WAEC SSCE - Chemistry - 2006

Bayanin Amsa

Bayanin Amsa

The substance that is an ore of iron is Haematite. Haematite (Fe2O3) is a mineral form of iron oxide and is a primary source of iron ore. It is found in abundance throughout the world and is mined to extract iron metal. When iron is needed for manufacturing purposes, haematite is processed into pig iron, which is the basic building block of steel. Bauxite is an ore of aluminium, Cassiterite is an ore of tin, and steel is not an ore but an alloy composed mainly of iron with carbon and other elements.

Bayanin Amsa

Bayanin Amsa

Isotopes are atoms of the same element with different numbers of neutrons, and they have the same atomic number but different mass numbers. The relative abundance of an isotope is the proportion of that isotope in a sample of the element. In this question, element X has two isotopes with mass numbers of 6 and 7, and their relative abundances are given as 1 and 12.5, respectively. The relative atomic mass (RAM) of X can be calculated using the following formula: RAM = (mass of isotope 1 x abundance of isotope 1 + mass of isotope 2 x abundance of isotope 2) / total abundance Plugging in the given values, we get: RAM = (6 x 1 + 7 x 12.5) / (1 + 12.5) Simplifying the expression, we get: RAM = (6 + 87.5) / 13.5 RAM = 93.5 / 13.5 RAM ≈ 6.93 Therefore, the relative atomic mass of X is approximately 6.93. Hence, the correct option is "6.9".

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Metals can be extracted from their ores by a process called reduction. In this process, a chemical reaction is used to remove the metal from its ore. Reduction involves adding a reducing agent to the ore, which reacts with the metal to form a pure, metallic substance. The reducing agent takes away the oxygen or other elements from the ore, leaving behind the pure metal. This process is commonly used for metals like iron, copper, and aluminum, which are all extracted from their ores through reduction. Oxidation, hydrolysis, and decomposition are not commonly used to extract metals from their ores.

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The reason why waste plastics accumulate in the soil and pollute the environment is that plastic materials are non-biodegradable. This means that they cannot be broken down by the natural processes that decompose other organic materials. As a result, plastics persist in the environment for hundreds of years, slowly breaking down into smaller and smaller pieces called microplastics. These microplastics can contaminate soil, water, and air, and harm wildlife and human health. Therefore, it is important to reduce plastic waste by recycling, reusing, and reducing our plastic consumption.

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An alpha particle consists of two protons and two neutrons, and it is symbolized by the Greek letter alpha (α). When an atom emits an alpha particle, it loses two protons and two neutrons, which means that its atomic number decreases by 2 and its mass number decreases by 4. In the given question, the parent atom is X with a mass number of 238 and atomic number of 92. It decays by alpha particle emission, which means it loses two protons and two neutrons. Therefore, the daughter atom Y will have a mass number of 238-4=234 and an atomic number of 92-2=90. So, the atomic number and mass number of the daughter atom Y are 90 and 234, respectively. Therefore, the correct option is: "90 and 234 respectively".

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Chlorine is not used in the manufacturing of common salt. Common salt, also known as table salt or sodium chloride, is typically obtained through the evaporation of saltwater or brine. Chlorine is not needed in this process because saltwater and brine naturally contain high concentrations of sodium chloride. Chlorine is used in the other three processes listed: - Production of polyvinylchloride (PVC): Chlorine is used to create vinyl chloride monomer, which is then polymerized to form PVC. - Manufacturing of hydrochloric acid: Chlorine is used to react with hydrogen gas to produce hydrochloric acid. - Manufacturing of domestic bleach: Chlorine is used as the active ingredient in many types of bleach, including household bleach.

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The d-sub shell is one of the electron shells in an atom. It can hold a maximum of 10 electrons, which are distributed among its orbitals. Each orbital in the d-sub shell can hold 2 electrons. There are 5 different orbitals in the d-sub shell, labeled as dxy, dxz, dyz, dz2, and dx2-y2. These orbitals have different shapes and orientations in space, and they can accommodate electrons with different energies. Therefore, the answer is 5.

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Bayanin Amsa

The compound that is an alkanoate is CH₃COOCH₃. An alkanoate is a salt or an ester of an organic acid that contains a carbonyl group (-CO-) bonded to an oxygen atom (-O-) that is in turn bonded to an alkyl group. CH₃COOCH₃ is an ester of acetic acid (CH₃COOH) and methanol (CH₃OH), which is formed by the condensation reaction between the carboxylic acid and alcohol. This ester has the characteristic -COO- functional group and an alkyl group, which makes it an alkanoate. CH₃COOH is acetic acid, a carboxylic acid. CH₃CH₂OH is ethanol, a primary alcohol. CH₃CH₂COOH is propanoic acid, a carboxylic acid.

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Bayanin Amsa

The greenhouse effect is a natural process that occurs when certain gases in the Earth's atmosphere, such as carbon dioxide, methane, and water vapor, trap heat from the sun and keep the planet warm. However, human activities, such as burning fossil fuels, deforestation, and industrial processes, have increased the concentration of these greenhouse gases in the atmosphere, leading to an enhanced greenhouse effect and causing global warming and climate change. To reduce the greenhouse effect, it is necessary to reduce the concentration of these greenhouse gases in the atmosphere. This can be achieved by controlling the human activities that contribute to their emissions. Some effective ways to control the greenhouse effect are: - Reducing the use of fossil fuels: Fossil fuels are the main source of carbon dioxide emissions. Therefore, reducing the use of fossil fuels, such as oil, coal, and gas, can help reduce the concentration of carbon dioxide in the atmosphere. - Planting more trees: Trees absorb carbon dioxide from the atmosphere through photosynthesis, so planting more trees can help reduce the concentration of carbon dioxide in the atmosphere. - Improving energy efficiency: Using energy-efficient appliances, insulating homes and buildings, and reducing energy waste can help reduce the use of fossil fuels and, therefore, reduce greenhouse gas emissions. - Promoting the use of renewable energy: Renewable energy sources, such as solar, wind, and hydropower, do not emit greenhouse gases and can help reduce greenhouse gas emissions. - Reducing waste and promoting recycling: Landfills and incineration of waste produce methane and carbon dioxide, which are greenhouse gases. Therefore, reducing waste and promoting recycling can help reduce greenhouse gas emissions. In conclusion, the greenhouse effect can be reduced by controlling human activities that contribute to the emissions of greenhouse gases. By implementing measures such as reducing the use of fossil fuels, planting more trees, improving energy efficiency, promoting renewable energy, and reducing waste, we can reduce the concentration of greenhouse gases in the atmosphere and mitigate the effects of global warming and climate change.

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The reaction between vegetable oil and plant ash extract is a saponification reaction. Saponification is a process that involves the reaction of an ester, such as vegetable oil, with an alkali, such as plant ash extract, to form soap and glycerol. In the saponification reaction, the ester bond in the vegetable oil is broken, and the alkali reacts with the fatty acids in the oil to form soap molecules and glycerol. The plant ash extract provides the necessary alkali, usually in the form of potassium or sodium hydroxide. The reaction is exothermic, meaning it releases heat, and it is typically carried out in a heated vessel. The resulting mixture is a crude soap that contains impurities such as excess alkali, glycerol, and unsaponified fats. To produce a purer soap, the crude soap is dissolved in water and the impurities are removed through a process called salting out. The purified soap can then be used for cleaning or other purposes. In conclusion, the reaction between vegetable oil and plant ash extract is a saponification reaction, which produces soap and glycerol. This reaction has been used for centuries to produce soap and is still used today in traditional soap-making processes.

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The solution that forms a white precipitate when aqueous ammonia is added, which dissolves in excess ammonia, is the solution of zinc chloride (ZnCl_2(aq)). The addition of aqueous ammonia to zinc chloride solution forms a white precipitate of zinc hydroxide (Zn(OH)_2(s)), which dissolves in excess ammonia to form a colorless, complex ion called tetraamminezinc(II) (Zn(NH₃)₄²⁺). This reaction can be represented by the following equation: ZnCl_2(aq) + 2NH_3(aq) + H₂O(l) → Zn(OH)_2(s) + 2NH₄Cl(aq) The formation of a white precipitate that dissolves in excess ammonia indicates the presence of a metal cation that can form a complex ion with ammonia. Zinc ions (Zn²⁺) have a small size and high charge density, making them highly polarizing and able to form complex ions with ammonia. In contrast, lead ions (Pb²⁺), copper ions (Cu²⁺), and iron ions (Fe²⁺) can also form white precipitates with aqueous ammonia, but these precipitates do not dissolve in excess ammonia due to the formation of insoluble hydroxides.

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In the iodate ion (IO₃^-), the overall charge of the ion is -1. Since oxygen atoms typically have an oxidation state of -2 in ionic compounds, and there are three oxygen atoms in the ion, their total charge contribution is -6. Therefore, the oxidation state of iodine can be calculated by equating the total charge of the ion (-1) with the sum of the oxidation states of iodine and oxygen (-6 + x, where x is the oxidation state of iodine). Solving for x, we get: -6 + x = -1 x = +5 Therefore, the oxidation state of iodine in the iodate ion (IO₃^-) is +5.

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The given reaction involves the combination of hydrogen ions (H⁺) and hydroxide ions (OH^-) to form water (H₂O). The reaction releases energy in the form of heat, which is an indication that the reaction is exothermic. When a substance is formed from its constituent elements, the energy change taking place is called the enthalpy of formation. However, in this case, no new substance is being formed from its constituent elements. When an ionic compound dissolves in water, the energy change taking place is called the enthalpy of solution. However, in this case, no ionic compound is dissolving in water. When an acid and a base react to form a salt and water, the energy change taking place is called the enthalpy of neutralization. Since the given reaction is an acid-base reaction, and water is formed as one of the products, the energy change taking place is the enthalpy of neutralization. Thus, the answer is neutralization.

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The indicator used to neutralize CH₃COOH and NaOH solution is called phenolphthalein, and its pH range is 8-10. When an acid, like CH₃COOH, is mixed with a base, like NaOH, a neutralization reaction occurs, and the resulting solution is neither acidic nor basic but neutral. Phenolphthalein is a chemical compound that changes color depending on the pH of the solution it is in. It is colorless in acidic solutions with a pH below 8 and pink in basic solutions with a pH above 10. Therefore, when phenolphthalein is added to the CH₃COOH and NaOH solution, it will remain colorless as long as the solution is acidic but will turn pink once the solution becomes basic. Since the pH range of phenolphthalein is 8-10, it is suitable for use in neutralizing the CH₃COOH and NaOH solution, as the resulting solution will have a pH within that range, indicating that the acid and base have reacted completely to form a neutral solution.

Bayanin Amsa

Bayanin Amsa

In the given chemical reaction, the oxidation state of Fe changes from +2 to +3, which means that Fe has lost an electron. Similarly, the oxidation state of Mn changes from +7 in MnO^-4 to +2 in Mn²⁺, which means that Mn has gained five electrons. Therefore, Fe is undergoing oxidation and MnO^-4 is undergoing reduction. Reduction is a process in which a species gains electrons, leading to a decrease in its oxidation state. In this case, MnO^-4 is gaining electrons, so it is being reduced. Thus, the answer is MnO^-4.

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The ammonium compound used in the manufacture of dry cells is (NH4)Cl, which is also known as Ammonium Chloride. Dry cells are a type of electrochemical cell that converts chemical energy into electrical energy. They are commonly used in household appliances such as flashlights and remote controls. Ammonium chloride is used in the manufacture of dry cells as an electrolyte. An electrolyte is a substance that conducts electricity when dissolved in water. In dry cells, the ammonium chloride electrolyte helps to facilitate the flow of electrons between the electrodes, allowing the cell to generate an electric current. Out of the options provided, (NH4)Cl is the only ammonium compound. NH4NO3 is ammonium nitrate, which is commonly used as a fertilizer and explosive. (NH4)SO4 is ammonium sulfate, which is also used as a fertilizer. (NH4)2CO3 is ammonium carbonate, which is used in the manufacturing of ceramics and as a cleaning agent. NH4Cl is the only option that is commonly used in the manufacture of dry cells as an electrolyte.

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The aqueous method that cannot be used to distinguish between a strong acid and a weak acid is the action on starch iodide paper. The reason for this is that both strong acids and weak acids are capable of reacting with the starch-iodide paper to produce a blue-black color. This reaction occurs because the iodide ions in the paper react with the hydrogen ions released by the acids to form hydrogen iodide gas. The hydrogen iodide gas then reacts with the starch in the paper to produce the blue-black color. Conductivity measurement and measurement of pH are useful methods to distinguish between strong acids and weak acids. Strong acids completely dissociate in water, producing a high concentration of ions that conduct electricity well and have a very low pH. Weak acids only partially dissociate in water, producing a lower concentration of ions that conduct electricity less efficiently and have a higher pH. Measurement of heat of neutralization is also a useful method to distinguish between strong acids and weak acids. Strong acids release more heat during neutralization than weak acids because they produce more hydrogen ions when they react with a base.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

(a)(i) Differences between solids and gases

Solid	Gas
It has a definite shape and a definite volume.	It has no definite shape or volume and fills its container.
Its particles are closely packed and it is not easily compressed.	Its particles are far apart and it is highly compressible.

(a)(ii) X is freezing (solidification); Y is sublimation; Z is condensation (liquefaction).

(b)(i) Charles' Law

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

Thus, \(V \propto T\), and \(\dfrac{V}{T}=\text{constant}\).

(b)(ii) Graphical illustration of Charles' Law

The straight line passes through the origin. For example, its gradient is \(\dfrac{160-40}{400-100}=0.40\ \text{cm}^3\text{K}^{-1}\), showing that \(V/T\) is constant.

(c) Relative molecular mass of G

By Graham's law, rate of diffusion is inversely proportional to the square root of relative molecular mass:

\[\frac{r_{H_2}}{r_G}=\sqrt{\frac{M_G}{M_{H_2}}}\]

Since equal volumes diffuse,

\[\frac{r_{H_2}}{r_G}=\frac{t_G}{t_{H_2}}=\frac{37.4}{10}=3.74\]

\[3.74=\sqrt{\frac{M_G}{2}}\]

\[M_G=2(3.74)^2=27.98\]

Therefore, the relative molecular mass of G is \(\boxed{28}\).

(d)(i) Assumptions of the kinetic theory of gases

1. A gas consists of tiny particles which are in continuous rapid random motion.
2. The particles are very far apart compared with their sizes, and forces of attraction between them are negligible.

(d)(ii) Effect on the rate of formation of HCl

\[H_{2(g)}+Cl_{2(g)}\rightarrow 2HCl_{(g)}\]

1. Increase in temperature: The molecules gain kinetic energy and move faster. They collide more frequently and with greater energy, so the number of effective collisions increases. Therefore, the rate of formation of HCl increases.
2. Decrease in pressure: The gas molecules become farther apart. Collision frequency and hence the number of effective collisions decrease. Therefore, the rate of formation of HCl decreases.

(e) Metals which produce hydrogen

1. With dilute mineral acid: zinc, for example \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\).
2. With cold water: sodium, for example \(2Na+2H_2O\rightarrow2NaOH+H_2\).
3. With steam: iron, for example \(3Fe+4H_2O\rightarrow Fe_3O_4+4H_2\).
4. With hot concentrated alkali: aluminium, for example \(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\).

Bayanin Amsa

(a)(i) Law of constant composition: a pure compound always contains the same elements combined in the same fixed proportion by mass.
(ii) Law of multiple proportions: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.

(b) For X: oxygen \(= 1.535 - 1.365 = 0.170\) g. For Y: oxygen \(= 1.450 - 1.160 = 0.290\) g.

(i) X: Cu \(= \dfrac{1.365}{63.5} = 0.0215\), O \(= \dfrac{0.170}{16} = 0.0106\); ratio Cu:O \(\approx 2:1\), so X = Cu₂O.
Y: Cu \(= \dfrac{1.160}{63.5} = 0.0183\), O \(= \dfrac{0.290}{16} = 0.0181\); ratio Cu:O \(\approx 1:1\), so Y = CuO.

(ii) Mass of copper combining with 0.500 g oxygen:
I. In X: \(\dfrac{1.365}{0.170} \times 0.500 = \mathbf{4.01\ g}\).
II. In Y: \(\dfrac{1.160}{0.290} \times 0.500 = \mathbf{2.00\ g}\).

(iii) The result (4.01 g and 2.00 g of copper combining with the same 0.500 g of oxygen, a 2:1 ratio) illustrates the law of multiple proportions.

(c)(i) The fragrant odour is from the ester ethyl butanoate: CH₃CH₂CH₂COOCH₂CH₃.
(ii) The white solid left is sodium propan-2-oxide (sodium isopropoxide): (CH₃)₂CHONa, i.e. CH₃CH(ONa)CH₃.

(d)(i) CH₃CH₂COOCH₂CH₃ is ethyl propanoate.
(ii) Structural formula: CH₃CH₂COOCH₂CH₃.
(iii) Reagents/conditions: propanoic acid and ethanol, warmed with a little concentrated H₂SO₄ as catalyst (esterification, under reflux).

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(a)(i) Energy profile diagram

The reaction is exothermic, since \(\Delta H=-13\ \text{kJ mol}^{-1}\). Therefore, the products are at a lower energy level than the reactants.

(ii) Rate of reaction

\[\text{Rate}=\frac{\text{change in concentration of HI}}{\text{time taken}}=\frac{0.001-0}{50}=2.0\times10^{-5}\ \text{mol dm}^{-3}\text{s}^{-1}.\]

(b) Types of salts

1. \(\mathrm{K_4Fe(CN)_6}\): complex salt.
2. \(\mathrm{(NH_4)_2Fe(SO_4)_2\cdot6H_2O}\): double salt.
3. \(\mathrm{Mg(OH)NO_3}\): basic salt.
4. \(\mathrm{NaH_2PO_4}\): acid salt.

(c) Action of carbon(IV) oxide on lime water

Initially, carbon(IV) oxide reacts with lime water to form insoluble calcium trioxocarbonate(IV), \(\mathrm{CaCO_3}\), which is a white precipitate. This makes the lime water milky.

\[\mathrm{Ca(OH)_2(aq)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)}\]

On passing excess carbon(IV) oxide, the calcium trioxocarbonate(IV) dissolves to form soluble calcium hydrogentrioxocarbonate(IV). The solution therefore becomes clear.

\[\mathrm{CaCO_3(s)+CO_2(g)+H_2O(l)\rightarrow Ca(HCO_3)_2(aq)}\]

(d)(i) Uses

1. \(\mathrm{CaCO_3}\): manufacture of cement.
2. \(\mathrm{CaSO_4}\): manufacture of plaster of Paris for casts.
3. \(\mathrm{NaHCO_3}\): used in baking powder.

(d)(ii) Suitable drying agents

1. \(\mathrm{NH_3}\): calcium oxide, \(\mathrm{CaO}\).
2. \(\mathrm{HCl}\): concentrated tetraoxosulphate(VI) acid, \(\mathrm{H_2SO_4}\).
3. \(\mathrm{SO_2}\): concentrated tetraoxosulphate(VI) acid, \(\mathrm{H_2SO_4}\).

(d)(iii) Ammonia as a reducing agent

Ammonia reduces copper(II) oxide to copper:

\[\mathrm{2NH_3(g)+3CuO(s)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)}\]

(e) Formula of the industrial raw material

Constituent	Mass in 100 g	Moles	Mole ratio
Fe	28.1 g	\(28.1/56=0.502\)	\(0.502/0.502=1\)
Cl	35.7 g	\(35.7/35.5=1.006\)	\(1.006/0.502=2\)
\(\mathrm{H_2O}\)	36.2 g	\(36.2/18=2.011\)	\(2.011/0.502=4\)

Hence, the mole ratio is \(\mathrm{Fe:Cl:H_2O}=1:2:4\).

\[\boxed{\mathrm{FeCl_2\cdot4H_2O}}\]

Bayanin Amsa

(a) Electrolysis of CuSO₄ between pure copper electrodes

(i) Mass of cathode after the experiment

With copper electrodes the copper dissolved at the anode is deposited on the cathode, so mass lost by the anode equals mass gained by the cathode.

Mass dissolved at anode \( = 7.20 - 4.00 = 3.20\text{ g} \)

Mass of cathode after \( = 5.75 + 3.20 = \mathbf{8.95\text{ g}} \)

(ii) Electrode equations

I. Anode (oxidation): \[ Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^- \]

II. Cathode (reduction): \[ Cu^{2+}_{(aq)} + 2e^- \to Cu_{(s)} \]

(iii) The colour of the solution does not change. The blue Cu²⁺ ions removed at the cathode are exactly replaced by Cu²⁺ ions formed as the copper anode dissolves, so the concentration of the blue Cu²⁺ ions stays constant.

(iv) Value of the Faraday

Time \( = 1\text{ h }20\text{ min} = 80 \times 60 = 4800\text{ s} \)

Quantity of electricity \( Q = It = 2.0 \times 4800 = 9600\text{ C} \)

Moles of Cu deposited \( = \dfrac{3.20}{63.5} = 0.0504\text{ mol} \)

Each Cu²⁺ needs 2 electrons, so moles of electrons \( = 2 \times 0.0504 = 0.1008\text{ mol} \)

Faraday \( = \dfrac{9600}{0.1008} = \mathbf{9.52 \times 10^{4}\ C\,mol^{-1}} \) (about 95 200 C, close to the accepted 96 500 C).

(b) MnO₄^- + I^- + H⁺ reaction

(i) Oxidation: \[ 2I^- \to I_2 + 2e^- \]

(ii) Reduction: \[ MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O \]

(c) Extraction of tin

(i) Tin occurs chiefly as cassiterite, SnO₂. The ore is concentrated by froth flotation or gravity separation, roasted in air to burn off sulphur and arsenic impurities, then the oxide is reduced with carbon (coke) in a furnace:

\[ SnO_2 + 2C \to Sn + 2CO \]

The molten tin is run off and refined.

(ii) Use of tin: coating steel to make tin-plate for food cans (or making solder and alloys such as bronze).

(iii) Property: tin is resistant to corrosion and non-toxic, so it protects the steel and does not contaminate food.

(d) Pollution

(i) Pollution is the introduction of harmful or unwanted substances (pollutants) into the environment, causing damage to living things or the surroundings.

(ii) A generator undergoes incomplete combustion in a closed room and produces the poisonous gas carbon monoxide, CO. Carbon monoxide combines irreversibly with the haemoglobin of the blood (forming carboxyhaemoglobin), preventing the blood from carrying oxygen. This leads to suffocation and death.

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(a)(i) Laboratory preparation and collection of dry chlorine gas

Chlorine is prepared by warming manganese(IV) oxide with concentrated hydrochloric acid:

\[\mathrm{MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)}\]

The gas is passed through saturated sodium chloride solution to remove hydrogen chloride, then through concentrated sulphuric acid to remove water vapour. Since chlorine is denser than air, it is collected by downward delivery, that is, upward displacement of air. It is not collected over water because it dissolves and reacts in water.

(a)(ii) Uses of chlorine

1. It is used to sterilise drinking water and swimming pools.
2. It is used as a bleaching agent for paper pulp and textiles.

(b)(i) River water flowing through an industrial town may contain poisonous industrial effluents, such as acids, oils and heavy-metal salts, as well as sewage and disease-causing microorganisms. It is therefore unsafe for drinking without treatment.

(b)(ii) Uses of substances in water treatment

Substance	Use
Sand	Filtration: removal of suspended solid particles.
Chlorine	Disinfection: kills germs and bacteria.
Calcium oxide	Neutralises acidic water and raises its pH.
Alum	Coagulates fine colloidal particles so that they settle or can be filtered off.

(c) Volume of hydrogen chloride gas

\[\mathrm{2NaCl(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2HCl(g)}\]

Molar mass of \(\mathrm{NaCl}=23+35.5=58.5\ \mathrm{g\ mol^{-1}}\).

\[\text{Moles of NaCl}=\frac{5.85}{58.5}=0.100\ \mathrm{mol}\]

From the equation, \(2\) mol of \(\mathrm{NaCl}\) produce \(2\) mol of \(\mathrm{HCl}\). Therefore, moles of \(\mathrm{HCl}=0.100\ \mathrm{mol}\).

\[\text{Volume of HCl at s.t.p.}=0.100\times22.4=\boxed{2.24\ \mathrm{dm^3}}\]

(d)

1. Mercury is a metal that is liquid at room temperature.
2. Bromine is a non-metal that is liquid at room temperature.
3. Argon is a monatomic gas at room temperature.

(e) Differences between metals and non-metals

Property	Metals	Non-metals
Physical	Good conductors of heat and electricity; malleable and ductile.	Usually poor conductors of heat and electricity; brittle when solid.
Chemical	Lose electrons to form positive ions and generally form basic oxides.	Gain or share electrons and generally form acidic oxides.

Bayanin Amsa

None