH\(_3\))CH\(_2\)OH;
III. CH\(_3\)CH\(_2\)CH(CH\(_3\))OH.
(b) (i) Write the formulae of the products formed in the following reactions:
I. CH\(_3\)CH\(_2\)COOH \(\frac{K_{(s)}}{}\)
II. CH\(_3\)CH\(_2\)COOH. \(\frac{C_4H_6OH, heat}{Conc.H_2SO_4}\)
III. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)OH \(\frac{H^+/KMnO_4}{(excess)}\)
(ii) Name the major product(s) of each of the reactions in (b)(i).
(c) A gaseous hydrocarbon R of mass 7.0 g occupies a volume of 2.24 dm\(^3\) at s. t.p. If the percentage composition by mass of hydrogen is 14.3, determine its:
(i) empirical formula;
(ii) molecular formula. [ H = 1.00, C = 12.0, Molar volume of gas at s.t.p, = 22.4 dm\(^3\) ]
(d) Define structural isomerism.
(a)(i) Structures of the isomers of the alkene C\(_4\)H\(_8\)
The alkene of molecular formula C\(_4\)H\(_8\) exists as the following isomers (three structural isomers, of which but-2-ene shows an additional pair of geometric isomers):
Thus the structural isomers are but-1-ene, but-2-ene and 2-methylprop-1-ene; but-2-ene occurs as cis- and trans-forms.
(a)(ii) Class of alkanol
| Compound | Structure | Class |
| I | CH\(_3\)C(CH\(_3\))\(_2\)OH (the \(-\)OH carbon is joined to three other carbons) | Tertiary (3°) alkanol |
| II | CH\(_3\)CH(CH\(_3\))CH\(_2\)OH (the \(-\)OH carbon is joined to one other carbon) | Primary (1°) alkanol |
| III | CH\(_3\)CH\(_2\)CH(CH\(_3\))OH (the \(-\)OH carbon is joined to two other carbons) | Secondary (2°) alkanol |
(b)(i) Formulae of the products
I. Propanoic acid with potassium metal:
\[2CH_3CH_2COOH + 2K \rightarrow 2CH_3CH_2COOK + H_2\]Products: CH\(_3\)CH\(_2\)COOK and H\(_2\).
II. Propanoic acid with butanol (C\(_4\)H\(_9\)OH) and concentrated H\(_2\)SO\(_4\), heat (esterification):
\[CH_3CH_2COOH + C_4H_9OH \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3CH_2COOC_4H_9 + H_2O\]Products: CH\(_3\)CH\(_2\)COOC\(_4\)H\(_9\) and H\(_2\)O.
III. Butan-1-ol oxidised by acidified KMnO\(_4\) (excess) — the primary alkanol is oxidised fully to the carboxylic acid:
\[CH_3CH_2CH_2CH_2OH \xrightarrow{H^+/KMnO_4\,(excess)} CH_3CH_2CH_2COOH\]Product: CH\(_3\)CH\(_2\)CH\(_2\)COOH.
(b)(ii) Names of the major product(s)
- I. Potassium propanoate (and hydrogen gas)
- II. Butyl propanoate (and water)
- III. Butanoic acid
(c) Empirical and molecular formula of hydrocarbon R
Given: mass of R = 7.0 g, volume = 2.24 dm\(^3\) at s.t.p., % H by mass = 14.3, molar volume = 22.4 dm\(^3\), H = 1.00, C = 12.0.
(i) Empirical formula
Since R is a hydrocarbon, % carbon = \(100 - 14.3 = 85.7\%\).
| Element | C | H |
| % by mass | 85.7 | 14.3 |
| Moles = %÷A\(_r\) | \(\frac{85.7}{12.0}=7.14\) | \(\frac{14.3}{1.0}=14.3\) |
| Divide by smaller (7.14) | \(\frac{7.14}{7.14}=1\) | \(\frac{14.3}{7.14}=2\) |
\[\therefore \text{Empirical formula} = CH_2\]
(ii) Molecular formula
Number of moles of R:
\[n = \frac{\text{volume at s.t.p.}}{\text{molar volume}} = \frac{2.24}{22.4} = 0.1\ \text{mol}\]Molar mass:
\[M_r = \frac{\text{mass}}{\text{moles}} = \frac{7.0}{0.1} = 70\ \text{g mol}^{-1}\]Let the molecular formula be \((CH_2)_n\). The empirical formula mass \(= 12.0 + 2(1.0) = 14\).
\[14n = 70 \quad\Rightarrow\quad n = \frac{70}{14} = 5\]\[\therefore \text{Molecular formula} = C_5H_{10}\](d) Structural isomerism
Structural isomerism is the phenomenon in which two or more compounds have the same molecular formula but different structural (arrangement of atoms) formulae, and hence different properties.
Bayanin Amsa
(a)(i) Structures of the isomers of the alkene C\(_4\)H\(_8\)
The alkene of molecular formula C\(_4\)H\(_8\) exists as the following isomers (three structural isomers, of which but-2-ene shows an additional pair of geometric isomers):
Thus the structural isomers are but-1-ene, but-2-ene and 2-methylprop-1-ene; but-2-ene occurs as cis- and trans-forms.
(a)(ii) Class of alkanol
| Compound | Structure | Class |
| I | CH\(_3\)C(CH\(_3\))\(_2\)OH (the \(-\)OH carbon is joined to three other carbons) | Tertiary (3°) alkanol |
| II | CH\(_3\)CH(CH\(_3\))CH\(_2\)OH (the \(-\)OH carbon is joined to one other carbon) | Primary (1°) alkanol |
| III | CH\(_3\)CH\(_2\)CH(CH\(_3\))OH (the \(-\)OH carbon is joined to two other carbons) | Secondary (2°) alkanol |
(b)(i) Formulae of the products
I. Propanoic acid with potassium metal:
\[2CH_3CH_2COOH + 2K \rightarrow 2CH_3CH_2COOK + H_2\]Products: CH\(_3\)CH\(_2\)COOK and H\(_2\).
II. Propanoic acid with butanol (C\(_4\)H\(_9\)OH) and concentrated H\(_2\)SO\(_4\), heat (esterification):
\[CH_3CH_2COOH + C_4H_9OH \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3CH_2COOC_4H_9 + H_2O\]Products: CH\(_3\)CH\(_2\)COOC\(_4\)H\(_9\) and H\(_2\)O.
III. Butan-1-ol oxidised by acidified KMnO\(_4\) (excess) — the primary alkanol is oxidised fully to the carboxylic acid:
\[CH_3CH_2CH_2CH_2OH \xrightarrow{H^+/KMnO_4\,(excess)} CH_3CH_2CH_2COOH\]Product: CH\(_3\)CH\(_2\)CH\(_2\)COOH.
(b)(ii) Names of the major product(s)
- I. Potassium propanoate (and hydrogen gas)
- II. Butyl propanoate (and water)
- III. Butanoic acid
(c) Empirical and molecular formula of hydrocarbon R
Given: mass of R = 7.0 g, volume = 2.24 dm\(^3\) at s.t.p., % H by mass = 14.3, molar volume = 22.4 dm\(^3\), H = 1.00, C = 12.0.
(i) Empirical formula
Since R is a hydrocarbon, % carbon = \(100 - 14.3 = 85.7\%\).
| Element | C | H |
| % by mass | 85.7 | 14.3 |
| Moles = %÷A\(_r\) | \(\frac{85.7}{12.0}=7.14\) | \(\frac{14.3}{1.0}=14.3\) |
| Divide by smaller (7.14) | \(\frac{7.14}{7.14}=1\) | \(\frac{14.3}{7.14}=2\) |
\[\therefore \text{Empirical formula} = CH_2\]
(ii) Molecular formula
Number of moles of R:
\[n = \frac{\text{volume at s.t.p.}}{\text{molar volume}} = \frac{2.24}{22.4} = 0.1\ \text{mol}\]Molar mass:
\[M_r = \frac{\text{mass}}{\text{moles}} = \frac{7.0}{0.1} = 70\ \text{g mol}^{-1}\]Let the molecular formula be \((CH_2)_n\). The empirical formula mass \(= 12.0 + 2(1.0) = 14\).
\[14n = 70 \quad\Rightarrow\quad n = \frac{70}{14} = 5\]\[\therefore \text{Molecular formula} = C_5H_{10}\](d) Structural isomerism
Structural isomerism is the phenomenon in which two or more compounds have the same molecular formula but different structural (arrangement of atoms) formulae, and hence different properties.
Tambaya 55 Rahoto
(a)(i) Determine the oxidation number of sulphur in Na\(_2\)S\(_2\)O\(_3\)
(ii) Name the allotropes of sulphur.
(iii) State two ways in which the structure of graphite and diamond are similar.
(b)(i) Name two green-house gases.
(ii) State one effect of an increased level of green-house gases on the environment.
(iii) State one source from which nitrogen (I) oxide is released into the environment.
(iv) Write a chemical equation to show the effect of heat on each of the following compounds: I. KNO\(_{3(s)}\) II. AgNO\(_{3(s)}\)
(C)(i)Describe briefly how pure crystals of calcium chloride could be obtained from a solution of calcium chloride
ii) Explain briefly each of the following observations:
I. ammonia gas is highly soluble in water;
II. boiling ploint of chlorine is lower than that of iodine
(d) Consider the reaction represented by the following equation:- 2NaCl + H\(_2\)SO\(_{4(s)}\) \(\to\) Na\(_2\)SO\(_{4(s)}\) + 2HCl\(_{(g)}\)
Calculate the volume of HCl gas that can be obtained at s.t.p. from 5.85 g of sodium. chloride. [ Na = 23.0, Cl = 35.5, Molar volume of gas at s.t.p. = 22.4 dm\(^3\)]
(a)(i) Oxidation number of sulphur in \(Na_2S_2O_3\). Let it be \(x\); Na is \(+1\), O is \(-2\), and the compound is neutral:
\[2(+1) + 2x + 3(-2) = 0\]
\[2 + 2x - 6 = 0 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = +2\]
Sulphur has an average oxidation number of +2.
(ii) The allotropes of sulphur are rhombic sulphur and monoclinic sulphur (plastic sulphur is also sometimes listed).
(iii) Two ways in which graphite and diamond are similar: both are allotropes of carbon (made up of carbon atoms only), and both are giant covalent (macromolecular) structures with very high melting points.
(b)(i) Two greenhouse gases: carbon(IV) oxide (\(CO_2\)) and methane (\(CH_4\)). (Water vapour and CFCs are also acceptable.)
(ii) One effect of an increased level of greenhouse gases: global warming (a rise in the average atmospheric temperature, leading to melting of ice caps and rising sea levels).
(iii) One source of nitrogen(I) oxide, \(N_2O\): bacterial action on nitrogenous fertilizers in the soil (or from vehicle exhausts).
(iv) Effect of heat:
I. \[2KNO_3 \rightarrow 2KNO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) To obtain pure crystals of calcium chloride from its solution: evaporate the solution carefully to concentrate it to the point of crystallization, then allow it to cool so that crystals form; filter off the crystals and dry them (calcium chloride is deliquescent, so drying should be done quickly in a desiccator).
(ii) I. Ammonia is highly soluble in water because ammonia molecules are polar and form hydrogen bonds with water (and react with water to form ammonium hydroxide).
II. The boiling point of chlorine is lower than that of iodine because chlorine molecules are smaller and lighter, so the van der Waals (intermolecular) forces between them are weaker than the stronger van der Waals forces between the larger iodine molecules; less energy is needed to separate chlorine molecules.
(d) \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\). From 5.85 g of NaCl:
Molar mass of \(NaCl = 23 + 35.5 = 58.5\,g\,mol^{-1}\)
\[\text{moles of } NaCl = \frac{5.85}{58.5} = 0.1\,mol\]
From the equation, 2 moles NaCl give 2 moles HCl (1:1), so moles of \(HCl = 0.1\,mol\).
\[\text{volume at s.t.p.} = 0.1 \times 22.4 = 2.24\,dm^3\]
Bayanin Amsa
(a)(i) Oxidation number of sulphur in \(Na_2S_2O_3\). Let it be \(x\); Na is \(+1\), O is \(-2\), and the compound is neutral:
\[2(+1) + 2x + 3(-2) = 0\]
\[2 + 2x - 6 = 0 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = +2\]
Sulphur has an average oxidation number of +2.
(ii) The allotropes of sulphur are rhombic sulphur and monoclinic sulphur (plastic sulphur is also sometimes listed).
(iii) Two ways in which graphite and diamond are similar: both are allotropes of carbon (made up of carbon atoms only), and both are giant covalent (macromolecular) structures with very high melting points.
(b)(i) Two greenhouse gases: carbon(IV) oxide (\(CO_2\)) and methane (\(CH_4\)). (Water vapour and CFCs are also acceptable.)
(ii) One effect of an increased level of greenhouse gases: global warming (a rise in the average atmospheric temperature, leading to melting of ice caps and rising sea levels).
(iii) One source of nitrogen(I) oxide, \(N_2O\): bacterial action on nitrogenous fertilizers in the soil (or from vehicle exhausts).
(iv) Effect of heat:
I. \[2KNO_3 \rightarrow 2KNO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) To obtain pure crystals of calcium chloride from its solution: evaporate the solution carefully to concentrate it to the point of crystallization, then allow it to cool so that crystals form; filter off the crystals and dry them (calcium chloride is deliquescent, so drying should be done quickly in a desiccator).
(ii) I. Ammonia is highly soluble in water because ammonia molecules are polar and form hydrogen bonds with water (and react with water to form ammonium hydroxide).
II. The boiling point of chlorine is lower than that of iodine because chlorine molecules are smaller and lighter, so the van der Waals (intermolecular) forces between them are weaker than the stronger van der Waals forces between the larger iodine molecules; less energy is needed to separate chlorine molecules.
(d) \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\). From 5.85 g of NaCl:
Molar mass of \(NaCl = 23 + 35.5 = 58.5\,g\,mol^{-1}\)
\[\text{moles of } NaCl = \frac{5.85}{58.5} = 0.1\,mol\]
From the equation, 2 moles NaCl give 2 moles HCl (1:1), so moles of \(HCl = 0.1\,mol\).
\[\text{volume at s.t.p.} = 0.1 \times 22.4 = 2.24\,dm^3\]
