WAEC SSCE - Chemistry - 2011
Tambaya 1 Rahoto
When NH 4CI was dissolved in water, the container was cold to touch. This implies that
Bayanin Amsa
When NH4CI is dissolved in water, it undergoes an endothermic process. This is because the dissolution of NH4CI in water requires energy in the form of heat to break the bonds between the NH4CI molecules and allow the NH4CI to mix with the water. This energy is absorbed from the surrounding environment, resulting in a decrease in temperature and the container feeling cold to the touch. Therefore, the correct option is "the process is endothermic".
Tambaya 2 Rahoto
The compound that makes palm wine taste sour after exposure to the air for few days is
Tambaya 3 Rahoto
Which of the following separation techniques would show that black ink is a mixture of chemical compounds?
Bayanin Amsa
Chromatography would show that black ink is a mixture of chemical compounds. Chromatography is a separation technique that separates different components of a mixture based on their ability to move through a medium, such as paper or a column, at different rates. Since black ink is not a pure substance, but rather a mixture of different chemical compounds, chromatography would be able to separate the different compounds and show that they are present in the ink. By comparing the separation patterns of the black ink to those of known compounds, the different components of the ink can be identified.
Tambaya 4 Rahoto
Which of the following statements about a molar solution is correct? it
Bayanin Amsa
A molar solution is a solution that contains one mole of solute per liter of solution. This means that the correct statement about a molar solution is: "contains one mole of the solute in 1 dm3 of solution". Option A, "is a supersaturated solution", is incorrect because a supersaturated solution contains more solute than it can normally dissolve at a given temperature. Option B, "cannot dissolve more of the solute at that temperature", is incorrect because a saturated solution is the one that cannot dissolve more solute at a given temperature, not a molar solution. Option C, "contains any amount of solute in a given volume of solution", is incorrect because it does not describe the specific concentration of a molar solution, which has a defined amount of solute per liter of solution.
Tambaya 5 Rahoto
The energy change that accompanies the addition of an electron to an isolated gaseous atom is
Bayanin Amsa
The energy change that occurs when an electron is added to an isolated gaseous atom is called "electron affinity." This process involves adding an electron to a neutral atom, which then becomes a negatively charged ion. Electron affinity is a measure of how much energy is released or absorbed when this happens. If energy is released, the electron affinity is negative, meaning the atom is more likely to attract electrons. If energy is absorbed, the electron affinity is positive, meaning the atom is less likely to attract electrons. The electron affinity is related to the ionization energy, which is the energy required to remove an electron from an atom.
Tambaya 7 Rahoto
What type of reaction is represented by the following equation?\(\frac{2}{1}\)D\(\frac{3}{1}\)H → \(\frac{4}{4}\)He + \(\frac{1}{0}\)n + energy
Bayanin Amsa
The given equation represents a nuclear reaction, where the nuclei of two hydrogen isotopes, deuterium (D) and tritium (H), combine to form a helium nucleus and a neutron. This process is called nuclear fusion. Nuclear fusion is a type of nuclear reaction in which two atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. In this reaction, the two hydrogen isotopes, deuterium (D) and tritium (H), combine to form helium-4 (He-4) and a neutron (n), with the release of a large amount of energy. So, the type of reaction represented by the given equation is Nuclear Fusion.
Tambaya 8 Rahoto
Aluminium is extracted electrolysis from
Bayanin Amsa
Aluminium is extracted from bauxite using the process of electrolysis. Bauxite is a naturally occurring mineral that contains a mixture of aluminium, iron, titanium, and silica. The first step in the extraction of aluminium from bauxite involves the purification of bauxite by removing any impurities. This is done by treating the bauxite with sodium hydroxide solution to produce a solution of sodium aluminate. Next, the purified solution of sodium aluminate is subjected to electrolysis in a cell called Hall-Heroult cell. In this cell, the purified solution of sodium aluminate is mixed with molten cryolite, which acts as a solvent to lower the melting point of the aluminium oxide. The mixture is then electrolyzed at high temperatures to separate the aluminium from the other elements present in the solution. During the electrolysis process, aluminium ions are reduced at the cathode, forming liquid aluminium metal, while oxygen is produced at the anode. The liquid aluminium is then siphoned off from the bottom of the cell and sent for further processing to obtain pure aluminium metal. Therefore, the correct answer is bauxite.
Tambaya 9 Rahoto
The initial volume of a gas at 300 K was 220 cm 3 Determine its temperature if the volume became 250 cm 3
Bayanin Amsa
This is an application of Charles's Law, which states that at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature. Mathematically, V/T = k, where V is the volume of the gas, T is its absolute temperature, and k is a constant. So, if the initial volume of the gas is 220 cm 3 at a temperature of 300 K, we can use the above equation to find the value of k. V/T = k 220/300 = k k = 0.7333 (rounded to four decimal places) Now, if the volume of the gas changes to 250 cm 3, we can use the same equation to find its new temperature, T. V/T = k 250/T = 0.7333 T = 250/0.7333 T = 341 K (rounded to the nearest whole number) Therefore, the temperature of the gas when its volume changes from 220 cm 3 to 250 cm 3 is 341 K. Hence, the correct answer is.
Tambaya 10 Rahoto
Which of the following compounds determines the octane rating of petrol?
Bayanin Amsa
The compound that determines the octane rating of petrol is 2,2,4-trimethylpentane. Octane rating is a measure of a fuel's ability to resist "knocking" during combustion, caused by the air/fuel mixture detonating spontaneously in the engine. The higher the octane rating, the more resistant the fuel is to knocking. The octane rating of a fuel is determined by comparing its performance to that of a mixture of 2,2,4-trimethylpentane and n-heptane, which are assigned octane ratings of 100 and 0, respectively. Therefore, 2,2,4-trimethylpentane is used as a standard reference for octane rating, and is one of the compounds that determines the octane rating of petrol.
Tambaya 11 Rahoto
How many isotopes has hydrogen?
Bayanin Amsa
Hydrogen has three isotopes - protium, deuterium, and tritium. Protium is the most common and has one proton and no neutrons, deuterium has one proton and one neutron, and tritium has one proton and two neutrons. Therefore, the correct answer is (2) three isotopes.
Tambaya 12 Rahoto
Which of the following statements about displacement reaction is correct?
Bayanin Amsa
A displacement reaction is a type of chemical reaction where an element or a group of atoms of a molecule is replaced by another element or group of atoms. The correct statement about a displacement reaction is that a more electropositive element displaces a less electropositive one. This is because an element with higher electropositivity has a greater tendency to lose electrons and form positive ions, making it more reactive. Therefore, it can easily displace a less electropositive element from its compound by forming a new compound with it. The position of elements in the reactivity series plays a crucial role in determining the outcome of a displacement reaction. The higher an element is in the reactivity series, the more likely it is to displace a lower element. Finally, displacement reactions can occur at any point during the reaction, and they do not require the reaction to be at equilibrium.
Tambaya 13 Rahoto
Which of the following hydrohalic acids is the weakest?
Tambaya 14 Rahoto
What mass of copper would be formed when a current of 10.0 A is passed through a solution of CuSO4 for 1 hour? [Cu = 63.5; 1F = 96500C]
Bayanin Amsa
The mass of copper that would be formed can be calculated using Faraday's law of electrolysis. According to the law, the amount of substance produced at an electrode is directly proportional to the amount of electric charge passed through the electrolyte. Mathematically, the amount of substance produced (in moles) is given by: moles of substance = (electric charge passed) / (Faraday constant * ionic charge) In this case, copper ions (Cu2+) are being reduced at the cathode, so the ionic charge is 2+. The Faraday constant is 96500 C mol-1, which means that one mole of electrons corresponds to a charge of 96500 C. Now, we need to determine the amount of electric charge that is passed through the solution of CuSO4. We are given that a current of 10.0 A is passed through the solution for 1 hour. This means that the total electric charge passed is: electric charge passed = current x time electric charge passed = 10.0 A x 3600 s electric charge passed = 36000 C Substituting the values of electric charge, Faraday constant, and ionic charge into the above equation, we get: moles of Cu = (36000 C) / (96500 C mol-1 x 2) moles of Cu = 0.1869 mol Finally, we can calculate the mass of copper (Cu) that is formed using its molar mass: mass of Cu = moles of Cu x molar mass of Cu mass of Cu = 0.1869 mol x 63.5 g/mol mass of Cu = 11.86 g (rounded to two decimal places) Therefore, the mass of copper that would be formed when a current of 10.0 A is passed through a solution of CuSO4 for 1 hour is 11.86 g. Hence, the correct answer is.
Tambaya 15 Rahoto
On evaporation to dryness, 250 cm3 of saturated solution of salt X with relative molar mass 101 gave 50.5g of the. What is the soluility of the salt?
Tambaya 16 Rahoto
Rare gases are stable because they
Tambaya 17 Rahoto
Which of the following elements can conveniently be placed in two groups in the periodic table ?
Bayanin Amsa
The element that can conveniently be placed in two groups in the periodic table is hydrogen. Hydrogen is the first element in the periodic table and has an atomic number of 1. It is located in group 1 of the periodic table because it has only one valence electron and shares many chemical properties with the alkali metals, such as lithium and sodium. However, hydrogen also shares some chemical properties with the halogens, such as fluorine and chlorine, which are located in group 17 of the periodic table. Like the halogens, hydrogen can form a single covalent bond by sharing one electron with another atom, and it can also form negative ions called hydrides, which have similar properties to halides. Therefore, hydrogen can be placed in two groups in the periodic table: group 1 and group 17. However, its properties are more similar to the alkali metals than the halogens, so it is generally classified as a nonmetal or a metalloid rather than a halogen. In conclusion, hydrogen is the element that can conveniently be placed in two groups in the periodic table: group 1 and group 17.
Tambaya 18 Rahoto
An element with the electron configuration of 1s 22s 22p 6 would have a combining power of
Bayanin Amsa
The electron configuration 1s22s22p6 belongs to the noble gas neon (Ne). Elements in the noble gas family have a full outer shell of electrons, which makes them stable and unreactive. Combining power, also known as valency, is a measure of how many chemical bonds an element can form with other atoms. Noble gases have a combining power of 0 because they already have a full outer shell of electrons and do not need to gain or lose any electrons to achieve stability. Therefore, the answer is 0.
Tambaya 20 Rahoto
Which of the following metals could be used as sacrifical anode for preventing the corrosion of iron?
Bayanin Amsa
To prevent the corrosion of iron, a more reactive metal can be used as a sacrificial anode. This metal will undergo oxidation instead of iron, and hence, protect it from rusting. Out of the given options, magnesium is the most reactive metal, followed by zinc and aluminium. Therefore, magnesium can be used as a sacrificial anode to protect iron from corrosion. Copper, lead, and silver are less reactive and cannot be used as sacrificial anodes for preventing the corrosion of iron.
Tambaya 21 Rahoto
Consider the following electrochemical cell notation: M(s) / M2+ (aq)// H+ (aq)/ H2(g). The value of the electrode potential is positive when?
Bayanin Amsa
The electrode potential is positive when electrons flow from the metal electrode, M(s), to the hydrogen electrode, H2(g). The electrochemical cell notation M(s) / M2+ (aq) // H+ (aq) / H2(g) represents a cell with a metal electrode M(s) in contact with a solution containing M2+ ions and a hydrogen electrode in contact with a solution containing H+ ions. The double vertical lines (||) represent the salt bridge or porous membrane used to separate the two solutions while allowing the flow of ions to maintain electrical neutrality. The electrode potential of a cell is a measure of the tendency of electrons to flow from one electrode to another. In this case, the electrode potential is positive when electrons flow from the metal electrode, M(s), to the hydrogen electrode, H2(g). This means that the metal electrode has a higher tendency to lose electrons and become oxidized, while the hydrogen electrode has a lower tendency to gain electrons and become reduced. When electrons flow from the metal electrode to the hydrogen electrode, a current is generated and the flow of electrons is accompanied by the flow of positive ions from the metal solution to the hydrogen solution, and vice versa. The flow of current is related to the electrode potential and the resistance of the circuit, and can be used to do work such as powering a light bulb or charging a battery. In conclusion, the electrode potential is positive when electrons flow from the metal electrode, M(s), to the hydrogen electrode, H2(g), in an electrochemical cell. This is because the metal electrode has a higher tendency to lose electrons and become oxidized, while the hydrogen electrode has a lower tendency to gain electrons and become reduced.
Tambaya 22 Rahoto
Which of the following elements would produce coloured ions in aqueous solution?
Bayanin Amsa
Iron is the element that would produce colored ions in aqueous solution. This is because iron has variable oxidation states, which means that it can lose different numbers of electrons to form different ions. When iron ions are in solution, they can absorb specific wavelengths of light and reflect others, resulting in the appearance of color. The color of the ion depends on the oxidation state of iron and the ligands (molecules or ions that surround the metal ion) attached to it. For example, Fe2+ ions can form a complex with the ligand SCN- to give a blood-red color, while Fe3+ ions can form a complex with the ligand EDTA to give a blue color. In contrast, calcium, magnesium, and sodium ions do not usually produce colored ions in aqueous solution.
Tambaya 23 Rahoto
A gas that is collected by upward delivery is likely to be
Bayanin Amsa
A gas that is collected by upward delivery is likely to be lighter than air. This is because upward delivery involves collecting the gas in an inverted container with its open end downwards in a trough filled with water. The gas being collected is less dense than air, and thus it rises upwards and displaces the air in the container, eventually filling it up. The principle of this method is based on the fact that gases with lower density rise upwards, and gases with higher density sink downwards. Therefore, the correct option is that the gas is lighter than air.
Tambaya 24 Rahoto
The colour of phenolphthalein indicator in alkaline solution at the end-point of an acid-base titration is?
Bayanin Amsa
Phenolphthalein is a commonly used acid-base indicator. It is colourless in acidic solutions and pink in basic solutions. Therefore, at the end-point of an acid-base titration, when the solution has become basic, the phenolphthalein indicator will turn pink. Thus, the answer is pink.
Tambaya 25 Rahoto
Which of the following statements about enthalpy of neutralization is correct? It
Bayanin Amsa
Enthalpy of neutralization is the heat released or absorbed when an acid and a base react together to form a salt and water. The enthalpy of neutralization depends on the strength of the acid and the base involved in the reaction. The correct statement about enthalpy of neutralization is that it is constant for a strong acid and a strong base. This is because, in a strong acid-strong base neutralization reaction, the acid and the base completely ionize in solution, forming H+ and OH- ions respectively. The reaction between H+ and OH- ions is highly exothermic and releases a constant amount of heat. This constant value of heat released per mole of acid-base reaction is known as the enthalpy of neutralization. Therefore, the correct option is: "enthalpy of neutralization is constant for a strong acid and a strong base."
Tambaya 26 Rahoto
The following substances are ores of metals except
Bayanin Amsa
Out of the four options given, graphite is the substance that is not an ore of metal. Graphite is a form of carbon and is not considered as an ore of metal. Bauxite is an ore of aluminum, cuprite is an ore of copper, and cassiterite is an ore of tin. Ores are naturally occurring minerals or rocks that contain metal or metal compounds in sufficient amounts to make it economically feasible to extract the metal.
Tambaya 27 Rahoto
The bond formed when two electrons that are shared between two atoms are donated by only one of the atoms is
Bayanin Amsa
The bond formed when two electrons that are shared between two atoms are donated by only one of the atoms is called a dative bond. In this type of bond, also known as a coordinate covalent bond, one atom donates a pair of electrons to be shared with another atom, which accepts the electrons. The atom that donates the electrons is called the donor or Lewis base, while the atom that accepts the electrons is called the acceptor or Lewis acid. Dative bonds are commonly found in coordination compounds, where a metal ion accepts electron pairs from surrounding ligands to form a complex.
Tambaya 28 Rahoto
Bubbling excess carbon (IV) oxide into calcium hydroxide solution results in the formaton of
Tambaya 29 Rahoto
The volume occupied by 17g of H2S at s.t.p. is [H = 1.00, S = 32.0, Molar volume = 22.4 dm3]
Bayanin Amsa
To solve this problem, we need to use the ideal gas law, which relates the volume, pressure, temperature, and number of moles of a gas. At standard temperature and pressure (STP), which is defined as 0°C and 1 atmosphere of pressure, one mole of any gas occupies 22.4 dm³ of volume. First, we need to calculate the number of moles of H2S present in 17g of the gas. To do this, we divide the mass of the gas by its molar mass. Molar mass of H2S = (2 × atomic mass of H) + atomic mass of S = (2 × 1.00 g/mol) + 32.0 g/mol = 34.0 g/mol Number of moles of H2S = mass of H2S ÷ molar mass of H2S = 17g ÷ 34.0 g/mol = 0.5 mol Now, we can use the ideal gas law to calculate the volume of 0.5 mol of H2S at STP. PV = nRT where P = 1 atm, V is the volume we want to find, n = 0.5 mol, R is the gas constant (0.0821 L atm/K mol), and T = 273 K. Solving for V, we get: V = nRT/P = (0.5 mol) × (0.0821 L atm/K mol) × (273 K) ÷ (1 atm) = 11.2 L or dm³ Therefore, the volume occupied by 17g of H2S at STP is 11.2 dm³. Answer: 11.2 dm³.
Tambaya 30 Rahoto
What is the amount of magnesium that would contain 1.20 x 1024 particles? [Mg = 24, Avogadro's constant = 6.02 x 1023]
Bayanin Amsa
Avogadro's constant is the number of particles in one mole of a substance, which is approximately equal to 6.02 x 10^23 particles. Therefore, if we have 1.20 x 10^24 particles of magnesium, we can calculate the number of moles by dividing the number of particles by Avogadro's constant: n = (1.20 x 10^24 particles) / (6.02 x 10^23 particles/mole) n = 2.0 moles Therefore, the amount of magnesium that would contain 1.20 x 10^24 particles is 2.0 moles. So, the correct option is "2.0 moles".
Tambaya 31 Rahoto
When bromine is added to ethene at room temperature, the compound formed is
Bayanin Amsa
When bromine is added to ethene at room temperature, the compound formed is 1,2-dibromoethane. This reaction is an example of an addition reaction in which the double bond of the ethene molecule breaks, and the two bromine atoms are added to the carbon atoms of the ethene molecule. The product formed in this reaction is 1,2-dibromoethane because the two bromine atoms are added to adjacent carbon atoms of the ethene molecule.
Tambaya 32 Rahoto
When element 20A combines with element 8Y,
Bayanin Amsa
The combination of elements with atomic numbers 20 and 8 (A and Y, respectively) would result in the formation of an ionic compound, AY. Element 20 (A) is calcium, a metal that easily loses two electrons to form a stable Ca2+ ion. Element 8 (Y) is oxygen, a non-metal that readily accepts two electrons to form a stable O2- ion. When calcium (A) reacts with oxygen (Y), calcium loses two electrons to form Ca2+, and oxygen gains two electrons to form O2-. The oppositely charged ions, Ca2+ and O2-, are then held together by electrostatic forces of attraction to form the ionic compound, AY. Therefore, the correct option is: "an ionic compound, AY is formed."
Tambaya 33 Rahoto
The number of atoms in one mole of a substance is equal to the
Bayanin Amsa
The number of atoms in one mole of a substance is equal to Avogadro's number. Avogadro's number is a constant, denoted as 'N\(_A\)', which represents the number of particles (atoms, molecules, ions, etc.) in one mole of a substance. The value of Avogadro's number is approximately 6.022 x 10\(^{23}\) particles per mole. Therefore, if we know the molar mass of a substance (in grams per mole), we can calculate the number of atoms (or molecules) in one mole of that substance by multiplying the molar mass by Avogadro's number. This relationship is commonly used in chemistry to relate the mass of a substance to the number of particles it contains.
Tambaya 34 Rahoto
Which of the following ions has the electron configuration 2,8,8?
Bayanin Amsa
The electron configuration 2,8,8 represents an atom with 18 electrons, which corresponds to the noble gas argon (Ar). To determine which ion has the same electron configuration, we need to look at the number of electrons that each ion has lost or gained compared to the neutral atom. Na+ has lost one electron, which means it now has 10 electrons, with a configuration of 2,8. This is not the same as 2,8,8, so Na+ is not the correct ion. Mg2+ has lost two electrons, which means it now has 10 electrons, with a configuration of 2,8. This is also not the same as 2,8,8, so Mg2+ is not the correct ion. F- has gained one electron, which means it now has 10 electrons, with a configuration of 2,8. This is not the same as 2,8,8, so F- is not the correct ion. Cl- has gained one electron, which means it now has 18 electrons, with a configuration of 2,8,8. Therefore, the correct answer is Cl-.
Tambaya 35 Rahoto
Which of the following metallic oxides is amphoteric?
Bayanin Amsa
An amphoteric oxide is one that can react as both an acid and a base. This means that it can react with both acids and bases to form salts. The only oxide listed that can do this is aluminum oxide (Al2O3). Therefore, the answer is AI2O3. MgO and Na2O are basic oxides, which means they react with acids to form salts. Fe2O3 is an acidic oxide, which means it reacts with bases to form salts.
Tambaya 36 Rahoto
Consider the reaction represented by the following equation: xKMnO 4(aq) + ySO2(g) + zH2O(I) → K 2SO4(AQ) + 2MnSO4(aq) + 2H2SO x,y and z are respectively
Tambaya 37 Rahoto
The equation P = \(\frac{K}{V}\) illustrates
Bayanin Amsa
The equation P = \(\frac{K}{V}\) illustrates the relationship between pressure and volume of a gas at constant temperature. This is known as Boyle's law, which states that the pressure of a gas is inversely proportional to its volume when temperature is kept constant. In other words, as the volume of a gas decreases, its pressure will increase, and vice versa, as long as the temperature is held constant.
Tambaya 38 Rahoto
In metallic colids, the forces of attraction are between the mobile valence electrons and
Bayanin Amsa
In metallic colids, the forces of attraction are between the mobile valence electrons and the positively charged nuclei. In a metallic solid, the outermost valence electrons of metal atoms are not tightly bound to their respective atoms, but instead, they move freely throughout the entire crystal structure. These electrons are referred to as delocalized electrons. The positive metal ions are arranged in a lattice structure and are surrounded by these delocalized electrons. These electrons are attracted to the positively charged nuclei, which keeps the metal ions together in the lattice structure. Therefore, the forces of attraction in metallic colids are between the mobile valence electrons and the positively charged nuclei.
Tambaya 39 Rahoto
Which of the following processes does not involve the use of limestone?
Tambaya 40 Rahoto
Which of the following compounds would react with ethanoic acid to give a sweet smelling liquid?
Bayanin Amsa
The compound that would react with ethanoic acid to give a sweet smelling liquid is an alkanol. An alkanol is an organic compound that has a hydroxyl (-OH) group attached to an alkyl group. When an alkanol such as ethanol (C2H5OH) reacts with ethanoic acid (CH3COOH), the reaction is an esterification reaction, which produces an ester and water. The reaction is catalyzed by the presence of an acid, such as sulfuric acid (H2SO4). The general equation for this reaction is: alcohol + carboxylic acid ⇌ ester + water In this case, the reaction between ethanol and ethanoic acid produces the ester ethyl acetate, which has a sweet, fruity odor. This sweet smelling liquid is used in the production of perfumes, artificial flavors, and solvents. Therefore, the compound that would react with ethanoic acid to give a sweet smelling liquid is an alkanol, which has a hydroxyl group (-OH) attached to an alkyl group.
Tambaya 41 Rahoto
Which of the following equimolar solutions has the highest conductivity?
Tambaya 42 Rahoto
Consider the following reaction equation: X(g)
+ Y(g) ⇌ XY(g); ∆H = + 220KJ mol-1 If the temperature of the system is increased, the
Bayanin Amsa
The reaction equation given is X(g) + Y(g) ⇌ XY(g); ∆H = +220KJ mol-1. The ∆H value is positive which indicates that the reaction is endothermic, meaning that heat is absorbed when the reaction takes place. According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in temperature, pressure, or concentration of reactants/products, the system will respond in a way that tends to counteract that change. In this case, if the temperature of the system is increased, the equilibrium position will shift in the direction that absorbs heat. Since the reaction is endothermic, the forward reaction will absorb heat, so the system will favor the forward reaction to absorb the excess heat. Therefore, the forward reaction would be favored if the temperature of the system is increased. Hence, the correct option is "forward reaction would be favored."
Tambaya 44 Rahoto
Which of the following arrangements is in order of increasing metallic property?
Tambaya 45 Rahoto
What is the value of n in the following equation? Cr2O72- + 14H+ + ne- → 2Cr 3+ + 7H2O
Bayanin Amsa
In the given chemical equation, the electrons (ne-) appear on the left-hand side of the equation as a reactant. This indicates that the equation is showing a reduction half-reaction, where electrons are gained. To balance the electrons gained in the reduction half-reaction, we need to identify the number of electrons that are being transferred. In this case, we can see that 6 electrons are being gained by the reactants (Cr2O72- and 14H+) to form 2 moles of Cr3+ and 7 moles of H2O. Therefore, the value of n in the equation is 6.
Tambaya 46 Rahoto
Which of the following electrons configurations correctly represents an inert element?
Bayanin Amsa
An inert element is an element that has a stable electron configuration, which means it is not likely to lose, gain or share electrons. It is also known as a noble gas. The correct electron configuration for an inert element is a full outermost energy level, which is represented as an electron configuration with eight valence electrons. Therefore, the electron configuration that correctly represents an inert element is 1s22s22p6. This configuration represents the noble gas, neon (Ne), which has a full outermost energy level and does not easily react with other elements.
Tambaya 47 Rahoto
(a)Define the term solubility.
(b) The table below gives the solubility of salt Z at various mperatures
|
|
0 | 10 | 20 | 30 | 40 | 50 | 60 |
| Solubility (mol dm\(^{-3}\)) |
0.13 | 0.21 | 0.31 | 0.45 | 0.63 | 0.85 | 1.10 |
(i) Plot a graph of solubility against temperature.
(ii) From the graph determine the solubility of salt Z at 35°C.
(iii) If 100cm\(^{-3}\) of the saturated solution is cooled from 55°C to 35°C, calculate the mass of salt Z that would crystallize out. [Molar mass of salt Z = 100 g]
(c)(i) Write a balanced equation to illustrate the reaction of AI\(_2\)O\(_3\) with dilute I. HCI; II. NaOH.
(ii) What is the name given to an oxide that exhibits both acidic and basic properties?
(iii) Give one metallic oxide which exhibite these properties.
(d)(i) Determine the oxidation number of: I. Al in [Al (H\(_2\)O)\(_6\)]\(^{3+}\); II. H in NaH.
(ii) Give the IUPAC name of each of the following substances; I. CuSO\(_4\).5H\(_2\)O; II. CaCO\(_3\); Ill. KMnO\(_4\).
Bayanin Amsa
None
Tambaya 48 Rahoto
(a)(i) What is a functional group?
(ii) State the functional group in each of the following compounds: I. CH\(_3\)CH\(_2\)CH(CH\(_3\))OH; II. CH\(_3\)CH\(_2\)CH\(_2\)COOH.
(b)(i) Complete the following equations:
(ii) Draw the structure(s) of the product(s) formed in (b)(i) above.
(c)(i).Write an equation for the prepzration of butan-2-ol from butene.
(d) (i) Give the reagents required for the following conversions to take place: I. CH\(_2\) = CH\(_2\) to CH\(_3\)CH\(_2\)OH; II. CH\(_3\)CH\(_2\)OH to CH\(_3\)COOH; Ill. CH\(_3\)COOH to CH\(_3\)COOCH\(_2\)CH\(_3\).
(e) Consider the following organic structure; CH\(_3\)OHCHCH = CHCOOH
(i) State what would be observed when the organic compound is treated with each of the following reagents: I. cold NaHCO\(_{3(aq)}\); II. hot solution of I\(_2\) in NaOH\(_{(aq)}\); Ill. bromine water.
None
Bayanin Amsa
None
Tambaya 49 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
(a) A colourless gas P was given off when dilute tetraoxosulphate (VI) acid was added to zinc salt Q. On bubbling the gas through lime water, a white precipitate R was formed. Identify P, Q and R.
(b) Name a suitable apparatus that could be used to perform each of the following activities in the laboratory
(i) storage of dilute silver trioxonitrate (V):
(ii) heating copper metal;
(iii) separation of a mixture of water and Kerosene. Give one reason for each of your answers in (b)
None
Bayanin Amsa
None
Tambaya 50 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
Credit will be given for strict adherence to the instructions, for observations precisely record and for accurate inferences. All tests, observations and inferences must be clearly entered in your answer book in ink, at the time they are made.
C is a mixture of two salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion drawn from the result of each fest
(a) Put all of C into a boiling tube and add about 5cm\(^3\) of distilled water. Stir thoroughly and filter. Keep both the residue and the filtrate.
(b) To about 2 cm\(^3\) of the filtrate, add few drops of Pb(NO\(_3\))\(_{(aq)}\). Boil the mixture and then allow to cool.
(c)(i) Put the residue in a test tube and add dilute HNO\(_{3}\). Shake the mixture and divide the solution into two portions
(ii) to the first portion from (c)(i), add NaOH\(_{(aq)}\) in drops acid then in excess.
(iii) To the second portion from (c)(ii), add aqueous ammonia in drops and then in excess
Bayanin Amsa
None
Tambaya 51 Rahoto
The following table gives the atomic numbers of elements V, W, X, Y and Z.
| Element |
V | W | X | Y | Z |
| Atomic number |
11 | 16 | 18 | 19 | 24 |
(a)Which of the elements: (i) belong(s) to group 1?
(ii) is/are riotle gas(es)?
(iii) form(s) coloured compound(s)?
(iv) form(s) an anion?
(v) react(s) with water to liberate hydrogen?
(vi) react(s) with water to form alkaline solution?
(b) What is the:
(i) charge on the ion formed in
(a)(iv) above?
(ii) group of the element(s) in (a)(i) above?
(c)(i) Write the formula of the compound formed between element V and element W.
(ii) Sate the type of bond formed in (c)(i) above. Explain your answer.
(d)(i) What is a covalent compound?
(ii) Give two factors that influence covalent bonding.
(iii) State the type of bond that exists in each of the following substances: MgO, NH\(_3\) and Fe.
(iv) What are intermolecular forces?
Tambaya 52 Rahoto
(a) A compound X reacts with excess HNO\(_{3(aq)}\) to give carbon (IV) oxide and another compound Y. A solution of Y reacts with NaOH\(_{(aq)}\) to form a white precipitate which is insoluble in excess NaOH\(_{(aq)}\). Identify X and Y.
(b) (i) Write a balanced equation to illustrate the reducing property of ammonia in its reaction with CuO.
(ii) Explain why it is not advisable to heat ammonium dioxonitrate (III) directly.
(iii) Give two uses nitrogen.
(c) Give the reason why (i) dilute H\(_{2}\)SO\(_{4}\) is not suitable for the preparation of CO\(_{2(g)}\) from CaCO\(_{3(s)}\) (ii) concentrated H\(_{2}\)SO\(_{4}\) cannot be used to dry ammonia gas.
(d) State two: (i) physical properties; (ii) chemical properties of metals.
(e) What is the oxidation number of: (i) chlorine in I. Cl\(_{2}\). II. ClO\(_{-(3)}\)
(ii) vanadium in V\(_{2}\)O\(_{5}\)
(f)(i) Explain the term half-life. (ii) Two radioactive elements, P and Q have half-life of 1200 seconds and 3600 seconds respectively.
I. Which of the elements is more stable? II. Give a reason for your answer.
Bayanin Amsa
None
