JAMB UTME - Chemistry - 1990

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Smoke is a mixture of small particles and gases that are released when materials burn. The composition of smoke depends on the material that is burning, but in general, smoke consists of solid or liquid particles dispersed in gas. When materials burn, they release gases such as carbon monoxide, carbon dioxide, and nitrogen oxides. These gases mix with air and can form small particles, which become suspended in the air as smoke. The particles in smoke can be very small, often less than 1 micron in diameter, and can be inhaled deep into the lungs, which can be harmful to human health. Therefore, the correct option is "solid or liquid particles dispersed in gas." Smoke can contain a variety of particles, including ash, soot, and other chemicals, depending on what is being burned.

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The given chemical equation indicates that 1 mole of sodium oxalate (Na2C2O4) reacts with 1 mole of calcium chloride (CaCl2) to produce 1 mole of calcium oxalate (CaC2O4) and 2 moles of sodium chloride (NaCl). First, we need to calculate the number of moles of sodium oxalate in the given solution. The molecular weight of Na2C2O4 is (2x23) + (2x12) + (4x16) = 134 g/mol. Therefore, 1.9 g of Na2C2O4 is equal to 1.9/134 = 0.01418 moles. According to the equation, 1 mole of Na2C2O4 reacts with 1 mole of CaCl2 to produce 1 mole of CaC2O4. So, the number of moles of CaC2O4 produced will also be 0.01418 moles. Now, we can use the molarity formula, M = moles/volume (in liters), to calculate the minimum volume of 0.1 M CaC2O4 solution required. Rearranging the formula, we get the equation, volume = moles/M. Substituting the values, volume = 0.01418/0.1 = 0.1418 L = 141.8 mL. Therefore, the minimum volume of 0.1 M calcium oxalate solution required is 141.8 mL, which is equivalent to 1.40 x 10^2 cm3. The correct option is: 1.40 χ 102 dm3.

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A physical change is a change in which the chemical composition of a substance remains the same, but there is a change in its physical state or appearance. Out of the options provided, the dissolution of sodium chloride (NaCl) in water is a physical change. This is because the chemical composition of NaCl remains the same before and after the dissolving process. Sodium chloride, which is a solid in its pure form, dissolves in water to form a clear and colorless solution, but the chemical composition of NaCl remains the same. The bubbling of chlorine into water and the passing of steam over heated iron are examples of chemical changes because a new substance is formed with a different chemical composition. The bubbling of chlorine into a jar containing hydrogen is also a chemical change because it results in the formation of hydrogen chloride gas (HCl), which has different chemical properties than the reactants (chlorine and hydrogen). Hence, the correct option is the dissolution of sodium chloride in water.

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This is a question on acid-base titration. From the given data, we know that 25.00 cm³ of the monobasic acid requires 20.00 cm³ of 0.1 M NaOH solution for complete neutralization. Therefore, the number of moles of NaOH required to neutralize the acid is: 0.1 mol/L × 0.020 L = 0.002 mol Since the acid is monobasic, the number of moles of the acid is also 0.002 mol. We can use this information to calculate the molar mass of the acid. The molar mass is equal to the mass of the acid divided by the number of moles of the acid: Molar mass = Mass of acid / Number of moles of acid We are given that 2.0 g of the acid was made up to 250 cm³. Therefore, the concentration of the acid is: Concentration = Mass of acid / Volume of solution = 2.0 g / 250 cm³ = 0.008 mol/L Now, we can calculate the number of moles of the acid present in 25.00 cm³ of the solution: Number of moles of acid = Concentration × Volume = 0.008 mol/L × 0.0250 L = 0.0002 mol Finally, we can calculate the molar mass of the acid: Molar mass = Mass of acid / Number of moles of acid = 2.0 g / 0.0002 mol = 100 g/mol Therefore, the molar mass of the acid is 100 g/mol. Answer option (C) is correct.

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The pH of a solution is defined as the negative logarithm of the concentration of hydrogen ions (H+) in moles per liter (pH = -log[H+]). To determine the concentration of H+ in moles per dm3 (equivalent to liters), we can rearrange the formula: [H+] = 10^(-pH). Therefore, for a sodium solution with pH 4.398, the concentration of H+ is: [H+] = 10^(-4.398) = 3.98 x 10^(-5) moles per dm3. The correct answer is option A: 4.0 x 10^(-5). Note: A dm3 is the same as a liter.

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24mg = 1 mole of magnesium and

32g of O2 = mole of O2

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The compound that represents the polymerization product of ethyne is option A. Polymerization is a process by which small molecules, called monomers, combine to form a larger molecule called a polymer. In the case of ethyne, also known as acetylene, the monomer contains two carbon atoms and two hydrogen atoms with a triple bond between them. When multiple ethyne molecules undergo polymerization, the triple bonds break and the carbon atoms link together to form a long chain of carbon atoms. The resulting polymer is called polyacetylene and has a structure similar to that of option A, which shows a long chain of carbon atoms with alternating single and double bonds. Options B, C, and D do not represent the correct structure for the polymerization product of ethyne. Option B shows a cyclic structure, which is not typically formed in the polymerization of ethyne. Option C shows a branched structure, which is also not typically formed in this process. Option D shows a linear structure with all single bonds, which does not reflect the alternating single and double bonds present in polyacetylene.

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The question is asking which of the gases listed can be collected by the upward displacement of air. When a gas is collected by the upward displacement of air, it means that the gas is lighter than air and will rise up when released into the atmosphere. The air that is displaced by the gas will be pushed down and out of the container. To determine which gas is lighter than air, we need to look at their molecular weights. The lighter the gas, the more likely it is to be collected by the upward displacement of air. - NO: Nitrogen monoxide has a molecular weight of 30 g/mol, which is lighter than air (which has a molecular weight of 28.8 g/mol). Therefore, it can be collected by the upward displacement of air. - H2: Hydrogen gas has a molecular weight of 2 g/mol, which is much lighter than air. Therefore, it can be collected by the upward displacement of air. - NH3: Ammonia gas has a molecular weight of 17 g/mol, which is lighter than air. Therefore, it can be collected by the upward displacement of air. - Cl2: Chlorine gas has a molecular weight of 71 g/mol, which is much heavier than air. Therefore, it cannot be collected by the upward displacement of air. Therefore, the gases that can be collected by the upward displacement of air are NO, H2, and NH3.

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The kinetic theory of gases states that the molecules of a gas are in constant motion and move in straight lines until they collide with other molecules or the walls of their container. Therefore, the postulate that the molecules of a gas move in straight lines between collisions implies that collisions between gas molecules are elastic. In other words, no energy is lost during the collision and the total kinetic energy of the system remains constant. Thus, the correct option is "collisions are perfectly elastic." The other options are not directly related to the given postulate.

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V1 = 146cm3 T1 = 273 + 118 = 291 k

P1 = 0.971 atmosphere

V2 = ? T2 = 273K P2 = 1 atom

V2 = (0.971 atom χ 146cm3 χ 273k)/(1atm χ 291k) = 132997 0r = 133cm3

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Ag+ + e-
H = 108 g Ag
= 0.11 10.8 Ag
4OH → O2 + 2H2O + 4e
4F ↔ 1mole O2
4F = 22.4 dm3 χ 0.1 = 0.56 dm3

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In the reaction: SnO2 + 2CO
From the equation above
2(12 g) carbon = 151 g SnO2
24 g carbon =151 g SnO2
= 0.302 kg SnO2 = 24/151g χ 0.0302 kg carbon = 0.048 kg carbon

Mass of coke is required not mass of carbon,so it is (0.048$\times$100)$÷$80 = 0.06

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Chlorine has two isotopes with mass numbers 35 and 37. The atomic mass of chlorine is 35.5. This means that the average mass of all chlorine atoms is 35.5 atomic mass units (amu). Since the atomic mass is a weighted average of the masses of all isotopes of chlorine, we can set up an equation to find the relative abundance of the isotope with mass number 37: (35 x % abundance of 35) + (37 x % abundance of 37) = 35.5 We can simplify this equation by letting x be the % abundance of the isotope with mass number 37: (35 x (100 - x)) + (37 x x) = 35.5 x 100 Solving for x gives us x = 24.24, which is approximately 25%. Therefore, the relative abundance of the isotope with mass number 37 in chlorine is 25%. So, the correct answer is 25.

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The given information states that 0.1 faraday of electricity deposited 2.95 g of nickel during electrolysis of an aqueous solution. So, we can calculate the amount of nickel deposited per unit of electricity (per faraday) as follows: Amount of nickel deposited per faraday = 2.95 g / 0.1 F = 29.5 g/F Now, to calculate the number of moles of nickel deposited by 0.4 faraday, we can use the following formula: Amount of substance (in moles) = Mass / Molar mass where Mass is the amount of nickel deposited and Molar mass is the atomic weight of nickel (Ni = 58.7 g/mol). So, the number of moles of nickel deposited by 0.4 faraday can be calculated as follows: Amount of nickel deposited by 0.4 F = 29.5 g/F × 0.4 F = 11.8 g Number of moles of nickel deposited by 0.4 F = 11.8 g / 58.7 g/mol = 0.20 mol Therefore, the answer is 0.20. In summary, 0.1 faraday of electricity deposited 2.95 g of nickel, which means that 1 faraday will deposit 29.5 g of nickel. To calculate the number of moles of nickel deposited by 0.4 faraday, we multiply the amount of nickel deposited per faraday (29.5 g/F) by 0.4 F, which gives us 11.8 g of nickel. Then, we divide this mass by the molar mass of nickel (58.7 g/mol) to get the number of moles of nickel deposited, which is 0.20 mol.

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Sodium hydrogen, also known as sodium hydride, is a highly reactive substance that readily reacts with water to produce hydrogen gas. Therefore, it should be stored in properly closed containers to prevent it from coming into contact with moisture and air, which could lead to dangerous reactions. When sodium hydrogen is exposed to air, it readily absorbs water vapor from the air, which can cause it to react violently, producing hydrogen gas and heat. This reaction can be explosive if the concentration of hydrogen gas becomes too high. Therefore, the correct option is "readily absorbs water vapor from the air." Sodium hydrogen should be stored in an airtight container with a desiccant to absorb any moisture that may be present. This helps to prevent the absorption of water vapor from the air, which could lead to dangerous reactions.

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The diagram above shows a process where a solid (X) is heated to form a gas (Y), which is then cooled to form a liquid (Z). The changes in physical state are sublimation (X to Y), and condensation (Y to Z). The solid changes directly to gas without passing through the liquid phase during sublimation, and the gas changes to liquid during condensation. Therefore, the answer is (A) sublimation, condensation and freezing.

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The question is asking for the solubility of CuSO4 in water at a certain temperature. The given information is that 20g of CuSO4 is dissolved in 100g of water at 180°C. To solve this problem, we need to first calculate the molar mass of CuSO4, which is: Cu: 1 x 64 = 64 g/mol S: 1 x 32 = 32 g/mol O: 4 x 16 = 64 g/mol Total: 160 g/mol Next, we need to calculate the number of moles of CuSO4 that were dissolved in the water. We can do this using the formula: moles = mass/molar mass mass = 20g molar mass = 160 g/mol moles = 20/160 = 0.125 mol Finally, we can calculate the solubility of CuSO4 in water at this temperature using the formula: solubility = moles/volume We are given that the volume of the solution is 100g, but we need to convert this to dm3. Since the density of water is 1 g/cm3, we can assume that 100g of water has a volume of 100 cm3 or 0.1 dm3. solubility = 0.125/0.1 = 1.25 mol dm-3 Therefore, the solubility of CuSO4 in water at 180°C is 1.25 mol dm-3, which corresponds to.

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The cycloalkane with the molecular formula C5H10 can have three isomers. This is because cycloalkanes with five carbon atoms can only form a cyclopentane ring, and there are three possible ways to arrange the carbon atoms in a cyclopentane ring that result in different isomers. These three isomers are: 1. Cyclopentane 2. Methylcyclobutane 3. Dimethylcyclopropane Therefore, the answer is "three isomers."

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The problem can be solved by using the ideal gas equation, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas: PV = nRT Where R is the universal gas constant. At STP (standard temperature and pressure), the temperature is 273.15 K and the pressure is 1 atm. Using the given volume of 500 cm3, we can convert it to dm3 by dividing by 1000: V = 500 cm3 = 0.5 dm3 Substituting these values into the ideal gas equation, we get: (1 atm) (0.5 dm3) = n (0.08206 L.atm/mol.K) (273.15 K) Simplifying this equation, we can solve for n: n = (1 atm) (0.5 dm3) / (0.08206 L.atm/mol.K) (273.15 K) n = 0.0200 mol Now we can use the formula for the density of a gas to find its relative molecular mass (M): density = mass / volume M = (mass of gas) / (number of moles of gas) We are given the mass of the gas (1.58 g) and the volume of the gas (0.5 dm3), so we can substitute these values into the density equation: density = (1.58 g) / (0.5 dm3) density = 3.16 g/dm3 Finally, we can solve for the relative molecular mass by rearranging the density equation: M = (density) (GMV) / (RT) Substituting the known values, we get: M = (3.16 g/dm3) (22.40 dm3/mol) / (0.08206 L.atm/mol.K) (273.15 K) M = 71.0 g/mol Therefore, the relative molecular mass of the gas is 71 g/mol, which corresponds to.

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The octane number of a fuel is a measure of its ability to resist "knocking" or detonation during combustion in an internal combustion engine. A fuel with a higher octane number is more resistant to knocking, and therefore is typically used in high-performance engines that require high compression ratios or high operating temperatures. The octane number of a fuel is determined by comparing its performance to a mixture of 2,2,4-trimethylpentane (iso-octane) and n-heptane. This mixture is assigned an octane rating of 100 because it has very high resistance to knocking. Fuels that perform as well as the 2,2,4-trimethylpentane/n-heptane mixture are assigned higher octane numbers, while fuels that perform worse are assigned lower numbers. In this problem, we are given that the performance of the fuel is the same as a mixture of 55 g of 2,2,4-trimethylpentane and 45 g of n-heptane. We can use this information to determine the octane number of the fuel by comparing its performance to the 55/45 mixture. The octane number of the fuel can be calculated as follows: octane number = (volume percentage of iso-octane in the mixture) x 100 To find the volume percentage of iso-octane in the mixture, we need to calculate the moles of each component in the mixture, using their respective molecular weights: moles of iso-octane = 55 g / 114.23 g/mol = 0.481 moles moles of n-heptane = 45 g / 100.21 g/mol = 0.449 moles The volume percentage of iso-octane in the mixture can be calculated using their respective densities and molecular weights: volume percentage of iso-octane = (0.481 mol x 114.23 g/mol) / (0.481 mol x 114.23 g/mol + 0.449 mol x 100.21 g/mol) x 100% = 55.96% Therefore, the octane number of the fuel is approximately 56, which is closest to option (A) 45. Note that the octane number of a fuel is typically reported as an integer, so the actual octane number of the fuel in this problem would likely be rounded down to 55.

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The test that can completely identify any of sulfur (VI) oxide, hydrogen, carbon (IV) oxide, and nitrogen (II) oxide is passing each gas into lime water. This is because the reaction of lime water with each gas is specific and characteristic. When sulfur (VI) oxide reacts with lime water, it forms sulfuric acid which turns lime water milky. When hydrogen reacts with lime water, it forms calcium hydroxide and water, which does not produce any noticeable change in the lime water. When carbon (IV) oxide reacts with lime water, it forms calcium carbonate which turns lime water cloudy. When nitrogen (II) oxide reacts with lime water, it forms calcium nitrite which turns lime water brown.

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The term that indicates the number of bonds that can be formed by an atom is "Valence". Valence refers to the outermost shell of electrons in an atom. These electrons are involved in chemical bonding with other atoms to form molecules. The number of electrons in the valence shell determines the number of bonds that an atom can form. For example, the element carbon has four valence electrons, which means it can form up to four bonds with other atoms. Similarly, oxygen has six valence electrons and can form up to two bonds. Therefore, the term that indicates the number of bonds that can be formed by an atom is "Valence".

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To calculate the volume of 11.0 M hydrochloric acid needed to obtain 1 dm^3 of 0.05 M acid, we can use the formula: M1V1 = M2V2 where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. We can rearrange the formula to solve for the unknown volume V1: V1 = (M2 x V2) / M1 Substituting the values given in the question: M1 = 11.0 M M2 = 0.05 M V2 = 1 dm^3 Plugging in the values: V1 = (0.05 M x 1 dm^3) / 11.0 M V1 = 0.0045 dm^3 = 0.045 L = 45 mL Therefore, the volume of 11.0 M hydrochloric acid that must be diluted to obtain 1 dm^3 of 0.05 M acid is 45 mL. The answer is not listed in mL or L, but rather in dm^3. 1 dm^3 is equivalent to 1000 mL or 1 L. So, 0.045 L is equal to 0.045 dm^3, which is the same as the listed answer of 0.05 dm^3.

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An oxide that reacts with both HCl (hydrochloric acid) and NaOH (sodium hydroxide) to give salt and water only is classified as an amphoteric oxide. Amphoteric oxides are substances that can act as either an acid or a base depending on the reaction conditions. They have the ability to donate or accept a proton (H+) depending on the nature of the other reactant. In the case of HCl, which is an acid, the amphoteric oxide will act as a base and accept the proton to form salt and water. In the case of NaOH, which is a base, the amphoteric oxide will act as an acid and donate a proton to form salt and water. Examples of amphoteric oxides include aluminum oxide (Al2O3) and zinc oxide (ZnO). Therefore, the answer is "an amphoteric oxide".

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Pb(NO3)2 + H2S → PbS + 2HNO3
34 g H2S → 239 g pbs
5.02g pbs → (34)/(237g) χ 5.02 g = 0.714 g
34g → 22.4 dm3
=0.714 → (22.4)/34g) χ 100 = 4.70

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The question is asking which metal oxide is added to the reaction mixture to make colored glasses. The metals added are those which form colored silicates. Potassium, barium, zinc, and copper are all metals, but only copper oxide (CuO) forms colored silicates. Therefore, the correct answer is copper.

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The metal that will liberate hydrogen from steam or dilute acid is iron. When iron reacts with steam or dilute acid, it forms iron oxide or iron (II) salt and hydrogen gas. This is because iron is more reactive than copper, lead, and mercury, and it can displace hydrogen from water or acid. Copper, lead, and mercury are less reactive than iron, so they cannot displace hydrogen from steam or dilute acid.

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The catalyst commonly used in the Haber process for the manufacture of ammonia is iron. The Haber process is a chemical reaction that combines nitrogen and hydrogen gases to produce ammonia gas. This reaction is slow and requires high pressure and temperature to proceed efficiently. The presence of a catalyst speeds up the reaction rate without being consumed in the process. Iron is preferred as the catalyst in the Haber process because it is readily available, inexpensive, and effective in promoting the reaction. Additionally, iron can withstand the harsh reaction conditions of high pressure and temperature without degrading. The iron catalyst is usually in the form of small pellets or shavings, which provide a large surface area for the reactant gases to interact with the catalyst, increasing the reaction rate.