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Pergunta 1 Relatório
Pipe-borne water is usually chlorinated in order to
Detalhes da Resposta
Pipe-borne water is usually chlorinated in order to kill harmful bacteria in the water. Chlorine is a strong oxidizing agent that can destroy the cell walls and enzymes of bacteria, viruses and other microorganisms that might be present in the water. By killing these microorganisms, chlorination helps to prevent the spread of waterborne diseases and make the water safe for consumption.
Pergunta 2 Relatório
When alkanols react with sodium, the ga evolved is
Detalhes da Resposta
When alkanols react with sodium, the gas evolved is hydrogen. This is because sodium is more reactive than hydrogen and reacts with alkanols to liberate hydrogen gas. The reaction is as follows: R-OH + Na → R-ONa + 1/2H2 where R represents the alkyl group. The reaction is a redox reaction, in which sodium is oxidized and alkanols are reduced. The hydrogen gas evolved can be tested using the 'squeaky pop' test, which involves holding a lighted splint near the gas. The hydrogen gas ignites with a squeaky pop sound due to its high reactivity with oxygen.
Pergunta 3 Relatório
Which of the following can undergo both addition reactions and substitution reactions?
Detalhes da Resposta
Benzene is the only compound that can undergo both addition reactions and substitution reactions. This is because of its unique structure, which consists of a ring of six carbon atoms with alternating double bonds. The double bonds in benzene are considered to be "delocalized," meaning that the electrons involved in the bonds are spread out over the entire ring rather than being localized between specific carbon atoms. This delocalization makes benzene relatively stable and resistant to addition reactions, but it also makes it susceptible to substitution reactions. In a substitution reaction, one or more of the hydrogen atoms in the benzene ring can be replaced with another atom or group of atoms. In an addition reaction, on the other hand, two molecules combine to form a single molecule, usually with the addition of a double or triple bond. Because benzene already has a stable ring structure with delocalized double bonds, it is not easily able to undergo addition reactions, but it can undergo substitution reactions.
Pergunta 4 Relatório
AN AQUEOUS SOLUTION IS ACIDIC IF
Detalhes da Resposta
An aqueous solution is acidic if it contains more H3O- ions than OH- ions. This is because the concentration of H3O- ions determines the acidity or basicity of a solution. In an acidic solution, there is an excess of H3O- ions, which can donate a proton (H+) to other substances, making it corrosive and able to change the color of litmus to red. A high pH value indicates that the solution is basic, not acidic. A bitter taste is not a reliable indicator of acidity, as many substances can have a bitter taste without being acidic.
Pergunta 5 Relatório
Which of the following methods is suitable for the preparation of an insoluble salt?
Detalhes da Resposta
The method suitable for the preparation of an insoluble salt is the Double decomposition method. In this method, two soluble salts are mixed together to form an insoluble salt, which is then separated from the mixture through filtration. This method is particularly useful for preparing salts that are insoluble in water, as it does not require the use of a precipitating agent.
Pergunta 6 Relatório
Which of the following is not a property of trioxonitrate (V) salts?
Detalhes da Resposta
Pergunta 7 Relatório
What is the function of sulphur (IV) oxide in the reaction represented by the following equation? SO 2(aq) + 2H2S(g) → 3S (s) + 2H 2O (l)
Detalhes da Resposta
In the given reaction, sulphur (IV) oxide (SO2) acts as an oxidizing agent. It oxidizes hydrogen sulphide (H2S) gas to form solid sulfur (S) and water (H2O). The oxidation state of sulphur in SO2 is +4, while in H2S, it is -2. During the reaction, sulphur (IV) oxide is reduced to sulphur (S) with a decrease in its oxidation state, while hydrogen sulphide is oxidized to sulphur (S) with an increase in its oxidation state. Therefore, sulphur (IV) oxide acts as an oxidizing agent in the given reaction.
Pergunta 9 Relatório
Which of the following compounds is used as a gaseous fuel?
Detalhes da Resposta
The compound that is used as a gaseous fuel is CH 3-CH 2-CH 2-CH 3. This is the chemical formula for butane, a hydrocarbon that is commonly used as a fuel in lighters and camping stoves. Butane is a gas at room temperature and pressure, which makes it ideal for use as a gaseous fuel. It burns cleanly and efficiently, producing mainly carbon dioxide and water vapor as byproducts.
Pergunta 10 Relatório
Which of the following will produce oxygen and hydrogen during its electrolysis using platinum electrode?
Detalhes da Resposta
Out of the given options, dilute sodium chloride solution is the one that will produce oxygen and hydrogen during its electrolysis using platinum electrodes. During electrolysis, the ions present in the solution are attracted to the electrodes and undergo reduction or oxidation, depending on their charge. In the case of dilute sodium chloride solution, it contains Na+ and Cl- ions. At the cathode (negative electrode), the Na+ ions are attracted and undergo reduction to form sodium atoms and hydrogen gas: 2Na+ + 2e- → 2Na 2H+ + 2e- → H2(g) At the anode (positive electrode), the Cl- ions are attracted and undergo oxidation to form chlorine gas and release oxygen gas: 2Cl- → Cl2(g) + 2e- 2H2O → O2(g) + 4H+ + 4e- Since platinum electrodes are inert, they do not participate in the chemical reactions and do not affect the electrolysis process.
Pergunta 11 Relatório
Substances which absorb water from the atmosphere without dissolving in it are sid to be
Detalhes da Resposta
Substances which absorb water from the atmosphere without dissolving in it are said to be hygroscopic. This means that they have the ability to attract and hold water molecules from the surrounding environment due to their physical and chemical properties. Examples of hygroscopic substances include salt, sugar, and silica gel. Hygroscopic substances are commonly used as desiccants, which are materials that are used to absorb moisture from the air and keep products dry. They are also used in certain industrial processes to remove water from various substances.
Pergunta 12 Relatório
Two metallic ions commonly present in hard water are
Detalhes da Resposta
The two metallic ions commonly present in hard water are Mg2+ and Ca2+. Hard water is water that contains high levels of dissolved minerals, such as calcium and magnesium. These minerals come from the ground and rock formations that the water flows through. When hard water is heated or evaporates, it leaves mineral deposits behind that can build up in pipes, appliances, and other surfaces. Mg2+ and Ca2+ ions are the most common minerals that contribute to the hardness of water. They form insoluble precipitates with soaps, reducing their lathering ability, and can also cause scaling in pipes and boilers. Therefore, it is important to remove or reduce these ions from water for many domestic and industrial applications.
Pergunta 13 Relatório
How many electrons are removed from Cr2- when it is oxidized to CrO42-?
Detalhes da Resposta
When Cr2- is oxidized to CrO42-, it gains oxygen atoms and loses electrons. To determine how many electrons are lost, we need to look at the change in oxidation state of chromium in the reaction. In Cr2-, the oxidation state of chromium is +2. In CrO42-, the oxidation state of chromium is +6. Therefore, chromium has lost 4 electrons in the reaction (going from +2 to +6). So, the correct answer is 4 electrons.
Pergunta 14 Relatório
Petrol consists mainly of
Detalhes da Resposta
Petrol consists mainly of hydrocarbons. Hydrocarbons are organic compounds made up of carbon and hydrogen atoms only. In the case of petrol, these hydrocarbons are typically made up of between 5 and 12 carbon atoms per molecule, with some additional small amounts of other compounds. When petrol is burned in an engine, the hydrocarbons react with oxygen in the air to produce energy, carbon dioxide, and water vapor.
Pergunta 15 Relatório
Dehydration of ethanol produces
Detalhes da Resposta
Dehydration of ethanol produces ethene. During the reaction, a molecule of water is eliminated from ethanol, and the remaining atoms combine to form ethene. This is an example of an elimination reaction, where a small molecule is removed from a larger molecule to form a new product. The reaction is typically carried out at high temperatures in the presence of a catalyst such as concentrated sulfuric acid or alumina. Ethene is an important industrial chemical that is used as a feedstock for the production of many other organic chemicals.
Pergunta 16 Relatório
The spontaneity of a chemical reaction is determined by the change in the
Detalhes da Resposta
The spontaneity of a chemical reaction is determined by the change in the free energy of the system. Free energy is the energy available to do work, and a chemical reaction is spontaneous when the change in free energy is negative. In other words, the products of the reaction have lower free energy than the reactants, and the reaction can occur without the input of additional energy. Concentration, temperature, pressure, and volume can affect the spontaneity of a reaction, but the driving force behind the spontaneity is the change in free energy.
Pergunta 17 Relatório
What mass of copper would be deposited by a current of 1. 0 ampere passing for 965 seconds through copper (ll) tetraoxosulphate (IV) solution? [Cu = 63.5; 1F = 96500C]
Detalhes da Resposta
To solve this problem, we can use the formula: mass of substance deposited = (current × time × molar mass) ÷ (Faraday’s constant × number of electrons transferred) First, we need to find the number of moles of copper deposited by dividing the charge passed by the Faraday’s constant: Q = I × t = 1.0 A × 965 s = 965 C n = Q ÷ (Faraday’s constant × number of electrons transferred) n = 965 C ÷ (96500 C/mol × 2) n = 0.005 mol Next, we can use the number of moles to calculate the mass of copper using its molar mass: mass = n × molar mass mass = 0.005 mol × 63.5 g/mol mass = 0.318 g Therefore, the mass of copper deposited by the current is 0.318 g. The correct option is (a) 0.318g.
Pergunta 18 Relatório
The liquid product of the destructive distillation of coal is
Detalhes da Resposta
The liquid product of the destructive distillation of coal is ammoniacal liquor. This is because during the destructive distillation of coal, coal is heated in the absence of air to produce several by-products such as coke, coal tar, and coal gas. Ammoniacal liquor is a by-product of coal gas production. It is a brown liquid that contains ammonia, carbon dioxide, and other impurities. The ammonia in ammoniacal liquor is usually recovered through a process called scrubbing, which involves passing the gas through water to dissolve the ammonia.
Pergunta 19 Relatório
Which of the following can be added to dissolve the precipitate formed in the reaction represented by the following equation? CaCl2(aq) + H2C2O 4(s) ⇌ CaC 2O2
Detalhes da Resposta
Pergunta 20 Relatório
A positive reaction to Fehling's test indicates the presence of
Detalhes da Resposta
A positive reaction to Fehling's test indicates the presence of reducing sugars. Fehling's test is a chemical test used to detect the presence of reducing sugars such as glucose, fructose, and maltose in a solution. The test involves adding Fehling's reagent, which is a mixture of copper sulfate and potassium tartrate, to the solution being tested and heating it. If reducing sugars are present, they will react with the copper ions in the Fehling's reagent, causing a red precipitate of copper(I) oxide to form. Therefore, the presence of a red precipitate after the Fehling's test indicates the presence of reducing sugars in the solution.
Pergunta 21 Relatório
A device used in the laboratory for intermittent production of gases without heating is
Detalhes da Resposta
The device used in the laboratory for intermittent production of gases without heating is Kipp's apparatus. The Kipp's apparatus is a laboratory equipment used for generating gases such as hydrogen, carbon dioxide, and sulfur dioxide, etc. without heating. It is made up of a glass bottle with three parts: a bulb-shaped bottom, a cylindrical middle section, and a funnel-shaped top. The middle section contains a tube with a tap or stopcock, while the bottom is filled with the reactants needed to generate the gas. When the tap is opened, water or another liquid enters the bottom bulb and reacts with the solid to generate the gas, which then bubbles up and is collected in the top funnel. Kipp's apparatus is often used for qualitative analysis in the laboratory.
Pergunta 22 Relatório
Aluminium is extracted from
Detalhes da Resposta
Aluminium is extracted from bauxite. Bauxite is an ore that is rich in aluminium oxide (Al2O3). The process of extracting aluminium involves the following steps: first, the bauxite ore is purified to remove any impurities, then it is converted into alumina (Al2O3) by the Bayer process. Next, the alumina is electrolyzed in molten cryolite (Na3AlF6) to produce pure aluminium metal. The process of extracting aluminium is energy-intensive and requires large amounts of electricity.
Pergunta 23 Relatório
Which of the following alkanes has the highest boiling point?
Detalhes da Resposta
Hexane has the highest boiling point among the given options. Boiling point increases with increasing molecular size and branching decreases boiling point. Hexane has the highest molecular size and the least branching compared to the other options, making it have the highest boiling point.
Pergunta 24 Relatório
The products of the reaction between CH3CH 2COOH and CH3OH are H 2
O and
Detalhes da Resposta
The products of the reaction between CH3CH 2COOH (propanoic acid) and CH3OH (methanol) are CH 3CH2COOCH3 (methyl propanoate) and H2O (water). This is an esterification reaction where the carboxylic acid (propanoic acid) reacts with an alcohol (methanol) in the presence of an acid catalyst (such as sulfuric acid) to form an ester (methyl propanoate) and water. The general equation for this reaction is: R-COOH + R'-OH ⇌ R-COOR' + H2O In this specific case, the reaction between propanoic acid and methanol produces methyl propanoate and water as the products.
Pergunta 25 Relatório
Metals are said to be malleable because they
Detalhes da Resposta
Metals are said to be malleable because they can be beaten into thin sheets without breaking. This is due to the nature of metallic bonding, where positively charged metal ions are surrounded by a sea of delocalized electrons that are free to move around. When a force is applied to a metal, these delocalized electrons act as a lubricant between the metal ions, allowing them to slide past each other and change shape without breaking. This property is important for the shaping of metals into various forms and structures for use in construction, manufacturing, and other applications.
Pergunta 26 Relatório
Rusting of iron is an example of
Detalhes da Resposta
Rusting of iron is an example of a redox reaction. It involves the oxidation of iron, where it loses electrons and reacts with oxygen to form iron oxide. The process is also accompanied by reduction, where oxygen gains electrons to form hydroxide ions. The overall reaction is: Fe + 2H2O + O2 → Fe(OH)2 2Fe(OH)2 + 1/2O2 → Fe2O3.H2O (rust) Therefore, rusting of iron involves both oxidation and reduction, which is the basis of redox reactions.
Pergunta 27 Relatório
Xg of a pure sample of iron (II) sulphide reacted completely with excess dilute hydrochloric acid to give 3.20g of iron (II) chloride according to the following equation: FeS\(_{(s)}\) + 2HCI\(_{(aq)}\) ---> FeCl\(_{2(aq)}\) + H\(_2\)S\(_{(g)}\).
(a) Mention one method apart from heating by which the reaction can be made to proceed faster
(b) Calculate the value of X. [CI = 35.5, Fe = 56; FeS = 88g mol\(^{-1}\)]
(a) One method (apart from heating) by which the reaction can be made faster: increase the surface area of the iron(II) sulphide by using it in a finely powdered form (using a more concentrated acid is also acceptable).
(b) Calculating X. Equation: \[FeS_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_2S_{(g)}\]
Molar mass of \(FeCl_2 = 56 + (2 \times 35.5) = 56 + 71 = 127\,g\,mol^{-1}\)
\[\text{moles of } FeCl_2 = \frac{3.20}{127} = 0.0252\,mol\]
From the equation, 1 mole \(FeS\) gives 1 mole \(FeCl_2\), so moles of \(FeS = 0.0252\,mol\).
\[X = \text{mass of } FeS = 0.0252 \times 88 = 2.22\,g\]
So X = 2.22 g (approximately).
Detalhes da Resposta
(a) One method (apart from heating) by which the reaction can be made faster: increase the surface area of the iron(II) sulphide by using it in a finely powdered form (using a more concentrated acid is also acceptable).
(b) Calculating X. Equation: \[FeS_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_2S_{(g)}\]
Molar mass of \(FeCl_2 = 56 + (2 \times 35.5) = 56 + 71 = 127\,g\,mol^{-1}\)
\[\text{moles of } FeCl_2 = \frac{3.20}{127} = 0.0252\,mol\]
From the equation, 1 mole \(FeS\) gives 1 mole \(FeCl_2\), so moles of \(FeS = 0.0252\,mol\).
\[X = \text{mass of } FeS = 0.0252 \times 88 = 2.22\,g\]
So X = 2.22 g (approximately).
Pergunta 28 Relatório
(a)(i) What is meant by the activation energy of a reaction?
(ii) State the effect of a catalyst on activation energy.
(b) What substance serves as a catalyst in each of the following?
(I) Hydrogenation of oils
(ii) Biochemical reactions.
(a)(i) The activation energy of a reaction is the minimum amount of energy that the reacting particles must possess before a collision between them can lead to a reaction (i.e. the minimum energy needed for an effective collision).
(ii) A catalyst lowers (reduces) the activation energy of the reaction by providing an alternative reaction pathway of lower activation energy, so more colliding particles have the required energy and the reaction proceeds faster.
(b) Catalyst in each case:
Detalhes da Resposta
(a)(i) The activation energy of a reaction is the minimum amount of energy that the reacting particles must possess before a collision between them can lead to a reaction (i.e. the minimum energy needed for an effective collision).
(ii) A catalyst lowers (reduces) the activation energy of the reaction by providing an alternative reaction pathway of lower activation energy, so more colliding particles have the required energy and the reaction proceeds faster.
(b) Catalyst in each case:
Pergunta 29 Relatório
(a) List two uses of sodium trioxocarbonate (IV).
(b) Sodium trioxocarbonate (IV) solution is alkaline.
(i) What phenomenon is responsible for this observation?
(ii) Name the product obtained on passing carbon (IV) oxide into saturated sodium trioxocarbonate (IV) solution.
(a) Two uses of sodium trioxocarbonate(IV), \(Na_2CO_3\) (washing soda): in the manufacture of glass and in softening hard water. (It is also used in the manufacture of soap, detergents and paper.)
(b)(i) The phenomenon responsible for the alkalinity of sodium trioxocarbonate(IV) solution is salt hydrolysis. Being the salt of a strong base (NaOH) and a weak acid (trioxocarbonate(IV) acid), the carbonate ion reacts with water to release hydroxide ions, making the solution alkaline:
\[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\]
(ii) When carbon(IV) oxide is passed into a saturated sodium trioxocarbonate(IV) solution, the product obtained is sodium hydrogen trioxocarbonate(IV) (sodium hydrogencarbonate), \(NaHCO_3\):
\[Na_2CO_3 + CO_2 + H_2O \rightarrow 2NaHCO_3\]
Detalhes da Resposta
(a) Two uses of sodium trioxocarbonate(IV), \(Na_2CO_3\) (washing soda): in the manufacture of glass and in softening hard water. (It is also used in the manufacture of soap, detergents and paper.)
(b)(i) The phenomenon responsible for the alkalinity of sodium trioxocarbonate(IV) solution is salt hydrolysis. Being the salt of a strong base (NaOH) and a weak acid (trioxocarbonate(IV) acid), the carbonate ion reacts with water to release hydroxide ions, making the solution alkaline:
\[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\]
(ii) When carbon(IV) oxide is passed into a saturated sodium trioxocarbonate(IV) solution, the product obtained is sodium hydrogen trioxocarbonate(IV) (sodium hydrogencarbonate), \(NaHCO_3\):
\[Na_2CO_3 + CO_2 + H_2O \rightarrow 2NaHCO_3\]
Pergunta 30 Relatório
(a)(i) Define heat of neutralization
(ii) Give the reason why copper (II) chloride can be prepared by neutralization, unlike lead (II) chloride.
(b)(i) Describe in outline, the manufacture of trioxonitrate (V) acid by the catalytic oxidation of ammonia, giving equations where appropriate.
(ii) What are the products obtained when sodium tioxonitrate (V) is heated strongly?
(c) When powdered magnesium is heated to redness in a stream of nitrogen, magnesium nitride (Mg\(_3\)N\(_2\)) is formed.
(i) Write an equation for the reaction
(ii) Hence, calculate the amount (in mole) of magnesium nitride that can be obtained from 3.0g of magnesium [Mg = 24].
(a)(i) Heat of neutralization is the heat change when one mole of hydrogen ions (H+) from an acid reacts completely with one mole of hydroxide ions (OH-) from a base to form one mole of water under standard conditions.
(ii) Copper(II) chloride is soluble in water, so it can be made in solution by neutralizing an acid with a base and then crystallized. Lead(II) chloride is insoluble in cold water, so a neutralization in solution would simply precipitate it; it is therefore prepared by precipitation (double decomposition), not by neutralization.
(b)(i) Manufacture of trioxonitrate(V) acid (Ostwald process)
(ii) Sodium trioxonitrate(V) on strong heating gives sodium trioxonitrate(III) and oxygen:
\[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
(c)(i) \[3Mg + N_2 \rightarrow Mg_3N_2\]
(ii) Amount of Mg \(= \dfrac{3.0}{24} = 0.125\ \text{mol}\).
From the equation, 3 mol Mg give 1 mol Mg3N2, so
amount of Mg3N2 \(= \dfrac{0.125}{3} = 0.0417\ \text{mol} \approx 4.17 \times 10^{-2}\ \text{mol}\).
Detalhes da Resposta
(a)(i) Heat of neutralization is the heat change when one mole of hydrogen ions (H+) from an acid reacts completely with one mole of hydroxide ions (OH-) from a base to form one mole of water under standard conditions.
(ii) Copper(II) chloride is soluble in water, so it can be made in solution by neutralizing an acid with a base and then crystallized. Lead(II) chloride is insoluble in cold water, so a neutralization in solution would simply precipitate it; it is therefore prepared by precipitation (double decomposition), not by neutralization.
(b)(i) Manufacture of trioxonitrate(V) acid (Ostwald process)
(ii) Sodium trioxonitrate(V) on strong heating gives sodium trioxonitrate(III) and oxygen:
\[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
(c)(i) \[3Mg + N_2 \rightarrow Mg_3N_2\]
(ii) Amount of Mg \(= \dfrac{3.0}{24} = 0.125\ \text{mol}\).
From the equation, 3 mol Mg give 1 mol Mg3N2, so
amount of Mg3N2 \(= \dfrac{0.125}{3} = 0.0417\ \text{mol} \approx 4.17 \times 10^{-2}\ \text{mol}\).
Pergunta 31 Relatório
(a) Write an equation for the reaction of chlorine with
(i) potassium iodide solution;
(ii) zinc on heating
(b) Identify the product Q in the following reaction: Cl\(_2\) + 2NaOH \(\to\) NaCl + H\(_2\)O + Q.
(a) Reactions of chlorine:
(b) In \(Cl_2 + 2NaOH \rightarrow NaCl + H_2O + Q\) (cold, dilute sodium hydroxide), the product Q is sodium chlorate(I) (sodium oxochlorate(I) / sodium hypochlorite), \(NaOCl\).
The balanced equation is: \[Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O\]
Detalhes da Resposta
(a) Reactions of chlorine:
(b) In \(Cl_2 + 2NaOH \rightarrow NaCl + H_2O + Q\) (cold, dilute sodium hydroxide), the product Q is sodium chlorate(I) (sodium oxochlorate(I) / sodium hypochlorite), \(NaOCl\).
The balanced equation is: \[Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O\]
Pergunta 32 Relatório
Copy and complete the following table
| Element | Number of Neutrons | Electronic Configuration | Group in the periodic Table |
| \(^{23}_{11}Na\) | --- | \(1s^{2}2s^{2}2p^{6}3s^{1}\) | 1 |
| \(^4_2He\) | 2 | ---- | --- |
| ---- | 7 | \(1s^{2}2s^{2}2p^{2}\) | ---- |
Column 1 Column 2 Column 3 Data Data Data Data Data Data Data Data Data
| Element | Number of Neutrons | Electronic Configuration | Group in the Periodic Table |
| \(^{23}_{11}Na\) | 12 | \(1s^{2}2s^{2}2p^{6}3s^{1}\) | 1 |
| \(^{4}_{2}He\) | 2 | \(1s^{2}\) | 0 |
| \(^{13}_{6}C\) | 7 | \(1s^{2}2s^{2}2p^{2}\) | IV or 4 |
Detalhes da Resposta
Column 1 Column 2 Column 3 Data Data Data Data Data Data Data Data Data
| Element | Number of Neutrons | Electronic Configuration | Group in the Periodic Table |
| \(^{23}_{11}Na\) | 12 | \(1s^{2}2s^{2}2p^{6}3s^{1}\) | 1 |
| \(^{4}_{2}He\) | 2 | \(1s^{2}\) | 0 |
| \(^{13}_{6}C\) | 7 | \(1s^{2}2s^{2}2p^{2}\) | IV or 4 |
Pergunta 33 Relatório
(a)(i) What type of reaction is involved in each of the conversion processes indicated as I to V below?
(ii) Name one isomer of glucose
(iii) Explain why palmwine becomes sour on prolonged exposure to air.
(b)(i) List the reagents and the reaction condition necessary for ethanoic acid to form an alkanoate.
(ii) Give two uses of alkanoates
(c)(i) What is the lUPAC name of the following compound?
(ii) Outline one chemical test to distinguish between methane and the compound in (c)(i) above.
(iii) Write an equation for the combustion of ethene in excess oxygen.
(a)(i) Type of reaction in each conversion (I to V)
| Step | Type of reaction |
|---|---|
| I | Hydrolysis |
| II | Fermentation (decomposition) |
| III | Dehydration |
| IV | Polymerization |
| V | Oxidation |
(a)(ii) One isomer of glucose: fructose (galactose is also acceptable), each of formula C6H12O6.
(a)(iii) Why palm wine becomes sour on prolonged exposure to air. Palm wine contains ethanol. On exposure to air, bacterial/atmospheric oxidation converts the ethanol to ethanoic (acetic) acid, and it is this acid that gives the drink its sour taste:
\[ CH_3CH_2OH + O_2 \rightarrow CH_3COOH + H_2O \]
(b)(i) Reagents and conditions for ethanoic acid to form an alkanoate
\[ CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{conc.\,H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O \]
(b)(ii) Two uses of alkanoates (esters)
(c)(i) IUPAC name of the compound: but-2-yne, \( CH_3-C{\equiv}C-CH_3 \).
(c)(ii) Chemical test to distinguish methane from but-2-yne. Pass each gas in turn into bromine water (or acidified KMnO4). But-2-yne, being unsaturated, rapidly decolourizes the reagent; methane, being saturated, gives no visible reaction and the colour remains.
(c)(iii) Combustion of ethene in excess oxygen
\[ C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)} \]
Detalhes da Resposta
(a)(i) Type of reaction in each conversion (I to V)
| Step | Type of reaction |
|---|---|
| I | Hydrolysis |
| II | Fermentation (decomposition) |
| III | Dehydration |
| IV | Polymerization |
| V | Oxidation |
(a)(ii) One isomer of glucose: fructose (galactose is also acceptable), each of formula C6H12O6.
(a)(iii) Why palm wine becomes sour on prolonged exposure to air. Palm wine contains ethanol. On exposure to air, bacterial/atmospheric oxidation converts the ethanol to ethanoic (acetic) acid, and it is this acid that gives the drink its sour taste:
\[ CH_3CH_2OH + O_2 \rightarrow CH_3COOH + H_2O \]
(b)(i) Reagents and conditions for ethanoic acid to form an alkanoate
\[ CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{conc.\,H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O \]
(b)(ii) Two uses of alkanoates (esters)
(c)(i) IUPAC name of the compound: but-2-yne, \( CH_3-C{\equiv}C-CH_3 \).
(c)(ii) Chemical test to distinguish methane from but-2-yne. Pass each gas in turn into bromine water (or acidified KMnO4). But-2-yne, being unsaturated, rapidly decolourizes the reagent; methane, being saturated, gives no visible reaction and the colour remains.
(c)(iii) Combustion of ethene in excess oxygen
\[ C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)} \]
Pergunta 34 Relatório
(a)(i) What is the general formula for alkanoic acids?
(ii) State two chemical properties of ethanoic acid.
(b) Which of propene, butane and pentane
(i) will decolorize acidified KMnO\(_4\) solution?
(ii) can be easily polymerized?
(iii) is an isomer of methylpropane?
(iv) can be obtained from an alkanol by dehydration?
(a)(i) The general formula for alkanoic acids is \(C_nH_{2n+1}COOH\) (equivalently \(C_nH_{2n}O_2\)).
(ii) Two chemical properties of ethanoic acid, \(CH_3COOH\):
(b) Of propene, butane and pentane:
Detalhes da Resposta
(a)(i) The general formula for alkanoic acids is \(C_nH_{2n+1}COOH\) (equivalently \(C_nH_{2n}O_2\)).
(ii) Two chemical properties of ethanoic acid, \(CH_3COOH\):
(b) Of propene, butane and pentane:
Pergunta 35 Relatório
(a) State two postulates of the kinetic theory of gases.
(b) Write two chemical properties that are common to both carbon (IV) oxide and sulphur (IV) oxide.
(a) Two postulates of the kinetic theory of gases:
(b) Two chemical properties common to both carbon(IV) oxide, \(CO_2\), and sulphur(IV) oxide, \(SO_2\):
Detalhes da Resposta
(a) Two postulates of the kinetic theory of gases:
(b) Two chemical properties common to both carbon(IV) oxide, \(CO_2\), and sulphur(IV) oxide, \(SO_2\):
Pergunta 36 Relatório
(a) If a steel spoon were to be plated with silver, state what would be suitable for use as the;
(i) anode. (ii) cathode; (iii) electrolyte
(b)(i) Write an equation for one of the reactions involved in the purification of bauxite.
(ii) Give the reason why the carbon anodes are changed at intervals during the electrolysis of pure alumina solution in molten cryolite.
(a) To electroplate a steel spoon with silver:
(b)(i) An equation for one reaction in the purification of bauxite (dissolving the amphoteric aluminium oxide in hot sodium hydroxide, leaving impurities behind):
\[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\]
(ii) The carbon anodes must be changed (replaced) at intervals because the oxygen liberated at the anode reacts with the hot carbon to form carbon(IV) oxide (and carbon(II) oxide), so the carbon anodes gradually burn away (are oxidized) and become too small to use.
Detalhes da Resposta
(a) To electroplate a steel spoon with silver:
(b)(i) An equation for one reaction in the purification of bauxite (dissolving the amphoteric aluminium oxide in hot sodium hydroxide, leaving impurities behind):
\[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\]
(ii) The carbon anodes must be changed (replaced) at intervals because the oxygen liberated at the anode reacts with the hot carbon to form carbon(IV) oxide (and carbon(II) oxide), so the carbon anodes gradually burn away (are oxidized) and become too small to use.
Pergunta 37 Relatório
(a)(i) What is an electrolyte?
(ii) Classify each of the following as strong electrolyte/weak, electrolyte/non-electrolyte. Potassium chloride; sodium ethanoate, aqueous ammonia; cane sugar
(b)(i) Write half-cell equations for the reactions in the Daniel cell .
(ii) Why is the Daniel cell classified as an electrochemical cell?
(iii) Give two other examples of electrochemical cell.
(c) Explain the following observations.
(i) Graphite conducts electricity, unlike most non-metals
(ii) In the electrolysis of copper (II) tetraoxosulphate (VI) solution, the blue colour fades with platinum electrodes while the colour intensity is unaffected with copper electrodes (equations required)
(iii) A solution of dry hydrogen chloride in methylbenzene (toluene) does not conduct electricity whereas hydrochloric acid does.
(a)(i) An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it because it contains free mobile ions.
(ii)
(b)(i) Daniell cell half-cell reactions
Anode (oxidation): \[Zn \rightarrow Zn^{2+} + 2e^-\]
Cathode (reduction): \[Cu^{2+} + 2e^- \rightarrow Cu\]
(ii) It is an electrochemical cell because it converts chemical energy (from the spontaneous redox reaction) into electrical energy.
(iii) Other examples: the dry (Leclanche) cell and the lead-acid accumulator.
(c)(i) In graphite, each carbon atom bonds to only three others, using three of its four valence electrons. The fourth electron of every atom is delocalized between the layers and is free to move, so graphite conducts electricity.
(ii) With platinum electrodes, Cu2+ ions are discharged at the cathode and removed from solution, so the blue colour fades:
\[Cu^{2+} + 2e^- \rightarrow Cu\]
at the anode oxygen is evolved: \[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
With copper electrodes, the copper anode dissolves and replaces the Cu2+ removed at the cathode, so the concentration and blue colour stay the same:
\[Cu \rightarrow Cu^{2+} + 2e^-\]
(iii) In methylbenzene, hydrogen chloride exists as un-ionized covalent molecules, so there are no mobile ions and it does not conduct. In water, HCl ionizes into mobile H+ and Cl- ions, so hydrochloric acid conducts electricity.
Detalhes da Resposta
(a)(i) An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it because it contains free mobile ions.
(ii)
(b)(i) Daniell cell half-cell reactions
Anode (oxidation): \[Zn \rightarrow Zn^{2+} + 2e^-\]
Cathode (reduction): \[Cu^{2+} + 2e^- \rightarrow Cu\]
(ii) It is an electrochemical cell because it converts chemical energy (from the spontaneous redox reaction) into electrical energy.
(iii) Other examples: the dry (Leclanche) cell and the lead-acid accumulator.
(c)(i) In graphite, each carbon atom bonds to only three others, using three of its four valence electrons. The fourth electron of every atom is delocalized between the layers and is free to move, so graphite conducts electricity.
(ii) With platinum electrodes, Cu2+ ions are discharged at the cathode and removed from solution, so the blue colour fades:
\[Cu^{2+} + 2e^- \rightarrow Cu\]
at the anode oxygen is evolved: \[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
With copper electrodes, the copper anode dissolves and replaces the Cu2+ removed at the cathode, so the concentration and blue colour stay the same:
\[Cu \rightarrow Cu^{2+} + 2e^-\]
(iii) In methylbenzene, hydrogen chloride exists as un-ionized covalent molecules, so there are no mobile ions and it does not conduct. In water, HCl ionizes into mobile H+ and Cl- ions, so hydrochloric acid conducts electricity.
Pergunta 38 Relatório
(a) What is the IUPAC name of Fe\(_2\)(SO\(_4\))\(_3\)?
(b)(i) Write an equation to represent the reaction of hydrogen sulphide with iron (III) chloride solution.
(ii) Mention one change observed during the reaction in (b)(i) above.
(a) IUPAC name of \(Fe_2(SO_4)_3\)
The compound contains the iron(III) cation, \(Fe^{3+}\), and the sulphate anion, \(SO_4^{2-}\). Because iron shows more than one oxidation state, the oxidation number of the metal must be shown in Roman numerals, and the anion is named in full as the tetraoxosulphate(VI) ion. The IUPAC name is therefore iron(III) tetraoxosulphate(VI) (commonly, iron(III) sulphate).
(b)(i) Reaction of hydrogen sulphide with iron(III) chloride solution
Hydrogen sulphide is a reducing agent. It reduces iron(III) to iron(II), and is itself oxidised, so that its sulphur is deposited as free (elemental) sulphur:
\[2FeCl_3(aq) + H_2S(g) \rightarrow 2FeCl_2(aq) + 2HCl(aq) + S(s)\]In ionic terms the essential change is \(2Fe^{3+} + H_2S \rightarrow 2Fe^{2+} + 2H^+ + S\), which shows clearly that iron is reduced from the \(+3\) to the \(+2\) state while sulphide sulphur (\(-2\)) is oxidised to sulphur (\(0\)).
(b)(ii) One observed change
The yellow-brown iron(III) chloride solution turns pale green, showing that iron(III) has been reduced to iron(II). At the same time a pale yellow deposit (turbidity) of sulphur forms in the mixture. Either of these observations is acceptable: the colour change from yellow-brown to pale green, or the appearance of the yellow solid sulphur.
Detalhes da Resposta
(a) IUPAC name of \(Fe_2(SO_4)_3\)
The compound contains the iron(III) cation, \(Fe^{3+}\), and the sulphate anion, \(SO_4^{2-}\). Because iron shows more than one oxidation state, the oxidation number of the metal must be shown in Roman numerals, and the anion is named in full as the tetraoxosulphate(VI) ion. The IUPAC name is therefore iron(III) tetraoxosulphate(VI) (commonly, iron(III) sulphate).
(b)(i) Reaction of hydrogen sulphide with iron(III) chloride solution
Hydrogen sulphide is a reducing agent. It reduces iron(III) to iron(II), and is itself oxidised, so that its sulphur is deposited as free (elemental) sulphur:
\[2FeCl_3(aq) + H_2S(g) \rightarrow 2FeCl_2(aq) + 2HCl(aq) + S(s)\]In ionic terms the essential change is \(2Fe^{3+} + H_2S \rightarrow 2Fe^{2+} + 2H^+ + S\), which shows clearly that iron is reduced from the \(+3\) to the \(+2\) state while sulphide sulphur (\(-2\)) is oxidised to sulphur (\(0\)).
(b)(ii) One observed change
The yellow-brown iron(III) chloride solution turns pale green, showing that iron(III) has been reduced to iron(II). At the same time a pale yellow deposit (turbidity) of sulphur forms in the mixture. Either of these observations is acceptable: the colour change from yellow-brown to pale green, or the appearance of the yellow solid sulphur.
Pergunta 39 Relatório
(a)(i) Arrange the following elements in the order of increasing reactivity. Iron, Lead, Magnessium, Aluminium.
(ii) Which of the following elements in (a)(i) above reacts with sodium hydroxide to give hydrogen?
(b) What property of tetraoxosulphate (VI) acid does each of the following reactions illustrate?
(I) S + 2H\(_2\)SO\(_4\) ---> 3SO\(_4\) + 2H\(_2\)O
(ii) MgO + H\(_2\)SO\(_4\) ---> MgSO\(_4\) + H\(_2\)O
(iii) C\(_{12}\)H\(_{22}\)C\(_{11}\) + H\(_2\)SO\(_4\) ---> 12C + H\(_2\)SO\(_4\) + 11H\(_2\)O
(a)(i) In increasing order of reactivity:
Lead < Iron < Aluminium < Magnesium.
(ii) Aluminium reacts with sodium hydroxide solution to give hydrogen gas.
(b) Tetraoxosulphate(VI) acid behaves as:
(i) an oxidising agent in its reaction with sulphur.
(ii) a typical acid in the neutralisation of magnesium oxide.
(iii) a dehydrating agent in its reaction with sugar, \(C_{12}H_{22}O_{11}\).
Detalhes da Resposta
(a)(i) In increasing order of reactivity:
Lead < Iron < Aluminium < Magnesium.
(ii) Aluminium reacts with sodium hydroxide solution to give hydrogen gas.
(b) Tetraoxosulphate(VI) acid behaves as:
(i) an oxidising agent in its reaction with sulphur.
(ii) a typical acid in the neutralisation of magnesium oxide.
(iii) a dehydrating agent in its reaction with sugar, \(C_{12}H_{22}O_{11}\).
Pergunta 40 Relatório
(a)(i) List the quantum number that are assigned to an electron in an atom.
(ii) What is the maximum number of electrons that can occupy the 3d orbital?
(b) An element represented as P has the following electronic configuration: 1s\(^2\)2s\(^2\)2p\(^6\)2s\(^2\)
(i) Write the electronic configuration of the ion of P.
(ii) Without identifying P, write the likely formula of its chloride.
(iii) State with reason, whether P will be a good oxidizing or reducing agent.
(c)(i) What is electron affinity?
(ii) Explain briefly why ammonia can precipitate in dative bonding.
(d)(i) If an element in Group IV loses an alpha particle, to which group would the product belong?
(ii) Two equally toxic substances X and Y which decay to non-toxic products, were absorbed through the skin. If their half-lives are 8 minutes and 2 months respectively, which of them constitutes the greater health hazard? Explain your answer
(a)(i) Quantum numbers assigned to an electron in an atom are:
(ii) The maximum number of electrons that can occupy the 3d subshell is 10.
(b) The configuration is taken as \(1s^2\,2s^2\,2p^6\,3s^2\).
(i) P loses two electrons to form P2+. Its electronic configuration is:
\(1s^2\,2s^2\,2p^6\)
(ii) The likely formula of the chloride of P is PCl2.
(iii) P is a good reducing agent because it readily loses two outer electrons to attain a stable noble-gas configuration. Thus, it donates electrons to other substances.
(c)(i) Electron affinity is the energy evolved or absorbed when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous negative ions.
(ii) Ammonia can participate in dative bonding because the nitrogen atom has a lone pair of electrons which it can donate to an electron-deficient atom or ion for sharing.
(d)(i) On losing an alpha particle, the atomic number decreases by 2. Therefore, an element in Group IV will give a product in Group II.
(ii) X, with a half-life of 8 minutes, constitutes the greater health hazard. Its shorter half-life means that it decays more rapidly and releases radiation at a higher rate than Y within a given period.
Detalhes da Resposta
(a)(i) Quantum numbers assigned to an electron in an atom are:
(ii) The maximum number of electrons that can occupy the 3d subshell is 10.
(b) The configuration is taken as \(1s^2\,2s^2\,2p^6\,3s^2\).
(i) P loses two electrons to form P2+. Its electronic configuration is:
\(1s^2\,2s^2\,2p^6\)
(ii) The likely formula of the chloride of P is PCl2.
(iii) P is a good reducing agent because it readily loses two outer electrons to attain a stable noble-gas configuration. Thus, it donates electrons to other substances.
(c)(i) Electron affinity is the energy evolved or absorbed when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous negative ions.
(ii) Ammonia can participate in dative bonding because the nitrogen atom has a lone pair of electrons which it can donate to an electron-deficient atom or ion for sharing.
(d)(i) On losing an alpha particle, the atomic number decreases by 2. Therefore, an element in Group IV will give a product in Group II.
(ii) X, with a half-life of 8 minutes, constitutes the greater health hazard. Its shorter half-life means that it decays more rapidly and releases radiation at a higher rate than Y within a given period.
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