JAMB UTME - Chemistry - 1992

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The functional group that gives bubbles on treatment with a saturated solution of NaHCO3 is the carboxyl group. When an organic compound containing a carboxyl group reacts with NaHCO3, it undergoes a chemical reaction called an acid-base reaction or a neutralization reaction. During this reaction, the carboxylic acid reacts with NaHCO3 to form salt, carbon dioxide gas, and water. The carbon dioxide gas is responsible for the formation of bubbles. Therefore, option D, which is the carboxyl group, is the correct answer.

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An emulsion paint consists of liquid particles dispersed in liquid. Emulsion paints are water-based paints that are made up of a pigment (coloring material), binder (a film-forming substance), and water. In an emulsion paint, the pigment and binder are finely dispersed in water and held together by an emulsifier (a substance that keeps the particles from separating). When the paint is applied to a surface, the water evaporates, leaving behind a film of pigment and binder that adheres to the surface. The use of water as a solvent makes emulsion paints more environmentally friendly than solvent-based paints, which use harmful volatile organic compounds (VOCs).

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Mg + 2e → 24 gms
2f → 24 gms
2 x 96,500 → 24 gms
Q = (500 x 1440 x 60)
Q = 4320000 C
1 mole of mg = (1) / (2 x 96,500) x (43200000) / (1)
= (24)/(1) = (1036800000) / (193000) = 5372.020725 gms
in leg = (5372.020725) / (1000) = 5.4 leg

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To calculate the grams of H2SO4 needed for the preparation of 0.175 dm³ of 6.00 M H2SO4, we need to use the formula: Molarity (M) = moles (n) / volume (V) Rearranging the formula, we get: moles (n) = Molarity (M) x volume (V) We are given the volume (V) as 0.175 dm³ and the Molarity (M) as 6.00 M. So, moles of H2SO4 = 6.00 M x 0.175 dm³ = 1.05 moles of H2SO4 The molar mass of H2SO4 can be calculated as: 2(1.00 g/mol) + 32.06 g/mol + 4(16.00 g/mol) = 98.08 g/mol Therefore, 1.05 moles of H2SO4 weighs: 1.05 moles x 98.08 g/mol = 102.98 g So, approximately 103.0 grams of H2SO4 are necessary for the preparation of 0.175 dm³ of 6.00 M H2SO4. Therefore, the answer is: 103.0 g

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The measure of the average kinetic energy of the molecules of a substance is called temperature. Temperature is a measure of how hot or cold a substance is and is directly related to the kinetic energy of the particles in that substance. The higher the temperature, the more kinetic energy the particles have, and the faster they move. Conversely, the lower the temperature, the less kinetic energy the particles have, and the slower they move. Therefore, temperature is the correct option that measures the average kinetic energy of the molecules of a substance. Volume, mass, and pressure are not related to the average kinetic energy of the molecules of a substance.

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Element P has only 1 electron in its outermost shell, which is highly reactive, and it is easier for P to lose this electron than to gain 7 electrons to fill its outermost shell. Therefore, P is highly likely to be a metal. Element Q has 2 electrons in its outermost shell, and it is relatively easy for it to lose 2 electrons and form a cation. Therefore, Q is also highly likely to be a metal. Element R has 3 electrons in its outermost shell, and it is relatively easy for it to lose 3 electrons and form a cation. Therefore, R is also highly likely to be a metal. Element S has 7 electrons in its outermost shell, which is close to a full shell with 8 electrons. It is easier for S to gain 1 electron and form an anion rather than lose 7 electrons to form a cation. Therefore, S is unlikely to be a metal. Hence, the answer is S.

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The nucleus of the isotope tritium contains one proton and two neutrons, and no electrons. This means that the correct option is "two neutrons and one proton." The atomic number of an element is determined by the number of protons in the nucleus. Since tritium has one proton, its atomic number is 1, and it is a hydrogen isotope. The mass number of tritium is 3, which is the sum of its protons and neutrons.

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Entropy is a measure of the randomness or disorder of a system. A positive change in entropy means that the disorder or randomness of the system has increased. -: H2O(I) → H2O(g) - When water vaporizes, it goes from a more ordered, liquid state to a more random, gaseous state. So, the entropy change will be positive. -: Cu2+(aq) + Fe(s) → Fe2+(aq) + Cu(s) - This is a redox reaction where copper ions are reduced to copper atoms while iron is oxidized to iron ions. Since a more ordered solid is converted into a less ordered solution, the entropy change will be negative. -: N2(g) + 3H2(g) → 2NH3(g) - Here, two gases combine to form a single gas, so the disorder of the system decreases, and the entropy change will be negative. -: 2HCl(s) → H2(g) + Cl2(g) - In this reaction, two solids combine to form two gases. As a result, the disorder of the system increases, and the entropy change will be positive. Therefore, the option in which the entropy change is positive is.

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The central atom in a water molecule (H2O) is oxygen. Oxygen has six valence electrons, two of which are shared with two hydrogen atoms to form covalent bonds. Therefore, the oxygen in the water molecule has two pairs of electrons, one shared with each hydrogen atom, to form two single covalent bonds. These two pairs of electrons are known as bonding pairs. Since the oxygen atom has six valence electrons, it has two lone pairs of electrons. These are the two pairs of electrons that are not involved in the covalent bonding with hydrogen atoms. Lone pairs of electrons are also known as non-bonding pairs or unshared pairs. Therefore, the answer is - there are two lone pairs of electrons on the central atom of the H2O molecule.

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The rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases will effuse (or diffuse) faster than heavier gases under the same conditions. In this problem, we are given that 100 cm³ of oxygen pass through a porous plug in 50 seconds. We can use this information to determine the rate of effusion of oxygen. First, we need to convert the volume from cm³ to liters (L), since the units of gas rate are typically expressed in L/s or mL/s. 100 cm³ = 0.1 L Now we can calculate the rate of effusion of oxygen: Rate of effusion of oxygen = volume / time = 0.1 L / 50 s = 0.002 L/s Next, we need to use the rate of effusion of oxygen to determine the time it would take for the same volume of hydrogen to pass through the same porous plug. Since hydrogen has a lower molar mass than oxygen (1 g/mol versus 16 g/mol), we would expect it to effuse faster. In fact, the ratio of their effusion rates can be calculated using Graham's law of effusion: Rate of effusion of hydrogen / Rate of effusion of oxygen = sqrt (Molar mass of oxygen / Molar mass of hydrogen) Rate of effusion of hydrogen / 0.002 L/s = sqrt (16 g/mol / 1 g/mol) Rate of effusion of hydrogen / 0.002 L/s = 4 Rate of effusion of hydrogen = 4 x 0.002 L/s = 0.008 L/s Finally, we can use the rate of effusion of hydrogen to calculate the time it would take for the same volume (0.1 L) to pass through the porous plug: Time taken for hydrogen to pass through the porous plug = volume / rate of effusion of hydrogen Time taken for hydrogen to pass through the porous plug = 0.1 L / 0.008 L/s = 12.5 s Therefore, the answer is 12.5 s.

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The chemical formula for sulfuric acid is H2SO4, which dissociates into two H+ ions and one sulfate ion (SO42-) when dissolved in water. Therefore, in a 0.5 M solution of H2SO4, the concentration of H+ ions would be twice as much, i.e., 1.0 M. To calculate the number of moles of H+ ions in 1 dm3 of 1.0 M solution, we can use the formula: moles = concentration x volume Substituting the values, we get: moles = 1.0 mol/dm3 x 1 dm3 moles = 1.0 moles Therefore, there are 1.0 moles of H+ ions in 1 dm3 of 0.5 M solution of H2SO4. Hence, the correct answer is 1.0 moles.

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The process represented by the given chemical equation is known as slaking. Slaking is a chemical reaction that occurs when water is added to quicklime (CaO) to produce slaked lime or calcium hydroxide (Ca(OH)2). In the given equation, CaO reacts with H2O to form Ca(OH)2 with the evolution of heat (ΔH = -65 kJ). The reaction is exothermic, meaning that heat is released during the process. Slaked lime or calcium hydroxide is commonly used in construction, agriculture, and chemical industries. It is used as a building material, a pH adjuster, a flocculant in water treatment, and in the production of various chemicals. Therefore, the process represented by the given equation is known as slaking.

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During electrolysis, oxidation occurs at the anode while reduction occurs at the cathode. In this case, copper (II) tetraoxosulphate (VI) is being electrolyzed, so the copper ion (Cu2+) and the sulfate ion (SO42-) are present in the solution. At the anode, oxidation occurs, and the sulfate ion (SO42-) is likely to be oxidized to oxygen gas (O2) and water (H2O) since carbon is a good oxidizing agent. Thus, option B (oxygen and copper) is correct. At the cathode, reduction occurs, and copper (II) ions (Cu2+) are likely to be reduced to copper metal (Cu) since copper is more likely to be reduced than hydrogen. Therefore, option B (oxygen and copper) is the correct answer.

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Sulphur (IV) oxide gas dissolves in water vapour to produce acid rain during rainfall. When sulphur (IV) oxide gas dissolves in water vapour, it forms sulphurous acid, which is a weak acid. When it rains, this acid is brought down to the earth's surface, causing acid rain. Acid rain can damage buildings, plants, and aquatic life, and it can also make lakes and rivers too acidic for some fish and other aquatic organisms to survive. Therefore, it is important to control the emission of sulphur (IV) oxide gas into the atmosphere.

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The oxide of nitrogen that is unstable in air is NO (nitric oxide). Nitric oxide is a highly reactive gas and readily reacts with oxygen in the air to form nitrogen dioxide (NO2). Nitrogen dioxide is a brownish gas that is responsible for the brown haze seen over many cities. NO2 is relatively stable in air and is not as reactive as NO. On the other hand, NO2 can further react with oxygen to form nitrogen trioxide (N2O3) and nitrogen pentoxide (N2O5). Both of these oxides are relatively stable in air and are not as reactive as NO. Therefore, the unstable oxide of nitrogen in air is NO.

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The characteristic reaction of carbonyl compounds is addition. Carbonyl compounds have a C=O functional group, which is a polar group that attracts nucleophiles (electron-rich species). This makes the carbonyl carbon atom electrophilic (electron-poor), and it readily undergoes addition reactions with nucleophiles. The addition reaction occurs by the nucleophile attacking the electrophilic carbon, resulting in the formation of a new bond and breaking of the C=O double bond. Examples of addition reactions of carbonyl compounds include nucleophilic addition of alcohols to form hemiacetals or acetals, nucleophilic addition of amines to form imines or enamines, and nucleophilic addition of Grignard reagents to form alcohols.

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An increase in temperature causes an increase in pressure of a gas in a fixed volume due to an increase in the number of collisions between the gas molecules and the walls of the container. As the temperature of the gas increases, the kinetic energy of the gas molecules also increases, causing them to move around more rapidly and collide with the walls of the container more frequently and with greater force. This increased force of collisions leads to an increase in pressure within the container. Therefore, the correct answer is "number of collisions between the gas molecules and the walls of the container."

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The problem describes a gas with a constant pressure but a changing volume and temperature. According to Charles's Law, the volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature. That is, as the temperature of the gas increases, so does its volume, and vice versa. Using this law, we can set up a proportion between the initial and final temperatures and volumes: (initial volume) / (initial temperature) = (final volume) / (final temperature) We are given the initial volume (2 dm³) and temperature (300 K), and we are looking for the final temperature that will result in double the initial volume (4 dm³). (2 dm³) / (300 K) = (4 dm³) / (final temperature) Solving for the final temperature: final temperature = (4 dm³) x (300 K) / (2 dm³) = 600 K Therefore, the temperature at which the gas's volume will be doubled while keeping the pressure constant is 600 K. The correct answer is "600k".

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This is a problem in the field of thermodynamics, specifically dealing with the ideal gas law. The ideal gas law is a formula that relates the pressure, volume, and temperature of an ideal gas. It is written as follows: PV = nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas in Kelvin. In this problem, we are told that the mass of the gas is constant, so the number of moles of gas is also constant. Therefore, we can write the ideal gas law as: P1V1 = P2V2 where P1 and V1 are the initial pressure and volume of the gas, and P2 and V2 are the final pressure and volume of the gas. We are told that the pressure is constant, so we can simplify the equation to: V1/T1 = V2/T2 where T1 is the initial temperature of the gas and T2 is the final temperature of the gas. We are given that the initial volume of the gas is 2 dm^3, and we want to find the temperature at which its volume will be doubled. Therefore, we can set V2 = 4 dm^3. Plugging in the known values, we get: 2/300 = 4/T2 Simplifying this equation, we get: T2 = (300 x 4) / 2 = 600 K Therefore, the answer is option D: 600 K. To summarize, we used the ideal gas law and the fact that the number of moles of gas is constant to write an equation relating the initial and final temperature and volume of the gas. We then solved for the final temperature when the volume is doubled, given the initial volume and temperature of the gas.

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The large temperature interval between the melting point and boiling point of a metal is due to the fact that melting and boiling involve very different types of energy transfer. Melting is the process by which a solid substance is heated and transformed into a liquid. In order to melt a metal, heat energy must be added to break the metallic bonds between the atoms that hold the metal in a rigid, crystalline structure. The temperature at which this occurs is called the melting point. Boiling, on the other hand, is the process by which a liquid substance is heated and transformed into a gas. In order to boil a liquid metal, heat energy must be added to break the intermolecular forces between the atoms or molecules that hold the liquid together. The temperature at which this occurs is called the boiling point. Metals have a large temperature interval between their melting and boiling points because the metallic bonds that hold the atoms together in a solid are much stronger than the intermolecular forces that hold a liquid metal together. This means that it takes much more energy to break the metallic bonds and melt the metal than it does to break the intermolecular forces and boil the liquid metal. Therefore, option C is the correct answer: melting does not break the metallic bond but boiling does.

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Duralumin is a type of aluminum alloy that is made by mixing aluminum with copper and another metal. The options given are zinc and gold, lead and manganese, nickel and silver, and manganese and magnesium. Among these options, the correct one is manganese and magnesium. Therefore, duralumin consists of aluminum, copper, manganese, and magnesium. This alloy is known for its strength, hardness, and durability, and is commonly used in the manufacturing of aircraft and other high-stress applications.

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Tritium is an isotope of hydrogen, and like all hydrogen atoms, its nucleus contains only one proton. However, unlike the common hydrogen isotope, tritium also contains two neutrons in its nucleus. Therefore, the correct answer is: "two neutrons and one proton."

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Water for town supply is chlorinated to make it free from bacteria. Chlorine is a powerful disinfectant that kills most of the bacteria, viruses, and other microorganisms that can be present in water. By adding chlorine, the water becomes safe for drinking and other household uses. Chlorine also helps to remove bad odour from the water. However, it does not remove temporary or permanent hardness from water.

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To find the concentration of hydroxide ion in the orange juice, we need to use the equation for the ionization constant of water (Kw), which is Kw = [H+][OH-] = 1.0 x 10^-14 at 25°C. Since we know the pH of the orange juice is 3.80, we can find the concentration of hydrogen ions [H+] using the equation pH = -log[H+]. So, pH 3.80 is equivalent to [H+] = 1.58 x 10^-4 M. To find the concentration of hydroxide ions [OH-], we can rearrange the equation for Kw to get [OH-] = Kw/[H+]. Substituting the values we found, we get [OH-] = 6.3 x 10^-11 M. Therefore, the correct option is 6.3 x 10^-11.

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Polysaccharides are complex carbohydrates made up of multiple simple sugar units joined together by glycosidic bonds. Starch is an example of a polysaccharide. It is a carbohydrate consisting of a large number of glucose units joined by glycosidic bonds. It is commonly found in plants and serves as a storage form of energy. Therefore, the correct answer is option D, "starch."

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Carbon Hydrogen

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The solubility of a gaseous substance is dependent on the nature of solvent, temperature, and pressure. Therefore, the correct option is I, III and IV only. - The nature of the solvent affects the solubility of a gas as different gases have different solubility in different solvents. For example, oxygen is more soluble in water than in organic solvents. - The temperature affects the solubility of a gas in a solvent as the solubility of gases generally decreases with an increase in temperature. - The pressure also affects the solubility of a gas in a solvent as the solubility of gases generally increases with an increase in pressure.