WAEC SSCE - Chemistry - 2017
Tambaya 1 Rahoto
The oxidation number of nitrogen in \(Pb(NO_{3})_{2}\) is
Bayanin Amsa
The oxidation number of an element in a compound represents the charge that the element would have if the compound were a completely ionic compound. The oxidation number of nitrogen in Pb(NO3)2 is +5. This can be determined using the concept of oxidation states, which are assigned to elements based on a set of rules. One of the rules is that the sum of the oxidation states of all the elements in a neutral compound must equal zero. In Pb(NO3)2, the oxidation state of lead (Pb) is +2, since it is a metal and typically has a positive charge in ionic compounds. The sum of the oxidation states of nitrogen (N) and oxygen (O) must then equal -2 to balance out the charge of lead. Since there are three oxygen atoms, each with an oxidation state of -2, the sum of the oxidation states of oxygen is -6. The oxidation state of nitrogen must then be +5 to balance the equation. So, the oxidation number of nitrogen in Pb(NO3)2 is +5.
Tambaya 2 Rahoto
Which of the following arrangements indicates the correct decreasing order of ionization energy of period 3 elements?
Tambaya 3 Rahoto
Atomic orbital is
Bayanin Amsa
An atomic orbital is a region around the nucleus of an atom where electrons are most likely to be found. It is not a physical path that electrons travel around the nucleus, but rather a three-dimensional space that describes the probability of finding an electron at a given location within an atom. Atomic orbitals are characterized by their energy level, shape, and orientation in space. The number and arrangement of electrons in atomic orbitals determine the chemical and physical properties of an element.
Tambaya 4 Rahoto
Which of the following pairs of gases are pollutants from car exhaust?
Tambaya 5 Rahoto
The condensation of two units of glucose will produce a dissacharide with the formula \(C_{12}H_{22}O_{11}\) which is
Bayanin Amsa
The condensation of two glucose molecules produces a disaccharide known as maltose, with the chemical formula \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \). During this reaction, an -OH group from one glucose combines with an -H from another glucose, releasing a molecule of water (H₂O) and forming a glycosidic bond between the glucose units. This specific reaction distinguishes maltose from other disaccharides like sucrose, which is formed from glucose and fructose. Therefore, the correct answer is maltose.
Tambaya 6 Rahoto
Calculate the mass of copper deposited if a current of 0.45A flows through \(CuSO_{4}\) solution for 1hour 15mins. [Cu=64.0, S= 32.0, O =16.0, 1F= 96500C]
Bayanin Amsa
Tambaya 7 Rahoto
In a saturated solution at a given temperature, the undissolved solutes are in equilibrium with
Bayanin Amsa
In a saturated solution, the undissolved solute particles and the dissolved solute particles are in a dynamic equilibrium. This means that at a given temperature, the rate at which solute particles dissolve into the solvent is equal to the rate at which solute particles come out of solution and reform the solid. Therefore, the undissolved solute particles are in equilibrium with the saturated solution, which is the solution that contains the maximum amount of solute that can dissolve at a given temperature. This equilibrium occurs because the solute particles have reached their maximum solubility limit and are unable to dissolve any further. In summary, the undissolved solute particles are in equilibrium with the saturated solution, not with the solvent or the insoluble solute particles.
Tambaya 8 Rahoto
Which of the following factors characterize members of the same homologous series?
Bayanin Amsa
Members of the same homologous series are characterized by having similar physical properties and a gradual change in chemical properties throughout the series. The successive members also undergo changes in their molecular formula by adding a repeating unit of \(CH_{2}\). This means that although the members may have different molecular formulas, they have a similar structural basis and exhibit similar chemical behaviors.
Tambaya 9 Rahoto
Soluble salts can be prepared by the following methods except
Tambaya 10 Rahoto
The number of atoms present in 2.5moles of a triatomic gas is equivalent to [1mole = \(6.02\times 10^{23}\)]
Bayanin Amsa
One mole of a substance contains a specific number of particles, which is called Avogadro's number, represented by \(6.02\times 10^{23}\) particles per mole. Therefore, to find the number of particles present in 2.5 moles of a substance, we can simply multiply Avogadro's number by 2.5: Number of particles = 2.5 moles x Avogadro's number Number of particles = 2.5 x \(6.02\times 10^{23}\) Number of particles = \(1.505\times 10^{24}\) So the answer is option (C) \(1.5\times 10^{24}\), which represents the number of atoms present in 2.5 moles of the triatomic gas.
Tambaya 12 Rahoto
The properties of a good primary standard include the following except
Tambaya 13 Rahoto
The reason for the decrease in the atomic size of elements across a period is that
Tambaya 14 Rahoto
The oxidation number of X in \(XO_{4}^{3+}\) is
Bayanin Amsa
The oxidation number of an element in a chemical compound is a number that represents the charge that an atom of that element would have if the molecule was completely ionic. In a compound, the sum of all the oxidation numbers must equal the net charge on the molecule. In the case of \(XO_{4}^{3+}\), the oxidation number of X can be determined by using this rule. The net charge on the molecule is +3, which means that the sum of all the oxidation numbers of the atoms in the molecule must equal +3. Since the oxidation number of oxygen is always -2, the oxidation number of X can be calculated as follows: \[X + 4(-2) = +3\] Solving for X, we get: \[X = +5\] Therefore, the oxidation number of X in \(XO_{4}^{3+}\) is +5.
Tambaya 15 Rahoto
The common feature of reactions at the anode is that
Bayanin Amsa
The common feature of reactions at the anode is that oxidation is involved. This means that during the reaction, the anode loses electrons and becomes positively charged. As a result, either a neutral atom or a negatively charged ion is produced. In some cases, the electrode may also dissolve due to the reaction. However, the key feature is always oxidation, which is the process of losing electrons.
Tambaya 16 Rahoto
Which of the following gases is not acidic?
Bayanin Amsa
The gas that is not acidic is CO, or carbon monoxide. Acidity is determined by a substance's ability to donate hydrogen ions (protons), and thus, increase the hydrogen ion concentration in a solution. In general, gases that contain oxygen atoms, such as SO3 and NO2, are more likely to be acidic because oxygen has a greater tendency to form bonds with hydrogen atoms. CO, on the other hand, does not contain oxygen and is not acidic.
Tambaya 17 Rahoto
The real harmful effect of the release of chlorofluorocarbons into the atmosphere is that it eventually causes
Bayanin Amsa
The release of chlorofluorocarbons (CFCs) into the atmosphere causes a harmful effect by depleting the ozone layer. The ozone layer is a natural layer of gas in the Earth's stratosphere that protects life on Earth by absorbing harmful ultraviolet (UV) radiation from the sun. CFCs break down ozone molecules, causing the ozone layer to become thinner, which allows more UV radiation to reach the Earth's surface. This can lead to an increase in skin cancer, cataracts, and other harmful effects on plants and animals. Therefore, the real harmful effect of the release of CFCs is that it eventually causes excessive ultraviolet light from the sun to reach the Earth's surface.
Tambaya 18 Rahoto
The name of the compound \(C_{4}H_{9}COOC_{3}H_{7}\) is
Bayanin Amsa
The name of the compound \(C_{4}H_{9}COOC_{3}H_{7}\) is "pentyl propanoate".
In naming organic compounds, the longest carbon chain is used as the base name, followed by the name of the functional group. In this case, the longest carbon chain has 5 carbons, so the base name is "pentyl". The functional group is a carboxyl group (-COO-), so the ending is "oate". The base name and the ending are combined to get "pentyl propanoate".
Tambaya 21 Rahoto
Classification of alkanols is based on the
Tambaya 22 Rahoto
The species with no electron in the 3d-orbital is
Tambaya 23 Rahoto
Fats and oils can be obtained from any of the following sources except
Bayanin Amsa
Paraffin oil is not a source of fats and oils. It is a type of hydrocarbon that is used as a fuel or a lubricant and is not a food source. The other options - animals, groundnuts, and palm kernel - are all sources of fats and oils that can be used for cooking or in the production of various food products.
Tambaya 24 Rahoto
Which of the following statements is NOT a chemical property of an acid?
Tambaya 25 Rahoto
The IUPAC name for the compound \(ClCH_{2}CH_{2}OH\) is
Tambaya 26 Rahoto
Potassium Chloride solid does not conduct electricity because
Tambaya 27 Rahoto
Petroleum is a non-renewable source of energy because it
Bayanin Amsa
Petroleum is a non-renewable source of energy because it cannot be regenerated once it is used up. This means that when we use up all the available petroleum, we cannot create more. Petroleum is formed naturally over millions of years from the remains of plants and animals that were buried underground. Although it may be cheap compared to some other sources of energy, it is not a sustainable option for the future. Unlike renewable sources of energy like solar, wind, and hydro power, which can be replenished naturally, petroleum is finite and will eventually run out if we continue to use it at the current rate. While petroleum can be recycled after use, it is still a finite resource and will eventually be depleted. Therefore, it is important to find alternative sources of energy that are renewable and sustainable.
Tambaya 28 Rahoto
Chemical reactions occurs when the colliding reactants have energy that is
Bayanin Amsa
Chemical reactions occur when the colliding reactants have energy that is greater than the activation energy. Activation energy is the minimum amount of energy required for a chemical reaction to occur. If the colliding reactants do not have enough energy to meet or exceed the activation energy, the reaction will not take place. In other words, the activation energy acts as a barrier that needs to be overcome for the reaction to proceed. Once the reactants have enough energy to surpass this barrier, they can form new products with a lower energy state than the reactants, which releases energy in the form of heat, light, or other forms of energy. Therefore, only when the colliding reactants have energy greater than the activation energy can they overcome this barrier and form new products with a lower energy state, resulting in a chemical reaction.
Tambaya 30 Rahoto
Which of the following statements about endothermic reaction is correct?
Bayanin Amsa
The correct statement about endothermic reactions is that "heat energy is absorbed." Endothermic reactions are those in which energy is absorbed from the surroundings, usually in the form of heat. This causes the temperature of the surroundings to decrease. Activation energy refers to the energy required to start a chemical reaction, and it can be high or low for both endothermic and exothermic reactions. A catalyst is a substance that speeds up a chemical reaction, but it is not always required for an endothermic reaction. Endothermic reactions can occur reversibly or irreversibly, depending on the specific reaction and conditions. The reversibility of a reaction refers to whether it can proceed in both forward and backward directions.
Tambaya 31 Rahoto
When Magnesium reacts with oxygen, heat is
Bayanin Amsa
When magnesium reacts with oxygen, it produces magnesium oxide. During this reaction, energy is released in the form of heat. This means that the reaction is exothermic, and the correct answer is "heat is evolved and the reaction is exothermic."
Tambaya 32 Rahoto
The following glasswares are used to measure the volume of liquids EXCEPT
Bayanin Amsa
The glassware that is NOT used to measure the volume of liquids is a test tube. Test tubes are typically used for mixing and holding small amounts of liquids, but they do not have precise measurement markings like the other glasswares listed. Graduated beakers, pipettes, and burettes have measurement markings and are specifically designed for accurately measuring the volume of liquids.
Tambaya 33 Rahoto
The use of diamond in abrasives is due to its
Bayanin Amsa
The use of diamond in abrasives is due to its hardness. Diamond is one of the hardest naturally occurring substances on Earth, and its hardness makes it ideal for use in abrasive applications. When diamond is used as an abrasive, it can cut through and grind down even the toughest materials, including metals and ceramics. Additionally, diamond is incredibly durable, so it can withstand the wear and tear of abrasive applications better than many other materials.
Tambaya 34 Rahoto
The property of elements which increases down a group of the periodic table is
Bayanin Amsa
The property that increases down a group in the periodic table is the ionic radius. The ionic radius is the size of an atom, measured from the center of the nucleus to the outermost shell of electrons. As you move down a group in the periodic table, the number of electron shells increases, making the atom larger and the ionic radius larger. This is due to the fact that the outermost electrons are farther away from the nucleus and are not held as tightly as the inner electrons. This results in a larger overall size of the atom.
Tambaya 35 Rahoto
The mass of one mole of \((NH_{4})_{2}CO_{3}\) is [O=16.0, N=14.0, C=12.0, H=1.0]
Tambaya 36 Rahoto
If 0.2g of a salt is required to saturate 200\(cm^{3}\) of water at room temperature, what is the solubility of the salt?
Bayanin Amsa
Tambaya 37 Rahoto
Which of the following statements is correct?
Bayanin Amsa
The correct statement is "Ionic compounds in solution would conduct electricity." Explanation: 1. Covalent compounds are formed when two or more non-metal atoms share electrons to form a molecule. They do not readily ionize in solution because they do not contain ions. 2. Covalent compounds consist of molecules made up of atoms bonded together by sharing electrons. These molecules do not contain ions. 3. A hydrogen bond is a type of intermolecular force that occurs between a hydrogen atom and an electronegative atom (usually oxygen, nitrogen, or fluorine) in a different molecule. It is not formed between a metal and a non-metal. 4. Ionic compounds are made up of a metal and a non-metal. When an ionic compound dissolves in water, it dissociates into ions, which can carry an electric charge and conduct electricity. Covalent compounds do not dissociate into ions in solution and do not conduct electricity.
Tambaya 38 Rahoto
Pauli exclusion principle is related to
Tambaya 39 Rahoto
The bond angle in tetrahedral molecules such as tetrachloromethane is
Bayanin Amsa
The bond angle in tetrahedral molecules like tetrachloromethane is 109.5°. This is because the molecule has a tetrahedral geometry, which means it has four atoms bonded to a central atom, with the angle between any two adjacent bonds being 109.5°. This angle is also known as the tetrahedral angle and is determined by the repulsion between the bonding pairs of electrons around the central atom. The four bonding pairs of electrons in tetrachloromethane repel each other equally, causing them to arrange themselves in a tetrahedral shape with the bond angle of 109.5°.
Tambaya 40 Rahoto
A hydrocarbon contained 14.3% hydrogen. The empirical formula for the hydrocarbon is [H= 1.0, C= 12.0]
Bayanin Amsa
The empirical formula of a compound gives the simplest whole number ratio of the elements in the compound. To find the empirical formula of a hydrocarbon, we need to know the percentage of hydrogen in it. In this case, the hydrocarbon contains 14.3% hydrogen. To convert this to a whole number ratio, we divide 14.3 by the atomic weight of hydrogen (1.0), which gives us 14.3. This means that for every 12 atoms of carbon, there are 14.3 atoms of hydrogen. To simplify this to a whole number ratio, we divide both 14.3 and 12 by the greatest common divisor, which is 2. This gives us a ratio of 7:6 for hydrogen to carbon atoms. The empirical formula for the hydrocarbon is therefore \(CH_{2}\), which represents the simplest whole number ratio of hydrogen to carbon atoms in the compound.
Tambaya 41 Rahoto
Consider the following reaction equation: \(2NO_{(g)} + O_{{2}{(g)}} \to 2NO_{{2}{(g)}}\). What is the change in the oxidation number of nitrogen?
Bayanin Amsa
The oxidation number is a number that represents the number of electrons an atom has gained or lost in a molecule. In the given reaction, the oxidation number of nitrogen changes from +2 to +4. At the beginning of the reaction, nitrogen in NO has an oxidation number of +2, while in NO2 it has an oxidation number of +4. This means that nitrogen has gained two electrons and undergone oxidation. To understand why this is happening, we need to look at the reaction. The reactants are NO and O2, and the product is NO2. The oxygen atoms in the reactants have an oxidation number of 0, while in the product they have an oxidation number of -2. This means that each oxygen atom has gained two electrons and undergone reduction. Since nitrogen has not gained or lost any electrons in the reaction, it is not undergoing reduction or oxidation, but it is participating in the reaction by forming bonds with the oxygen atoms. The change in oxidation number of nitrogen from +2 to +4 is a result of sharing two electrons with each oxygen atom in the product. Therefore, the correct answer is "+2 to +4".
Tambaya 42 Rahoto
Which of the following metals has the strongest metallic bond?
Tambaya 43 Rahoto
The property of a chemical reaction that is affected by the presence of a catalyst is
Bayanin Amsa
The property of a chemical reaction that is affected by the presence of a catalyst is the activation energy. Activation energy is the energy required for a chemical reaction to occur. Catalysts are substances that can speed up a chemical reaction by lowering the activation energy required for the reaction to happen. This means that the same chemical reaction that would otherwise require a lot of energy to start, can happen more easily and with less energy input in the presence of a catalyst. Imagine trying to push a heavy object up a hill - it takes a lot of effort and energy to get it started. But if there's a ramp, it becomes much easier to push the object up the hill, and less energy is required. In the same way, a catalyst acts like a ramp for a chemical reaction, making it easier for the reaction to occur with less energy input.
Tambaya 44 Rahoto
Halogens generally react with metals to form
Bayanin Amsa
Halogens are a group of elements that include fluorine, chlorine, bromine, iodine, and astatine. When halogens react with metals, they tend to form salts. Salts are a type of compound that is formed when an acid and a base react with each other. In the case of halogens and metals, the halogen acts like an acid and the metal acts like a base. The halogen donates an electron to the metal, which forms a positively charged ion, and the halogen becomes negatively charged. These two oppositely charged ions then combine to form a salt. For example, when chlorine reacts with sodium metal, they form sodium chloride, which is table salt.
Tambaya 45 Rahoto
When water was added to a white anhydrous substance X, the colour changed to blue. X is
Bayanin Amsa
The white anhydrous substance X changes color to blue when water is added. This is an indication of a chemical reaction taking place, and the resulting blue color is likely due to the formation of a complex compound. Out of the given options, CuSO4 is a blue compound, also known as copper(II) sulfate, and is commonly used as a colorant and in various industrial applications. When it is dissolved in water, it forms a blue solution due to the hydration of copper ions. Therefore, it is most likely that the anhydrous substance X is CuSO4 because the addition of water resulted in a blue color, which is consistent with the characteristic blue color of CuSO4.
Tambaya 46 Rahoto
The purity of a solid sample can best be determined by its
Bayanin Amsa
The melting point of a solid sample is the most reliable way to determine its purity. The melting point of a pure substance is a fixed temperature, which is well defined and does not change with time. If there are impurities present in the sample, the melting point will be lower and will not be a fixed temperature. By measuring the melting point, we can determine the purity of the sample. If the melting point is lower than expected, it indicates the presence of impurities in the sample. This is because impurities lower the energy needed to break the bonds holding the solid together, allowing it to melt at a lower temperature.
Tambaya 47 Rahoto
Which of the following species determines the chemical properties of an atom?
Bayanin Amsa
The species that determines the chemical properties of an atom is the electron. This is because electrons are responsible for the bonding of atoms with each other, which determines the chemical properties of a substance. Electrons are negatively charged particles that orbit around the nucleus of an atom. The number of electrons in the outermost shell (valence shell) of an atom determines how it will interact with other atoms, forming chemical bonds. The chemical reactivity and behavior of an atom depend on its valence electrons, which determine the atom's ability to form chemical bonds with other atoms. Therefore, electrons play a crucial role in determining the chemical properties of an atom.
Tambaya 48 Rahoto
A 0.1\(moldm^{-3}\) solution of sodium hydroxide was diluted with distilled water to 0.001\(moldm^{-3}\). What is the dilution factor?
Bayanin Amsa
Dilution factor
= molarconc.ofdilutedmolarconc.ofundiluted
= 0.0010.1=0.01
Tambaya 49 Rahoto
The seperation of Nitrogen from oxygen using fractional distillation of air is possible because
Bayanin Amsa
The separation of Nitrogen from Oxygen using fractional distillation of air is possible because of the difference in their boiling points. Nitrogen has a boiling point of -196°C while oxygen has a boiling point of -183°C. This means that at a certain temperature, oxygen will boil and turn into a gas before nitrogen. By carefully controlling the temperature, we can separate the two gases as they boil at different times. This is known as fractional distillation, which is a process that separates a mixture into its different components based on their boiling points. Therefore, the key factor in separating nitrogen from oxygen is their difference in boiling points, not their density, reactivity, or position in the periodic table.
Tambaya 50 Rahoto
A primary alkanol has a molecular mass of 60. What is the structural formula of the compound? [C= 12.0, H= 1.0, O= 16.0)
Tambaya 51 Rahoto
(a)(i) What is an acid-base indicator?
(ii) Give one example of an acid-base indicator.
(b) State the property exhibited by nitrogen(IV) oxide in each of the following equations:,
(i) 4Cu + 2NO\(_2\) -> 4CuO + N\(_2\) (ii) H\(_2\)O + 2NO\(_2\) --> HNO\(_3\) + HNO\(_2\)
(c)(i) Define enthalpy of combustion..
(ii) State why the enthalpy of combustion is always negative.
(d)(i) Distinguish between a primary cell and a secondary cell.
(ii) Give an example of each of the cells stated in I (d)(i).
(e) Define the term mole.
(f) Calculate the amount of hydrochloric acid in 40.0 cm\(^3\) of 0.40 moldm\(^{-3}\) dilute HCl.
(g) Name two substances which can be used as electrodes during the electroylsis of acidified water.
(h) List two forces of attraction that can exist between covalent molecules.
(i) Name the products formed when butane undergoes incomplete combustion.
(j) Write the electron configuration of \(_{26}\)Fe\(^{3+}\)
(a)(i) An acid-base indicator is a substance (usually a weak acid or weak base) that shows different colours in acidic and alkaline media and is used to detect the end point of a neutralization.
(ii) Example: litmus (or methyl orange, or phenolphthalein).
(b)(i) In \(4Cu + 2NO_2 \rightarrow 4CuO + N_2\), nitrogen(IV) oxide acts as an oxidizing agent (it is reduced and oxidizes the copper).
(ii) In \(H_2O + 2NO_2 \rightarrow HNO_3 + HNO_2\), nitrogen(IV) oxide undergoes disproportionation, acting as both an oxidizing and a reducing agent (N goes from +4 to +5 and to +3).
(c)(i) Enthalpy of combustion is the heat change when one mole of a substance is completely burnt in excess oxygen under standard conditions.
(ii) It is always negative because combustion is exothermic - heat is released to the surroundings.
(d)(i) A primary cell cannot be recharged; its chemical reaction is irreversible. A secondary cell can be recharged because its reaction is reversible.
(ii) Primary cell: dry (Leclanche) cell. Secondary cell: lead-acid accumulator.
(e) A mole is the amount of a substance that contains as many elementary entities as there are atoms in 12 g of carbon-12 (that is, \(6.02 \times 10^{23}\) entities).
(f) Amount of HCl \(= C \times V = 0.40 \times \dfrac{40.0}{1000} = 0.016\ \text{mol}\).
(g) Platinum and carbon (graphite) - inert electrodes.
(h) Van der Waals (dispersion) forces and hydrogen bonds.
(i) Carbon(II) oxide (carbon monoxide) and water (with some carbon/soot).
(j) \(_{26}\)Fe is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\); Fe3+ loses the two 4s and one 3d electron:
\[1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\]
Bayanin Amsa
(a)(i) An acid-base indicator is a substance (usually a weak acid or weak base) that shows different colours in acidic and alkaline media and is used to detect the end point of a neutralization.
(ii) Example: litmus (or methyl orange, or phenolphthalein).
(b)(i) In \(4Cu + 2NO_2 \rightarrow 4CuO + N_2\), nitrogen(IV) oxide acts as an oxidizing agent (it is reduced and oxidizes the copper).
(ii) In \(H_2O + 2NO_2 \rightarrow HNO_3 + HNO_2\), nitrogen(IV) oxide undergoes disproportionation, acting as both an oxidizing and a reducing agent (N goes from +4 to +5 and to +3).
(c)(i) Enthalpy of combustion is the heat change when one mole of a substance is completely burnt in excess oxygen under standard conditions.
(ii) It is always negative because combustion is exothermic - heat is released to the surroundings.
(d)(i) A primary cell cannot be recharged; its chemical reaction is irreversible. A secondary cell can be recharged because its reaction is reversible.
(ii) Primary cell: dry (Leclanche) cell. Secondary cell: lead-acid accumulator.
(e) A mole is the amount of a substance that contains as many elementary entities as there are atoms in 12 g of carbon-12 (that is, \(6.02 \times 10^{23}\) entities).
(f) Amount of HCl \(= C \times V = 0.40 \times \dfrac{40.0}{1000} = 0.016\ \text{mol}\).
(g) Platinum and carbon (graphite) - inert electrodes.
(h) Van der Waals (dispersion) forces and hydrogen bonds.
(i) Carbon(II) oxide (carbon monoxide) and water (with some carbon/soot).
(j) \(_{26}\)Fe is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\); Fe3+ loses the two 4s and one 3d electron:
\[1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\]
Tambaya 52 Rahoto
(a)(i) 1. State the periodic law.
2. What is meant by the term periodic property of elements?
(ii) List three properties of an element which show periodicity.
(iii) Explain briefly how each of the properties listed in (a)(i) in varies across the period.
(b) Defulle relative atomic mass.
(c)(i) What phenomenon is exhibited by an element Z which exist as \(^{35}_{17}Z\) and \(^{37}_{17}X\)
(ii) What accounts for the difference in the mass numbers of the element Z?
(iii) Calculate the relative atomic mass of Z if the percentage abundance of \(^{37}_{17}Z\) is 75%
(d)(i) State the method used for collecting each of the following gases: I. CO II. HCI III. H\(_2\)
(ii) Give a reason for your answer stated in (d)(i) I and II
Bayanin Amsa
None
Tambaya 53 Rahoto
(a)(i) Describe briefly the industrial preparation of ammonia.
(ii) Write a balanced equation for the reaction in (a)(i).
(iii) State one way of increasing the yield of ammonia in 4(a)(i).
(iv) State two uses of ammonia.
(b) Describe briefly, one chemical test for each of the following gases in the laboratory: (i) hydrogen; (ii) carbon (IV) oxide; (iii) oxygen.
(c)(i) State the composition of water gas.
(ii) List two uses of water gas.
(d) Describe briefly a simple experiment to determine the type of hardness in a sample of water.
(a)(i) Industrial preparation of ammonia (Haber process): nitrogen (from air) and hydrogen (from natural gas) are mixed in the ratio 1:3 and passed over a finely divided iron catalyst at about 450 °C and about 200 atmospheres pressure.
(ii) \[N_2 + 3H_2 \rightleftharpoons 2NH_3\]
(iii) Increase the yield by increasing the pressure (or by lowering the temperature, or removing ammonia as it forms).
(iv) Uses of ammonia: manufacture of fertilizers (ammonium salts) and manufacture of trioxonitrate(V) acid.
(b) Chemical tests
- (i) Hydrogen: a lighted splint applied to the gas burns it with a squeaky "pop" sound.
- (ii) Carbon(IV) oxide: the gas turns limewater (calcium hydroxide solution) milky.
- (iii) Oxygen: the gas rekindles (relights) a glowing splint.
(c)(i) Water gas is a mixture of carbon(II) oxide and hydrogen (CO + H2).
(ii) Uses of water gas: as a fuel; as a source of hydrogen in industry (for example in the manufacture of methanol/ammonia).
(d) Type of hardness: boil a sample of the water and then add soap solution. If the water now lathers readily (hardness removed by boiling), the hardness is temporary. If it still does not lather easily after boiling, the hardness is permanent. Comparing lathering before and after boiling distinguishes the two.
Bayanin Amsa
(a)(i) Industrial preparation of ammonia (Haber process): nitrogen (from air) and hydrogen (from natural gas) are mixed in the ratio 1:3 and passed over a finely divided iron catalyst at about 450 °C and about 200 atmospheres pressure.
(ii) \[N_2 + 3H_2 \rightleftharpoons 2NH_3\]
(iii) Increase the yield by increasing the pressure (or by lowering the temperature, or removing ammonia as it forms).
(iv) Uses of ammonia: manufacture of fertilizers (ammonium salts) and manufacture of trioxonitrate(V) acid.
(b) Chemical tests
- (i) Hydrogen: a lighted splint applied to the gas burns it with a squeaky "pop" sound.
- (ii) Carbon(IV) oxide: the gas turns limewater (calcium hydroxide solution) milky.
- (iii) Oxygen: the gas rekindles (relights) a glowing splint.
(c)(i) Water gas is a mixture of carbon(II) oxide and hydrogen (CO + H2).
(ii) Uses of water gas: as a fuel; as a source of hydrogen in industry (for example in the manufacture of methanol/ammonia).
(d) Type of hardness: boil a sample of the water and then add soap solution. If the water now lathers readily (hardness removed by boiling), the hardness is temporary. If it still does not lather easily after boiling, the hardness is permanent. Comparing lathering before and after boiling distinguishes the two.
Tambaya 54 Rahoto
(a) Write the molecular formula of X.
(i) What type of reaction is represented by the equation?
(ii) Consider the following reaction equation: \( \mathrm{C_{12}H_2 \to X + C_8H_{18}} \)
(iii) Draw the structure of two isomers of X.
(iv) Name the isomers drawn in (a)(iii).
(v) Write a balanced equation for the reaction between X and hydrogen.
(b) Describe one test for fats.
(c) Sulphur (IV) oxide is converted to tetraoxosulphate (VI) acid according to the following equation: \( \mathrm{2SO_{2(g)} + O_{2(g)} + 2H_2O_{(l)} \to 2H_2OSO_{4(aq)}} \). If 1.5 moles of oxgen reacts with sulphur (IV) oxide, calculate the mass of tetraoxosulphate (VI) acid produced. [H = 1.0; O = 16.0; S = 32.0].
(d) Consider the following neutralization reaction:
\( \mathrm{CH_3COOH + NaOH \to CH_3COONa + H_2O;} \ \bigtriangleup H_1 \)
\( \mathrm{CH_3COOH + NH_4OH \to CH_3COONH_4 + H_2O;} \ \bigtriangleup H_2 \)
\( \mathrm{NaOH + HCl \to NaCl + H_2O} \ \bigtriangleup H_3 \)
(i) Arrange the enthalphy changes for the reactions in order of increasing magnitude.
(ii) Explain briefly your order in (d)(i).
(e) Consider the following substances. \( \mathrm{Cu_{(s)}} \), \( \mathrm{BeCl_2} \), \( \mathrm{NaH_{(s)}} \), \( \mathrm{HF_{(s)}} \)and \( \mathrm{CCl_{4(l)}} \). State the substance(s) which;
(i) can conduct electricity;
(ii) is/are soluble in water.
The equation in (a)(ii) is the catalytic cracking of dodecane, \(C_{12}H_{26} \to X + C_8H_{18}\). Balancing atoms: carbon \(12-8=4\); hydrogen \(26-18=8\). Therefore X is butene, \(C_4H_8\), an alkene.
(a) Molecular formula of X
\[ X = C_4H_8 \]
(a)(i) Type of reaction
It is catalytic cracking (thermal decomposition of a long-chain alkane into a shorter alkane and an alkene).
(a)(iii) Structures of two isomers of X (\(C_4H_8\))
Two chain/positional isomers of butene are drawn below, showing every bond and hydrogen atom:
(a)(iv) Names of the isomers
Isomer A is but-1-ene (\(CH_2=CH{-}CH_2{-}CH_3\)) and Isomer B is but-2-ene (\(CH_3{-}CH=CH{-}CH_3\)).
(a)(v) Reaction of X with hydrogen
Butene undergoes catalytic addition (hydrogenation) with hydrogen to give butane:
\[ C_4H_{8} + H_2 \xrightarrow{\;Ni\;} C_4H_{10} \]
(b) One test for fats
Translucent (grease) spot test: rub a small quantity of the substance onto a piece of filter paper and allow it to dry (warm gently if solid). A permanent translucent (oily) spot that does not disappear on drying confirms the presence of fat. (Alternatively, adding Sudan III reagent gives a red-stained oily layer, confirming fat.)
(c) Mass of tetraoxosulphate (VI) acid produced
Equation: \(2SO_{2(g)} + O_{2(g)} + 2H_2O_{(l)} \to 2H_2SO_{4(aq)}\)
Molar mass of \(H_2SO_4 = (2\times1.0) + 32.0 + (4\times16.0) = 2 + 32 + 64 = 98\ \text{g mol}^{-1}\).
From the equation, \(1\) mole of \(O_2\) produces \(2\) moles of \(H_2SO_4\).
\[ 1.5\ \text{mol } O_2 \Rightarrow 2 \times 1.5 = 3\ \text{mol } H_2SO_4 \]
\[ \text{Mass} = \text{moles} \times \text{molar mass} = 3 \times 98 = \mathbf{294\ g} \]
(d)(i) Enthalpy changes in order of increasing magnitude
\[ \Delta H_2 < \Delta H_1 < \Delta H_3 \]
(d)(ii) Explanation
The enthalpy of neutralisation depends on the energy released when \(H^+\) and \(OH^-\) ions combine to form water; the more completely the acid and base ionise, the more \(H^+\) and \(OH^-\) are available and the greater the heat evolved. \(\Delta H_3\) (strong acid HCl + strong base NaOH) is largest because both are fully ionised. \(\Delta H_1\) (weak acid \(CH_3COOH\) + strong base NaOH) is smaller because some energy is absorbed to ionise the weak acid. \(\Delta H_2\) (weak acid + weak base \(NH_4OH\)) is smallest because energy is absorbed to ionise both the weak acid and the weak base.
(e)(i) Substance(s) that can conduct electricity
\(Cu_{(s)}\) only, because it is a metal with a sea of mobile (delocalised) electrons.
(e)(ii) Substance(s) soluble in water
\(BeCl_2\), \(NaH_{(s)}\) and \(HF\) dissolve in water (they are ionic/polar and interact with the polar water molecules), whereas the non-polar covalent \(CCl_4\) is insoluble.
Bayanin Amsa
The equation in (a)(ii) is the catalytic cracking of dodecane, \(C_{12}H_{26} \to X + C_8H_{18}\). Balancing atoms: carbon \(12-8=4\); hydrogen \(26-18=8\). Therefore X is butene, \(C_4H_8\), an alkene.
(a) Molecular formula of X
\[ X = C_4H_8 \]
(a)(i) Type of reaction
It is catalytic cracking (thermal decomposition of a long-chain alkane into a shorter alkane and an alkene).
(a)(iii) Structures of two isomers of X (\(C_4H_8\))
Two chain/positional isomers of butene are drawn below, showing every bond and hydrogen atom:
(a)(iv) Names of the isomers
Isomer A is but-1-ene (\(CH_2=CH{-}CH_2{-}CH_3\)) and Isomer B is but-2-ene (\(CH_3{-}CH=CH{-}CH_3\)).
(a)(v) Reaction of X with hydrogen
Butene undergoes catalytic addition (hydrogenation) with hydrogen to give butane:
\[ C_4H_{8} + H_2 \xrightarrow{\;Ni\;} C_4H_{10} \]
(b) One test for fats
Translucent (grease) spot test: rub a small quantity of the substance onto a piece of filter paper and allow it to dry (warm gently if solid). A permanent translucent (oily) spot that does not disappear on drying confirms the presence of fat. (Alternatively, adding Sudan III reagent gives a red-stained oily layer, confirming fat.)
(c) Mass of tetraoxosulphate (VI) acid produced
Equation: \(2SO_{2(g)} + O_{2(g)} + 2H_2O_{(l)} \to 2H_2SO_{4(aq)}\)
Molar mass of \(H_2SO_4 = (2\times1.0) + 32.0 + (4\times16.0) = 2 + 32 + 64 = 98\ \text{g mol}^{-1}\).
From the equation, \(1\) mole of \(O_2\) produces \(2\) moles of \(H_2SO_4\).
\[ 1.5\ \text{mol } O_2 \Rightarrow 2 \times 1.5 = 3\ \text{mol } H_2SO_4 \]
\[ \text{Mass} = \text{moles} \times \text{molar mass} = 3 \times 98 = \mathbf{294\ g} \]
(d)(i) Enthalpy changes in order of increasing magnitude
\[ \Delta H_2 < \Delta H_1 < \Delta H_3 \]
(d)(ii) Explanation
The enthalpy of neutralisation depends on the energy released when \(H^+\) and \(OH^-\) ions combine to form water; the more completely the acid and base ionise, the more \(H^+\) and \(OH^-\) are available and the greater the heat evolved. \(\Delta H_3\) (strong acid HCl + strong base NaOH) is largest because both are fully ionised. \(\Delta H_1\) (weak acid \(CH_3COOH\) + strong base NaOH) is smaller because some energy is absorbed to ionise the weak acid. \(\Delta H_2\) (weak acid + weak base \(NH_4OH\)) is smallest because energy is absorbed to ionise both the weak acid and the weak base.
(e)(i) Substance(s) that can conduct electricity
\(Cu_{(s)}\) only, because it is a metal with a sea of mobile (delocalised) electrons.
(e)(ii) Substance(s) soluble in water
\(BeCl_2\), \(NaH_{(s)}\) and \(HF\) dissolve in water (they are ionic/polar and interact with the polar water molecules), whereas the non-polar covalent \(CCl_4\) is insoluble.
Tambaya 55 Rahoto
(a)(i) Describe briefly how trioxonitrate (V) ions could be tested for in the laboratory.
(ii) State two uses of each of the following compounds: I. sodium chloride; II. sodium trioxocarbonate (IV).
(b) Write balanced equations for the reactions involved in the extraction of iron in the blast furnace.
(ii) State Faraday's first law of electrolysis.
(iii) State two applications of electrolysis.
(c) Concentrated tetraoxosulphate (VI) acid is added to sugar crystals in a beaker. State what would be observed. Explain briefly your answer.
(d) Write an equation for the reaction of zinc powder with:
(i) dilute tetraoxosulphate (VI) acid;
(ii) concentrated tetraoxosulphate (VI) acid.
(e) What property of concentrated tetraoxosulphate (VI) acid is shown in (d)(ii)
(a)(i) Test for trioxonitrate(V) (nitrate) ion - brown ring test: add freshly prepared iron(II) sulphate solution to the test solution, then carefully pour concentrated tetraoxosulphate(VI) acid down the side of the tube so it forms a layer below. A brown ring forms at the junction of the two liquids.
(ii)
- I. Sodium chloride: seasoning/preserving food; raw material for making chlorine, sodium hydroxide and sodium trioxocarbonate(IV).
- II. Sodium trioxocarbonate(IV): softening hard water; manufacture of glass and soap.
(b)(i) Extraction of iron in the blast furnace
\[C + O_2 \rightarrow CO_2\]
\[CO_2 + C \rightarrow 2CO\]
\[Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\]
\[CaCO_3 \rightarrow CaO + CO_2\]
\[CaO + SiO_2 \rightarrow CaSiO_3 \text{ (slag)}\]
(ii) Faraday's first law: the mass of a substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
(iii) Applications: electroplating; purification (refining) of metals; extraction of reactive metals.
(c) The white sugar turns black (a spongy black mass of carbon rises up), with much heat and steam given off. Concentrated tetraoxosulphate(VI) acid is a powerful dehydrating agent: it removes hydrogen and oxygen from the sucrose as water, leaving carbon:
\[C_{12}H_{22}O_{11} \xrightarrow{conc.\ H_2SO_4} 12C + 11H_2O\]
(d)(i) \[Zn + H_2SO_4\ (dilute) \rightarrow ZnSO_4 + H_2\]
(ii) \[Zn + 2H_2SO_4\ (conc.,\ hot) \rightarrow ZnSO_4 + SO_2 + 2H_2O\]
(e) In (d)(ii) the concentrated acid acts as an oxidizing agent.
Bayanin Amsa
(a)(i) Test for trioxonitrate(V) (nitrate) ion - brown ring test: add freshly prepared iron(II) sulphate solution to the test solution, then carefully pour concentrated tetraoxosulphate(VI) acid down the side of the tube so it forms a layer below. A brown ring forms at the junction of the two liquids.
(ii)
- I. Sodium chloride: seasoning/preserving food; raw material for making chlorine, sodium hydroxide and sodium trioxocarbonate(IV).
- II. Sodium trioxocarbonate(IV): softening hard water; manufacture of glass and soap.
(b)(i) Extraction of iron in the blast furnace
\[C + O_2 \rightarrow CO_2\]
\[CO_2 + C \rightarrow 2CO\]
\[Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\]
\[CaCO_3 \rightarrow CaO + CO_2\]
\[CaO + SiO_2 \rightarrow CaSiO_3 \text{ (slag)}\]
(ii) Faraday's first law: the mass of a substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
(iii) Applications: electroplating; purification (refining) of metals; extraction of reactive metals.
(c) The white sugar turns black (a spongy black mass of carbon rises up), with much heat and steam given off. Concentrated tetraoxosulphate(VI) acid is a powerful dehydrating agent: it removes hydrogen and oxygen from the sucrose as water, leaving carbon:
\[C_{12}H_{22}O_{11} \xrightarrow{conc.\ H_2SO_4} 12C + 11H_2O\]
(d)(i) \[Zn + H_2SO_4\ (dilute) \rightarrow ZnSO_4 + H_2\]
(ii) \[Zn + 2H_2SO_4\ (conc.,\ hot) \rightarrow ZnSO_4 + SO_2 + 2H_2O\]
(e) In (d)(ii) the concentrated acid acts as an oxidizing agent.
