WAEC SSCE - Chemistry - 1997
Pergunta 1 Relatório
Separation of mixture of solids by physical methods can be based on differences in the following except
Detalhes da Resposta
Physical methods of separation involve the use of physical properties to separate mixtures of substances. Some physical properties that can be used for separation include melting point, solubility, particle size, molar mass, and magnetic property. However, molar mass is not typically used as a basis for separation of mixtures of solids by physical methods, as it is a property that relates to the composition of individual substances, rather than the mixture as a whole. Therefore, the answer is "molar mass."
Pergunta 2 Relatório
Which of the following Group 1 elements has the highest ionization energy?
Detalhes da Resposta
Ionization energy is the energy required to remove an electron from an atom. The higher the ionization energy, the more difficult it is to remove an electron. As we move down Group 1 of the periodic table, the atomic radius increases, and the outermost electron is farther away from the positively charged nucleus. As a result, the ionization energy decreases. Therefore, the element with the smallest atomic radius in Group 1, which is Li, will have the highest ionization energy. Therefore, the answer is: 3Li.
Pergunta 3 Relatório
A given volume of methane diffuses in 20 seconds. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions? [CH4 = 16; SO2 = 64]
Detalhes da Resposta
The rate of diffusion of a gas is inversely proportional to the square root of its molar mass. This means that the heavier the gas, the slower it diffuses. The molar mass of methane (CH4) is 16, while the molar mass of sulphur (IV) oxide (SO2) is 64. Therefore, the ratio of the square root of the molar masses of methane to sulphur (IV) oxide is: √(64/16) = √4 = 2 This means that sulphur (IV) oxide diffuses at a rate that is 2 times slower than methane under the same conditions. Since it takes methane 20 seconds to diffuse, it will take sulphur (IV) oxide 2 times longer, which is: 20 x 2 = 40 seconds Therefore, the answer is 40 seconds.
Pergunta 4 Relatório
How much heat will be liberated if 8g of hydrogen burns in excess oxygen according to the following equation? 2H2 + O2→ 2H2O,∆H = -571.7KJ (H = 1)
Detalhes da Resposta
Pergunta 6 Relatório
Which of the following is the relative molecular mass of a compound which has empirical formula CH2O? (H = 1; C = 12; O = 16)
Detalhes da Resposta
Pergunta 7 Relatório
Radon is used as a tracer in medical research because it
Detalhes da Resposta
Radon is used as a tracer in medical research because it is radioactive. Radon is a noble gas and has a complete valence shell, but its radioactivity makes it useful as a tracer. Radioactive substances can be traced and monitored through the body, which can be useful in medical research to study biological processes or diagnose certain medical conditions. Radon is also advantageous because it is a gas and can easily flow through small spaces, allowing it to be easily inhaled or ingested by the patient being studied.
Pergunta 8 Relatório
A mixture of NaCI(s) and CaCO3(s) is best separated by
Detalhes da Resposta
The best method to separate a mixture of NaCl and CaCO3 is dissolution followed by filtration. This is because NaCl is soluble in water, while CaCO3 is not. By dissolving the mixture in water, NaCl will dissolve, while CaCO3 remains undissolved. Filtration can then be used to separate the insoluble CaCO3 from the solution containing the dissolved NaCl. The other options are not suitable because sublimation is not effective for separating NaCl and CaCO3, and dissolution followed by crystallization and evaporation will only work for the soluble component (NaCl) and not the insoluble component (CaCO3).
Pergunta 9 Relatório
Which of the following is a general method of preparing acids?
Detalhes da Resposta
The general method of preparing acids is the reaction between an anhydride and water. An anhydride is a compound that has had water removed from it, usually by heating. When water is added to an anhydride, it undergoes a reaction called hydration, which results in the formation of an acid. This is a general method for preparing acids because many different types of anhydrides can be used, resulting in a wide variety of acids being produced.
Pergunta 10 Relatório
If the value of ∆H for a reaction is negative, it means that the reaction is
Detalhes da Resposta
If the value of ∆H for a reaction is negative, it means that the reaction is exothermic. An exothermic reaction releases heat to the surroundings. ∆H is the change in enthalpy during a reaction, and when ∆H is negative, it means that the products have lower enthalpy than the reactants. This implies that energy is released during the reaction, which results in the increase in temperature of the surroundings. Exothermic reactions are spontaneous, and they occur on their own without any external energy input.
Pergunta 11 Relatório
The loss of molecules of water of crystallization to the atmosphere by some crystalline salts is known as
Detalhes da Resposta
Efflorescence is the loss of water of crystallization to the atmosphere by some crystalline salts. This occurs when the salt is exposed to dry air, causing the water molecules to evaporate and leaving the salt in a dry and powdery form. The process is the opposite of hygroscopy, which is the ability of a substance to absorb moisture from the atmosphere. The word "efflorescence" comes from the Latin word "efflorescere," which means "to flower out."
Pergunta 12 Relatório
The valence electrons of element 12Mg are in the
Detalhes da Resposta
The valence electrons of element 12 (Mg) are in the 3s orbital. Valence electrons are the electrons in the outermost energy level (or shell) of an atom. In the case of magnesium, which has an atomic number of 12, the electron configuration is 1s2 2s2 2p6 3s2. This means that magnesium has 2 electrons in its 3s orbital, which is its outermost energy level. The 3s orbital is located in the third energy level (or shell) of the magnesium atom. The 1s and 2s orbitals are located in the first and second energy levels, respectively, and the 2px and 2py orbitals are also located in the second energy level. Valence electrons are important because they are involved in chemical bonding and reactions. In the case of magnesium, having 2 valence electrons in its 3s orbital allows it to form chemical bonds with other atoms by either losing or sharing those electrons. Therefore, the valence electrons of element 12 (Mg) are in the 3s orbital, which is located in the outermost energy level of the atom.
Pergunta 13 Relatório
Which of the following is used widely in the manufacture of flavours and perfumes?
Detalhes da Resposta
Alkanoates are widely used in the manufacture of flavours and perfumes. These are organic compounds that contain the functional group -COO- and are also known as esters. They are often used in the food industry to create fruity or sweet flavours and in the perfume industry to create floral or fruity scents. Alkanones, alkanes, complex sugars, and aliphatic compounds are not commonly used in the manufacture of flavours and perfumes.
Pergunta 14 Relatório
Element X, with electronic configuration 2, 8, 2 and element Y with electronic configuration 2, 8, 7 are like to combine by
Detalhes da Resposta
The electronic configuration of an element determines its reactivity and bonding behavior. Element X has two valence electrons in its outermost shell while Element Y has seven valence electrons. To achieve stability, Element X needs to lose its two valence electrons to form a cation with a +2 charge, while Element Y needs to gain one electron to form an anion with a -1 charge. This creates an electrostatic attraction between the oppositely charged ions, resulting in an ionic bond. Therefore, the most likely type of bond that Element X and Y will form is an electrovalent or ionic bond.
Pergunta 15 Relatório
What is the basicity of the acid in the followed reaction? NaCO + 2CHCOOH → 2CH COONa + H2O + CO2
Detalhes da Resposta
Pergunta 16 Relatório
What quantity of electrons (in mole) is lost when one mole of iron (ll) ions is oxidized to iron (lll) ions?
Detalhes da Resposta
When iron (II) ions (Fe²⁺) are oxidized to iron (III) ions (Fe³⁺), they lose one electron each. So, for one mole of Fe²⁺, there are 6.022 x 10²³ ions, and for each ion, one electron is lost during oxidation. Therefore, one mole of Fe²⁺ ions will lose one mole of electrons during oxidation. Therefore, the answer is 1 mole.
Pergunta 17 Relatório
Which of the following when heated strongly in air will leave a metal as residue?
Detalhes da Resposta
Pergunta 18 Relatório
How many unpaired electrons are there in an atom of an element with the following electronic configuration? 1s22s22p6
Detalhes da Resposta
An atom with the electronic configuration 1s22s22p6 has a completely filled valence shell, which means that all of its electrons are paired. Therefore, the number of unpaired electrons in this atom is zero (0).
Pergunta 19 Relatório
The following oxides react with both acids and bases to form salts except
Detalhes da Resposta
The oxide that does not react with both acids and bases to form salts is carbon (IV) oxide. Carbon (IV) oxide is a neutral oxide, meaning it does not have acidic or basic properties. It does not ionize in water to form H+ or OH- ions and therefore does not react with acids or bases to form salts. Zinc oxide, lead (II) oxide, aluminium oxide and tin (IV) oxide are all basic oxides, which means they react with acids to form salts and with bases to form complex salts or basic salts.
Pergunta 20 Relatório
What is the value of x in the following equation? MnO- 4 + 4H+ + xe- → MnO 2 + 2H2O
Detalhes da Resposta
Pergunta 21 Relatório
Which of the following metals is not extracted by electrolysis ?
Detalhes da Resposta
Pergunta 22 Relatório
The phenomenon observed when dust particles collide randomly in a beam of sunlight is known as
Detalhes da Resposta
The phenomenon observed when dust particles collide randomly in a beam of sunlight is known as the Tyndall effect. When a beam of light passes through a transparent medium, such as air, the path of the light appears straight and unobstructed. However, when the light passes through a medium containing particles, such as dust or smoke, the path of the light becomes visible and appears as a beam. This phenomenon is known as the Tyndall effect. The Tyndall effect occurs because the particles in the medium scatter the light in all directions, making the beam of light visible. When the scattered light enters the eye, it creates a visible path that appears as a beam of light. In the case of dust particles in a beam of sunlight, the dust particles act as the scattering medium, causing the light to scatter in all directions and making the beam of light visible. This can be observed in everyday situations, such as when dust particles are present in a room and sunlight enters through a window. Therefore, the phenomenon observed when dust particles collide randomly in a beam of sunlight is known as the Tyndall effect, which is caused by the scattering of light by particles in a medium.
Pergunta 23 Relatório
Ethene undergoes mainly addition reactions because it is
Detalhes da Resposta
Ethene undergoes mainly addition reactions because it is unsaturated. This means that it contains a double bond between two carbon atoms, which is a reactive site for the addition of other molecules. In an addition reaction, the double bond is broken and new atoms or groups are added to the carbon atoms of the ethene molecule. This results in the formation of a new compound that contains all the atoms of the original reactants. Addition reactions are characteristic of unsaturated compounds like ethene, which have reactive double or triple bonds that can be easily broken to form new bonds.
Pergunta 24 Relatório
Which of the following is an ore of aluminium?
Detalhes da Resposta
Bauxite is an ore of aluminium. It is a reddish-brown rock that is mainly composed of aluminum hydroxide minerals. Bauxite is the primary source of aluminium metal, which is used in a variety of industrial applications due to its strength, lightness, and resistance to corrosion. Therefore, option D (Bauxite) is the correct answer.
Pergunta 25 Relatório
The reaction between alkanoic acids and alkanols in the presence of a mineral acid is known as
Detalhes da Resposta
The reaction between alkanoic acids and alkanols in the presence of a mineral acid is known as esterification. In this reaction, an ester is formed as a product. The process involves the removal of a water molecule from the carboxylic acid and the alcohol, forming an ester linkage. The mineral acid acts as a catalyst and helps to speed up the reaction. This reaction is commonly used in the production of fragrances, flavors, and plastics.
Pergunta 26 Relatório
If an element R belongs to the same group as sodium, an aqueous solution of ROH will
Detalhes da Resposta
If an element R belongs to the same group as sodium, then it is likely to have similar properties as sodium. Sodium hydroxide (NaOH) is an alkali and a base. When a group 1 metal hydroxide dissolves in water, it forms an alkaline solution. Hence, if an aqueous solution of ROH is similar to NaOH, it will also form an alkaline solution. An alkaline solution has a pH greater than 7 and turns red litmus paper blue. Therefore, the correct option is "have pH greater than 7".
Pergunta 27 Relatório
An atom 17 protons, 17 electrons and 18 neutrons has a mass number of
Detalhes da Resposta
The mass number of an atom is the sum of the number of protons and the number of neutrons in the nucleus of the atom. Therefore, the mass number of the atom with 17 protons and 18 neutrons is: Mass number = number of protons + number of neutrons = 17 + 18 = 35 Hence, the correct option is 35.
Pergunta 28 Relatório
The by-product of the fermentation of sugar to ethanol is
Detalhes da Resposta
The by-product of the fermentation of sugar to ethanol is carbon dioxide (CO2). During fermentation, sugar is converted into ethanol and carbon dioxide gas. The carbon dioxide gas is a by-product of the reaction and is responsible for the bubbling seen in the fermentation process.
Pergunta 29 Relatório
Which of the following ions will migrate to the cathode during electrolysis?
Detalhes da Resposta
During electrolysis, positively charged ions (cations) migrate towards the negative electrode (cathode), whereas negatively charged ions (anions) migrate towards the positive electrode (anode). Therefore, the ion that will migrate to the cathode depends on the nature of the electrolyte solution and the electrode potential. Out of the options given, zinc ions (Zn²⁺) and tetraoxosulphate (VI) ions (SO₄²⁻) are both cations, which means they will migrate towards the cathode during electrolysis. Chloride ions (Cl⁻), sulphide ions (S²⁻), and trioxonitrate (V) ions (NO₃⁻) are all anions and will move towards the anode during electrolysis. Therefore, the answer to the question is either Zinc ions (Zn²⁺) or Tetraoxosulphate (VI) ions (SO₄²⁻), depending on the specific electrolyte solution and conditions of the electrolysis.
Pergunta 31 Relatório
When chlorine is added to slaked lime, the product obtained is
Detalhes da Resposta
When chlorine is added to slaked lime, the product obtained is bleaching powder. Bleaching powder is a pale yellow powder with the chemical formula CaOCl2. It is used as a bleaching agent, disinfectant, and for water treatment. Chlorine gas reacts with slaked lime (calcium hydroxide) to form calcium hypochlorite, which is the main component of bleaching powder. The reaction can be represented as follows: Ca(OH)2 + Cl2 → CaOCl2 + H2O Hence, the correct option is (a) bleaching powder.
Pergunta 32 Relatório
From the following equations, what is the order of decreasing reactivity of X, Y and Z? Y (s) + X2+(aq) → Y2+(aq) + X(s) X(s) + Z2+(aq) → X2+(aq) + Z(s)
Detalhes da Resposta
Pergunta 33 Relatório
In linear molecules, the bond angle is
Detalhes da Resposta
In linear molecules, the bond angle is 180o. Linear molecules are those in which all the atoms attached to the central atom are in a straight line. The bond angle is the angle between the two chemical bonds involving the central atom. Since there is a straight line, the angle between the two bonds is 180o.
Pergunta 34 Relatório
What is the most probable group of an element which is a soft, silvery white solid and reacts violently with water?
Detalhes da Resposta
The element described is likely to be an alkali metal because it is a soft, silvery white solid and reacts violently with water. Alkali metals are located in Group 1 of the periodic table, so the answer is Group 1. Group 1 elements are known for their low densities, low melting points, and high reactivity, particularly with water.
Pergunta 35 Relatório
Alkanes react with the halogens mainly by
Detalhes da Resposta
Alkanes react with halogens mainly by substitution. In this type of reaction, one or more atoms or groups in a molecule are replaced by other atoms or groups. In the case of alkanes and halogens, a halogen atom substitutes a hydrogen atom in the alkane molecule, forming a halogenoalkane. The substitution occurs due to the greater reactivity of halogens compared to the alkane's carbon-hydrogen bonds. The process usually requires energy to initiate, such as heat or light. The reaction mechanism involves a free radical mechanism where the halogen molecule breaks into two halogen radicals, which react with alkane molecules to form alkyl radicals and hydrogen halides. The alkyl radicals react with halogen molecules to form halogenoalkanes and regenerate the halogen radicals, which continue the reaction.
Pergunta 36 Relatório
The mass number of an element is the sum of its
Detalhes da Resposta
The mass number of an element is the sum of its protons and neutrons. This is because protons and neutrons are found in the nucleus of an atom, and together they contribute most of the mass of an atom. Electrons, on the other hand, are found in the electron shells outside of the nucleus and have negligible mass. Therefore, the mass number is determined by the number of protons and neutrons in an atom, and it helps to identify different isotopes of the same element.
Pergunta 37 Relatório
Which of the substance lettered P to V in the equation below would be left on the filter paper if the reaction mixture were filtered? P (aq) + Q(aq) → V(s) + R(aq) + T(g)
Detalhes da Resposta
The substance that would be left on the filter paper if the reaction mixture were filtered is V. The letter "s" in the formula V(s) means that it is a solid. During filtration, solids are separated from liquids. The other substances, P, Q, R, and T, are either in aqueous or gaseous form, and they would pass through the filter paper. Therefore, V is the only substance that would be left on the filter paper.
Pergunta 38 Relatório
An increase in the pressure of a gas results in a decrease in its
Detalhes da Resposta
An increase in the pressure of a gas results in a decrease in its volume. This is known as Boyle's law, which states that at a constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure. In other words, as the pressure increases, the volume of the gas decreases and vice versa. This can be explained by the fact that increasing the pressure of a gas means that the gas particles are compressed closer together, resulting in less space between the particles and a smaller volume. Conversely, decreasing the pressure of a gas means that the gas particles are allowed to expand and take up more space, resulting in a larger volume.
Pergunta 39 Relatório
If the atomic number of an element X is 11 and that of nitrogen is 7, the most likely formula of the nitride of X is
Detalhes da Resposta
Pergunta 40 Relatório
Which of the following is a general method of preparing acids?
Detalhes da Resposta
Pergunta 41 Relatório
The position of equilibrium in a reversible reaction is affected by
Detalhes da Resposta
Pergunta 42 Relatório
The common feature of reactions at the anode is that
Detalhes da Resposta
The common feature of reactions at the anode is that oxidation is involved. This means that the substance being oxidized loses electrons, resulting in an increase in its oxidation state. An anode is the positively charged electrode in an electrochemical cell, where oxidation takes place. During the process, the anode becomes negatively charged as electrons are released, leaving behind positively charged ions. Hence, the correct answer is oxidation is involved.
Pergunta 43 Relatório
An acid is a substance which in the presence of water produces
Detalhes da Resposta
An acid is a substance which in the presence of water produces hydroxonium ions. When an acid is dissolved in water, it dissociates into positively charged hydrogen ions (H+) and negatively charged ions. However, since the hydrogen ions are highly reactive, they immediately bond with water molecules to form hydroxonium ions (H3O+). This is why the presence of water is essential for acids to exhibit their acidic properties. The concentration of hydroxonium ions determines the acidity of a solution, and the more hydroxonium ions present, the stronger the acid. Salts, oxygen, effervescence, and hydrogen gas are not typically produced by the reaction between an acid and water.
Pergunta 44 Relatório
What volume of oxygen at s.t.p would react with carbon to form 4.40g of CO2 according to the following equation? C(s) + O2(g) → CO2(g) [O = 16; C =12; 1 mole of a gas occupies 22.4dm3 at s.t.p]
Detalhes da Resposta
The balanced chemical equation is C(s) + O2(g) → CO2(g). From the equation, we can see that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. The molar mass of CO2 is 12 + 2(16) = 44 g/mol. Given that 4.40 g of CO2 is formed, the number of moles of CO2 can be calculated as: number of moles = mass / molar mass = 4.40 g / 44 g/mol = 0.1 mol Since 1 mole of oxygen reacts with 1 mole of carbon to produce 1 mole of CO2, the number of moles of oxygen required is also 0.1 mol. Using the ideal gas equation PV = nRT, we can find the volume of oxygen required at standard temperature and pressure (s.t.p), where P = 1 atm, V = ?, n = 0.1 mol, R = 0.0821 L.atm/K.mol, and T = 273 K: V = nRT / P = 0.1 mol x 0.0821 L.atm/K.mol x 273 K / 1 atm = 2.24 L 1 mole of any gas occupies 22.4 L at s.t.p, so the volume of oxygen required is 2.24 L, which is.
Pergunta 46 Relatório
The use of diamond in abrasives is due ti its
Detalhes da Resposta
Diamond is used in abrasives due to its hardness. Hardness is the ability of a material to resist scratches and abrasions. Diamond is the hardest naturally occurring substance, meaning that it is extremely difficult to scratch or wear down. Therefore, it is an excellent abrasive material, and it is used in cutting, grinding, and polishing tools.
Pergunta 47 Relatório
Which of the following statements is correct about sodium chloride in the solid state?
Detalhes da Resposta
Pergunta 48 Relatório
Lead and tin are the components of
Detalhes da Resposta
Lead and tin are the components of "soft solder". Soft solder is a metal alloy that is used to join two metal pieces together, typically in electronics or plumbing applications. The most common composition of soft solder is 60% tin and 40% lead, although there are other compositions that can be used as well. The lead in soft solder helps to reduce the melting point of the alloy, making it easier to work with, while the tin provides strength and durability to the joint. Soft solder is called "soft" because it has a low melting point, usually around 180-190°C, which means it can be melted and applied with a soldering iron or torch. This makes it a useful material for a variety of applications where a strong, durable joint is needed, such as in electrical wiring or plumbing systems.
Pergunta 49 Relatório
(a)(i) List two reactants for the laboratory preparation of ammonia
(ii) State three physical properties of ammonia
(iii) Describe in outline, the manufacture of ammonia by the Haber process.
(b) Write an equation in each case to show the:
(i) reaction between ammonia gas and heated copper (II) oxide
(ii) action of heat on ammonium trioxocarbonate (IV)
(c)(i) Which industrial process is used for convey ammonia to trioxonitrate (V) acid?
(ii) Give the reason why electropositive metals do not generally are off hydrogen with dilute trioxonitrate (V) acid
(d) Give one example in each case, to show how trioxonitrate (V) acid reacts generally with: (i) bases (ii) non-metals.
(a)(i) Two reactants for laboratory preparation of ammonia: an ammonium salt (e.g. ammonium chloride, NH4Cl) and a base (e.g. calcium hydroxide, slaked lime).
(ii) Three physical properties of ammonia: colourless gas; pungent (choking) smell; less dense than air; very soluble in water.
(iii) Haber process: Nitrogen (from air) and hydrogen (from natural gas/water) are mixed in the ratio 1 : 3 by volume and passed over a finely divided iron catalyst at about 450°C and a pressure of about 200-250 atm:
N2(g) + 3H2(g) ⇌ 2NH3(g)
The ammonia is cooled and liquefied, and the unreacted gases are recycled.
(b)(i) 3CuO + 2NH3 → 3Cu + N2 + 3H2O
(ii) (NH4)2CO3 → 2NH3 + H2O + CO2
(c)(i) Ammonia is converted to trioxonitrate(V) acid by the Ostwald process.
(ii) Electropositive metals do not liberate hydrogen with dilute HNO3 because HNO3 is a strong oxidizing agent: it oxidizes any hydrogen formed to water while it is itself reduced to oxides of nitrogen (NO or NO2), so oxides of nitrogen are evolved instead of hydrogen.
(d)(i) With bases (neutralization): HNO3 + NaOH → NaNO3 + H2O.
(ii) With non-metals (oxidation): C + 4HNO3 → CO2 + 4NO2 + 2H2O.
Detalhes da Resposta
(a)(i) Two reactants for laboratory preparation of ammonia: an ammonium salt (e.g. ammonium chloride, NH4Cl) and a base (e.g. calcium hydroxide, slaked lime).
(ii) Three physical properties of ammonia: colourless gas; pungent (choking) smell; less dense than air; very soluble in water.
(iii) Haber process: Nitrogen (from air) and hydrogen (from natural gas/water) are mixed in the ratio 1 : 3 by volume and passed over a finely divided iron catalyst at about 450°C and a pressure of about 200-250 atm:
N2(g) + 3H2(g) ⇌ 2NH3(g)
The ammonia is cooled and liquefied, and the unreacted gases are recycled.
(b)(i) 3CuO + 2NH3 → 3Cu + N2 + 3H2O
(ii) (NH4)2CO3 → 2NH3 + H2O + CO2
(c)(i) Ammonia is converted to trioxonitrate(V) acid by the Ostwald process.
(ii) Electropositive metals do not liberate hydrogen with dilute HNO3 because HNO3 is a strong oxidizing agent: it oxidizes any hydrogen formed to water while it is itself reduced to oxides of nitrogen (NO or NO2), so oxides of nitrogen are evolved instead of hydrogen.
(d)(i) With bases (neutralization): HNO3 + NaOH → NaNO3 + H2O.
(ii) With non-metals (oxidation): C + 4HNO3 → CO2 + 4NO2 + 2H2O.
Pergunta 50 Relatório
(a) State two properties of alkalis
(b)(i) Why is tetraoxosulphate (VI) acid able to produce two types of salt?
(ii) Write an equation for the reactior involved on heating sodium trioxosulphate (IV) with hydrochloric acid.
(a) Two properties of alkalis: they turn red litmus blue (pH greater than 7); they have a soapy/slippery feel. (They also neutralize acids to give a salt and water, and taste bitter.)
(b)(i) Tetraoxosulphate(VI) acid, H2SO4, produces two types of salt because it is dibasic: it contains two replaceable (ionizable) hydrogen atoms. When only one H is replaced by a metal it forms an acid salt (e.g. NaHSO4); when both are replaced it forms a normal salt (e.g. Na2SO4).
(ii) Heating sodium trioxosulphate(IV), Na2SO3, with hydrochloric acid:
Na2SO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g)
Detalhes da Resposta
(a) Two properties of alkalis: they turn red litmus blue (pH greater than 7); they have a soapy/slippery feel. (They also neutralize acids to give a salt and water, and taste bitter.)
(b)(i) Tetraoxosulphate(VI) acid, H2SO4, produces two types of salt because it is dibasic: it contains two replaceable (ionizable) hydrogen atoms. When only one H is replaced by a metal it forms an acid salt (e.g. NaHSO4); when both are replaced it forms a normal salt (e.g. Na2SO4).
(ii) Heating sodium trioxosulphate(IV), Na2SO3, with hydrochloric acid:
Na2SO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g)
Pergunta 51 Relatório
(a)(i) State two general methods of preparing soluble salts.
(ii) Mention three pieces of apparatus required for determining the solubility of a salt at a given temperature.
(b) The solubilities of two salts represented as K and L were determined at various temperatures. The results are shown in the table below:
| Temperature \( (^{o}\mathrm{C}) \) | 0 | 20 | 40 | 60 | 80 | 90 |
| Solubility of K \( (\mathrm{mol.\ dm}^{-3}) \) | 0.38 | 0.46 | 0.54 | 0.62 | 0.69 | 0.73 |
| Solubility of L \( (\mathrm{mol.\ dm}^{-3}) \) | 0.12 | 0.34 | 0.64 | 1.08 | 1.64 | 2.00 |
(i) Plot the solubility curves of K and L on the same graph. Use the curves to answer questions (ii) - (iv) below.
(ii) What is the solubility of K at 50°C?
(iii) At what temperature is the solubility of L equal to \(1.0\mathrm{mol.\ dm}^{-3}\)?
(iv) Over what temperature range is K more soluble that L?
(v) Given that the molar mass of L is 101g, determine whether a solution containing 3.4g of L per \(250\mathrm{cm}^{3}\) at 20°C is saturated or unsaturated.
(a)(i) General methods of preparing soluble salts
- Neutralisation of an acid by an alkali or base.
- Direct combination of the constituent elements.
(a)(ii) Apparatus required
A thermometer, a water bath and a glass stirrer.
(b)(i) Solubility curves
The solubility curves of K and L are plotted below on the same axes.
(b)(ii) From the curve for K, at 50°C:
Solubility of K = 0.58 mol dm−3.
(b)(iii) From the curve for L, a solubility of 1.0 mol dm−3 is obtained at approximately 57°C.
(b)(iv) K is more soluble than L from 0°C to about 31.5°C.
(b)(v)
Mass concentration of L in the solution is:
\[\frac{3.4\ \text{g}}{250\ \text{cm}^3}\times 1000 = 13.6\ \text{g dm}^{-3}\]
At 20°C, the solubility of L is 0.34 mol dm−3. In mass units:
\[0.34\times 101 = 34.34\ \text{g dm}^{-3}\]
Since 13.6 g dm−3 is less than 34.34 g dm−3, the solution is unsaturated.
Detalhes da Resposta
(a)(i) General methods of preparing soluble salts
- Neutralisation of an acid by an alkali or base.
- Direct combination of the constituent elements.
(a)(ii) Apparatus required
A thermometer, a water bath and a glass stirrer.
(b)(i) Solubility curves
The solubility curves of K and L are plotted below on the same axes.
(b)(ii) From the curve for K, at 50°C:
Solubility of K = 0.58 mol dm−3.
(b)(iii) From the curve for L, a solubility of 1.0 mol dm−3 is obtained at approximately 57°C.
(b)(iv) K is more soluble than L from 0°C to about 31.5°C.
(b)(v)
Mass concentration of L in the solution is:
\[\frac{3.4\ \text{g}}{250\ \text{cm}^3}\times 1000 = 13.6\ \text{g dm}^{-3}\]
At 20°C, the solubility of L is 0.34 mol dm−3. In mass units:
\[0.34\times 101 = 34.34\ \text{g dm}^{-3}\]
Since 13.6 g dm−3 is less than 34.34 g dm−3, the solution is unsaturated.
Pergunta 52 Relatório
(a) List two products obtained when crude oil is refined
(b)(i) What is the general formula for alkanols?
(ii) State the type of reaction involved in the conversion of ethanol to ethanoic acid
(iii) Write an equation to show how ethanoic acid reacts with sodium trioxocarbonate (IV).
(a) Two products from refining crude oil: petrol (gasoline) and kerosene (also diesel, bitumen, lubricating oil).
(b)(i) General formula for alkanols: CnH2n+1OH.
(ii) Converting ethanol to ethanoic acid is an oxidation reaction.
(iii) Ethanoic acid with sodium trioxocarbonate(IV):
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
Detalhes da Resposta
(a) Two products from refining crude oil: petrol (gasoline) and kerosene (also diesel, bitumen, lubricating oil).
(b)(i) General formula for alkanols: CnH2n+1OH.
(ii) Converting ethanol to ethanoic acid is an oxidation reaction.
(iii) Ethanoic acid with sodium trioxocarbonate(IV):
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
Pergunta 53 Relatório
(a) Give one example of a metal which:
(i) can displace hydrogen from cold water
(ii) at red-heat, reacts reversibly with steam
(iii) does not react with dilute hydrochloric acid
(b) When dry hydrogen was passed over lead (II) oxide, a greyish solid J was obtained.
(i) Identify J
(ii) What type of reaction was involved in the formation of J from lead (II) oxide?
(a)(i) Sodium (Na).
(ii) Copper (Cu).
(iii) Magnesium (Mg).
(b)(i) J is lead (Pb).
(ii) Reduction.
Detalhes da Resposta
(a)(i) Sodium (Na).
(ii) Copper (Cu).
(iii) Magnesium (Mg).
(b)(i) J is lead (Pb).
(ii) Reduction.
Pergunta 54 Relatório
Consider the reaction represented by the following Q equation: N\(_2\)O\(_{4(g)}\) \(\rightleftharpoons\) 2NO\(_{2(g)}\) \(\Delta\)H = +57.2KJmol\(^{-1}\)
(a) When is the reaction said to be at equilibrium?
(b) Mention two conditions that can favour the forward reaction
(c) Name the principle involved in (b) above.
(a) The reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction, so that the concentrations of N2O4 and NO2 remain constant (a dynamic equilibrium).
(b) The forward reaction is endothermic (ΔH = +57.2 kJ mol-1) and produces more moles of gas (1 → 2). Two conditions that favour it are:
- Increasing the temperature.
- Decreasing the pressure (increasing the volume).
(Removing NO2 as it forms would also favour it.)
(c) The principle involved is Le Chatelier's principle.
Detalhes da Resposta
(a) The reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction, so that the concentrations of N2O4 and NO2 remain constant (a dynamic equilibrium).
(b) The forward reaction is endothermic (ΔH = +57.2 kJ mol-1) and produces more moles of gas (1 → 2). Two conditions that favour it are:
- Increasing the temperature.
- Decreasing the pressure (increasing the volume).
(Removing NO2 as it forms would also favour it.)
(c) The principle involved is Le Chatelier's principle.
Pergunta 55 Relatório
(a) State Gay Lussac's law of combining volumes
(b) Hydrogen reacts with oxygen according to the following equation;
2H\(_{2(g)}\) + O\(_{2(g)}\) ---> 2H\(_2\)O\(_{(g)}\). If 50cm\(^3\) of hydrogen were sparked with 30cm\(^{3}\) of oxygen, calculate the volume of unused oxygen after cooling to the initial temperature and pressure.
(a) Gay-Lussac's law of combining volumes: When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volume of any gaseous product, provided the temperature and pressure remain constant.
(b) 2H2 + O2 → 2H2O. Hydrogen and oxygen react in the volume ratio 2 : 1.
50 cm3 of H2 requires \(\dfrac{50}{2}=25\ \text{cm}^3\) of O2.
Oxygen supplied = 30 cm3; oxygen used = 25 cm3.
\[ \text{Unused oxygen}=30-25=5\ \text{cm}^3 \]Detalhes da Resposta
(a) Gay-Lussac's law of combining volumes: When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volume of any gaseous product, provided the temperature and pressure remain constant.
(b) 2H2 + O2 → 2H2O. Hydrogen and oxygen react in the volume ratio 2 : 1.
50 cm3 of H2 requires \(\dfrac{50}{2}=25\ \text{cm}^3\) of O2.
Oxygen supplied = 30 cm3; oxygen used = 25 cm3.
\[ \text{Unused oxygen}=30-25=5\ \text{cm}^3 \]Pergunta 56 Relatório
(a)(i) What is meant by cracking of petroleum fractions?
(ii) Write an equation for the laboratory preparation of ethene from ethanol.
(iii) Give one chemical test to distinguish between ethane and ethene.
(b)(i) Name the class of carbohydrates to which starch and cellulose belong.
(ii) What process is used for isolating ethanol from the other products of fermentation of sugar?
(iii) Name the organic product of the reaction between ethanol and sodium
(iv). Write the structural formula of 2-chloroethanol.
(c) State the reason why:
(i) benzene produces more soot than ethene on burning in excess air;
(ii) ethanoic acid has a higher boiling point than methanoic acid;
(iii) sodium chloride is used during the manufacture of soap.
(d) Give one use of: (i) ethyne (ii) coal (iii) carbon black
(a)(i) Cracking is the breaking down of large, long-chain hydrocarbon molecules from heavier petroleum fractions into smaller, more useful molecules (such as petrol and alkenes) using heat and/or a catalyst.
(ii) C2H5OH → C2H4 + H2O (heated with concentrated H2SO4 at about 170°C).
(iii) Add bromine water: ethene rapidly decolourizes the red-brown bromine water, while ethane does not.
(b)(i) Starch and cellulose are polysaccharides. (ii) Ethanol is isolated by fractional distillation. (iii) The organic product of ethanol + sodium is sodium ethoxide (C2H5ONa). (iv) 2-chloroethanol: ClCH2CH2OH.
(c)(i) Benzene produces more soot because it has a higher proportion of carbon (high carbon-to-hydrogen ratio), so it undergoes more incomplete combustion, releasing unburnt carbon. (ii) Ethanoic acid has a higher relative molecular mass and larger molecule than methanoic acid, giving stronger van der Waals forces (in addition to hydrogen bonding), so more energy is needed to boil it. (iii) Sodium chloride is added to precipitate (salt out) the soap, separating it from the glycerol and water.
(d)(i) Ethyne: oxy-acetylene welding/cutting. (ii) Coal: as a fuel (and a source of coke and coal tar). (iii) Carbon black: as a reinforcing filler in motor tyres (and a pigment in ink).
Detalhes da Resposta
(a)(i) Cracking is the breaking down of large, long-chain hydrocarbon molecules from heavier petroleum fractions into smaller, more useful molecules (such as petrol and alkenes) using heat and/or a catalyst.
(ii) C2H5OH → C2H4 + H2O (heated with concentrated H2SO4 at about 170°C).
(iii) Add bromine water: ethene rapidly decolourizes the red-brown bromine water, while ethane does not.
(b)(i) Starch and cellulose are polysaccharides. (ii) Ethanol is isolated by fractional distillation. (iii) The organic product of ethanol + sodium is sodium ethoxide (C2H5ONa). (iv) 2-chloroethanol: ClCH2CH2OH.
(c)(i) Benzene produces more soot because it has a higher proportion of carbon (high carbon-to-hydrogen ratio), so it undergoes more incomplete combustion, releasing unburnt carbon. (ii) Ethanoic acid has a higher relative molecular mass and larger molecule than methanoic acid, giving stronger van der Waals forces (in addition to hydrogen bonding), so more energy is needed to boil it. (iii) Sodium chloride is added to precipitate (salt out) the soap, separating it from the glycerol and water.
(d)(i) Ethyne: oxy-acetylene welding/cutting. (ii) Coal: as a fuel (and a source of coke and coal tar). (iii) Carbon black: as a reinforcing filler in motor tyres (and a pigment in ink).
Pergunta 57 Relatório
(a)(i) State the two main processes involved in the manufacture of oxygen from air
(ii) Name the type of chemical bonding which exists between oxygen atoms in a molecule of oxygen
(b) What term is used to describe the relationship between oxygen and ozone (O\(_3\))?
(a)(i) The two main processes in obtaining oxygen from air are the liquefaction of air (compressing and cooling air until it liquefies) followed by the fractional distillation of the liquid air (nitrogen boils off first at -196°C, leaving oxygen).
(ii) The bonding between the two oxygen atoms in an O2 molecule is a covalent bond (a double covalent bond).
(b) The relationship between oxygen (O2) and ozone (O3) is allotropy; they are allotropes of the element oxygen.
Detalhes da Resposta
(a)(i) The two main processes in obtaining oxygen from air are the liquefaction of air (compressing and cooling air until it liquefies) followed by the fractional distillation of the liquid air (nitrogen boils off first at -196°C, leaving oxygen).
(ii) The bonding between the two oxygen atoms in an O2 molecule is a covalent bond (a double covalent bond).
(b) The relationship between oxygen (O2) and ozone (O3) is allotropy; they are allotropes of the element oxygen.
Pergunta 58 Relatório
(a)(i) List three characteristic properties of transition metals
(ii) 0.45g of a metal M was deposited when a current of 1.8 amperes was passed for 12.5 minutes through a solution containing M\(^{2+}\). Calculate the relative atomic mass of M. [1 Faraday = 96500 C]
(iii) Give the reason why copper-plated iron corrodes easily when the surface is scratched.
(b)(i) State the law of definite proportions (constant composition).
(ii) Describe in outline, an experimental procedure for determining the proportion of oxygen in a given sample of copper(II) oxide.
(iii) Write an equation to show how copper (II) oxide can be obtained directly from copper (II) trioxonitrate (V)
(a)(i) Three characteristic properties of transition metals: they show variable oxidation states; they form coloured ions/compounds; they act as catalysts. (They also form complex ions.)
(ii) Current = 1.8 A, time = 12.5 min = 750 s, M2+ (so 2 electrons per atom).
\[ Q=1.8\times750=1350\ \text{C};\quad n(e^-)=\frac{1350}{96500}=0.01399\ \text{mol} \] \[ n(M)=\frac{0.01399}{2}=6.995\times10^{-3}\ \text{mol} \] \[ \text{Relative atomic mass}=\frac{0.45}{6.995\times10^{-3}}=64.3\approx 64 \](iii) When the copper coat is scratched, the more reactive (more electropositive) iron is exposed. Iron then acts as the anode and copper as the cathode in an electrochemical cell, so the iron loses electrons and corrodes faster than it would on its own.
(b)(i) Law of definite proportions: A given chemical compound always contains the same elements combined together in the same fixed proportion by mass, whatever its source or method of preparation.
(ii) Weigh a known mass of dry copper(II) oxide in a weighed boat/tube. Pass dry hydrogen over it while heating; the hydrogen reduces the oxide to copper (CuO + H2 → Cu + H2O). Heat to constant mass, cool in the stream of hydrogen, and reweigh. The loss in mass is the mass of oxygen, so the percentage of oxygen = (loss in mass / mass of CuO) × 100.
(iii) 2Cu(NO3)2 → 2CuO + 4NO2 + O2
Detalhes da Resposta
(a)(i) Three characteristic properties of transition metals: they show variable oxidation states; they form coloured ions/compounds; they act as catalysts. (They also form complex ions.)
(ii) Current = 1.8 A, time = 12.5 min = 750 s, M2+ (so 2 electrons per atom).
\[ Q=1.8\times750=1350\ \text{C};\quad n(e^-)=\frac{1350}{96500}=0.01399\ \text{mol} \] \[ n(M)=\frac{0.01399}{2}=6.995\times10^{-3}\ \text{mol} \] \[ \text{Relative atomic mass}=\frac{0.45}{6.995\times10^{-3}}=64.3\approx 64 \](iii) When the copper coat is scratched, the more reactive (more electropositive) iron is exposed. Iron then acts as the anode and copper as the cathode in an electrochemical cell, so the iron loses electrons and corrodes faster than it would on its own.
(b)(i) Law of definite proportions: A given chemical compound always contains the same elements combined together in the same fixed proportion by mass, whatever its source or method of preparation.
(ii) Weigh a known mass of dry copper(II) oxide in a weighed boat/tube. Pass dry hydrogen over it while heating; the hydrogen reduces the oxide to copper (CuO + H2 → Cu + H2O). Heat to constant mass, cool in the stream of hydrogen, and reweigh. The loss in mass is the mass of oxygen, so the percentage of oxygen = (loss in mass / mass of CuO) × 100.
(iii) 2Cu(NO3)2 → 2CuO + 4NO2 + O2
Pergunta 59 Relatório
(a)(i) Define entropy
(ii) What term is used to describe a reaction in which heat is absorbed from the surrounding?
(b) State two conditions that can lead to ineffective collisions during a chemical reaction.
(a)(i) Entropy is a measure of the degree of disorder or randomness of a system.
(ii) A reaction in which heat is absorbed from the surroundings is described as endothermic.
(b) Two conditions that lead to ineffective collisions:
- The colliding particles possess energy less than the activation energy.
- The particles collide with the wrong (unfavourable) orientation.
Detalhes da Resposta
(a)(i) Entropy is a measure of the degree of disorder or randomness of a system.
(ii) A reaction in which heat is absorbed from the surroundings is described as endothermic.
(b) Two conditions that lead to ineffective collisions:
- The colliding particles possess energy less than the activation energy.
- The particles collide with the wrong (unfavourable) orientation.
Pergunta 60 Relatório
(a) Mention three chemical properties of chlorine
(b) What type of reaction is represented by each of the following equations?
(i) Mg\(_{(s)}\) + 2HCI\(_{(aq)}\) --> Mg\(_2\)Cl\(_{2(aq)}\) + H\(_{2(g)}\)
(ii) Ag\(^+_{aq}\) + Cl\(^-\)\(_{(aq)}\) ----> AgCl\(_{(s)}\)
Detalhes da Resposta
None
Pergunta 61 Relatório
(a) List two industrial uses of concentrated tetraoxosulphate (VI) acid
(b) The diagram below represents the set-up for the electrolysis of dilute tetraoxosulphate (VI) acid. Use it to answer Questions (i) to (iii).
(i) Which letter on the diagram represents the battery?
(ii) Write an equation for the reaction occurring at the cathode
(iii) What product does V represent?
Detalhes da Resposta
None
Pergunta 62 Relatório
(a) List two properties of alpha particles.
(b) X is an element which exists as an isotopic mixture containing 90% of \(^{39}_{19} X\) and 10% of \(^{41}_{19}X\).
(i) How many neutrons are present in the isotope \(^{41}_{19}X\)?
(ii) Calculate the mean relative atomic mass of X.
(a) Two properties of alpha particles: they are positively charged (each is a helium nucleus, He2+); they have low penetrating power (stopped by paper or the skin). (They are also relatively heavy and cause intense ionization.)
(b)(i) Neutrons in 4119X = mass number - atomic number = 41 - 19 = 22 neutrons.
(ii) Mean relative atomic mass
\[ \bar{A}=\frac{(90\times39)+(10\times41)}{100}=\frac{3510+410}{100}=39.2 \]Detalhes da Resposta
(a) Two properties of alpha particles: they are positively charged (each is a helium nucleus, He2+); they have low penetrating power (stopped by paper or the skin). (They are also relatively heavy and cause intense ionization.)
(b)(i) Neutrons in 4119X = mass number - atomic number = 41 - 19 = 22 neutrons.
(ii) Mean relative atomic mass
\[ \bar{A}=\frac{(90\times39)+(10\times41)}{100}=\frac{3510+410}{100}=39.2 \]
