WAEC SSCE - Chemistry - 1995
Pergunta 1 Relatório
If a solid has a low melting point and dissolves readily in benzene, it would probably
Detalhes da Resposta
If a solid has a low melting point and dissolves readily in benzene, it would probably have covalent bonding. Covalent compounds are those that are held together by sharing of electrons between atoms, and they typically have low melting points and do not conduct electricity in either the solid or molten state. Benzene, a nonpolar solvent, readily dissolves covalent compounds, which explains why the solid in question dissolves readily in it. Strong electrostatic forces of attraction are associated with ionic compounds, which typically have high melting points and are soluble in polar solvents like water.
Pergunta 2 Relatório
H3O+(aq) + OH-(aq) → 2H2O(l). The heat change accompanying the process represented by the equation above is the heat of
Detalhes da Resposta
The heat change accompanying the given process is the heat of neutralization. This is because the reaction involves the combination of hydrogen ions (H+) from an acid (represented by the H3O+ ion) and hydroxide ions (OH-) from a base to form water molecules (H2O). This process of combining an acid and a base to form a salt and water is called neutralization. The heat released or absorbed during this process is known as the heat of neutralization.
Pergunta 3 Relatório
Metals which react with steam only when they are red-hot include
Detalhes da Resposta
Pergunta 4 Relatório
A positive brown ring test indicates the presence of
Detalhes da Resposta
The positive brown ring test indicates the presence of nitrate ions (NO3). The test is done by adding freshly prepared iron(II) sulfate (FeSO4) solution and concentrated sulfuric acid (H2SO4) to the test solution. If nitrate ions are present in the solution, a brown ring will form at the junction of the two liquids. This occurs because nitrate ions reduce Fe2+ ions to Fe3+ ions, which then react with the sulfuric acid to form Fe(HSO4)2 which appears as the brown ring.
Pergunta 5 Relatório
The following statements about graphite are correct except that it
Detalhes da Resposta
Pergunta 6 Relatório
What mass of copper will be deposited by the liberation of Cu2+ when 0.1F of electricity flows through an aqueous solution of a copper (ll) salt? [Cu = 64]
Detalhes da Resposta
Pergunta 7 Relatório
The following salts dissolve readily in cold water except
Detalhes da Resposta
Among the given options, the salt that does not dissolve readily in cold water is PbSO4. This is because PbSO4 is an ionic compound with a high lattice energy, meaning that its ions are held together by strong electrostatic forces of attraction. These forces of attraction make it difficult for the PbSO4 crystals to break apart and dissolve in water. Therefore, PbSO4 is considered insoluble in water. The other options, such as CaCI2, (NH4)2SO4, Na2CO3, and Na2SO3, readily dissolve in cold water due to their ionic nature and low lattice energy.
Pergunta 8 Relatório
What is the amount (in mole) of hydrogen gas that would be produced if 0.6 mole of hydrochloric acid reacted with excess zinc according to the following equation? Zn(s) + 2HCI(aq) → ZnCI2(aq) + H2(g)
Detalhes da Resposta
According to the balanced chemical equation, 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. Zn(s) + 2HCI(aq) → ZnCI2(aq) + H2(g) Therefore, if 0.6 mole of hydrochloric acid is used, it will react with 0.3 mole of zinc to produce 0.3 mole of hydrogen gas. Thus, the amount of hydrogen gas produced is 0.3 mole. Therefore, the correct option is: 0.3 mole.
Pergunta 9 Relatório
Copper (ll) tetraoxosulphate (VI) is often added to swimming pools because it
Detalhes da Resposta
Copper (II) tetraoxosulphate (VI) is often added to swimming pools because it prevents the growth of algae. Algae can grow in the warm, moist environment of swimming pools and can make the water appear green and murky. Copper (II) tetraoxosulphate (VI) is a powerful algicide that can prevent the growth of algae in swimming pools. When added to the water, it releases copper ions which are toxic to algae and prevent their growth. Copper (II) tetraoxosulphate (VI) is a safe and effective way to keep swimming pools clean and clear.
Pergunta 10 Relatório
Sodium chloride cannot conduct electricity in the solid state because it
Detalhes da Resposta
Sodium chloride (NaCl) cannot conduct electricity in the solid state because it does not contain mobile ions. In the solid state, the sodium (Na) and chloride (Cl) ions are held tightly in a lattice structure, and cannot move freely to conduct an electric current. When dissolved in water, NaCl dissociates into its ions, which are free to move and conduct electricity. Therefore, NaCl is an example of an ionic compound that conducts electricity only when dissolved in water or melted.
Pergunta 11 Relatório
Which of the following is an alloy of mercury?
Detalhes da Resposta
The alloy of mercury is amalgam. An amalgam is a type of alloy that consists of mercury as one of the components. Amalgams are usually used in dentistry to fill cavities. When mercury is mixed with other metals like silver, tin, or copper, it forms a soft, pliable material that can be easily shaped to fill the cavity. Once the amalgam hardens, it becomes a durable and long-lasting filling. Therefore, the correct answer is Amalgam.
Pergunta 12 Relatório
A WEAK ACID IS ONE WHICH
Detalhes da Resposta
A weak acid is one which is slightly ionized in water. In other words, when a weak acid is dissolved in water, only a small fraction of its molecules dissociate into ions. This is in contrast to a strong acid, which dissociates almost completely into ions when dissolved in water. As a result of its weak ionization, a weak acid has a lower acidity compared to a strong acid.
Pergunta 13 Relatório
Which of the following cannot be deduced from the electronic configuration of a transition metal?
Detalhes da Resposta
Pergunta 14 Relatório
Alums are classified as
Detalhes da Resposta
Alums are classified as double salts. Double salts are compounds that contain two different cations and one anion, held together by ionic bonds. In the case of alums, they are double salts of a monovalent cation (such as potassium or ammonium), a trivalent cation (such as aluminum or chromium), and a sulfate anion. The general formula for alums is M^+M^3+(SO4)2·12H2O, where M^+ is a monovalent cation and M^3+ is a trivalent cation. Alums are commonly used as mordants in dyeing and in various industrial processes.
Pergunta 15 Relatório
Fructose and glucose are classified as carbohydrates because they
Detalhes da Resposta
Pergunta 16 Relatório
What is the percentage by mass of copper in copper in copper (l) oxide (CU2O)? [O = 16; CU =64]
Detalhes da Resposta
To determine the percentage by mass of copper in copper (I) oxide (Cu2O), we need to calculate the molar mass of Cu2O, which is the sum of the atomic masses of the elements in the compound. Molar mass of Cu2O = 2Cu + O = (2 x 64) + 16 = 144 g/mol The molar mass of copper in Cu2O is 2 x 64 = 128 g/mol The percentage by mass of copper in Cu2O is: (128 g Cu / 144 g Cu2O) x 100% = 88.9% Therefore, the correct answer is 88.9%.
Pergunta 17 Relatório
Which of the following species is always present in acidified water?
Detalhes da Resposta
The species that is always present in acidified water is H3O+. When water is mixed with an acid, it produces H+ ions, which react with water to form H3O+ ions. These ions are responsible for the acidic properties of the solution. Therefore, out of the given options, H3O+ is the correct choice.
Pergunta 18 Relatório
The function of limestone in the extraction of iron in the blast furnace is
Detalhes da Resposta
The function of limestone in the extraction of iron in the blast furnace is to remove impurities from the iron ore. During the extraction process, iron ore is heated with coke (a form of carbon) and air to produce molten iron, which is then drained from the bottom of the furnace. However, the iron ore contains impurities such as silicon dioxide (SiO2), which is acidic in nature and can corrode the lining of the furnace. Limestone (calcium carbonate, CaCO3) is added to the blast furnace to act as a flux, which neutralizes the acidic impurities and forms a slag, a waste material that can be easily removed from the furnace. Thus, the function of limestone is to purify the iron ore by removing the impurities and prevent the corrosion of the furnace lining.
Pergunta 19 Relatório
Which of the following is hydrocarbons is unsaturated?
Detalhes da Resposta
The hydrocarbon that is unsaturated is benzene. A hydrocarbon is considered unsaturated when it contains at least one double or triple bond between carbon atoms, which means that it has fewer hydrogen atoms per carbon atom than a saturated hydrocarbon. Benzene is an unsaturated hydrocarbon with a ring structure and alternating double bonds between carbon atoms. The other options, ethane, propane, 2-methylbutane, and 2,2,4-trimethylpentane, are all saturated hydrocarbons, which means they only have single bonds between carbon atoms and are fully saturated with hydrogen atoms.
Pergunta 20 Relatório
What is the amount (in mole) of hydrogen gas that would be produced if 0.6 mole of hydrochloric acid reacted with excess zinc according to the following equation? Zn(s) + 2HCI(aq) → ZnCI2(aq) + H2(g)
Detalhes da Resposta
From the balanced chemical equation, it is observed that 1 mole of Zinc reacts with 2 moles of Hydrochloric acid to produce 1 mole of Hydrogen gas. Therefore, 0.6 mole of Hydrochloric acid would react with 0.3 mole of Zinc and produce 0.3 mole of Hydrogen gas. So the answer is (c) 0.3 mole.
Pergunta 21 Relatório
The most suitable method for preparing lead (ll) chloride is by
Detalhes da Resposta
The most suitable method for preparing lead (II) chloride is by mixing aqueous solutions of Pb(NO3)2 and NaCI. When aqueous solutions of lead nitrate and sodium chloride are mixed, a white precipitate of lead (II) chloride is formed. The other options mentioned in the question are not suitable for preparing lead (II) chloride. For example, the action of dilute HCl on PbSO4 will produce lead (II) sulfate, not lead (II) chloride. Similarly, the action of dilute HCl on lead will produce lead (II) chloride, but it is not the most suitable method for preparing lead (II) chloride. Therefore, the correct answer is "mixing aqueous solutions Pb(NO3)2 and NaCI."
Pergunta 22 Relatório
Which of the following is not correct of the reaction represented by the equation below? 2C(s) + O2(g) → 2CO(g); ∆H = - xKJ
Detalhes da Resposta
Pergunta 23 Relatório
In which of the following processes are larger molecules broken down into smaller molecules?
Detalhes da Resposta
The process in which larger molecules are broken down into smaller molecules is called "degradation" or "cracking". Out of the given options, the process that involves larger molecules being broken down into smaller molecules is "hydrolysis of starch". Hydrolysis is a chemical reaction that involves the breaking of a bond in a molecule using water. In the case of starch, it is broken down into smaller glucose molecules by the addition of water molecules, resulting in a simpler and smaller molecule. The other processes listed involve the formation or combination of molecules rather than their breakdown.
Pergunta 24 Relatório
The following can be obtained directly from the destructive distillation of coal except
Detalhes da Resposta
Destructive distillation is the process of heating coal in the absence of air to obtain useful products. During this process, coal is broken down into various components. Ammoniacal liquor, coal gas, coke, and coal tar are some of the products obtained from the destructive distillation of coal. However, producer gas is not obtained directly from the destructive distillation of coal. It is obtained by passing air over coke at a high temperature. Therefore, the correct option is "producer gas."
Pergunta 25 Relatório
What mass of copper will be deposited by the liberation of Cu2+ when 0.1F of electricity flows through an aqueous solution of a copper (ll) salt? [Cu = 64]
Detalhes da Resposta
The amount of substance that is deposited during an electrolysis reaction can be calculated using Faraday's laws of electrolysis. According to Faraday's laws, the amount of substance deposited is directly proportional to the amount of electricity that flows through the solution. The formula that relates the amount of substance to the amount of electricity is: Amount of substance = (Electricity in Coulombs) / (Faraday constant x Valency) where the Faraday constant is the amount of electric charge per mole of electrons, and the valency is the number of electrons involved in the redox reaction. For the given problem, the copper (II) ion has a valency of 2, and we are given that 0.1 F of electricity flows through the solution. The Faraday constant is 96,485 Coulombs per mole of electrons. Therefore, the amount of copper that will be deposited can be calculated as: Amount of substance = (0.1 F) / (2 x 96,485 C/mol) = 5.18 x 10^-5 mol The molar mass of copper is 64 g/mol, so the mass of copper that will be deposited is: Mass of copper = Amount of substance x Molar mass = 5.18 x 10^-5 mol x 64 g/mol = 0.00331 g ≈ 3.2 g (rounded to one decimal place) Therefore, the answer is 3.2g.
Pergunta 26 Relatório
The following alkanols will yield alkanoic acids on reacting with excess acidified K2Cr2O7 except
Detalhes da Resposta
The alkanols that will yield alkanoic acids on reacting with excess acidified K2Cr2O7 are primary and secondary alkanols. Tertiary alkanols do not undergo oxidation. Therefore, the correct answer is (CH3)3COH, which is a tertiary alcohol and cannot be oxidized to an alkanoic acid.
Pergunta 27 Relatório
Which of the following molecules is linear in shape?
Detalhes da Resposta
The shape of a molecule is determined by its electronic geometry, which depends on the arrangement of the atoms and the number of electron pairs around the central atom. A molecule is considered linear if it has two atoms and no lone pairs of electrons on the central atom. Based on this definition, the molecule that is linear in shape is CI2 because it has two chlorine atoms and no lone pairs of electrons on the central atom. All the other molecules listed have lone pairs or more than two atoms, which means they are not linear in shape.
Pergunta 28 Relatório
Consider the reaction represented by the equation below: XO + YO → X + YO2 In the reaction, YO acts as
Detalhes da Resposta
In the given chemical equation, XO is being reduced, and YO is being oxidized. An oxidizing agent is a substance that causes another substance to be oxidized, while a reducing agent is a substance that causes another substance to be reduced. YO is being reduced, which means it is causing the reduction of XO. Thus, YO is acting as a reducing agent.
Pergunta 29 Relatório
The gas produced when a mixture of sodium propanoate and soda lime is heated is
Detalhes da Resposta
Pergunta 30 Relatório
(a) Mention the chemical substance manufactured starting from each of the foirownc sets of materials:
(i) sugar and yeast;
(ii) ammonia, air and water;
(iii) vegetable oil and caustic alkali.
(b) State one air pollutant generated during the manufacture of fertilizers.
(a) Substances manufactured
- (i) Sugar and yeast: Ethanol (ethyl alcohol). The yeast supplies the enzyme zymase which ferments the sugar: \[C_6H_{12}O_{6} \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2\]
- (ii) Ammonia, air and water: Trioxonitrate(V) acid (nitric acid, \(HNO_3\)). In the Ostwald process ammonia is catalytically oxidised to nitrogen(II) oxide, further oxidised by air to nitrogen(IV) oxide, then dissolved in water: \[4NH_3 + 5O_2 \to 4NO + 6H_2O\]\[2NO + O_2 \to 2NO_2\]\[3NO_2 + H_2O \to 2HNO_3 + NO\]
- (iii) Vegetable oil and caustic alkali: Soap (by saponification, the alkaline hydrolysis of the oil).
(b) Air pollutant from fertilizer manufacture
One example is sulphur(IV) oxide, \(SO_2\) (released where sulphuric acid is used to make ammonium tetraoxosulphate(VI) fertilizer). Oxides of nitrogen or escaped ammonia are also acceptable.
Detalhes da Resposta
(a) Substances manufactured
- (i) Sugar and yeast: Ethanol (ethyl alcohol). The yeast supplies the enzyme zymase which ferments the sugar: \[C_6H_{12}O_{6} \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2\]
- (ii) Ammonia, air and water: Trioxonitrate(V) acid (nitric acid, \(HNO_3\)). In the Ostwald process ammonia is catalytically oxidised to nitrogen(II) oxide, further oxidised by air to nitrogen(IV) oxide, then dissolved in water: \[4NH_3 + 5O_2 \to 4NO + 6H_2O\]\[2NO + O_2 \to 2NO_2\]\[3NO_2 + H_2O \to 2HNO_3 + NO\]
- (iii) Vegetable oil and caustic alkali: Soap (by saponification, the alkaline hydrolysis of the oil).
(b) Air pollutant from fertilizer manufacture
One example is sulphur(IV) oxide, \(SO_2\) (released where sulphuric acid is used to make ammonium tetraoxosulphate(VI) fertilizer). Oxides of nitrogen or escaped ammonia are also acceptable.
Pergunta 31 Relatório
(a)(i) List the three types of particles present in atoms.
(ii) name the element which does not contain all the three particles in its atom. Mention the particle that is not present.
(b) Give the reason why:
(i) the relative atomic masses of some elements are not whole number;
(ii) relative atomic masses are used instead of the actual masses of atoms in grams;
(iii) metals are good conductors of electricity.
(c)(i) Name the type of bond present in the oxonium ion,
(ii) State one effect of the existence of intermolecular hydrogen bonding on the physical properties of ethanol.
(d)(i) Explain what is meant by water of crystallization.
(ii) When 5.0g of a compound Y was heated to constant mass, 1.8g of water vapour was given off. Determine the number of molecules of water of crystallization in one molecule of Y, given that the molar mass of its anhydrous form is 160g. [H = 1, 0 = 16]
(a)(i) Three types of particles in atoms
- Protons (positively charged, in the nucleus)
- Neutrons (uncharged, in the nucleus)
- Electrons (negatively charged, in the shells)
(a)(ii) The element is hydrogen (the ordinary isotope, \(^{1}_{1}H\)). The particle absent is the neutron (its atom has 1 proton and 1 electron only).
(b) Reasons
- (i) The relative atomic masses of some elements are not whole numbers because such elements exist as a mixture of isotopes; the relative atomic mass is the weighted average of the masses of the isotopes according to their relative abundances.
- (ii) Relative atomic masses are used instead of actual masses in grams because the actual mass of a single atom is extremely small (of the order of \(10^{-23}\) g) and inconvenient to handle, whereas relative values are simple comparative numbers.
- (iii) Metals are good conductors of electricity because they contain mobile (delocalised) free electrons that drift through the metallic lattice and carry charge.
(c)(i) The oxonium ion \(H_3O^+\) contains ordinary covalent bonds and one coordinate (dative) covalent bond (the oxygen lone pair is donated to the extra proton).
(c)(ii) Intermolecular hydrogen bonding raises the boiling point of ethanol (it boils much higher than expected for its molecular mass because extra energy is needed to break the hydrogen bonds). It also makes ethanol fully miscible with water.
(d)(i) Water of crystallisation is the fixed number of water molecules chemically combined with one formula unit of a salt in its crystal structure.
(d)(ii) Mass of anhydrous salt \(= 5.0 - 1.8 = 3.2\) g.
\[n(\text{anhydrous}) = \frac{3.2}{160} = 0.02\ \text{mol}\]\[n(H_2O) = \frac{1.8}{18} = 0.10\ \text{mol}\]\[\text{ratio } H_2O : \text{salt} = \frac{0.10}{0.02} = 5\]Therefore there are 5 molecules of water of crystallisation in one molecule of Y.
Detalhes da Resposta
(a)(i) Three types of particles in atoms
- Protons (positively charged, in the nucleus)
- Neutrons (uncharged, in the nucleus)
- Electrons (negatively charged, in the shells)
(a)(ii) The element is hydrogen (the ordinary isotope, \(^{1}_{1}H\)). The particle absent is the neutron (its atom has 1 proton and 1 electron only).
(b) Reasons
- (i) The relative atomic masses of some elements are not whole numbers because such elements exist as a mixture of isotopes; the relative atomic mass is the weighted average of the masses of the isotopes according to their relative abundances.
- (ii) Relative atomic masses are used instead of actual masses in grams because the actual mass of a single atom is extremely small (of the order of \(10^{-23}\) g) and inconvenient to handle, whereas relative values are simple comparative numbers.
- (iii) Metals are good conductors of electricity because they contain mobile (delocalised) free electrons that drift through the metallic lattice and carry charge.
(c)(i) The oxonium ion \(H_3O^+\) contains ordinary covalent bonds and one coordinate (dative) covalent bond (the oxygen lone pair is donated to the extra proton).
(c)(ii) Intermolecular hydrogen bonding raises the boiling point of ethanol (it boils much higher than expected for its molecular mass because extra energy is needed to break the hydrogen bonds). It also makes ethanol fully miscible with water.
(d)(i) Water of crystallisation is the fixed number of water molecules chemically combined with one formula unit of a salt in its crystal structure.
(d)(ii) Mass of anhydrous salt \(= 5.0 - 1.8 = 3.2\) g.
\[n(\text{anhydrous}) = \frac{3.2}{160} = 0.02\ \text{mol}\]\[n(H_2O) = \frac{1.8}{18} = 0.10\ \text{mol}\]\[\text{ratio } H_2O : \text{salt} = \frac{0.10}{0.02} = 5\]Therefore there are 5 molecules of water of crystallisation in one molecule of Y.
Pergunta 32 Relatório
(a)(i) Explain the term pH
(ii) If sodium hydroxide solution were added to a solution of a strong acid, what would happen to the pH of the solution?
(b) Give one example of each of the following:
(i) acidic oxide
(ii) acid salt.
(a)(i) The pH of a solution is a measure of its acidity or alkalinity; it is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii) Adding sodium hydroxide to a strong acid neutralizes some of the H+ ions, so the pH increases (the solution becomes less acidic and moves toward, then above, 7).
(b)(i) Acidic oxide: carbon(IV) oxide, CO2 (or SO2, SO3).
(ii) Acid salt: sodium hydrogen trioxocarbonate(IV), NaHCO3 (or sodium hydrogen tetraoxosulphate(VI), NaHSO4).
Detalhes da Resposta
(a)(i) The pH of a solution is a measure of its acidity or alkalinity; it is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii) Adding sodium hydroxide to a strong acid neutralizes some of the H+ ions, so the pH increases (the solution becomes less acidic and moves toward, then above, 7).
(b)(i) Acidic oxide: carbon(IV) oxide, CO2 (or SO2, SO3).
(ii) Acid salt: sodium hydrogen trioxocarbonate(IV), NaHCO3 (or sodium hydrogen tetraoxosulphate(VI), NaHSO4).
Pergunta 33 Relatório
(a) List three properties of a system that is in a state of chemical equilibrium.
(b) Consider reaction represented by the following equation: 3H\(_{2(g)}\) + N\(_{2(g)}\) \(\rightleftharpoons\) 2NH\(_{3(g)}\); H = 92KJ
(i) Explain the effect of increasing the temperature of the reaction on the yield of ammonia
(ii) Uses of energy profile diagram to illustrate the effect of a positive catalyst on the rate of either the forward reaction or the reverse reaction.
(c) In the extraction of aluminium from bauxite:
(i) outline the procedure used for purifying the ore;
(ii) write equation for the reaction at each electrode, during the electrolysis of the pure alumina;
(iii) state the function of molten cryolite in the electrolytic cell for the extraction.
(a) Three properties of a system in a state of chemical equilibrium
- The change is dynamic: the forward and reverse reactions continue simultaneously and the rate of the forward reaction equals the rate of the reverse reaction.
- The macroscopic properties remain constant with time: the concentrations of reactants and products (and observable properties such as colour and pressure) do not change once equilibrium is reached.
- It is reached only in a closed system, and at equilibrium the free-energy change of the system is zero (\(\Delta G = 0\)).
(b) \(3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)};\ \Delta H = -92\ \text{kJ}\) (the forward reaction is exothermic).
(i) Effect of increasing the temperature on the yield of ammonia
Since the forward reaction is exothermic, by Le Chatelier's principle an increase in temperature favours the endothermic (reverse) direction. The equilibrium therefore shifts to the left, so the yield of ammonia decreases as the temperature is raised.
(ii) Effect of a positive catalyst (energy profile diagram)
A positive catalyst provides an alternative reaction pathway of lower activation energy. On the energy profile below, the peak of the catalysed curve (green, dashed) is lower than that of the uncatalysed curve (red, solid), while the energy levels of the reactants and of the products are unchanged. Because the activation-energy barrier is lowered by the same amount for both directions, the catalyst speeds up the forward and the reverse reactions equally; equilibrium is reached faster, but the position of equilibrium (and hence the yield) is unchanged.
(c) Extraction of aluminium from bauxite
(i) Purifying the ore (Bayer process): the powdered bauxite is dissolved, under pressure, in hot concentrated sodium hydroxide solution. The amphoteric aluminium oxide dissolves as sodium aluminate, while insoluble impurities such as \(Fe_2O_3\) and \(SiO_2\) do not dissolve and are filtered off:
\[Al_2O_{3(s)} + 2NaOH_{(aq)} + 3H_2O_{(l)} \rightarrow 2NaAl(OH)_{4(aq)}\]
The filtered sodium aluminate solution is then diluted and seeded so that aluminium hydroxide precipitates out:
\[NaAl(OH)_{4(aq)} \rightarrow Al(OH)_{3(s)} + NaOH_{(aq)}\]
The aluminium hydroxide is filtered off, washed, dried and heated strongly (calcined) to give pure alumina:
\[2Al(OH)_{3(s)} \xrightarrow{\text{heat}} Al_2O_{3(s)} + 3H_2O_{(g)}\]
(ii) Electrode reactions during the electrolysis of the pure molten alumina:
\[\text{Cathode (reduction): } Al^{3+} + 3e^- \rightarrow Al\]
\[\text{Anode (oxidation): } 2O^{2-} \rightarrow O_2 + 4e^-\]
(iii) Function of molten cryolite: it acts as a solvent (flux) for the alumina and lowers the melting point of the electrolyte from about 2045 °C to about 950 °C, so that electrolysis can be carried out at a much lower temperature while also increasing the electrical conductivity of the melt.
Detalhes da Resposta
(a) Three properties of a system in a state of chemical equilibrium
- The change is dynamic: the forward and reverse reactions continue simultaneously and the rate of the forward reaction equals the rate of the reverse reaction.
- The macroscopic properties remain constant with time: the concentrations of reactants and products (and observable properties such as colour and pressure) do not change once equilibrium is reached.
- It is reached only in a closed system, and at equilibrium the free-energy change of the system is zero (\(\Delta G = 0\)).
(b) \(3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)};\ \Delta H = -92\ \text{kJ}\) (the forward reaction is exothermic).
(i) Effect of increasing the temperature on the yield of ammonia
Since the forward reaction is exothermic, by Le Chatelier's principle an increase in temperature favours the endothermic (reverse) direction. The equilibrium therefore shifts to the left, so the yield of ammonia decreases as the temperature is raised.
(ii) Effect of a positive catalyst (energy profile diagram)
A positive catalyst provides an alternative reaction pathway of lower activation energy. On the energy profile below, the peak of the catalysed curve (green, dashed) is lower than that of the uncatalysed curve (red, solid), while the energy levels of the reactants and of the products are unchanged. Because the activation-energy barrier is lowered by the same amount for both directions, the catalyst speeds up the forward and the reverse reactions equally; equilibrium is reached faster, but the position of equilibrium (and hence the yield) is unchanged.
(c) Extraction of aluminium from bauxite
(i) Purifying the ore (Bayer process): the powdered bauxite is dissolved, under pressure, in hot concentrated sodium hydroxide solution. The amphoteric aluminium oxide dissolves as sodium aluminate, while insoluble impurities such as \(Fe_2O_3\) and \(SiO_2\) do not dissolve and are filtered off:
\[Al_2O_{3(s)} + 2NaOH_{(aq)} + 3H_2O_{(l)} \rightarrow 2NaAl(OH)_{4(aq)}\]
The filtered sodium aluminate solution is then diluted and seeded so that aluminium hydroxide precipitates out:
\[NaAl(OH)_{4(aq)} \rightarrow Al(OH)_{3(s)} + NaOH_{(aq)}\]
The aluminium hydroxide is filtered off, washed, dried and heated strongly (calcined) to give pure alumina:
\[2Al(OH)_{3(s)} \xrightarrow{\text{heat}} Al_2O_{3(s)} + 3H_2O_{(g)}\]
(ii) Electrode reactions during the electrolysis of the pure molten alumina:
\[\text{Cathode (reduction): } Al^{3+} + 3e^- \rightarrow Al\]
\[\text{Anode (oxidation): } 2O^{2-} \rightarrow O_2 + 4e^-\]
(iii) Function of molten cryolite: it acts as a solvent (flux) for the alumina and lowers the melting point of the electrolyte from about 2045 °C to about 950 °C, so that electrolysis can be carried out at a much lower temperature while also increasing the electrical conductivity of the melt.
Pergunta 34 Relatório
(a)(i) Sketch a graph to illustrate Charles' law.
(ii) A gas occupies 500 cm\(^{ 3}\) at 2TC. calculate its volume at 40°C constant pressure.
(b) List two gases that are used as refrigerant
(a)(i) Charles' law states that at constant pressure the volume of a fixed mass of gas is directly proportional to its absolute (kelvin) temperature. A graph of volume against absolute temperature is therefore a straight line passing through the origin:
The line is straight and, when produced, passes through the origin (0 K, 0 cm\(^3\)), showing \(V \propto T\) at constant pressure. If the same volume is plotted against temperature in \(^{\circ}\)C, the straight line is displaced and cuts the temperature axis at \(-273\ ^{\circ}\text{C}\) (absolute zero).
(a)(ii) Convert both temperatures to kelvin:
\[T_1 = 27 + 273 = 300\ \text{K}, \qquad T_2 = 40 + 273 = 313\ \text{K}, \qquad V_1 = 500\ \text{cm}^3\]At constant pressure, Charles' law gives:
\[\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad\Rightarrow\quad \frac{500}{300} = \frac{V_2}{313}\] \[V_2 = \frac{500 \times 313}{300} = 521.7 \approx 522\ \text{cm}^3\]The gas occupies 522 cm\(^3\) at 40 \(^{\circ}\)C.
(b) Two gases used as refrigerants:
- Ammonia (NH\(_3\))
- Carbon dioxide (CO\(_2\))
Detalhes da Resposta
(a)(i) Charles' law states that at constant pressure the volume of a fixed mass of gas is directly proportional to its absolute (kelvin) temperature. A graph of volume against absolute temperature is therefore a straight line passing through the origin:
The line is straight and, when produced, passes through the origin (0 K, 0 cm\(^3\)), showing \(V \propto T\) at constant pressure. If the same volume is plotted against temperature in \(^{\circ}\)C, the straight line is displaced and cuts the temperature axis at \(-273\ ^{\circ}\text{C}\) (absolute zero).
(a)(ii) Convert both temperatures to kelvin:
\[T_1 = 27 + 273 = 300\ \text{K}, \qquad T_2 = 40 + 273 = 313\ \text{K}, \qquad V_1 = 500\ \text{cm}^3\]At constant pressure, Charles' law gives:
\[\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad\Rightarrow\quad \frac{500}{300} = \frac{V_2}{313}\] \[V_2 = \frac{500 \times 313}{300} = 521.7 \approx 522\ \text{cm}^3\]The gas occupies 522 cm\(^3\) at 40 \(^{\circ}\)C.
(b) Two gases used as refrigerants:
- Ammonia (NH\(_3\))
- Carbon dioxide (CO\(_2\))
Pergunta 35 Relatório
(a) Mention one process apart from respiration, which increases the amount of carbon (IV) oxide in the atmosphere.
(b)(i) State one use of sodium hydrogentrioxocarbonate (IV).
(ii) Write an equation to show the action of heat on sodium hydrogentrioxocarbonate(IV).
(a) Process (other than respiration) that increases atmospheric \(CO_2\)
Combustion of fuels (the burning of wood, coal, petroleum and other fossil fuels). The decay of dead organic matter by bacteria is also acceptable.
(b)(i) Use of sodium hydrogentrioxocarbonate(IV), \(NaHCO_3\)
It is used as baking powder in bread and cake making (it releases \(CO_2\) which makes the dough rise). It is also used as an antacid to relieve stomach acidity and as a component of dry-powder fire extinguishers.
(b)(ii) Action of heat on \(NaHCO_3\)
On heating it decomposes to give sodium trioxocarbonate(IV), water and carbon(IV) oxide:
\[2NaHCO_3 \xrightarrow{\text{heat}} Na_2CO_3 + H_2O + CO_2\]Detalhes da Resposta
(a) Process (other than respiration) that increases atmospheric \(CO_2\)
Combustion of fuels (the burning of wood, coal, petroleum and other fossil fuels). The decay of dead organic matter by bacteria is also acceptable.
(b)(i) Use of sodium hydrogentrioxocarbonate(IV), \(NaHCO_3\)
It is used as baking powder in bread and cake making (it releases \(CO_2\) which makes the dough rise). It is also used as an antacid to relieve stomach acidity and as a component of dry-powder fire extinguishers.
(b)(ii) Action of heat on \(NaHCO_3\)
On heating it decomposes to give sodium trioxocarbonate(IV), water and carbon(IV) oxide:
\[2NaHCO_3 \xrightarrow{\text{heat}} Na_2CO_3 + H_2O + CO_2\]Pergunta 36 Relatório
(a) Name the device used for producing an electric current from a chemical
(b) Copy and complete the table below.
| Electrolyte | Product at the anode (carbon) | Product at the cathode (carbon) |
| Dilute \( \mathrm{NaCl}_{(aq)} \) Concentrated \( \mathrm{NaCI}_{(aq)} \) |
(a) The device used for producing an electric current from a chemical reaction is an electrochemical cell (also called a galvanic or voltaic cell, or simply a chemical cell).
(b) Completed table
Reasoning:
- In dilute \( NaCl_{(aq)} \) the solution is mostly water. At the anode, \( OH^- \) is discharged in preference to the dilute \( Cl^- \), giving oxygen. At the cathode, \( H^+ \) is discharged in preference to \( Na^+ \) (sodium is high in the electrochemical series), giving hydrogen.
- In concentrated \( NaCl_{(aq)} \) the high concentration of \( Cl^- \) causes chloride to be discharged at the anode instead of \( OH^- \), giving chlorine. Hydrogen is still liberated at the cathode.
| Electrolyte | Product at the anode (carbon) | Product at the cathode (carbon) |
|---|---|---|
| Dilute \( NaCl_{(aq)} \) | Oxygen, \( O_2 \) | Hydrogen, \( H_2 \) |
| Concentrated \( NaCl_{(aq)} \) | Chlorine, \( Cl_2 \) | Hydrogen, \( H_2 \) |
Electrode half-equations: at the cathode \( 2H^+ + 2e^- \rightarrow H_2 \); at the dilute anode \( 4OH^- \rightarrow O_2 + 2H_2O + 4e^- \); at the concentrated anode \( 2Cl^- \rightarrow Cl_2 + 2e^- \).
Detalhes da Resposta
(a) The device used for producing an electric current from a chemical reaction is an electrochemical cell (also called a galvanic or voltaic cell, or simply a chemical cell).
(b) Completed table
Reasoning:
- In dilute \( NaCl_{(aq)} \) the solution is mostly water. At the anode, \( OH^- \) is discharged in preference to the dilute \( Cl^- \), giving oxygen. At the cathode, \( H^+ \) is discharged in preference to \( Na^+ \) (sodium is high in the electrochemical series), giving hydrogen.
- In concentrated \( NaCl_{(aq)} \) the high concentration of \( Cl^- \) causes chloride to be discharged at the anode instead of \( OH^- \), giving chlorine. Hydrogen is still liberated at the cathode.
| Electrolyte | Product at the anode (carbon) | Product at the cathode (carbon) |
|---|---|---|
| Dilute \( NaCl_{(aq)} \) | Oxygen, \( O_2 \) | Hydrogen, \( H_2 \) |
| Concentrated \( NaCl_{(aq)} \) | Chlorine, \( Cl_2 \) | Hydrogen, \( H_2 \) |
Electrode half-equations: at the cathode \( 2H^+ + 2e^- \rightarrow H_2 \); at the dilute anode \( 4OH^- \rightarrow O_2 + 2H_2O + 4e^- \); at the concentrated anode \( 2Cl^- \rightarrow Cl_2 + 2e^- \).
Pergunta 37 Relatório
(a) List two differences between solids and liquids.
(b) The graph below is the heating curve for a solid X. Use the graph to answer Questions (i) — (iii) below.
(i) What is the melting point of X?
(ii) If the vapour of X is cooled, at what temperature will it start to condense?
(iii) (I) As X is heated, state what happens to the: I. frequency of collision of molecules of X;
(II) value of the entropy of the system.
(a) Two differences between solids and liquids
| Solids | Liquids |
|---|---|
| Have a definite (fixed) shape of their own. | Have no definite shape; they take the shape of the container. |
| Particles are closely packed and held in fixed positions, so solids are almost incompressible and do not flow. | Particles are more loosely held and can slide over one another, so liquids flow. |
(b) The heating curve for solid X is shown below. Each horizontal (flat) portion is a change of state, where the added heat is used to break the forces between particles instead of raising the temperature.
(i) Melting point of X
The melting point is the temperature of the first flat portion of the curve, where the solid is turning into liquid at constant temperature.
\[ \text{Melting point of X} = 60\,^{\circ}\text{C} \](ii) Temperature at which the vapour of X starts to condense
Condensation is the reverse of boiling, so the vapour condenses at the boiling point of X. This is the temperature of the second flat portion of the curve.
\[ \text{Condensation temperature} = 210\,^{\circ}\text{C} \](iii) As X is heated:
(I) Frequency of collision of molecules of X: it increases. Heating gives the molecules more kinetic energy, so they move faster and collide with one another more often.
(II) Value of the entropy of the system: it increases. As X changes from solid to liquid to gas the particles become more disordered and more freely arranged, so the entropy of the system rises.
Detalhes da Resposta
(a) Two differences between solids and liquids
| Solids | Liquids |
|---|---|
| Have a definite (fixed) shape of their own. | Have no definite shape; they take the shape of the container. |
| Particles are closely packed and held in fixed positions, so solids are almost incompressible and do not flow. | Particles are more loosely held and can slide over one another, so liquids flow. |
(b) The heating curve for solid X is shown below. Each horizontal (flat) portion is a change of state, where the added heat is used to break the forces between particles instead of raising the temperature.
(i) Melting point of X
The melting point is the temperature of the first flat portion of the curve, where the solid is turning into liquid at constant temperature.
\[ \text{Melting point of X} = 60\,^{\circ}\text{C} \](ii) Temperature at which the vapour of X starts to condense
Condensation is the reverse of boiling, so the vapour condenses at the boiling point of X. This is the temperature of the second flat portion of the curve.
\[ \text{Condensation temperature} = 210\,^{\circ}\text{C} \](iii) As X is heated:
(I) Frequency of collision of molecules of X: it increases. Heating gives the molecules more kinetic energy, so they move faster and collide with one another more often.
(II) Value of the entropy of the system: it increases. As X changes from solid to liquid to gas the particles become more disordered and more freely arranged, so the entropy of the system rises.
Pergunta 38 Relatório
What name is given to each of the following? The
(a) irregular random movement of smoke particles in air.
(b) existence of an element in various forms in the same physical-state
(c) disintegration of atomic nuclei, accompanied with radiation emission.
(d) conversion of a solid difitrunto vapour without melting.
(a) The irregular random movement of smoke particles in air - Brownian motion.
(b) The existence of an element in various forms in the same physical state - allotropy.
(c) The disintegration of atomic nuclei accompanied by the emission of radiation - radioactivity.
(d) The conversion of a solid directly into vapour without melting - sublimation.
Detalhes da Resposta
(a) The irregular random movement of smoke particles in air - Brownian motion.
(b) The existence of an element in various forms in the same physical state - allotropy.
(c) The disintegration of atomic nuclei accompanied by the emission of radiation - radioactivity.
(d) The conversion of a solid directly into vapour without melting - sublimation.
Pergunta 39 Relatório
(a) State the type of reaction involved in the conversion of:
(i) proteins.to amino acids:
(ii) ethanol to ethene;
(iii) benzene to bromobenzene
(b) Write an equation to show that ethene reasts with hydrogen in the presence of finely divided nickel.
(a)
- (i) Proteins to amino acids: hydrolysis.
- (ii) Ethanol to ethene: dehydration (an elimination reaction).
- (iii) Benzene to bromobenzene: substitution (halogenation).
(b) Ethene reacts with hydrogen over finely divided nickel to give ethane:
\[C_2H_4 + H_2 \xrightarrow{Ni} C_2H_6\]
Detalhes da Resposta
(a)
- (i) Proteins to amino acids: hydrolysis.
- (ii) Ethanol to ethene: dehydration (an elimination reaction).
- (iii) Benzene to bromobenzene: substitution (halogenation).
(b) Ethene reacts with hydrogen over finely divided nickel to give ethane:
\[C_2H_4 + H_2 \xrightarrow{Ni} C_2H_6\]
Pergunta 40 Relatório
(a)(i) What is isomerism?
(ii) Name the alkanol that is isomeric with methoxymethane (CH\(_3\)OCH\(_3\)).
(b)(i) Outline the laboratory:preparation of ethylethanoate. (Diagrams not required)
(ii) Write the structural formula of ethylethancqte
(iii) State two physical properties of ethylethanoate.
(c) When gas oil which consists of larger hydrocarbons was subjected to high temperature and pressure, the following reaction occurred.
C\(_{17}\)H\(_{36(l)}\) \(\to\) 3C\(_2\)H\(_{4(g)}\) + C\(_3\)H\(_{6(g)}\) + Q\(_{(l)}\)
(i) What name is given to the process indicated above?
(ii) State the importance of the process to the petroleum industry.
(iii) Find the formula of the product which Q represents in the equation above.
(iv) Mention one type of chemical industry that utilizes ethene as raw material.
(d) Consider the following compounds: CH\(_3\) — (CH\(_2\))\(_2\) —CH\(_3\); C\(_6\)H\(_5\) —CH = CH\(_2\); CH=C — CH\(_3\). State which of them:
(i) is used as a domestic fuel;
(ii) is an aromatic compound,
(iii) participates in such situation but not addition reactions;
(iv) would react with two moles of hydrogen per mole.
(a)(i) Isomerism is the existence of two or more compounds having the same molecular formula but different structural arrangements of their atoms (and hence different properties).
(a)(ii) Methoxymethane \(CH_3OCH_3\) has molecular formula \(C_2H_6O\). The isomeric alkanol is ethanol, \(C_2H_5OH\).
(b)(i) Laboratory preparation of ethyl ethanoate
- Warm a mixture of ethanoic acid and ethanol together with a little concentrated tetraoxosulphate(VI) acid (which acts as catalyst and dehydrating agent) under reflux.
- Esterification occurs: \[CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O\]
- Distil off the ester and purify it by shaking with sodium trioxocarbonate(IV) solution (to remove acid) followed by drying and redistillation.
(b)(ii) Structural formula: \(CH_3-CO-O-CH_2-CH_3\) (i.e. \(CH_3COOC_2H_5\)).
(b)(iii) Two physical properties: it is a colourless liquid with a sweet, fruity smell; it is only slightly soluble in water but volatile.
(c) \(C_{17}H_{36} \to 3C_2H_4 + C_3H_6 + Q\)
- (i) The process is cracking (thermal/catalytic cracking).
- (ii) It is important because it breaks large, less useful hydrocarbon molecules into smaller, more useful ones (petrol and gaseous alkenes), increasing the yield of petrol and supplying alkenes as raw materials.
- (iii) Balancing carbon: \(17 = (3\times2) + 3 + C_Q \Rightarrow C_Q = 8\). Balancing hydrogen: \(36 = (3\times4) + 6 + H_Q \Rightarrow H_Q = 18\). Therefore Q is \(C_8H_{18}\) (octane).
- (iv) The plastics (polymer) industry, which polymerises ethene to poly(ethene). The manufacture of ethanol is also acceptable.
(d) The compounds are butane \(CH_3(CH_2)_2CH_3\), phenylethene \(C_6H_5CH=CH_2\), and propyne \(CH\equiv C-CH_3\).
- (i) Domestic fuel: butane, \(CH_3(CH_2)_2CH_3\).
- (ii) Aromatic compound: phenylethene (styrene), \(C_6H_5CH=CH_2\).
- (iii) Undergoes substitution but not addition: butane (a saturated alkane).
- (iv) Reacts with two moles of hydrogen per mole: propyne, \(CH\equiv C-CH_3\) (the triple bond adds \(2H_2\)).
Detalhes da Resposta
(a)(i) Isomerism is the existence of two or more compounds having the same molecular formula but different structural arrangements of their atoms (and hence different properties).
(a)(ii) Methoxymethane \(CH_3OCH_3\) has molecular formula \(C_2H_6O\). The isomeric alkanol is ethanol, \(C_2H_5OH\).
(b)(i) Laboratory preparation of ethyl ethanoate
- Warm a mixture of ethanoic acid and ethanol together with a little concentrated tetraoxosulphate(VI) acid (which acts as catalyst and dehydrating agent) under reflux.
- Esterification occurs: \[CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O\]
- Distil off the ester and purify it by shaking with sodium trioxocarbonate(IV) solution (to remove acid) followed by drying and redistillation.
(b)(ii) Structural formula: \(CH_3-CO-O-CH_2-CH_3\) (i.e. \(CH_3COOC_2H_5\)).
(b)(iii) Two physical properties: it is a colourless liquid with a sweet, fruity smell; it is only slightly soluble in water but volatile.
(c) \(C_{17}H_{36} \to 3C_2H_4 + C_3H_6 + Q\)
- (i) The process is cracking (thermal/catalytic cracking).
- (ii) It is important because it breaks large, less useful hydrocarbon molecules into smaller, more useful ones (petrol and gaseous alkenes), increasing the yield of petrol and supplying alkenes as raw materials.
- (iii) Balancing carbon: \(17 = (3\times2) + 3 + C_Q \Rightarrow C_Q = 8\). Balancing hydrogen: \(36 = (3\times4) + 6 + H_Q \Rightarrow H_Q = 18\). Therefore Q is \(C_8H_{18}\) (octane).
- (iv) The plastics (polymer) industry, which polymerises ethene to poly(ethene). The manufacture of ethanol is also acceptable.
(d) The compounds are butane \(CH_3(CH_2)_2CH_3\), phenylethene \(C_6H_5CH=CH_2\), and propyne \(CH\equiv C-CH_3\).
- (i) Domestic fuel: butane, \(CH_3(CH_2)_2CH_3\).
- (ii) Aromatic compound: phenylethene (styrene), \(C_6H_5CH=CH_2\).
- (iii) Undergoes substitution but not addition: butane (a saturated alkane).
- (iv) Reacts with two moles of hydrogen per mole: propyne, \(CH\equiv C-CH_3\) (the triple bond adds \(2H_2\)).
Pergunta 41 Relatório
(a) What is the change in oxidation state of chromium in the reaction represented by the following equation?
3SO\(_2\) + Cr\(_2\)O\(^2_{-7}\) + 2H\(^+\) -> 3SO\(_4^{2-}\) + 2Cr\(^{3+}\) + H\(_2\)O
(b) Use the half equations given below to deduce the equation for the reaction between iron(II) ions and heptaoxodichromate (VI) ions in acidic solution.
Fe\(^{2+}\) --> Fe\(^{3+}\) + e\(^-\)
Cr\(_2\)O\(^{2-}_7\) + 14H\(^+\) + 6e\(^-\) ----> 2Cr\(^{3+}\) + 7H\(_2\)O.
Pergunta 42 Relatório
(a) State three reasons why air is classified as a mixture.
(b) List two methods that can be used to separate a mixture of iodine crystals and iron filings.
(a) Air is classified as a mixture because:
- Its components can be separated by physical means (fractional distillation of liquid air).
- Its components retain their own individual properties.
- Its composition is variable (the proportions are not fixed) and the components are not combined in any fixed ratio, and no chemical reaction or energy change occurs when they are together.
(b) A mixture of iodine crystals and iron filings can be separated by:
- Using a magnet to remove the iron filings (magnetic separation).
- Sublimation - warming the mixture so the iodine sublimes and is collected, leaving the iron filings behind.
Detalhes da Resposta
(a) Air is classified as a mixture because:
- Its components can be separated by physical means (fractional distillation of liquid air).
- Its components retain their own individual properties.
- Its composition is variable (the proportions are not fixed) and the components are not combined in any fixed ratio, and no chemical reaction or energy change occurs when they are together.
(b) A mixture of iodine crystals and iron filings can be separated by:
- Using a magnet to remove the iron filings (magnetic separation).
- Sublimation - warming the mixture so the iodine sublimes and is collected, leaving the iron filings behind.
Pergunta 43 Relatório
(a)(i) Explain what is meant by saturated solution
(ii) Describe in outline, a suitable procedure for preparing a saturated solution of sodium trioxonitrate(V) at 30°C.
(ii) State two techniques that can be used to recover crystals of sodium trioxonitrate(V) from its saturated solution.
(b) 1.0dm\(^3\) of an aqueous solution at 90°C contains 404g of potassium trioxonitrate(V) and 245g of potassium trioxochlorate (V).
(i) Determine which of the two salts will separate out when the solution is cooled to 60°C. N = 14. O = 16, CI = 35.5, K = 39; Solubility of KNO\(_3\) in water at 60\(^o\)C = 5.14 mol.dm\(^{-3}\), Solubility of KCIO\(_3\) in water at 60°C = 1.61 mol.dm\(^{-3}\)
(ii) Calculate the mass of salt that will separate out at 60°C
(c)(i) List two salts which cause hardness of water.
(ii) Explain why temporary hardness of water result in the furring of kettle.
(a)(i) A saturated solution is a solution that contains the maximum amount of dissolved solute it can hold at a given temperature, in the presence of (and in equilibrium with) undissolved solute.
(a)(ii) Preparing a saturated solution of \(NaNO_3\) at 30 C
- Warm some distilled water and keep it at 30 C in a thermostatic bath.
- Add sodium trioxonitrate(V) a little at a time, stirring, until no more will dissolve and some solid remains undissolved at 30 C.
- Filter off the excess undissolved solid; the filtrate is the saturated solution at 30 C.
(a)(iii) Two techniques to recover the crystals: evaporation of the solvent (evaporate to the point of crystallisation and allow to cool) and cooling/crystallisation of the hot saturated solution.
(b) Molar masses: \(KNO_3 = 39+14+48 = 101\ \text{g mol}^{-1}\); \(KClO_3 = 39+35.5+48 = 122.5\ \text{g mol}^{-1}\).
Amounts present in \(1.0\ dm^3\):
\[n(KNO_3) = \frac{404}{101} = 4.0\ \text{mol dm}^{-3}\]\[n(KClO_3) = \frac{245}{122.5} = 2.0\ \text{mol dm}^{-3}\](i) At 60 C the solubility of \(KNO_3\) is \(5.14\ \text{mol dm}^{-3}\); since \(4.0 < 5.14\), all the \(KNO_3\) stays in solution. The solubility of \(KClO_3\) is only \(1.61\ \text{mol dm}^{-3}\); since \(2.0 > 1.61\), the excess crystallises. Therefore potassium trioxochlorate(V), \(KClO_3\), separates out.
(ii) Amount separating \(= 2.0 - 1.61 = 0.39\ \text{mol}\).
\[\text{mass} = 0.39 \times 122.5 = 47.8\ \text{g}\](c)(i) Two salts causing hardness: calcium hydrogentrioxocarbonate(IV), \(Ca(HCO_3)_2\), and calcium tetraoxosulphate(VI), \(CaSO_4\) (magnesium salts also cause hardness).
(c)(ii) Temporary hardness furs a kettle because, on boiling, the dissolved \(Ca(HCO_3)_2\) decomposes to insoluble calcium trioxocarbonate(IV), which deposits as a hard scale (fur) on the kettle: \[Ca(HCO_3)_2 \xrightarrow{\text{heat}} CaCO_3 + H_2O + CO_2\]
Detalhes da Resposta
(a)(i) A saturated solution is a solution that contains the maximum amount of dissolved solute it can hold at a given temperature, in the presence of (and in equilibrium with) undissolved solute.
(a)(ii) Preparing a saturated solution of \(NaNO_3\) at 30 C
- Warm some distilled water and keep it at 30 C in a thermostatic bath.
- Add sodium trioxonitrate(V) a little at a time, stirring, until no more will dissolve and some solid remains undissolved at 30 C.
- Filter off the excess undissolved solid; the filtrate is the saturated solution at 30 C.
(a)(iii) Two techniques to recover the crystals: evaporation of the solvent (evaporate to the point of crystallisation and allow to cool) and cooling/crystallisation of the hot saturated solution.
(b) Molar masses: \(KNO_3 = 39+14+48 = 101\ \text{g mol}^{-1}\); \(KClO_3 = 39+35.5+48 = 122.5\ \text{g mol}^{-1}\).
Amounts present in \(1.0\ dm^3\):
\[n(KNO_3) = \frac{404}{101} = 4.0\ \text{mol dm}^{-3}\]\[n(KClO_3) = \frac{245}{122.5} = 2.0\ \text{mol dm}^{-3}\](i) At 60 C the solubility of \(KNO_3\) is \(5.14\ \text{mol dm}^{-3}\); since \(4.0 < 5.14\), all the \(KNO_3\) stays in solution. The solubility of \(KClO_3\) is only \(1.61\ \text{mol dm}^{-3}\); since \(2.0 > 1.61\), the excess crystallises. Therefore potassium trioxochlorate(V), \(KClO_3\), separates out.
(ii) Amount separating \(= 2.0 - 1.61 = 0.39\ \text{mol}\).
\[\text{mass} = 0.39 \times 122.5 = 47.8\ \text{g}\](c)(i) Two salts causing hardness: calcium hydrogentrioxocarbonate(IV), \(Ca(HCO_3)_2\), and calcium tetraoxosulphate(VI), \(CaSO_4\) (magnesium salts also cause hardness).
(c)(ii) Temporary hardness furs a kettle because, on boiling, the dissolved \(Ca(HCO_3)_2\) decomposes to insoluble calcium trioxocarbonate(IV), which deposits as a hard scale (fur) on the kettle: \[Ca(HCO_3)_2 \xrightarrow{\text{heat}} CaCO_3 + H_2O + CO_2\]
