WAEC SSCE - Chemistry - 2008
Tambaya 1 Rahoto
Diamond does not conduct electricity because it
Bayanin Amsa
Diamond does not conduct electricity because it has no free valence electrons. In diamond, all carbon atoms are bonded to four other carbon atoms in a tetrahedral arrangement, forming a giant covalent molecule. Each carbon atom has no free valence electron to move around and conduct electricity. Thus, diamond is an insulator and does not allow electricity to pass through it.
Tambaya 2 Rahoto
When ethanol is heated with excess concentrated tetraoxosulphate (VI) acid, the organic product formed is
Bayanin Amsa
When ethanol is heated with excess concentrated tetraoxosulphate (VI) acid (also known as sulfuric acid), the organic product formed is ethene. During the reaction, the concentrated sulfuric acid acts as a dehydrating agent, removing a water molecule from the ethanol molecule. This results in the formation of ethene, which is an unsaturated hydrocarbon with a double bond between two carbon atoms. The reaction can be represented by the following equation: CH3CH2OH + H2SO4 → CH2=CH2 + H2O + H2SO4 So, the correct answer is ethene.
Tambaya 3 Rahoto
Elements in the same group of the periodic table have
Bayanin Amsa
Elements in the same group of the periodic table have the same number of valence electrons. This is because elements in the same group have the same outermost electron configuration, which determines their chemical properties. For example, all elements in group 1 have one valence electron, which makes them highly reactive and likely to lose that electron to form a cation with a charge of +1. Similarly, all elements in group 17 have 7 valence electrons, which makes them highly reactive and likely to gain one electron to form an anion with a charge of -1. Therefore, elements in the same group have similar chemical behavior, but different atomic sizes and electron shell numbers.
Tambaya 4 Rahoto
Atom \(^{234} _{88} Q\) decay by alpha emission to give an atom R. The atomic number and mass number of atom R are respectively
Bayanin Amsa
Alpha decay occurs when a nucleus emits an alpha particle consisting of two protons and two neutrons, which reduces the atomic number by 2 and the mass number by 4. In this case, atom Q has an atomic number of 88 and a mass number of 234. After it decays by alpha emission, it will lose two protons and two neutrons, so the atomic number will be 88 - 2 = 86, and the mass number will be 234 - 4 = 230. Therefore, the correct answer is (a) 86 and 230.
Tambaya 5 Rahoto
A mixture of kerosene and diesel oil can be separated by
Bayanin Amsa
A mixture of kerosene and diesel oil can be separated by distillation. This is because the two liquids have different boiling points, and thus can be separated by heating the mixture to a temperature at which one of the liquids boils but the other remains a liquid. The vapor of the boiling liquid can then be collected and condensed back into a liquid, leaving the other liquid behind. In this case, kerosene has a lower boiling point than diesel oil, so heating the mixture will cause the kerosene to vaporize and be collected separately.
Tambaya 6 Rahoto
An aqueous solution of CaCI2 is
Bayanin Amsa
An aqueous solution of CaCl2 is a neutral solution. Calcium chloride (CaCl2) is a salt composed of a cation, calcium (Ca2+), and an anion, chloride (Cl-). When dissolved in water, the ions dissociate and become surrounded by water molecules. The dissociation of CaCl2 does not result in the formation of any H+ or OH- ions, so the solution does not have any excess of either acid or base. Hence, the solution remains neutral.
Tambaya 8 Rahoto
Water molecules are held together by
Bayanin Amsa
Water molecules are held together by hydrogen bonds. A hydrogen bond is a type of weak chemical bond that occurs between a hydrogen atom of one molecule and an electronegative atom of another molecule, such as oxygen or nitrogen. In the case of water, the hydrogen atoms are attracted to the oxygen atoms of neighboring water molecules, forming hydrogen bonds. This results in the unique properties of water, such as its high boiling and melting points, and its ability to dissolve a wide range of substances.
Tambaya 9 Rahoto
When 100 cm3 of a saturated solution of KCIO3 at 40oC is evaporated, 14 g of the salt is recovered what is the solubility of KCIO3 at 40oC? [KCIO3 = 122.5]
Bayanin Amsa
The solubility of a substance is the amount of the substance that can be dissolved in a given amount of solvent at a particular temperature. In this question, we are given that 100 cm3 of a saturated solution of KCIO3 at 40oC was evaporated to recover 14 g of the salt. This means that the remaining solution was no longer saturated and all of the excess KCIO3 had precipitated out. To find the solubility of KCIO3 at 40oC, we can use the formula: Solubility = (mass of solute recovered) / (volume of solution before evaporation) The mass of KCIO3 recovered is given as 14 g, and the volume of solution before evaporation is 100 cm3 or 0.1 dm3. We can calculate the number of moles of KCIO3 using the given molar mass of KCIO3: Number of moles = (mass of KCIO3) / (molar mass of KCIO3) = 14 / 122.5 = 0.114 moles Now, we can calculate the solubility of KCIO3: Solubility = (mass of solute recovered) / (volume of solution before evaporation) = (0.114 moles) / (0.1 dm3) = 1.14 mol dm-3 Therefore, the solubility of KCIO3 at 40oC is 1.14 mol dm-3. The correct option is: - 1.14 mol dm-3
Tambaya 10 Rahoto
If 5.00cm3 of 0.20 mol dm-3Na2CO3 was diluted with distilled water to obtain 250 cm3 solution, what is the concentration of the resulting solution?
Bayanin Amsa
To solve the problem, we need to use the formula: C1V1 = C2V2 Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. Plugging in the values we know: C1 = 0.20 mol dm-3 V1 = 5.00 cm3 = 0.005 dm3 V2 = 250 cm3 = 0.250 dm3 Solving for C2: C2 = (C1V1)/V2 C2 = (0.20 mol dm-3)(0.005 dm3)/(0.250 dm3) C2 = 0.004 mol dm-3 Therefore, the concentration of the resulting solution is 0.004 mol dm-3. Answer: 0.004 mol dm-3
Tambaya 11 Rahoto
Which of the following alloys is mainly used in making statues?
Bayanin Amsa
Bronze is mainly used in making statues. Bronze is an alloy of copper, which is a good conductor of heat and electricity, and tin, which strengthens copper. It is widely used in sculpture due to its strength, ductility, and resistance to corrosion. Bronze also has a low melting point, which allows for easier casting and shaping of the statue. In addition to being used for statues, bronze is also used in making bells, musical instruments, and decorative items.
Tambaya 12 Rahoto
A hydrocarbon containing 88.9% carbon has the empirical formula [H = 1 C = 12]
Bayanin Amsa
To find the molecular formula of the hydrocarbon, we need to first determine its empirical formula. The empirical formula represents the simplest whole number ratio of atoms present in a compound. From the given data, we know that the hydrocarbon contains 88.9% carbon. This means that the remainder of the compound is made up of hydrogen. To calculate the empirical formula, we assume a 100g sample of the hydrocarbon. Therefore, the sample would contain 88.9g of carbon and 11.1g of hydrogen. Next, we convert the mass of each element to the number of moles using their respective atomic masses. The atomic mass of carbon is 12 g/mol, and that of hydrogen is 1 g/mol. The number of moles of carbon can be calculated as follows: moles of carbon = mass of carbon/atomic mass of carbon moles of carbon = 88.9g/12 g/mol moles of carbon = 7.41 mol Similarly, the number of moles of hydrogen can be calculated as follows: moles of hydrogen = mass of hydrogen/atomic mass of hydrogen moles of hydrogen = 11.1g/1 g/mol moles of hydrogen = 11.1 mol The empirical formula is then calculated by dividing the number of moles of each element by the smallest number of moles obtained in the previous step. This gives the simplest whole number ratio of the elements. The empirical formula for the hydrocarbon is therefore: C : H = 7.41 mol : 11.1 mol Dividing by 7.41 gives: C : H = 1 : 1.5 Therefore, the empirical formula for the hydrocarbon is CH1.5. We can round this up to CH2. The molecular formula can be determined by multiplying the empirical formula by a whole number n, which represents the number of empirical formula units in the compound. However, since we don't have any additional information about the hydrocarbon, we cannot determine the molecular formula with certainty. Therefore, the answer is (B) CH2, which represents the empirical formula of the hydrocarbon.
Tambaya 13 Rahoto
Aluminium is a good roofing material because it
Bayanin Amsa
The answer is: "forms an oxide film over its surface which prevents corrosion". This is because when aluminium is exposed to air, it reacts with oxygen to form a thin layer of aluminium oxide on its surface. This oxide layer is tightly bound to the aluminium and acts as a barrier, protecting the underlying metal from further oxidation or corrosion. This makes aluminium a good roofing material, as it can resist the effects of weather and other corrosive elements over time.
Tambaya 14 Rahoto
Which of the following gases has the lowerst rate of diffusion under the same conditions of temperature and pressure? [H = 1, O = 16, Ne = 20, CI = 35]
Bayanin Amsa
Under the same conditions of temperature and pressure, gases with higher molar mass diffuse more slowly than those with lower molar mass. This is because heavier gas molecules have a slower average speed compared to lighter gas molecules. Therefore, among the given options, chlorine (CI) with a molar mass of 35 g/mol has the highest molar mass and hence the lowest rate of diffusion. In contrast, hydrogen (H) with a molar mass of 1 g/mol has the lowest molar mass and hence the highest rate of diffusion. Neon (Ne) and oxygen (O) have molar masses of 20 g/mol and 16 g/mol, respectively, and therefore have diffusion rates between those of hydrogen and chlorine.
Tambaya 16 Rahoto
Consider the reaction represented by the question below: KOH(aq) + HCI(aq) → KCI(aq) + H2O(l) What volume of 0.25 mol dm-3 KOH would be required to completely neutralize 40 cm3 of 0.10 mol dm -3 HCI?
Bayanin Amsa
The balanced equation for the reaction between KOH and HCl is: KOH + HCl → KCl + H2O From the equation, we can see that the mole ratio of KOH to HCl is 1:1. This means that for every 1 mole of HCl, 1 mole of KOH is required to neutralize it. To find the volume of 0.25 mol dm-3 KOH needed to neutralize 40 cm3 of 0.10 mol dm-3 HCl, we need to use the formula: nacid Vacid = nbase Vbase where n is the number of moles and V is the volume in dm3. Rearranging the formula to isolate Vbase, we get: Vbase = (nacid Vacid) / nbase Substituting the values we have: nacid = 0.10 mol dm-3 (concentration of HCl) Vacid = 40 cm3 = 0.04 dm3 (volume of HCl) nbase = 0.25 mol dm-3 (concentration of KOH) Vbase = (0.10 mol dm-3 x 0.04 dm3) / 0.25 mol dm-3 = 0.016 dm3 = 16 cm3 Therefore, the volume of 0.25 mol dm-3 KOH needed to completely neutralize 40 cm3 of 0.10 mol dm-3 HCl is 16 cm3. The answer is (d) 16cm3.
Tambaya 17 Rahoto
The colour of phenolphthalein indicator in dilute HNO<3(aq) is
Bayanin Amsa
In dilute HNO3(aq), phenolphthalein indicator will be colorless. Phenolphthalein is a pH indicator that changes color depending on the pH of the solution it is in. In acidic solutions (pH less than 7), phenolphthalein is colorless. HNO3(aq) is a strong acid, so it will completely dissociate in water to form H+ ions, resulting in a low pH and a colorless phenolphthalein indicator.
Tambaya 18 Rahoto
The mass spectrometer can be used to measure mass of
Bayanin Amsa
The mass spectrometer is an instrument used to determine the mass-to-charge ratio of ions. It works by ionizing a sample, accelerating the ions in an electric and/or magnetic field, and then separating the ions based on their mass-to-charge ratio. Therefore, it can be used to measure the mass of an ion, which can represent the mass of a molecule or atom. Therefore, the correct option is "an atom."
Tambaya 19 Rahoto
Which of the following statements is not true of halogens?
Bayanin Amsa
Halogens are a group of nonmetal elements consisting of fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Among the given options, the statement that is not true of halogens is "their ionic radii decrease down the group". In reality, as we move down the group, the ionic radii of halogens increase due to the addition of a new energy level in each successive element, which results in a larger atomic radius. The other statements are true of halogens: they exist in different physical states (F₂ and Cl₂ are gases, Br₂ is a liquid, and I₂ is a solid), they exist as diatomic molecules, and their melting and boiling points increase down the group due to stronger van der Waals forces between larger molecules.
Tambaya 20 Rahoto
Arrange the following compounds in order of increasing boiling point
l.CH 3CH2CH2OH
II.CH 3CH 2OH
III.CH3CH2CH2CH2OH
IV. CH3CH 2CH2CH 2CH 3
Bayanin Amsa
The boiling point of an organic compound is determined by the strength of intermolecular forces present between its molecules. These intermolecular forces are influenced by the molecular mass and polarity of the compound. The larger the molecular mass and polarity of the molecule, the stronger the intermolecular forces and thus the higher the boiling point. Among the given compounds, III has the largest molecular mass and also the maximum number of polar C-O bonds. Therefore, it has the highest boiling point. IV has a branched structure which decreases its surface area of contact and hence the intermolecular forces of attraction are weaker, thus it has the lowest boiling point among the given options. Therefore, the correct order of increasing boiling point would be: IV < II < I < III. Therefore, is the correct answer.
Tambaya 21 Rahoto
In dry Leclanche cell, manganese (IV) oxide acts as
Bayanin Amsa
In a dry Leclanche cell, manganese (IV) oxide acts as a depolarizer. The depolarizer helps to remove hydrogen gas, which is produced at the negative electrode when the cell is in use, by reacting with it to form water. Manganese (IV) oxide is a strong oxidizing agent and it reacts with the hydrogen gas, making it unavailable to react with the cell components, thereby preventing polarization of the cell. This ensures that the cell maintains a steady voltage output over a long period of time.
Tambaya 22 Rahoto
Which of the following set of equations could be used to represent the standard enthalpy of formation of glucose?
Bayanin Amsa
The equation that could be used to represent the standard enthalpy of formation of glucose is the second option: 6CO2(g) + 6H2O (l) → C6H12O6(s) + 6O 2(g) This is because the standard enthalpy of formation of a compound is the enthalpy change that occurs when one mole of the compound is formed from its constituent elements, with all substances in their standard states. In this case, glucose is formed from carbon dioxide, water, and oxygen gas, which are the standard states of their respective elements. This equation shows the balanced chemical reaction for the formation of glucose from these elements and the resulting enthalpy change.
Tambaya 23 Rahoto
Find the number of neutrons in an atom represented by 4521X.
Bayanin Amsa
The number 4521X represents an atom of an element with atomic number 21. The atomic number of an element represents the number of protons in its nucleus. Since the atom is neutral, the number of protons in its nucleus is equal to the number of electrons surrounding the nucleus. To determine the number of neutrons, we can use the formula: number of neutrons = mass number - atomic number The mass number of the atom is not given, so we cannot directly calculate the number of neutrons. However, we know that for a particular element, the number of protons plus the number of neutrons is equal to the mass number. Since the atomic number of the element is 21, we can look up its atomic mass on the periodic table, which is approximately 45. Therefore, the number of neutrons in the atom is: number of neutrons = mass number - atomic number number of neutrons = 45 - 21 number of neutrons = 24 Therefore, the correct answer is 24.
Tambaya 24 Rahoto
Arrange the following elements in order of increasing electron affinity: Br2CI2
I 2
F 2
Bayanin Amsa
Electron affinity refers to the energy released when an electron is added to an atom in the gas phase. The greater the electron affinity of an atom, the more energy is released when an electron is added to it. Based on this definition, the electron affinity of the elements listed can be arranged in the following order: F 2 < Cl2 < Br2 < I 2 Therefore, the correct arrangement in order of increasing electron affinity would be: I 2 < Br2 < CI2 < F 2
Tambaya 25 Rahoto
What is the atomic number of an element whose cation Y+ has the electronic configuration 1s22s22p6?
Tambaya 26 Rahoto
Consider the reaction represented by the following equation: Zn(s) + 2HCI(aq) → ZnCI2(aq) + H2(g) what volume of hydrogen gas is produced at s.t.p. when 3.25 g of zinc reacts with excess dilute HCI? [Zn = 65, Molar gas volume at s.t.p. = 22.4 dm-3]
Bayanin Amsa
First, we need to calculate the number of moles of zinc used in the reaction. Moles of Zn = mass ÷ molar mass Moles of Zn = 3.25 g ÷ 65 g/mol Moles of Zn = 0.05 mol From the balanced equation, we know that 1 mole of Zn produces 1 mole of H2 gas. Therefore, the number of moles of H2 gas produced is also 0.05 mol. At s.t.p (standard temperature and pressure), one mole of any gas occupies 22.4 dm3. So, the volume of hydrogen gas produced at s.t.p is: Volume = number of moles × molar gas volume at s.t.p Volume = 0.05 mol × 22.4 dm-3/mol Volume = 1.12 dm3 Therefore, the answer is 1.12 dm3.
Tambaya 27 Rahoto
Which of the following compounds with the corresponding masses contain the highest percentage of nitrogen? [N = 14]
Tambaya 29 Rahoto
The hydrolysis of groundnut oil by potassium hydroxide is known as
Bayanin Amsa
The correct answer is "saponification." Saponification is a chemical reaction that occurs when a fat or oil is hydrolyzed with a strong base, such as potassium hydroxide (KOH) or sodium hydroxide (NaOH), to form soap and glycerol. In the case of groundnut oil, when it reacts with potassium hydroxide, it breaks down into soap (potassium salt of fatty acids) and glycerol. The soap produced is water-soluble and can be used for cleaning or as a surfactant.
Tambaya 30 Rahoto
Chemicals that are produced in small quantities and with very high degree of purity are
Bayanin Amsa
Chemicals that are produced in small quantities and with very high degree of purity are called fine chemicals. Fine chemicals are usually produced for specific applications or purposes, such as pharmaceuticals, agrochemicals, or specialty chemicals. They are different from bulk chemicals, which are produced in large quantities and are typically less pure. Fine chemicals are also different from heavy and light chemicals, which are produced in large quantities and have a wide range of industrial and commercial applications.
Tambaya 31 Rahoto
The high solubility of ethanol in water is due to its
Bayanin Amsa
The high solubility of ethanol in water is due to its hydrogen bonding. Ethanol is a polar molecule, which means it has a partially positive end and a partially negative end. Water is also a polar molecule and it can form hydrogen bonds with ethanol molecules. These hydrogen bonds allow ethanol to dissolve readily in water, resulting in a high solubility. The other options are not correct because they do not explain why ethanol is soluble in water.
Tambaya 32 Rahoto
In a fixed volume of a gas, an increase in temperature results in an increase in pressure due to an increase in the
Bayanin Amsa
When the temperature of a gas is increased, the kinetic energy of the gas molecules increases. This means that the gas molecules move around faster and collide with the walls of the container more frequently and with greater force. This increase in the frequency and force of the collisions between the gas molecules and the walls of the container results in an increase in pressure. Therefore, an increase in temperature of a gas in a fixed volume results in an increase in pressure due to an increase in the number of collisions between the gas molecules and the walls of the container.
Tambaya 33 Rahoto
Which of the following compounds would dissolved in water to give a solution whose pH is less than 7?
Tambaya 34 Rahoto
A solution of a salt was acidified with HCI when a few drops of BaCI2 solution were added, a white precipitate was formed. Which of the following anions is present in the salt?
Tambaya 35 Rahoto
When concentrated H2
SO 4 is added to NaCI (s), the gas evolved
Tambaya 36 Rahoto
Which of the following compounds would not give a precipitate with ammoniacal AgNO3 solution?
Tambaya 37 Rahoto
Which of the following substances causes the depletion of the ozone layer in the atmosphere?
Bayanin Amsa
Chlorofluorocarbons (CFCs) cause depletion of the ozone layer in the atmosphere. CFCs are organic compounds that contain chlorine, fluorine, and carbon. When CFCs are released into the atmosphere, they rise up to the stratosphere, where they are broken down by ultraviolet radiation from the sun. This breakdown releases chlorine atoms, which react with ozone molecules to form oxygen gas and chlorine monoxide. Chlorine monoxide then reacts with another ozone molecule to form more oxygen gas and another chlorine atom. This cycle continues, with each chlorine atom destroying many thousands of ozone molecules. The reduction of the ozone layer can lead to increased levels of harmful ultraviolet radiation reaching the Earth's surface, which can have negative effects on human health and the environment.
Tambaya 38 Rahoto
Which of the following compounds with the corresponding masses contain the highest percentage of nitrogen? [N = 14]
Tambaya 39 Rahoto
Consider the equilibrium reaction represented by the equation: 2SO2(g) + O2(g) ⇌ 2SO3(g) ∆H = - 197 kJmol-1. At equilibrium, increase in the total pressure of the system brings about
Tambaya 40 Rahoto
Calculate the quantity of electricity passed when 0.4 A flows for 1 hour 20 minutes through an electrolytic cell
Bayanin Amsa
Tambaya 41 Rahoto
Consider the following elements and their electronic configurations
W: 1s22s2
X: 1s 22s22p6
Y: 1s22s22p63s2
Z: 1s22s22p63s23p2. Which of the elements belong to the same group in the periodic table?
Tambaya 42 Rahoto
What is the main type of reaction alkenes undergo?
Bayanin Amsa
Alkenes undergo addition reactions. Addition reactions occur when atoms or groups of atoms are added to the carbon-carbon double bond of the alkene, resulting in the formation of a single product with a single bond. The double bond in the alkene is converted into a single bond, and the added groups are attached to the carbon atoms that were originally involved in the double bond. This type of reaction is commonly observed in organic chemistry, and it is important in the synthesis of a wide range of compounds, including polymers, plastics, and pharmaceuticals. Examples of addition reactions of alkenes include hydrogenation, halogenation, and hydration.
Tambaya 43 Rahoto
The collision theory proposes that
Bayanin Amsa
The collision theory proposes that reactants must collide with a certain minimum amount of energy to form products. In other words, for a chemical reaction to occur, the reactant molecules must collide with sufficient energy and in the correct orientation. If the collision has insufficient energy, the reactants will bounce off each other and not form any products. Therefore, the higher the frequency of successful collisions between reactant molecules, the higher the rate of reaction. The collision theory also suggests that not all collisions are effective; only those that have sufficient energy and correct orientation will lead to product formation.
Tambaya 44 Rahoto
What is the mass of 6.02 x 1024 atoms of magnesium? [Mg = 24, Avogadro constant = 6.02 x 1023 atoms mol-1]
Bayanin Amsa
The molar mass of magnesium is 24 g mol-1 (as given). This means that the mass of one mole of magnesium atoms is 24 g. We are given the number of atoms of magnesium: 6.02 x 1024 atoms. To find the mass of these atoms, we need to convert the number of atoms to moles, and then multiply by the molar mass of magnesium. Number of moles of magnesium atoms = (6.02 x 1024 atoms) / (6.02 x 1023 atoms mol-1) = 10 mol Mass of 10 moles of magnesium atoms = 10 mol x 24 g mol-1 = 240 g Therefore, the mass of 6.02 x 1024 atoms of magnesium is 240g. Answer: 240g
Tambaya 45 Rahoto
(a) Consider the following reaction sequence.
(i) What process leads/tathe formation of K?
(ii) Write the formula of K.
(iii) Write the structural formula of L and name L.
(iv) Name A\(_n\)
(v) Write the structure of M and name M.
(b)(i) What are carbohydrates?
(ii) Give one example each of a I. monosaccharide; II. disaccharide; III. polysaccharide.
(c) Consider the following structure of a simple sugar :
(i) Which functional group makes the compound a reducing agent?
(ii) State what would C = O be observed when I. the compound is mixed with Fehling's solution and boiled;
I. few drops of concentrated H\(_2\)SO\(_4\) is added to the sample of the compound. H?C?OH
(iii) Write an equation for the reaction in (c)(ii)(II).
(d) A hydrocarbon Z with molecular mass 78 on combustion gave 3.385 g of CO\(_2\) and and 0.692 g of H\(_2\)O. Determine the molecular formula of Z. [ H = 1, C = 12, O = 16]
Tambaya 46 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
(a) In the laboratory preparation of crystals of CuSO\(_4\), a green powder Q was added to dilute H\(_2\)SO\(_4\), and stirred. Effervescence occurred and a gas R was given off which turned lime water milky. Excess Q was removed from the mixture. The solution of Cu\(_2\)SO\(_4\) was concentrated to half its original volume and allowed to stand.
(i) What is substance Q?
(ii) Name gas R
(ii) Why was excess Q used?
(iv) How would you know that the reaction is complete?
(v) What method was used to remove excess Q? (vi) Why was the solution of CusO\(_4\) not heated to dryness?
(b) Name the reagent(s) used for testing each of the following substances in the laboratory: (i) Water; (i) Primary alkanol.
Bayanin Amsa
None
Tambaya 47 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
Burette readings(initial and final) must be given to two decimal places. Volume of pipette user must also be recorded but on account of experimental procedure is required. All calculations must be done in your answer book. A is O.050 mol dm\(^{-3}\) of acid HX. Bis a solution of NaOH containing 0.025 moles per 250 solutions.
(a) Put A into the burette and titrate it against 20.00 cm\(^3\) or 25.00 cm\(^3\) portions B using phenolphthalein as indicator. Tabulate your readings and calculate the average volume or A used.
(b) your results and the information provided above, calculate the;
(i) amount of acid in the average
(ii) amount of base in 20.00 cm\(^3\) or 25.00 cm\(^3\);
(iii) mole ratio of acid to base
(c) Write a balanced chemical equation for the reaction between the acid H\(_y\)X and the base NaOH
(d) State the basicity of the acid H\(_y\)X.
Bayanin Amsa
None
Tambaya 48 Rahoto
(a)(1) Define covalent bond.
(ii) Give two properties of covalent compounds
(ii) With the aid of a diagram, show how ammonia molecule is formed
(iv) Illustrate with a diagram the formation of ammonium ion?
(v) What type of bond(s) exist(s) in I. ammonia, H. ammonium ion? (\(_1\)H\(_7\)N)
(b)(i) Write three subatomic particles with their corresponding relative masses. CH\(_2\)OH
(ii) Name the possible states in which water can exist.
(c) (i) State Graham's law of diffusion
(ii) Arrange the following gases, He, CH\(_4\) and N\(_2\) in order of increasing rates of diffusion. Give a reason for the order. [ H = 1, He = 4, C = 12, N = 14 ]
(d) Draw the structures of the following compounds:
(i) 2,3-dimethylbutane;
(ii) 1,4-dibromocyclohexane.
None
Bayanin Amsa
None
Tambaya 49 Rahoto
(a)(i) Define hard water
(ii) Name two substances responsible for hardness in water
(iii) Give two methods for the removal of hardness in water.
(b)(i) What are the raw materials require for the manufacture of tetraoxosulphate (VI) acid by the contact process?
(ii) Write an equation for tr reaction that requires a catalyst in the contact process (iii) State the catalyst used in (b)(ii).
(c)(i) Give two uses of sodium tetraoxosulphate (VI) Consider the reaction represented by the following equation:
Na\(_2\)SO\(_4\).10H\(_2\)O\(_{(s)}\) \(\to\) Na\(_2\)SO\(_4\).H\(_2\)O + 9H\(_2\)O\(_{(g)}\)
(ii) What name is given to this type of reaction?
(iii) Calculate the solubility of Na\(_2\)CO\(_3\) at 25°C. if 30.0 cm\(^3\) of its saturated solution at that temperature gave 1.80 g of the anhydrous salt. [C = 12, 0 = 16, Na = 23]
(d)(i) Define the term activated complex
(ii) State one reason why a collision may not produce a chemical reaction
(iii) The formation of water gas is represented by the following equation:
C\(_{(s)}\) + H\(_2\)O\(_{(g)}\) \(\to\) CO\(_{(g)}\) + H\(_{2(0)}\) \(\Delta\)H= + 131 kJmol\(^{-1}\)
Draw an energy profile diagram for the reaction showing the I. activated complex, II. enthalpy of reactants.
Bayanin Amsa
None
Tambaya 50 Rahoto
(a) State (i) Pauli's Excusion principle;
(ii) Hund's rule of maximum multiplicity.
(b)(i) Write the electronic configuration of each of the following ions of copper: I. Cu\(_+\) II. Cu\(_{(2+)}\) [\(_{29}Cu\)]
(ii) Give the number of unpaired-electrons in each of the ions in (b)(i) above.
(iii) State the type of reaction represented by the following equation:
2Cu\(^+_{(aq)}\) \(\to\) Cu\(^{2+}_{(aq)}\) + Cu\(_{(s)}\)
(iv) Write the formula of one compound of Cu+.
(c)(i) Name the type of radiation that will I. penetrate lead block; II. be stopped by thin paper
(ii) Give the charge on each of the radiations mentioned in I(c)(i) above.
(iii) What term is used to describe each of the following nuclear processes?
I. Combination of two lighter nuclei to form a heavy nucleus II. Splitting of a heavy nucleus into two or more lighter nuclei
III. Time required for one-half of the atoms of a radioactive substance to decay.
(d) Arrange the following ions in order of increasing size. Give a reason for your answer in each case. I. Li\(^{+}\), K\(^{+}\), Na\(^{+}\); II. O\(^{2-}\), F\(^{-}\), N\(^{3-}\)
(e) Determine the percentage composition of phosphorus and oxygen in phosphorus (V) oxide [ P = 31, O = 16 ]
