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Question 1 Rapport
What is the main impurity in haematite?
Détails de la réponse
The main impurity in haematite is typically SiO3, also known as silica. Haematite is an iron ore consisting primarily of iron oxide (Fe2O3) and is a common source of iron. However, impurities can be present in the ore, and silica is one of the most common impurities. Silica can affect the quality of the iron produced from the ore and can cause problems during the smelting process. Therefore, it is important to remove as much silica as possible during the processing of haematite.
Question 2 Rapport
Which of the following compound is used for removing impurities from bauxite?
Détails de la réponse
The compound used for removing impurities from bauxite is NaOH (sodium hydroxide). This process is called the Bayer's process. Bauxite contains impurities like iron oxide (Fe2O3), silicon dioxide (SiO2), and titanium dioxide (TiO2), which must be removed before the extraction of aluminum from bauxite. In the Bayer's process, crushed bauxite is treated with a hot, concentrated solution of NaOH (sodium hydroxide), which dissolves the aluminum oxide (Al2O3) present in bauxite, forming sodium aluminate (NaAlO2). The impurities such as iron oxide, silicon dioxide, and titanium dioxide are not soluble in NaOH and are filtered out. The filtered solution of sodium aluminate is then treated with carbon dioxide (CO2) to form aluminum hydroxide (Al(OH)3) which is then calcined to obtain pure aluminum oxide, which is used for the extraction of aluminum metal.
Question 3 Rapport
Which of the following processes involves neutralization?
Détails de la réponse
The process that involves neutralization is liming of soil. When soil is acidic, it can be treated with a base such as calcium carbonate (CaCO3) to raise its pH and reduce its acidity. This process is known as liming of soil and involves a neutralization reaction between the acidic components of the soil and the basic components of the calcium carbonate. The result is a more neutral soil pH which is more suitable for plant growth. In the other options, hardening of oils involves hydrogenation, souring of milk involves fermentation, and charring of sugar involves thermal decomposition.
Question 4 Rapport
Which of the following statements about dative bonding is not correct?
Détails de la réponse
The statement that "each participating atom contributes one electron" is not correct. In dative bonding, also known as coordinate bonding, one atom donates a pair of electrons to another atom. The atom that donates the electron pair is called the donor, and the atom that accepts the electron pair is called the acceptor. The donor atom contributes both electrons in the bond, while the acceptor atom does not contribute any electrons to the bond. Therefore, each participating atom does not contribute one electron.
Question 5 Rapport
When a sample of water was boiled, it lathered more readily with soap. it can be concluded that the sample most likely contained
Détails de la réponse
When water is boiled, it becomes more receptive to lathering with soap. This indicates that the sample contains calcium hydrogentrioxocarbonate (IV) (Ca(HCO3)2). Hard water contains calcium and magnesium ions which reacts with soap to form an insoluble precipitate (soap scum), which reduces the effectiveness of soap. Boiling hard water removes temporary hardness (calcium hydrogen-carbonate) by causing the bicarbonate ions to decompose and produce insoluble carbonate, which can be removed by filtration, leaving behind soft water that lathers more easily with soap.
Question 6 Rapport
Wine containing 8% to 17% ethanol can be converted to gin containing about 40% ethanol by
Détails de la réponse
Gin is a distilled alcoholic beverage that typically contains about 40% ethanol. To produce gin from wine, the wine must undergo a process called distillation. Distillation involves heating the wine to vaporize the ethanol, which has a lower boiling point than water, and then condensing the vapor to collect the ethanol. This results in a much higher concentration of ethanol in the final product. Fermentation, which is the process by which wine is produced, can only produce ethanol concentrations of up to about 17%. Therefore, distillation is necessary to increase the ethanol content of the wine to the level required for gin. Evaporation and oxidation are not used to produce gin from wine.
Question 7 Rapport
What are the two gases associated with the formation of acid rain?
Détails de la réponse
Question 8 Rapport
Which of the following statements is/are correct about tin plating? l The electrolyte must be a soluble salt of tin ll. the objective to be plated is made the cathode lll. The anode is made of tin lV. The negative electrode dissolves
Détails de la réponse
Question 9 Rapport
Sea shells contain mainly CaCO 3. Calcium oxide can be prepared from sea shells by
Détails de la réponse
Calcium oxide (CaO) can be prepared from sea shells which mainly contain calcium carbonate (CaCO3) by heating at a high temperature. This process is called thermal decomposition, where heat is applied to the sea shells causing the CaCO3 to break down into its constituent elements, calcium oxide (CaO) and carbon dioxide (CO2). The reaction can be represented as: CaCO3 → CaO + CO2 Therefore, heating the sea shells at a high temperature will yield calcium oxide (CaO).
Question 10 Rapport
Which of the following substances could be responsible for the banana taste of a food flavour?
Détails de la réponse
The substance responsible for the banana taste of a food flavour is Pentylethanoate. This is because Pentylethanoate, also known as amyl acetate, is a chemical compound that naturally occurs in bananas, as well as in other fruits such as apples and pears. It is commonly used as a flavoring agent in the food industry to replicate the taste and smell of these fruits, and is known for its distinct banana-like flavor and aroma. Therefore, Pentylethanoate is the substance that could be responsible for the banana taste of a food flavor.
Question 11 Rapport
An element X with electronic configuration 1s22s22p 63s 23p5combines with another element Y with the configuration 1s2s22p23s33p5. What is the formula of the compound formed?
Détails de la réponse
To determine the formula of the compound formed between elements X and Y, we need to find out the number of electrons that each element can gain or lose to form ions. Element X has 7 valence electrons and needs to gain one electron to achieve the stable configuration of the nearest noble gas, which is 3s23p6. Thus, element X can form an anion with a charge of -1, which is denoted by X-. Element Y has 5 valence electrons and needs to gain three electrons to achieve the stable configuration of the nearest noble gas, which is 3s23p6. Thus, element Y can form an anion with a charge of -3, which is denoted by Y3-. To form a compound, the total positive charge must balance the total negative charge. The least common multiple of 1 and 3 is 3, so we need three X- ions to balance two Y3- ions. Therefore, the formula of the compound formed is X3Y2. So, the correct option is: XY2.
Question 12 Rapport
Substance Q reacts with cold water to form a solution which turns red litmus blue. Substance Q could be
Détails de la réponse
Substance Q reacts with cold water to form a solution which turns red litmus blue. This indicates that the substance is basic or alkaline in nature. Out of the given options, sodium is the only substance that is basic in nature and reacts with cold water to form a solution which turns red litmus blue. Therefore, substance Q could be sodium.
Question 13 Rapport
Determine the amount of electrons needed to deposit 6.4g of copper during the electrolysis of copper (ll) tetraoxosulphate (IV) solution [Cu = 64]
Détails de la réponse
Question 14 Rapport
A given fuel has an octane number of 100. This means that
Détails de la réponse
The correct answer is that the knocking tendency of the fuel is low. The octane number of a fuel is a measure of its ability to resist "knocking" or "pinging" during combustion, caused by the air/fuel mixture detonating prematurely in the engine. A fuel with a higher octane number is more resistant to knocking and is generally considered to be of higher quality. An octane number of 100 indicates that the fuel has excellent resistance to knocking, and is therefore considered a high-quality fuel.
Question 15 Rapport
The equation for the reaction at the anode during the electrolysis of concentrated sodium chloride solution is
Détails de la réponse
The equation for the reaction at the anode during the electrolysis of concentrated sodium chloride solution is: 2CI- (aq) → CI2(g) + 2e- During the electrolysis of a concentrated sodium chloride solution, the chloride ions (CI-) are oxidized at the anode to form chlorine gas (CI2) and electrons. The resulting electrons flow to the cathode, where they reduce sodium ions (Na+) to form sodium metal (Na). Therefore, the overall reaction is: 2NaCl (aq) → 2Na(s) + CI2(g) Note that this equation does not include the reaction at the cathode.
Question 16 Rapport
Which of the following salts can be prepared by reacting the metal with dilute H2SO4?
Détails de la réponse
Question 17 Rapport
What is the organic product of the reaction of C2H 5OH with excess acidified K 2Cr 2O7(aq)?
Détails de la réponse
The reaction of C2H5OH with excess acidified K2Cr2O7(aq) is an oxidation reaction. In this reaction, the ethanol (C2H5OH) is oxidized to ethanoic acid (CH3COOH). Therefore, the organic product of this reaction is CH3COOH, which is acetic acid or ethanoic acid.
Question 18 Rapport
A given fuel has an octane number of 100. This means that
Détails de la réponse
The octane number is a measure of the ability of a fuel to resist knocking, which is the uncontrolled and explosive combustion of fuel in the engine's combustion chamber. A higher octane number means that the fuel has a greater ability to resist knocking. Therefore, if a fuel has an octane number of 100, it indicates that the knocking tendency of the fuel is low, and it can resist knocking well. Thus, the correct option is "the knocking tendency of the fuel is low."
Question 19 Rapport
Five metals represented by the letters V to Z are arranged in order of their reactivity as shown . V>W>X>Y>Z.
Which of the metals is most likely extracted by the electrolysis of its fused chloride?
Détails de la réponse
The electrolysis of a fused chloride involves the use of electricity to extract the metal from its molten chloride. The process involves the reduction of the metal ion to the metal at the cathode while the chloride ions are oxidized to chlorine gas at the anode. From the reactivity series, metals that are high up are more reactive than those below them. Therefore, the most reactive metal, V, is the most likely to be extracted by the electrolysis of its fused chloride. This is because it requires more energy to extract a metal that is higher up in the reactivity series compared to one that is lower down. Hence, V requires more energy in the form of electricity to extract it from its fused chloride compared to the other metals, making it the most likely to be extracted through this method.
Question 20 Rapport
What term is used to describe an oxide whose aqueous solution turns blue litmus red?
Détails de la réponse
The term used to describe an oxide whose aqueous solution turns blue litmus red is an acid anhydride. Acid anhydrides are oxides that react with water to form acidic solutions. They are usually non-metal oxides that react with water to form an acidic solution. When an acid anhydride dissolves in water, it reacts with the water molecules to produce an acid. The acid then turns blue litmus red, indicating the presence of an acidic solution. Therefore, option B, acid anhydride, is the correct answer.
Question 21 Rapport
The reaction of vegetable oil with a solution of wood ash is
Détails de la réponse
The reaction of vegetable oil with a solution of wood ash is called saponification. This is a chemical process that involves the conversion of a fat or oil into soap and glycerol. When vegetable oil is mixed with a solution of wood ash, the ash provides the hydroxide ions which react with the oil to form soap molecules and glycerol. The soap molecules are composed of a hydrophilic (water-loving) head and a hydrophobic (water-repelling) tail. This allows them to interact with both water and oil, making them effective at cleaning. Saponification is an important industrial process for producing soap, as well as for the production of biodiesel.
Question 22 Rapport
What volume of 0.20 mol dm -3 NaOH solution would yield 5.0 g of NaOH on evaporation to dryness? [NaOH = 40g mol-1]
Détails de la réponse
The first step is to calculate the number of moles of NaOH present in 5.0 g of the solution using the formula: Number of moles = mass ÷ molar mass The molar mass of NaOH is 40g mol-1. Therefore: Number of moles = 5.0 g ÷ 40g mol-1 = 0.125 mol Next, we use the formula: Number of moles = concentration x volume to calculate the volume of the NaOH solution needed to contain 0.125 mol of NaOH. Rearranging the formula gives: Volume = number of moles ÷ concentration Volume = 0.125 mol ÷ 0.20 mol dm-3 = 0.625 dm3 = 625 cm3 Therefore, the volume of 0.20 mol dm-3 NaOH solution needed to yield 5.0 g of NaOH on evaporation to dryness is 625 cm3. The correct option is (b) 625 cm3.
Question 23 Rapport
Which of the following quantities is the same for one mole of Br2(l) and one mole of He(g)?
Détails de la réponse
The quantity that is the same for one mole of Br2(l) and one mole of He(g) is the number of molecules. One mole of any substance contains the same number of particles, known as Avogadro's number (6.022 x 1023 particles per mole). Therefore, one mole of Br2(l) and one mole of He(g) both contain the same number of molecules. The number of atoms and total mass of particles in one mole of Br2(l) and one mole of He(g) are different, as Br2(l) is a molecular substance and contains two bromine atoms per molecule, while He(g) is an element and contains only one helium atom per particle. The volume occupied at s.t.p. by one mole of Br2(l) and one mole of He(g) is also different, as the volume depends on the size and shape of the particles, and the conditions of temperature and pressure.
Question 24 Rapport
Bees inject an acidic substance when they sting. Which of the following chemical could be used to treat bee stings?
Détails de la réponse
When bees sting, they inject an acidic substance into the skin, which causes pain and irritation. To treat the sting, one needs a base that can neutralize the acid. Among the options given, NaHCO3 (sodium bicarbonate) is a base that can effectively neutralize the acid and alleviate the pain and irritation caused by the sting. CH3COOH (acetic acid) is also acidic and cannot neutralize the acid. C2H5OH (ethanol) and H2O2 (hydrogen peroxide) are not bases and cannot neutralize the acid either.
Question 25 Rapport
Which of the following statements about an exothermic reaction is correct?
Détails de la réponse
An exothermic reaction is a type of chemical reaction in which energy is released from the reaction system into the surroundings. Hence, the correct statement about an exothermic reaction is that the products have less heat content than reactants. This means that the energy released during the reaction is given off to the surroundings, and the enthalpy change for the reaction is negative. In contrast, an endothermic reaction is a reaction that absorbs energy from the surroundings, and has a positive enthalpy change.
Question 26 Rapport
Consider the electrochemical cell represented below Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s). Which of the following equations describes the reaction occurring in the cell?
Détails de la réponse
The electrochemical cell is composed of two half-cells: Zn(s)/Zn2+(aq) and Cu2+(aq)/Cu(s). The anode (negative electrode) is the Zn(s)/Zn2+(aq) half-cell, where oxidation occurs, and the cathode (positive electrode) is the Cu2+(aq)/Cu(s) half-cell, where reduction occurs. During the oxidation process at the anode, solid zinc (Zn) is oxidized to form Zn2+ ions, which are released into the solution. This is represented by the half-equation: Zn(s) → Zn2+(aq) + 2e- During the reduction process at the cathode, Cu2+ ions in the solution are reduced to form solid copper (Cu). This is represented by the half-equation: Cu2+(aq) + 2e- → Cu(s) The overall equation for the cell reaction can be obtained by adding the two half-equations and canceling out the electrons: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Therefore, the correct equation that describes the reaction occurring in the cell is Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Answer (C).
Question 27 Rapport
What process does the following equation represent ? n(C\(_{6}\)H\(_{10}\)O\(_{5}\)) + nH\(_{2}\)O → nC\(_{6}\)H\(_{12}\)O\(_{6}\) ?
Détails de la réponse
The equation n(C6H10O5) + nH2O → nC6H12O6 represents the hydrolysis of starch. Hydrolysis is a chemical process that uses water to break the bonds between molecules. In this case, the bond between the glucose molecules in starch is broken down by the addition of a water molecule, resulting in the formation of glucose molecules. Therefore, "Hydrolysis of starch" is the correct answer.
Question 28 Rapport
Which of the following solutions will give a precipitate with dilute tetraoxosulphate (V) acid? l. Lead (ll) trioxonitrate (V) solution ll. Barium chloride solution lll. Sodium chloride solution
Détails de la réponse
Barium chloride solution (ll) will give a precipitate with dilute tetraoxosulphate (V) acid. This is because barium chloride is a salt of barium and chloride ions, while tetraoxosulphate (V) acid is a solution of sulphate ions and hydrogen ions. When barium chloride solution is added to dilute tetraoxosulphate (V) acid, the barium ions react with the sulphate ions to form a white insoluble solid called barium sulphate. This is the precipitate that is formed. Lead (ll) trioxonitrate (V) solution and sodium chloride solution will not give a precipitate with dilute tetraoxosulphate (V) acid.
Question 30 Rapport
Which of the following processes will pollute water?
Détails de la réponse
Discharge of industrial effluents into waterways can pollute water. When industries release their effluents, which are waste materials such as chemicals, heavy metals, or other pollutants, into waterways, they can contaminate the water and harm aquatic life. The pollutants can also make the water unfit for human consumption, agriculture or recreational purposes. Therefore, it is important to properly manage and treat industrial waste before discharging it into the environment to prevent water pollution. The other options do not directly cause water pollution: exposure of water to ultraviolet rays can help to disinfect the water, passage of river water through a sand bed can help to filter the water, and addition of a measured quantity of chlorine can help to disinfect the water as well.
Question 31 Rapport
The electronic configuration of an element is 1s 22s22p63s 2
3p3. Where is the element located in the periodic table?
Détails de la réponse
The electronic configuration of an element is 1s22s22p63s23p3 indicates that the element has two electrons in its 1s orbital, two electrons in its 2s orbital, six electrons in its 2p orbital, two electrons in its 3s orbital, and three electrons in its 3p orbital. Based on this configuration, we can locate the element on the periodic table by looking at the highest energy level filled with electrons, which is the 3rd energy level or period, and the number of valence electrons, which is 5. Thus, the element is located in Group V and Period 3 of the periodic table. Therefore, the correct option is "Group V, period 3".
Question 32 Rapport
Five metals represented by the letters V to Z are arranged in order of their reactivity as shown . V>W>X>Y>Z.
Which of the metals cannot displace Y from an aqueous solution of its salt?
Détails de la réponse
Based on the given order of reactivity, any metal placed before Y in the sequence can displace Y from its salt, but any metal after Y cannot. Therefore, the answer is Z because it is placed after Y in the sequence of reactivity. Thus, Z cannot displace Y from its aqueous salt solution.
Question 33 Rapport
Group O of elements are unreactive because
Détails de la réponse
Group O, also known as the noble gases, are unreactive because their highest occupied energy level is full, which means they have a stable electron configuration. They have no tendency to gain, lose or share electrons, as they already have the full complement of electrons in their outermost shell. This makes them extremely unreactive and inert, which is why they are often referred to as noble or inert gases. Therefore, the correct answer is "their highest occupied energy level is full".
Question 35 Rapport
Which of the following compounds has the same empirical formula as benzene?
Détails de la réponse
Question 36 Rapport
When excess ethene is shaken with acidified KMnO4 solution, the product obtained is
Détails de la réponse
When excess ethene is shaken with acidified KMnO4 solution, the product obtained is ethane-1,2-diol. This is because acidified KMnO4 solution is a strong oxidizing agent and can oxidize alkenes such as ethene. When ethene is reacted with KMnO4, it undergoes oxidative cleavage where the double bond is broken and two hydroxyl groups are added to the carbons of the double bond. The resulting product is a diol, specifically ethane-1,2-diol.
Question 37 Rapport
The pH of 0.50 mol dm-3 solutions of HCI, H2SO4 and CH3COOH were measured. which of the following statements about the values obtained is correct?
Détails de la réponse
The correct statement is that "the pH value of the CH3COOH was the highest." pH is a measure of how acidic or basic a solution is, and it is based on the concentration of hydrogen ions (H+) present in the solution. A lower pH indicates a higher concentration of H+ ions and a more acidic solution, while a higher pH indicates a lower concentration of H+ ions and a more basic solution. For the given solutions, HCl is a strong acid that completely dissociates in water to give H+ and Cl- ions. H2SO4 is also a strong acid that dissociates into two H+ ions and one SO42- ion. CH3COOH is a weak acid that only partially dissociates into H+ and CH3COO- ions. The pH of HCl and H2SO4 solutions will be lower (more acidic) than the pH of the CH3COOH solution because they are strong acids, and their concentration of H+ ions is higher. However, we cannot determine if they have the same pH values or not from the given information. The sum of pH values of these solutions will not be equal to 14, as this value is the pH of a neutral solution (where the concentration of H+ ions is equal to the concentration of OH- ions). The solutions given in the question are acidic, and their pH values will be less than 7. Therefore, the correct statement is that the pH value of the CH3COOH was the highest because it is a weak acid that only partially dissociates, resulting in a lower concentration of H+ ions and a higher pH value.
Question 38 Rapport
(a) State the role of each of the following substances in the treatment of river water for town supply (i) Sand bed. (ii) Alum (iii) Chlorine
(b)(i) Give three major uses of H\(_2\)SO\(_4\)
(ii) Explain the following observation: A strip of blue litmus paper dropped into concentrated H\(_2\)SO\(_4\) becomes charred whereas in dilute H\(_2\)SO\(_4\), it turns red and is not charred.
(iii) Write an equation to show how concentrated H\(_2\)SO\(_4\) reacts with zinc.
(c)(i) List two gaseous pollutants that can be generated by burning coal.
(ii) Explain why coal burns more easily when it is broken into pieces than when it is in lump form.
(ii) What gas is responsible for most of the explosions in coal mines?
(iv) Name: the non-volatile residue left behind after the destructive distlillation of coal
(d)(i) What is meant by allotropy?
(ii) Name two crystalline allotropes of carbon
(iii) Name two elements apart from carbon, which exhibit allotropy
(iv) It is now known that carbon has an allotropic form called fullerene, containing molecules of formula C\(_{60}\). Calculate the mass of one mole of these molecules [C = 12]
Détails de la réponse
None
Question 39 Rapport
(a) i) An organic compound X contains 40% carbon, 6.67% hydrogen, the rest being oxygen. If X has a relative molecular mass of 60, determine its
(i) empirical formula (ii) molecular formula. [H = 1 ; C = 12; O = 16]
(b) An alkanoic acid Y has a relative molecular mass of 74.
(i) State the functional group of Y
(ii) What t of reaction is involved when Y is converted to an alkanoate?
(iii) Determine the structural formula of Y.
(iv) Write an equation for the reaction between Y and sodium
(v) If X in (a) above boils at 118°C and belongs to the same homologous series as Y, state with reason, whether the boiling point of Y will be equal to, higher or lower than 118°C.
(c)(i) What is fermentation?
(ii) Write an equation for the fermentation of glucose.
(iii) What must be added to glucose solution to make it ferment?
(iv) Explain why a tightly corked grass bottle filled to the brim with fresh palm wine shatters on standing.
None
Détails de la réponse
None
Question 40 Rapport
(a) The electronic configuration of five elements represented by the letters P, Q, R, S and T are indicated below.
P - 1s\(_2\)2s\(_2\)2p\(_2\)
Q 1s\(_2\)2s\(_2\)2p\(_4\)
R 1s\(_2\)2s\(_2\)2p\(_6\)
S - 1s\(_2\)2s\(_2\)2p\(_6\)3s\(_2\)
T - 1s\(_2\)2s\(_2\)2p\(_6\)3s\(_2\)3p\(_5\)
without identifying the elements, state which of them
(i) belongs to group VI in the periodic table;
(ii) is strongly metallic in character;
(iii) readily ionizes by gaining one electron;
(iv) contains two unpaired electrons in the ground state atom
(v) readily loses two electrons during chemical bonding
(vi) Does not paricipates in chemical reactions?
(vii) is an s-block element.
(b)(i) Copy and complete the table below as appropriate
| Particle | Number of Protons | Number of Electrons | Number of Neutrons |
| \(^1_1H\) | 1 | 1 | |
| \(^{27}_{13}\)Al\(^{3+}\) | |||
| \(^{16}_{8}O^{2+}\) | 8 |
(ii) Give the reason why atomic radius increases down a group in the periodic table but decreases from left to right in a period.
(c)(i) What is meant by the half-life of a radioactive element?
(ii) The nuclide \(^{210}_{84}PO\) loses an alpha 4° particle to form lead. Write an equation for the reaction.
(d) State the type of chemical bonding which accounts for each of the following observations:
(i) Chlorine exists as discrete molecules
(ii) Sodium chloride dissolves readily in water;
(iii) CuSO\(_{4(aq)}\) forms a deep blue complex ion with excess NH\(_{3(aq)}\)
Question 41 Rapport
(a) Mention one oxide in each case, which
(i) used in bleaching
(ii) is a redish-brown gas
(iii) reacts with NaOH and also with HCI;
(iv) dissolves in water to give a solution with pH greater than 7;
(v) oxidizes hot, concentrated HCI to chlorine
(b)(i) State three methods that can be used to removo hardness in a sample of water that contains calcium hydrogentrioxocarbonate (IV).
(ii) Explain with the aid of appropriate equation, why it is not advisable to build a house with limestone in an environment polluted by sulphur (IV) oxide.
(c)(i) List two compounds of potassium which yield oxygen when heated strongly
(ii) Calculate the amount (in moles) of gas which occupies 250 cm\(^3\) at s.t.p. [1 mole of gas occupies 22.4 dm\(^3\) at s.t.p]
(iii) If 250 cm\(^3\) of a gas at s.t.p. is heated to 27°C at constant pressure, calculate its new volume.
(iv) Explain in terms of the collision theory what happens as a gas is heated at constant pressure.
Détails de la réponse
None
Question 42 Rapport
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book. A is a solution of HCl containing 7.30g dm\(^{-3}\), B is a solution of X\(_2\)CO\(_{3}\) containing 10.6 gdm\(^{-3}\)
(i) Put A into your burette and titrate readings against 20.0 cm\(^3\) or 25.0cm\(^3\) portions of B using methyl orange as indicator. Tabulate your burette reading and calculate the average volume of A used. The equation for the reaction involved in the titration is ;
X\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{(aq)}\) \(\to\) 2XCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(ii) From your results and the information provided above, calculate the (i) concentration or A in mol dm\(^{3-}\)
Question 43 Rapport
TEST OF PRACTICAL KNOWLEDGE QUESTION
State what you would see on;
(i) bubbling SO\(_2\) into acidified KMnO\(_4\) solution
(ii) mixing zinc dust with CuSO\(_4\) solution
(iii) adding concentrated HNO\(_3\) to freshly prepared FeSO\(_4\) solution.
(b) List two substances in case which, if added to dilute H\(_2\)SO\(_4\), would give you
(i) H\(_{2(g)}\)
(ii) ZnSO\(_{4(aq)}\)
(c)
The diagram labelled I to IV below illustrate different laboratory set-ups used in the separation of mixtures.
(i) Name the separation technique illustrated by each diagram
(ii) Which of the set-ups is used for concentrating dilute salt solutions, for the purpose of crystallization?
(iii) Which of the set-ups is used in obtaining clear water from muddy water?
(iv) Mention the set-up you would use to separate a polar solvent from a non-polar solvent
(v) State the modification you would make to the set-up labelled IV in order to use it for separating a mixture of NaCl and NH\(_4\)CI
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