WAEC SSCE - Chemistry - 1993
Pergunta 1 Relatório
Which of the following pieces of deposited if a current of 10A was passed through a solution of copper(ll) salt for 965 seconds? (1F = 96500 C)
Detalhes da Resposta
Pergunta 2 Relatório
Producer gas is a mixture of
Detalhes da Resposta
Producer gas is a mixture of nitrogen and carbon (II) oxide. Producer gas is a fuel gas that is produced by heating a carbon-based fuel with a limited amount of oxygen or air. This process, called gasification, produces a mixture of gases that are collectively known as producer gas. The primary components of producer gas are nitrogen, which makes up the majority of the gas, and carbon (II) oxide, which is the second most abundant component. Other minor components may include hydrogen, methane, and other gases, depending on the type of fuel and the gasification process used. So, in summary, producer gas is mainly made up of nitrogen and carbon (II) oxide, with some minor amounts of other gases.
Pergunta 3 Relatório
Alkanes are used mainly
Detalhes da Resposta
Alkanes are primarily used as domestic and industrial fuels. Alkanes are a group of hydrocarbons that contain only single bonds between carbon atoms and are therefore relatively inert. They are obtained from crude oil and natural gas and can be used as fuels for heating and cooking in households and industries. In addition to their use as fuels, alkanes are also used in the production of plastics, where they serve as starting materials for the synthesis of various types of plastics. They are also used in the hydrogenation of oils to produce margarine, and as fine chemicals, where they are used as solvents, lubricants, and as a starting material in the synthesis of pharmaceuticals and other specialty chemicals. However, these applications are not as significant as their use as fuels.
Pergunta 4 Relatório
Sulphur burns in air to form
Detalhes da Resposta
Sulphur burns in air to form an acidic oxide, which is sulfur dioxide (SO2). When sulfur burns in air, it reacts with oxygen to form sulfur dioxide gas. Sulfur dioxide is an acidic oxide because it reacts with water in the air to form sulfurous acid, which has a pH lower than 7. This means that it can react with bases to form salts and water. Sulfur dioxide is also a major air pollutant that can cause respiratory problems and acid rain when it dissolves in rainwater. In summary, when sulfur burns in air, it forms sulfur dioxide, which is an acidic oxide that can react with water to form sulfurous acid and can contribute to air pollution and acid rain.
Pergunta 5 Relatório
In which of the following changes is ∆S negative ?
Detalhes da Resposta
The change in entropy (∆S) is a measure of the disorder or randomness of a system. A negative ∆S indicates that there is a decrease in disorder or randomness. In the given options: - The first option is a phase change from liquid to solid, which is typically associated with a decrease in entropy. So, ∆S is negative. - The second option involves the formation of a gas, which is typically associated with an increase in entropy. So, ∆S is positive. - The third option involves the decomposition of a solid into gases, which is typically associated with an increase in entropy. So, ∆S is positive. - The fourth option is the dissociation of a salt into its ions, which is typically associated with an increase in entropy. So, ∆S is positive. - The fifth option involves the formation of a gas, which is typically associated with an increase in entropy. So, ∆S is positive. Therefore, the only change in which ∆S is negative is H2O(l) → H2O(s).
Pergunta 6 Relatório
When ammonium chloride is dissolved in water in a test tube, the tube feels colder showing that
Detalhes da Resposta
The process is endothermic when ammonium chloride is dissolved in water in a test tube. When ammonium chloride is dissolved in water, it absorbs heat from the surroundings to break apart the ionic bonds between the ammonium and chloride ions. This means that the process requires energy to take place and absorbs heat from the surrounding environment. As a result, the test tube containing the solution feels colder to the touch, indicating that heat is being transferred from the surroundings to the solution. In summary, the cooling effect observed when ammonium chloride is dissolved in water in a test tube is due to the endothermic process of breaking apart ionic bonds, which absorbs heat from the surroundings.
Pergunta 7 Relatório
Which of the following separation is routinely applied in the petroleum industry
Detalhes da Resposta
The separation technique routinely applied in the petroleum industry is fractional distillation. Fractional distillation is a separation technique that separates a mixture of liquids into its individual components based on their boiling points. In the petroleum industry, crude oil, which is a mixture of various hydrocarbons, is heated and passed through a column that has several trays. Each tray has a different temperature, with the temperature increasing from the bottom to the top of the column. The hydrocarbons with lower boiling points vaporize and rise up the column, while those with higher boiling points remain as liquids and collect at the bottom. The vaporized hydrocarbons are then condensed and collected as fractions at different levels of the column. This process allows for the separation of crude oil into its individual components, such as gasoline, diesel, and kerosene.
Pergunta 8 Relatório
Which of the following is correct about the electrons in the 1s orbital? They
Detalhes da Resposta
The correct option about the electrons in the 1s orbital is that they are closest to the nucleus. The 1s orbital is the lowest energy level in an atom, and it is located closest to the nucleus, which has a positive charge. The closer an electron is to the nucleus, the stronger the attractive force between them, so electrons in the 1s orbital are more strongly attracted to the nucleus than those in other orbitals. This makes them harder to remove from the atom, and they require more energy to be ionized. The other options are not correct: electrons in the 1s orbital do not necessarily have the same spin, they are not easier to remove than those in the 2p orbital, they do not have the same energy as those in the 2s orbital, and they do not have different magnetic quantum numbers.
Pergunta 9 Relatório
Which of the following solids has a network structure?
Detalhes da Resposta
Diamond has a network structure. In a network structure, all the atoms in the solid are covalently bonded to each other forming a giant molecule. In diamond, each carbon atom is covalently bonded to four other carbon atoms forming a tetrahedral structure, creating a three-dimensional network of carbon atoms bonded together. This network structure is responsible for diamond's unique physical and chemical properties such as high hardness and thermal conductivity.
Pergunta 10 Relatório
Which of the following pairs illustrates isotopy?
Detalhes da Resposta
The pair that illustrates isotopy is hydrogen and deuterium. Isotopy refers to the phenomenon where two atoms of different elements have the same atomic number but different atomic masses due to a different number of neutrons in their nuclei. Hydrogen and deuterium are both elements with atomic number 1, but deuterium has one neutron in its nucleus in addition to the proton, while hydrogen has no neutron. Therefore, they are isotopes of each other. In contrast, the other options are not examples of isotopy. But-l-ene and but-2-ene are structural isomers, ortho-hydrogen and para-hydrogen are different spin isomers, oxygen and ozone are allotropes of the same element, and alpha and beta particles are different types of subatomic particles.
Pergunta 11 Relatório
Some precious stones such as ruby and sapphire consist of aluminum oxide coloured by traces of the oxides of
Detalhes da Resposta
Precious stones such as ruby and sapphire consist of aluminum oxide (Al2O3) colored by traces of transition metal oxides. The transition metal ions, such as chromium (Cr) and iron (Fe), replace some of the aluminum ions in the crystal lattice of the aluminum oxide, and give the stones their characteristic colors. For example, chromium gives rubies their red color and iron gives sapphires their blue color. Therefore, the correct option is "transition metals".
Pergunta 12 Relatório
Starch undergoes complete hydrolysis to produce
Detalhes da Resposta
Starch is a polysaccharide, which is a type of carbohydrate made up of many glucose units linked together. Complete hydrolysis of starch breaks down these glucose units into their monomers, resulting in the production of glucose. Therefore, the correct answer is glucose.
Pergunta 13 Relatório
Zinc displaces copper from an aqueous solution of copper (ll) salt because
Detalhes da Resposta
Zinc displaces copper from an aqueous solution of copper (II) salt because zinc is more reactive than copper. Reactivity is the tendency of a metal to lose electrons and form positive ions. Metals that have a higher tendency to lose electrons are more reactive than those with a lower tendency. Zinc is more reactive than copper, which means it has a greater tendency to lose electrons and form positive ions. When zinc is added to an aqueous solution of copper (II) salt, such as copper (II) sulfate, the zinc atoms lose electrons to form zinc ions. These zinc ions then react with the copper ions in the solution by displacing them from the copper (II) sulfate. The displaced copper ions then form a solid layer of copper on the surface of the zinc metal, which can be observed as a red-brown coating on the zinc. Therefore, the reason why zinc displaces copper from an aqueous solution of copper (II) salt is that zinc is more reactive than copper, and it has a greater tendency to lose electrons and form positive ions.
Pergunta 14 Relatório
What volume of distilled water should be added to 400cm3 of 2.0 mole dm-3 \(H_{2}SO_{4}\) to obtain 0.20 mole dm-3 of solution>
Detalhes da Resposta
This is a dilution problem in chemistry. The formula for dilution is C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, respectively, and C2 and V2 are the final concentration and volume, respectively. We can rearrange the formula to solve for V2: V2 = (C1V1)/C2 Substituting the values given in the question, we get: V2 = (2.0 mol dm-3 x 400 cm3)/0.20 mol dm-3 Simplifying the expression, we get: V2 = 4,000 cm3 Therefore, 4,000cm3 of solution is needed. However, the question is asking for the volume of distilled water to add, so we need to subtract the initial volume from the final volume to get the volume of distilled water needed: Volume of distilled water = V2 - V1 = 4,000 cm3 - 400 cm3 = 3,600 cm3 Hence, the correct option is 3,600 cm3.
Pergunta 16 Relatório
Carbon is often deposited in the exhaust-pipe of cars because of the
Detalhes da Resposta
The carbon deposits in the exhaust pipe of cars are caused by the incomplete combustion of petrol. When petrol is burned, it reacts with oxygen to produce carbon dioxide and water vapor. However, if there is not enough oxygen, the petrol can only partially burn, leading to the formation of carbon particles (soot) that get carried through the exhaust system and eventually settle in the exhaust pipe. The presence of additives in petrol or contamination with diesel would not directly cause carbon deposits in the exhaust pipe.
Pergunta 17 Relatório
The following gases decolorize bromine water except
Detalhes da Resposta
Bromine water is a reddish-brown solution of bromine in water. When an unsaturated hydrocarbon gas (C2H4, C2H2, C3H4, C3H6) is passed through the bromine water, the double or triple bond of the hydrocarbon is broken, and the bromine is added across the double or triple bond. This results in the decolorization of the bromine water. On the other hand, when a saturated hydrocarbon gas such as ethane (C2H6) is passed through bromine water, there is no double or triple bond to break, and therefore no bromine is added to the molecule. Thus, bromine water is not decolorized by ethane. Therefore, the correct answer to the question is C2H6.
Pergunta 18 Relatório
In which of the following pairs are both substances deliquescent?
Detalhes da Resposta
Pergunta 19 Relatório
Consider the reaction represented by the following equation CaCI2 (aq) + H 2C2O 4(aq) ⇌ CaC 2O4(s) + 2HCI(aq). Which of the following would dissolve the precipitate of CaC 2O4 formed?
Detalhes da Resposta
The formation of a precipitate in this reaction indicates that CaC 2O4 is insoluble in water. In order to dissolve the precipitate, we need to find a substance that can react with it to form a soluble compound. , adding concentrated hydrochloric acid, is the correct answer. This is because the reaction between CaC 2O4 and HCl produces soluble calcium chloride (CaCl 2) and carbon dioxide (CO 2). The balanced chemical equation for this reaction is: CaC 2O4 + 4HCl → CaCl 2 + 2CO 2 + 2H 2O The carbon dioxide gas will escape as bubbles, leaving behind a solution of CaCl 2, which is soluble in water. The other options will not dissolve the CaC 2O4 precipitate. Stirring the mixture vigorously will not change the chemical nature of the substances in the solution. Adding more calcium chloride solution will not react with the CaC 2O4 precipitate since it is already present in excess. Increasing the concentration of ethanedioic acid will not react with CaC 2O4 either, as it is a weak acid and cannot displace the strong acid HCl. Adding distilled water will not dissolve the precipitate since CaC 2O4 is insoluble in water.
Pergunta 20 Relatório
Which of the following gases are diatomic except
Detalhes da Resposta
A diatomic molecule is a molecule composed of two atoms, whether they are of the same element or not. Therefore, the gases that are diatomic have molecules that consist of two atoms. Nitrogen, hydrogen, fluorine, and oxygen are diatomic gases, which means they exist as molecules composed of two atoms of the same element. On the other hand, helium is a monatomic gas, meaning it exists as individual atoms and not as molecules consisting of multiple atoms. Therefore, helium is the gas among the options given that is not diatomic.
Pergunta 21 Relatório
Which of the following indicates the correct increasing order of oxidizing power of the halogens?
Detalhes da Resposta
The oxidizing power of a halogen is related to its ability to gain electrons. Halogens have seven electrons in their outermost shell and require one more electron to complete their octet configuration. The process of gaining an electron involves reduction of the halogen. The halogens are arranged in the periodic table in order of increasing atomic number, so the trend in oxidizing power can be related to this arrangement. As we move down the group, the atomic radius of the halogens increases, and the valence electrons become farther away from the nucleus, which means it becomes harder to attract electrons and so the oxidizing power decreases. Therefore, the correct increasing order of oxidizing power of the halogens is: I < Br < Cl < F. Hence, is correct.
Pergunta 22 Relatório
Which of the following will not affect the degree of hardness in a sample of temporarily hard water?
Detalhes da Resposta
Pergunta 23 Relatório
Which of the following substances is a peroxide?
Detalhes da Resposta
The substance that is a peroxide is Na2O2. A peroxide is a compound that contains an oxygen-oxygen single bond, also known as a peroxide linkage. Na2O2 is a peroxide because it contains the peroxide linkage between the two oxygen atoms in the formula. In contrast, CuO, Pb3O4, Fe2O3, and AI2O3 do not contain a peroxide linkage and are not peroxides. In summary, Na2O2 is the only substance in the given options that is a peroxide because it contains an oxygen-oxygen single bond, which is a defining characteristic of peroxides.
Pergunta 24 Relatório
An alkene may be converted to an alkane by
Detalhes da Resposta
An alkene is a hydrocarbon that contains a carbon-carbon double bond, whereas an alkane is a hydrocarbon that contains only single bonds between the carbon atoms. The process that converts an alkene to an alkane is called hydrogenation, which involves the addition of hydrogen gas (H2) across the carbon-carbon double bond. In this reaction, the double bond is broken, and the two carbon atoms are each bonded to a hydrogen atom, forming a single bond. This reaction is commonly carried out in the presence of a metal catalyst, such as platinum, palladium, or nickel, which speeds up the reaction by facilitating the adsorption of hydrogen molecules onto the surface of the metal. Therefore, the correct answer is hydrogenation.
Pergunta 25 Relatório
Copper metal will react with concentrated trioxonitrate (V) acid to give?
Detalhes da Resposta
Pergunta 26 Relatório
Which of the following compounds of tin is a strong reducing agent?
Detalhes da Resposta
The compound SnH4 (tin(IV) hydride) is a strong reducing agent. This is because it can easily donate its hydrogen atoms, which act as reducing agents. In other words, SnH4 can easily give up electrons to other compounds, causing them to be reduced. This property of SnH4 makes it useful in a variety of chemical reactions, particularly in organic chemistry. On the other hand, SnCl2, SnCl4, SnO2, and Sn(OH)2 are not strong reducing agents. SnCl2 and SnCl4 are both tin(IV) compounds and do not contain any hydrogen atoms that can act as reducing agents. SnO2 and Sn(OH)2 contain oxygen and hydroxide groups respectively, which do not have a tendency to give up electrons and are not strong reducing agents.
Pergunta 27 Relatório
Trioxosulphate (lV) acid is not stored for a long period because it
Detalhes da Resposta
Trioxosulphate (lV) acid, also known as sulfuric acid, is not stored for a long period because it is an unstable and easily decomposed compound. This means that over time, it breaks down into other substances, making it less effective for its intended use. Additionally, the decomposition of sulfuric acid can lead to the formation of hazardous byproducts, which can be dangerous if not properly handled. Therefore, it is essential to use and store sulfuric acid with caution and care.
Pergunta 28 Relatório
When the trioxonitrate (IV) salt of an alkali metal Y is heated, the formula of the residue is
Detalhes da Resposta
Pergunta 29 Relatório
What does X atand for in the following expression? X = Log(10)(H+)
Detalhes da Resposta
In the expression X = Log(10)(H+), the variable X stands for the degree of acidity, commonly known as pH. The pH of a solution is a measure of its hydrogen ion concentration. The logarithmic function in this expression relates the hydrogen ion concentration to the pH scale, which ranges from 0 to 14. A pH of 7 is considered neutral, while values below 7 are acidic and values above 7 are basic. The pH scale is widely used in chemistry and biology to describe the acidity or basicity of a solution, and it is an important parameter in many chemical reactions and biological processes.
Pergunta 30 Relatório
Examples of polymers include the following except
Detalhes da Resposta
Polymers are large molecules made up of repeating subunits called monomers. They can be natural or synthetic. Starch, nylon, wool, and perspex are examples of polymers because they are made up of repeating units of monomers. However, glass is not a polymer because it is not made up of repeating units of monomers. Instead, it is an amorphous solid that is made up of silicon dioxide (SiO2) molecules that are linked together in a random arrangement. Therefore, the correct answer is glass.
Pergunta 31 Relatório
The product of the reaction between ethanol and excess acidified K2Cr2O2 is
Detalhes da Resposta
The product of the reaction between ethanol and excess acidified K2Cr2O2 is ethanoic acid. The reaction is an oxidation reaction where ethanol is oxidized to ethanoic acid and K2Cr2O2 is reduced to Cr3+. The acidified K2Cr2O2 acts as an oxidizing agent and removes hydrogen from the ethanol molecule, producing ethanoic acid. Therefore, the correct answer is ethanoic acid.
Pergunta 32 Relatório
The component of air that removed when air is bubbled into alkaline pyrogallol solution is
Detalhes da Resposta
The component of air that is removed when air is bubbled into alkaline pyrogallol solution is oxygen. Alkaline pyrogallol solution is a reagent that is used to absorb oxygen from a gas mixture. When air is bubbled into this solution, the oxygen present in the air reacts with pyrogallol and gets absorbed, forming a brown precipitate of the oxidized pyrogallol. The other components of air, such as nitrogen, carbon dioxide, water vapor, and other gases, remain unchanged and do not react with pyrogallol. Therefore, the correct answer is: - Oxygen
Pergunta 33 Relatório
Chlorine is is used in water treatment as
Detalhes da Resposta
Chlorine is used in water treatment as a germicide. Water treatment is a process of purifying water to make it safe for human consumption. One of the most common methods of water treatment involves the use of chlorine. Chlorine is added to water to kill harmful microorganisms such as bacteria and viruses that can cause waterborne diseases like cholera, typhoid fever, and dysentery. Chlorine works by disrupting the cellular function of microorganisms and breaking down their cell walls. This makes them unable to reproduce and eventually causes them to die off. Chlorine is effective against a wide range of microorganisms and is relatively inexpensive compared to other water treatment methods. In summary, chlorine is used in water treatment as a germicide because it is effective at killing harmful microorganisms in water and making it safe for human consumption.
Pergunta 34 Relatório
(a) X and Y belong to the same period in the Periodic Table is a group I element while Y belongs to group VII. State which of the elements would
(i) be a good oxidizing agent
(ii) have the smaller atomic volume
(iii) have the higher ionization potential
(b) Explain your answer in (a)(i) above.
X is a Group I element (a metal, e.g. sodium) and Y is a Group VII element (a non-metal, e.g. chlorine); both are in the same period.
(a)
- (i) A good oxidizing agent: Y (the Group VII element).
- (ii) Smaller atomic volume / smaller atomic radius: Y (across a period the nuclear charge rises and the atom contracts).
- (iii) Higher ionization potential (ionization energy): Y.
(b) Explanation for (a)(i): Y (Group VII) has 7 electrons in its outermost shell and needs to gain only one electron to attain a stable octet. It therefore has a strong tendency to accept (gain) electrons from other species; a species that gains electrons is an oxidizing agent. X (Group I) instead loses its single valence electron easily and acts as a reducing agent. Hence Y is the good oxidizing agent.
Detalhes da Resposta
X is a Group I element (a metal, e.g. sodium) and Y is a Group VII element (a non-metal, e.g. chlorine); both are in the same period.
(a)
- (i) A good oxidizing agent: Y (the Group VII element).
- (ii) Smaller atomic volume / smaller atomic radius: Y (across a period the nuclear charge rises and the atom contracts).
- (iii) Higher ionization potential (ionization energy): Y.
(b) Explanation for (a)(i): Y (Group VII) has 7 electrons in its outermost shell and needs to gain only one electron to attain a stable octet. It therefore has a strong tendency to accept (gain) electrons from other species; a species that gains electrons is an oxidizing agent. X (Group I) instead loses its single valence electron easily and acts as a reducing agent. Hence Y is the good oxidizing agent.
Pergunta 35 Relatório
(a)(i) List four characteristic properties of transition metals
(ii) Name two metals that can be extracted from their ore by electrolysis.
(b)(i) Determine the oxidation number of chromium in Cr\(_2\)O\(^{2-}_{7}\)
(ii) State the colour observed on adding a few drops of dilute tetraoxosulphate (VI) acid to the system representedby the following equation: Cr\(_2\)O\(^{2-}_{7(aq)}\) + H\(_2O_{(l)}\) \(\rightleftharpoons\) 2CrO\(^{2-}_{4(aq)}\) + 2H\(^+_{(aq)}\). Explain your answer.
(c)(i) State and explain what would be observed if hydrogen sulphide gas were bubbled into acidified K\(_2\)Cr\(_2\)O\(_7\). Write an equation for the reaction.
(ii) What precaution should be taken to avoid excessive exposure to hydrogen sulphide gas while it is being generated in the laboratory?
Detalhes da Resposta
None
Pergunta 36 Relatório
(a)(i) Define the term polymerization.
(ii) List the three conditions required for the polymerization on of ethene.
(iii) State the property which is common to compounds that can be easily polymerized
(b) Write appropriate equations to show how the following can be obtained from propan-1-ol in the labouratory
(i) propene;
(ii) propylmethanoate. State the type of reaction involved in each case.
(c)(i) A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Determine its molecular formula if its molar mass is 180 (H = 1, C = 12, O = 16)
(ii) Explain why ethanoic acid boils at a much higher temperature than butane even though their molar masses almost equal.
(a)(i) Polymerization: the chemical process in which many small molecules (monomers) join together to form a very large molecule (a polymer).
(a)(ii) Conditions for polymerization of ethene: high pressure, high temperature (moderately raised temperature), and a catalyst (e.g. a trace of oxygen or a Ziegler-Natta catalyst).
(a)(iii) The common property is that such compounds are unsaturated (they contain a carbon-carbon double bond, C=C).
(b) From propan-1-ol (CH3CH2CH2OH):
- (i) Propene, by dehydration with concentrated H2SO4 (or hot Al2O3): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O}\). Type: elimination (dehydration).
- (ii) Propyl methanoate, by esterification with methanoic acid: \(\text{HCOOH} + \text{C}_3\text{H}_7\text{OH} \rightleftharpoons \text{HCOOC}_3\text{H}_7 + \text{H}_2\text{O}\). Type: esterification (condensation).
(c)(i) Molecular formula from 40.0% C, 6.7% H, 53.3% O:
| Element | %/Ar | ratio |
|---|---|---|
| C | 40.0/12 = 3.33 | 1 |
| H | 6.7/1 = 6.7 | 2 |
| O | 53.3/16 = 3.33 | 1 |
Empirical formula = CH2O (mass = 30). \(n = \dfrac{180}{30} = 6\), so the molecular formula is C6H12O6.
(c)(ii) Ethanoic acid molecules contain O-H groups and form strong intermolecular hydrogen bonds (indeed dimers) between molecules, whereas butane, being non-polar, has only weak van der Waals forces. Much more energy is needed to overcome the hydrogen bonds, so ethanoic acid boils at a much higher temperature despite the similar molar mass.
Detalhes da Resposta
(a)(i) Polymerization: the chemical process in which many small molecules (monomers) join together to form a very large molecule (a polymer).
(a)(ii) Conditions for polymerization of ethene: high pressure, high temperature (moderately raised temperature), and a catalyst (e.g. a trace of oxygen or a Ziegler-Natta catalyst).
(a)(iii) The common property is that such compounds are unsaturated (they contain a carbon-carbon double bond, C=C).
(b) From propan-1-ol (CH3CH2CH2OH):
- (i) Propene, by dehydration with concentrated H2SO4 (or hot Al2O3): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O}\). Type: elimination (dehydration).
- (ii) Propyl methanoate, by esterification with methanoic acid: \(\text{HCOOH} + \text{C}_3\text{H}_7\text{OH} \rightleftharpoons \text{HCOOC}_3\text{H}_7 + \text{H}_2\text{O}\). Type: esterification (condensation).
(c)(i) Molecular formula from 40.0% C, 6.7% H, 53.3% O:
| Element | %/Ar | ratio |
|---|---|---|
| C | 40.0/12 = 3.33 | 1 |
| H | 6.7/1 = 6.7 | 2 |
| O | 53.3/16 = 3.33 | 1 |
Empirical formula = CH2O (mass = 30). \(n = \dfrac{180}{30} = 6\), so the molecular formula is C6H12O6.
(c)(ii) Ethanoic acid molecules contain O-H groups and form strong intermolecular hydrogen bonds (indeed dimers) between molecules, whereas butane, being non-polar, has only weak van der Waals forces. Much more energy is needed to overcome the hydrogen bonds, so ethanoic acid boils at a much higher temperature despite the similar molar mass.
Pergunta 37 Relatório
(a) Give one disadvantage of:
(i) hard water
(ii) soft water
(b) Explain why the degree of hardness in a sample of clear lime water is higher than in another sample has that been turned milky by carbon (IV) oxide.
(a) One disadvantage of:
- (i) Hard water: it wastes soap because it forms an insoluble scum with soap before it can lather (it can also fur/scale kettles and boilers).
- (ii) Soft water: it dissolves lead from lead pipes, giving water that can cause lead (plumbo) poisoning (soft water is more corrosive to pipes).
(b) Clear lime water contains dissolved calcium hydroxide, Ca(OH)2, which supplies Ca2+ ions responsible for hardness. When carbon(IV) oxide is bubbled through and the lime water turns milky, the Ca2+ ions are precipitated as insoluble calcium trioxocarbonate(IV):
\[ \text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O} \]Removing Ca2+ from solution as the insoluble carbonate lowers the concentration of hardness-causing ions. Hence the milky sample has a lower degree of hardness than the clear lime water.
Detalhes da Resposta
(a) One disadvantage of:
- (i) Hard water: it wastes soap because it forms an insoluble scum with soap before it can lather (it can also fur/scale kettles and boilers).
- (ii) Soft water: it dissolves lead from lead pipes, giving water that can cause lead (plumbo) poisoning (soft water is more corrosive to pipes).
(b) Clear lime water contains dissolved calcium hydroxide, Ca(OH)2, which supplies Ca2+ ions responsible for hardness. When carbon(IV) oxide is bubbled through and the lime water turns milky, the Ca2+ ions are precipitated as insoluble calcium trioxocarbonate(IV):
\[ \text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O} \]Removing Ca2+ from solution as the insoluble carbonate lowers the concentration of hardness-causing ions. Hence the milky sample has a lower degree of hardness than the clear lime water.
Pergunta 38 Relatório
(a)(i) Distinguish between a strong acid and a concentrated acid.
(ii) What is meant by amphoteric oxide? Give one example.
(b)(i) Describe the manufacture of tetraoxosulphate (VI) acid by contact process.
(ii) Write one equation each to show the action of tetraoxosulphate (VI) acid respectively as a dehydrating agent and as an oxidizing agent.
(iii) Give the reason why tetraoxosulphate classified as a heavy chemical.
(a)(i) Strong acid vs concentrated acid: a strong acid is one that ionises (dissociates) completely into ions in aqueous solution, whereas a concentrated acid is one containing a large amount of acid (solute) in a small amount of water (little water). Strength refers to degree of ionisation; concentration refers to amount of solute per unit volume.
(a)(ii) Amphoteric oxide: an oxide that can react with both acids and bases to form a salt and water, e.g. aluminium oxide, Al2O3 (or ZnO, PbO).
(b)(i) Contact process for H2SO4:
- Sulphur (or a sulphide ore) is burnt in air to give sulphur(IV) oxide: \(\text{S} + \text{O}_2 \to \text{SO}_2\).
- The purified, dry SO2 and excess air are passed over a catalyst of vanadium(V) oxide (V2O5) at about 450 °C and 1-2 atm, oxidising it to sulphur(VI) oxide: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\).
- The SO3 is absorbed in concentrated H2SO4 to form oleum (fuming sulphuric acid): \(\text{SO}_3 + \text{H}_2\text{SO}_4 \to \text{H}_2\text{S}_2\text{O}_7\).
- The oleum is diluted carefully with water to give concentrated H2SO4: \(\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \to 2\text{H}_2\text{SO}_4\).
(b)(ii) Equations:
- As a dehydrating agent (removing the elements of water from sugar): \(\text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{conc. H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O}\).
- As an oxidizing agent (hot concentrated acid with copper): \(\text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}\).
(b)(iii) Tetraoxosulphate(VI) acid is classified as a heavy chemical because it is manufactured and used in very large (industrial) quantities and is relatively cheap and impure.
Detalhes da Resposta
(a)(i) Strong acid vs concentrated acid: a strong acid is one that ionises (dissociates) completely into ions in aqueous solution, whereas a concentrated acid is one containing a large amount of acid (solute) in a small amount of water (little water). Strength refers to degree of ionisation; concentration refers to amount of solute per unit volume.
(a)(ii) Amphoteric oxide: an oxide that can react with both acids and bases to form a salt and water, e.g. aluminium oxide, Al2O3 (or ZnO, PbO).
(b)(i) Contact process for H2SO4:
- Sulphur (or a sulphide ore) is burnt in air to give sulphur(IV) oxide: \(\text{S} + \text{O}_2 \to \text{SO}_2\).
- The purified, dry SO2 and excess air are passed over a catalyst of vanadium(V) oxide (V2O5) at about 450 °C and 1-2 atm, oxidising it to sulphur(VI) oxide: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\).
- The SO3 is absorbed in concentrated H2SO4 to form oleum (fuming sulphuric acid): \(\text{SO}_3 + \text{H}_2\text{SO}_4 \to \text{H}_2\text{S}_2\text{O}_7\).
- The oleum is diluted carefully with water to give concentrated H2SO4: \(\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \to 2\text{H}_2\text{SO}_4\).
(b)(ii) Equations:
- As a dehydrating agent (removing the elements of water from sugar): \(\text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{conc. H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O}\).
- As an oxidizing agent (hot concentrated acid with copper): \(\text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}\).
(b)(iii) Tetraoxosulphate(VI) acid is classified as a heavy chemical because it is manufactured and used in very large (industrial) quantities and is relatively cheap and impure.
Pergunta 39 Relatório
Mention the respective properties of the following allotropes of carbon that account for their uses as indicated:
(a) diamond used for drilling rocks;
(b) diamond used as jewels;
(c) graphite used as electrodes;
(d) graphite used for slowing down neutrons in nuclear reactors;
(e) wood charcoal used in gas masks.
- (a) Diamond for drilling rocks: diamond is extremely hard (the hardest natural substance) because each carbon atom is covalently bonded to four others in a rigid three-dimensional structure, so it can cut and drill hard rocks.
- (b) Diamond as jewels: diamond has a very high refractive index and disperses light strongly, so it sparkles brilliantly when cut, making it valuable as a gem.
- (c) Graphite as electrodes: graphite is a good conductor of electricity (it has delocalised/mobile electrons between its layers) and has a high melting point, so it is used as an inert electrode.
- (d) Graphite for slowing down neutrons in nuclear reactors: graphite is a good moderator; its carbon atoms slow down fast neutrons without absorbing them, and it withstands the high temperatures involved.
- (e) Wood charcoal in gas masks: wood charcoal is highly porous with a very large surface area, so it adsorbs (takes onto its surface) poisonous gases, purifying the air breathed.
Detalhes da Resposta
- (a) Diamond for drilling rocks: diamond is extremely hard (the hardest natural substance) because each carbon atom is covalently bonded to four others in a rigid three-dimensional structure, so it can cut and drill hard rocks.
- (b) Diamond as jewels: diamond has a very high refractive index and disperses light strongly, so it sparkles brilliantly when cut, making it valuable as a gem.
- (c) Graphite as electrodes: graphite is a good conductor of electricity (it has delocalised/mobile electrons between its layers) and has a high melting point, so it is used as an inert electrode.
- (d) Graphite for slowing down neutrons in nuclear reactors: graphite is a good moderator; its carbon atoms slow down fast neutrons without absorbing them, and it withstands the high temperatures involved.
- (e) Wood charcoal in gas masks: wood charcoal is highly porous with a very large surface area, so it adsorbs (takes onto its surface) poisonous gases, purifying the air breathed.
Pergunta 40 Relatório
(a) State one air pollution that causes:
(i) blood poisoning
(ii) acid poisoning
(iii) blackening of the walls of buildings
(b) Mention one major chemical industry in each case which requires the following as raw materials:
(i) petrochemicals;
(ii) cellulose.
(a)
(i) Carbon(II) oxide (carbon monoxide) prevents the oxygenation of the blood.
(ii) Sulphur(IV) oxide dissolves in rain water in the atmosphere to form acid rain.
(iii) Hydrogen sulphide reacts with the lead content of paint to form black lead(II) sulphide, PbS.
(b)
(i) Petrochemical industry.
(ii) Paper industry.
Detalhes da Resposta
(a)
(i) Carbon(II) oxide (carbon monoxide) prevents the oxygenation of the blood.
(ii) Sulphur(IV) oxide dissolves in rain water in the atmosphere to form acid rain.
(iii) Hydrogen sulphide reacts with the lead content of paint to form black lead(II) sulphide, PbS.
(b)
(i) Petrochemical industry.
(ii) Paper industry.
Pergunta 41 Relatório
The compound whose formula is written below is a major component of a soft fatty substance:
(a) State the change that would be observed in the physical state of the fatty substance if hydrogen were bubbled through it for long time in the presence of finely divided nickel at about 180\(^o\)C
(c) Determine the amount (in mole) of hydrogen that would be consumed if one mole of the component reacted completely with hydrogen.
(c) State the product of the reaction of the fatty substance with hot concentrated sodium hydroxide solution
Pergunta 42 Relatório
Classify each of the following as physical change or a chemical change:
(a) fractional distillation of liquefied air;
(b) cracking of petroleum fractions;
(c) conversion of rhombic sulphur to monoclinic sulphur;
(d) chromatographic separation of chlorophyll.
- (a) Fractional distillation of liquefied air: physical change (the components are only separated by boiling point; no new substance is formed).
- (b) Cracking of petroleum fractions: chemical change (large molecules are broken into new, smaller molecules).
- (c) Conversion of rhombic sulphur to monoclinic sulphur: physical change (only a change in the crystalline form/allotrope; the substance is still sulphur).
- (d) Chromatographic separation of chlorophyll: physical change (the pigments are merely separated; no new substance is formed).
Detalhes da Resposta
- (a) Fractional distillation of liquefied air: physical change (the components are only separated by boiling point; no new substance is formed).
- (b) Cracking of petroleum fractions: chemical change (large molecules are broken into new, smaller molecules).
- (c) Conversion of rhombic sulphur to monoclinic sulphur: physical change (only a change in the crystalline form/allotrope; the substance is still sulphur).
- (d) Chromatographic separation of chlorophyll: physical change (the pigments are merely separated; no new substance is formed).
Pergunta 43 Relatório
(a) If L is the Avogadro constant and E° is standard cell potential, state what X and Y stand for in the following expressions
(i) X = \(\frac{\text{Mass of L molecules of gas or vapour}}{\text{ Mass of L molecules of hydrogen}}\)
(ii) Y = -nFE°
(b) State two differences between a primary cell and a secondary cell.
(a)
- (i) \(X = \dfrac{\text{mass of L molecules of gas or vapour}}{\text{mass of L molecules of hydrogen}}\) is the relative density (vapour density) of the gas. Because L molecules is one mole, this ratio equals the molar mass of the gas divided by the molar mass of hydrogen, i.e. relative density \(= \tfrac{1}{2}\times\) molar mass.
- (ii) \(Y = -nFE^\circ\) is the standard change in Gibbs (free) energy of the cell reaction, \(\Delta G^\circ\), where n is the number of moles of electrons transferred and F is the Faraday constant.
(b) Two differences between a primary cell and a secondary cell:
| Primary cell | Secondary cell |
|---|---|
| Cannot be recharged; used once and discarded. | Can be recharged and used repeatedly. |
| The cell reaction is irreversible. | The cell reaction is reversible. |
| Example: Leclanche (dry) cell. | Example: lead-acid accumulator. |
Detalhes da Resposta
(a)
- (i) \(X = \dfrac{\text{mass of L molecules of gas or vapour}}{\text{mass of L molecules of hydrogen}}\) is the relative density (vapour density) of the gas. Because L molecules is one mole, this ratio equals the molar mass of the gas divided by the molar mass of hydrogen, i.e. relative density \(= \tfrac{1}{2}\times\) molar mass.
- (ii) \(Y = -nFE^\circ\) is the standard change in Gibbs (free) energy of the cell reaction, \(\Delta G^\circ\), where n is the number of moles of electrons transferred and F is the Faraday constant.
(b) Two differences between a primary cell and a secondary cell:
| Primary cell | Secondary cell |
|---|---|
| Cannot be recharged; used once and discarded. | Can be recharged and used repeatedly. |
| The cell reaction is irreversible. | The cell reaction is reversible. |
| Example: Leclanche (dry) cell. | Example: lead-acid accumulator. |
Pergunta 44 Relatório
(a)(i) What is meant by the rate of a chemical reaction?
(ii) Explain in terms of the vision theory, the effect of temperature increase on reaction rate.
(b) When hydrogen peroxide is exposed to air, it decomposes
(i) Write an equation for the reaction.
(ii) Outline an experiment to illustrate that effect of a named catalyst on the rate of decomposition.
(iii) Sketch an energy profile diagram to show the effect of the catalyst on the reaction rate, given that the reaction is exothermic.
(c)(i) Explain why enthalpy data alone cannot be used to predict whether a reaction can occur spontaneously or not.
(a)(i) The rate of a chemical reaction is the change in concentration of a reactant or product per unit time.
(a)(ii) According to collision theory, reactant particles must collide effectively before products can form. On increasing the temperature, particles acquire greater kinetic energy. They move faster, collide more frequently, and a greater proportion of the collisions have energy equal to or greater than the activation energy. Hence, the number of effective collisions per second increases and the reaction rate increases.
(b)(i)
(b)(ii) Place equal volumes and concentrations of hydrogen peroxide solution in two identical conical flasks fitted with bungs and delivery tubes leading to gas syringes. Add the same measured mass of manganese(IV) oxide, \(\mathrm{MnO_2}\), to one flask and immediately replace the bung. Do not add \(\mathrm{MnO_2}\) to the other flask, which serves as the control. Record the volume of oxygen collected in each gas syringe at equal time intervals.
The flask containing \(\mathrm{MnO_2}\) produces oxygen much more rapidly, shown by a steeper volume of oxygen against time curve and by vigorous effervescence. The gas relights a glowing splint, confirming that it is oxygen. \(\mathrm{MnO_2}\) is a catalyst because it increases the rate without being used up chemically.
(b)(iii) The catalysed pathway has a lower activation energy, while the enthalpy change, \(\Delta H\), remains negative because the products are at a lower energy than the reactants.
(c)(i) Enthalpy change alone cannot predict spontaneity because spontaneity also depends on the entropy change and on temperature. At constant temperature and pressure, a reaction is spontaneous only when:
Thus, an endothermic reaction may be spontaneous if the increase in entropy is sufficiently large at a suitable temperature, whereas an exothermic reaction may be non-spontaneous if the entropy change is sufficiently unfavourable.
Detalhes da Resposta
(a)(i) The rate of a chemical reaction is the change in concentration of a reactant or product per unit time.
(a)(ii) According to collision theory, reactant particles must collide effectively before products can form. On increasing the temperature, particles acquire greater kinetic energy. They move faster, collide more frequently, and a greater proportion of the collisions have energy equal to or greater than the activation energy. Hence, the number of effective collisions per second increases and the reaction rate increases.
(b)(i)
(b)(ii) Place equal volumes and concentrations of hydrogen peroxide solution in two identical conical flasks fitted with bungs and delivery tubes leading to gas syringes. Add the same measured mass of manganese(IV) oxide, \(\mathrm{MnO_2}\), to one flask and immediately replace the bung. Do not add \(\mathrm{MnO_2}\) to the other flask, which serves as the control. Record the volume of oxygen collected in each gas syringe at equal time intervals.
The flask containing \(\mathrm{MnO_2}\) produces oxygen much more rapidly, shown by a steeper volume of oxygen against time curve and by vigorous effervescence. The gas relights a glowing splint, confirming that it is oxygen. \(\mathrm{MnO_2}\) is a catalyst because it increases the rate without being used up chemically.
(b)(iii) The catalysed pathway has a lower activation energy, while the enthalpy change, \(\Delta H\), remains negative because the products are at a lower energy than the reactants.
(c)(i) Enthalpy change alone cannot predict spontaneity because spontaneity also depends on the entropy change and on temperature. At constant temperature and pressure, a reaction is spontaneous only when:
Thus, an endothermic reaction may be spontaneous if the increase in entropy is sufficiently large at a suitable temperature, whereas an exothermic reaction may be non-spontaneous if the entropy change is sufficiently unfavourable.
Pergunta 45 Relatório
(a) Name the residue obtained on strongly heating the following:
(i) ZnCO\(_3\) in an open crucible;
(ii) CuSO\(_4\), 5H\(_2\)O and then allowing it to cool in a desiccator.
(b) State the colour changes observed on heating and cooling in each case in(a) above.
Pergunta 46 Relatório
(a) State the phenomenon illustrated by the:
(i) spreading of the smell of hydrogen sulphide gas in the laboratory;
(ii) existence of atoms of the same element having different mass numbers
(b) The atomic number of an element is 17. It has different atoms containing 18 neutrons and 20 neutrons, with a relative abundance of 75% and 25% respectively. Calculate the relative atomic mass of the element.
(a)
(i) The spreading of the smell of hydrogen sulphide gas is called diffusion, which is the movement of particles from a region of higher concentration to a region of lower concentration.
(ii) Isotopy.
(b)
Atomic number = 17
Mass numbers of the atoms:
17 + 18 = 35
17 + 20 = 37
Relative atomic mass = (35 × 0.75) + (37 × 0.25)
= 26.25 + 9.25
= 35.5
Detalhes da Resposta
(a)
(i) The spreading of the smell of hydrogen sulphide gas is called diffusion, which is the movement of particles from a region of higher concentration to a region of lower concentration.
(ii) Isotopy.
(b)
Atomic number = 17
Mass numbers of the atoms:
17 + 18 = 35
17 + 20 = 37
Relative atomic mass = (35 × 0.75) + (37 × 0.25)
= 26.25 + 9.25
= 35.5
Pergunta 47 Relatório
State which of the following can exhibit geometric isomerism:
Give reason for your answer
The four compounds shown are:
- (a) \(CH_3-CH{=}CH-CH_3\) (but-2-ene);
- (b) \(CH_3-CH_2-CHBr-CH_2Br\) (1,2-dibromobutane);
- (c) \(HOOC-CH(CH_3)-CH_2-COOH\) (2-methylbutanedioic acid);
- (d) \(HOOC-CH{=}CH-COOH\) (butenedioic acid).
Compounds that exhibit geometric (cis-trans) isomerism: (a) and (d).
Reason. Geometric isomerism requires two conditions to be met together:
- a carbon-to-carbon double bond (\(C{=}C\)), because the double bond prevents (restricts) free rotation about that bond; and
- each of the two doubly-bonded carbon atoms must carry two different groups.
In (a) each doubly-bonded carbon carries an \(H\) and a \(CH_3\) group (two different groups), so cis- and trans-but-2-ene exist. In (d) each doubly-bonded carbon carries an \(H\) and a \(-COOH\) group, so the cis form (maleic acid) and the trans form (fumaric acid) exist. Both conditions are satisfied, so (a) and (d) show geometric isomerism.
Compounds (b) and (c) contain only single (C-C) bonds. Single bonds allow free rotation, so the atoms cannot be locked into fixed cis/trans positions; therefore (b) and (c) do not exhibit geometric isomerism.
Detalhes da Resposta
The four compounds shown are:
- (a) \(CH_3-CH{=}CH-CH_3\) (but-2-ene);
- (b) \(CH_3-CH_2-CHBr-CH_2Br\) (1,2-dibromobutane);
- (c) \(HOOC-CH(CH_3)-CH_2-COOH\) (2-methylbutanedioic acid);
- (d) \(HOOC-CH{=}CH-COOH\) (butenedioic acid).
Compounds that exhibit geometric (cis-trans) isomerism: (a) and (d).
Reason. Geometric isomerism requires two conditions to be met together:
- a carbon-to-carbon double bond (\(C{=}C\)), because the double bond prevents (restricts) free rotation about that bond; and
- each of the two doubly-bonded carbon atoms must carry two different groups.
In (a) each doubly-bonded carbon carries an \(H\) and a \(CH_3\) group (two different groups), so cis- and trans-but-2-ene exist. In (d) each doubly-bonded carbon carries an \(H\) and a \(-COOH\) group, so the cis form (maleic acid) and the trans form (fumaric acid) exist. Both conditions are satisfied, so (a) and (d) show geometric isomerism.
Compounds (b) and (c) contain only single (C-C) bonds. Single bonds allow free rotation, so the atoms cannot be locked into fixed cis/trans positions; therefore (b) and (c) do not exhibit geometric isomerism.
