WAEC SSCE - Chemistry - 1995
Tambaya 1 Rahoto
In which of the following processes are larger molecules broken down into smaller molecules?
Bayanin Amsa
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.
Tambaya 2 Rahoto
Alums are classified as
Bayanin Amsa
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.
Tambaya 3 Rahoto
The following alkanols will yield alkanoic acids on reacting with excess acidified K2Cr2O7 except
Bayanin Amsa
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.
Tambaya 4 Rahoto
The function of limestone in the extraction of iron in the blast furnace is
Bayanin Amsa
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.
Tambaya 5 Rahoto
Which of the following cannot be deduced from the electronic configuration of a transition metal?
Bayanin Amsa
Tambaya 6 Rahoto
A WEAK ACID IS ONE WHICH
Bayanin Amsa
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.
Tambaya 7 Rahoto
Consider the reaction represented by the equation below: XO + YO → X + YO2 In the reaction, YO acts as
Bayanin Amsa
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.
Tambaya 8 Rahoto
The most suitable method for preparing lead (ll) chloride is by
Bayanin Amsa
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."
Tambaya 9 Rahoto
Which of the following is an alloy of mercury?
Bayanin Amsa
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.
Tambaya 10 Rahoto
Which of the following molecules is linear in shape?
Bayanin Amsa
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.
Tambaya 11 Rahoto
Metals which react with steam only when they are red-hot include
Tambaya 13 Rahoto
Which of the following is hydrocarbons is unsaturated?
Bayanin Amsa
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.
Tambaya 15 Rahoto
Sodium chloride cannot conduct electricity in the solid state because it
Bayanin Amsa
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.
Tambaya 16 Rahoto
The following salts dissolve readily in cold water except
Bayanin Amsa
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.
Tambaya 17 Rahoto
Which of the following is not correct of the reaction represented by the equation below? 2C(s) + O2(g) → 2CO(g); ∆H = - xKJ
Tambaya 18 Rahoto
If a solid has a low melting point and dissolves readily in benzene, it would probably
Bayanin Amsa
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.
Tambaya 19 Rahoto
The gas produced when a mixture of sodium propanoate and soda lime is heated is
Bayanin Amsa
Tambaya 20 Rahoto
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)
Bayanin Amsa
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.
Tambaya 21 Rahoto
What is the percentage by mass of copper in copper in copper (l) oxide (CU2O)? [O = 16; CU =64]
Bayanin Amsa
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%.
Tambaya 22 Rahoto
Copper (ll) tetraoxosulphate (VI) is often added to swimming pools because it
Bayanin Amsa
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.
Tambaya 23 Rahoto
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]
Bayanin Amsa
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.
Tambaya 24 Rahoto
The following can be obtained directly from the destructive distillation of coal except
Bayanin Amsa
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."
Tambaya 25 Rahoto
H3O+(aq) + OH-(aq) → 2H2O(l). The heat change accompanying the process represented by the equation above is the heat of
Bayanin Amsa
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.
Tambaya 26 Rahoto
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)
Bayanin Amsa
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.
Tambaya 27 Rahoto
Which of the following species is always present in acidified water?
Bayanin Amsa
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.
Tambaya 28 Rahoto
A positive brown ring test indicates the presence of
Bayanin Amsa
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.
Tambaya 29 Rahoto
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]
Bayanin Amsa
Tambaya 30 Rahoto
(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.
Bayanin Amsa
None
Tambaya 31 Rahoto
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.
Bayanin Amsa
None
Tambaya 32 Rahoto
(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.
Bayanin Amsa
None
Tambaya 33 Rahoto
(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.
Bayanin Amsa
None
Tambaya 34 Rahoto
(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.
Bayanin Amsa
None
Tambaya 35 Rahoto
(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 NaCl\(_{(aq)}\) Concentrated NaCI\(_{(aq)}\) |
Tambaya 36 Rahoto
(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.
Bayanin Amsa
None
Tambaya 37 Rahoto
(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.
Bayanin Amsa
None
Tambaya 38 Rahoto
(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]
None
Bayanin Amsa
None
Tambaya 39 Rahoto
(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.
Tambaya 40 Rahoto
(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
Tambaya 41 Rahoto
(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) Saturated solution: A saturated solution is a solution that has dissolved the maximum amount of solute at a given temperature and pressure, where any additional solute will remain undissolved at the bottom of the container.
a)(ii) Preparation of saturated solution: To prepare a saturated solution of sodium trioxonitrate(V) at 30°C, a known mass of the solute is added to a fixed volume of solvent (water) at 30°C, and stirred continuously until no more solute dissolves, and some amount remains undissolved.
a)(iii) Recovery techniques: Two techniques that can be used to recover crystals of sodium trioxonitrate(V) from its saturated solution are cooling the solution, which will cause the excess solute to crystallize out, and evaporation of the solvent, which will leave the solute behind as crystals.
b)(i) Separation of salts: To determine which of the two salts will separate out when the solution is cooled to 60°C, we need to compare the solubility of each salt at 60°C with the concentration of each salt in the solution at 90°C. The solubility of KNO3 in water at 60°C is higher than the solubility of KCIO3 in water at 60°C, so KNO3 will remain in solution while KCIO3 will separate out.
b)(ii) Calculation of mass: To calculate the mass of salt that will separate out at 60°C, we need to use the solubility data and the concentration of KCIO3 in the solution. The amount of KCIO3 in the solution at 90°C is 245g, which is equal to 245/122 = 2.01 mol. At 60°C, the solubility of KCIO3 is 1.61 mol.dm-3, so the maximum amount of KCIO3 that can remain in solution is 1.61 mol. Therefore, the amount of KCIO3 that will separate out is 2.01 - 1.61 = 0.40 mol. The molar mass of KCIO3 is (39+35.5+3x16) = 122g/mol, so the mass of KCIO3 that will separate out is 0.40 x 122 = 48.8g.
c)(i) Salts causing hardness of water: Two salts that cause hardness of water are calcium carbonate (CaCO3) and magnesium sulfate (MgSO4).
c)(ii) Temporary hardness of water: Temporary hardness of water is caused by the presence of dissolved calcium and magnesium bicarbonates, which can decompose upon heating, leading to the formation of insoluble calcium and magnesium carbonates. These carbonates deposit as a layer of scale (furring) on the inside of kettles and other hot water equipment.
Bayanin Amsa
a)(i) Saturated solution: A saturated solution is a solution that has dissolved the maximum amount of solute at a given temperature and pressure, where any additional solute will remain undissolved at the bottom of the container.
a)(ii) Preparation of saturated solution: To prepare a saturated solution of sodium trioxonitrate(V) at 30°C, a known mass of the solute is added to a fixed volume of solvent (water) at 30°C, and stirred continuously until no more solute dissolves, and some amount remains undissolved.
a)(iii) Recovery techniques: Two techniques that can be used to recover crystals of sodium trioxonitrate(V) from its saturated solution are cooling the solution, which will cause the excess solute to crystallize out, and evaporation of the solvent, which will leave the solute behind as crystals.
b)(i) Separation of salts: To determine which of the two salts will separate out when the solution is cooled to 60°C, we need to compare the solubility of each salt at 60°C with the concentration of each salt in the solution at 90°C. The solubility of KNO3 in water at 60°C is higher than the solubility of KCIO3 in water at 60°C, so KNO3 will remain in solution while KCIO3 will separate out.
b)(ii) Calculation of mass: To calculate the mass of salt that will separate out at 60°C, we need to use the solubility data and the concentration of KCIO3 in the solution. The amount of KCIO3 in the solution at 90°C is 245g, which is equal to 245/122 = 2.01 mol. At 60°C, the solubility of KCIO3 is 1.61 mol.dm-3, so the maximum amount of KCIO3 that can remain in solution is 1.61 mol. Therefore, the amount of KCIO3 that will separate out is 2.01 - 1.61 = 0.40 mol. The molar mass of KCIO3 is (39+35.5+3x16) = 122g/mol, so the mass of KCIO3 that will separate out is 0.40 x 122 = 48.8g.
c)(i) Salts causing hardness of water: Two salts that cause hardness of water are calcium carbonate (CaCO3) and magnesium sulfate (MgSO4).
c)(ii) Temporary hardness of water: Temporary hardness of water is caused by the presence of dissolved calcium and magnesium bicarbonates, which can decompose upon heating, leading to the formation of insoluble calcium and magnesium carbonates. These carbonates deposit as a layer of scale (furring) on the inside of kettles and other hot water equipment.
Tambaya 42 Rahoto
(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.
Bayanin Amsa
None
Tambaya 43 Rahoto
(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).
Bayanin Amsa
None
