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Ajụjụ 1 Ripọtì
Which of the following conducts electricity
Akọwa Nkọwa
Graphite is the option that conducts electricity.
Ajụjụ 2 Ripọtì
A colored gas that is known to be poisonous and can readily damage the mucous lining of the lungs is?
Akọwa Nkọwa
The colored gas that is known to be poisonous and can readily damage the mucous lining of the lungs is chlorine. Chlorine is a highly reactive chemical element that is used in the production of many everyday products, such as paper, textiles, and plastics. It is also used as a disinfectant in swimming pools and water treatment plants. Inhaling chlorine gas can cause severe respiratory problems, including coughing, chest pain, and difficulty breathing. Prolonged exposure to chlorine can cause lung damage, and in extreme cases, it can be fatal. Chlorine gas is also highly irritating to the eyes, skin, and mucous membranes. It is important to handle chlorine with caution and to use appropriate protective gear, such as gloves and respiratory masks, when working with it. Proper ventilation and monitoring of chlorine levels are also essential to prevent exposure to this toxic gas.
Ajụjụ 3 Ripọtì
An organic compound which liberate carbon(iv)oxide from trioxocarbonate(iv) solution is likely to be?
Akọwa Nkọwa
The organic compound that liberates carbon(iv)oxide from trioxocarbonate(iv) solution is CH3COOH (acetic acid). When acetic acid is added to a solution of trioxocarbonate(iv) (carbonate) it reacts to form carbon(iv)oxide gas, water and a salt. The balanced chemical equation for the reaction is: 2CH3COOH + Na2CO3 → CO2 + 2H2O + 2NaCH3COO The carbon(iv)oxide gas is released as bubbles, causing the solution to fizz. Therefore, CH3COOH is the organic compound that liberates carbon(iv)oxide from trioxocarbonate(iv) solution.
Ajụjụ 4 Ripọtì
A sample of gas with an initial volume of 2.5 dm3 is heated and then allowed to expand to 7.5 dm3 at constant at pressure. What is the ratio of the final temperature of the initial absolute temperature?
Akọwa Nkọwa
According to Charles's Law, the ratio of the initial and final temperatures is equal to the ratio of the initial and final volumes at constant pressure. The ratio of the final volume to the initial volume is: Vf / Vi = 7.5 dm3 / 2.5 dm3 = 3 Therefore, the ratio of the final temperature to the initial temperature is also 3: Tf / Ti = Vf / Vi = 3 So the answer is 3:1.
Ajụjụ 5 Ripọtì
Hard water is water with high concentrations of dissolved ions, in particular calcium and
Akọwa Nkọwa
Hard water is water that contains high amounts of dissolved minerals, specifically calcium and magnesium ions. These minerals come from the rocks and soil that the water flows through and can accumulate in the water as it travels to your home. When you use hard water, it can leave mineral deposits on your pipes, fixtures, and appliances, which can reduce their efficiency and lifespan. It can also make soap less effective and leave your skin feeling dry and itchy. Therefore, it is important to treat hard water if it is a problem in your area.
Ajụjụ 6 Ripọtì
The dehydration of CH3 CH2 CH2 CH2 OH will give?
Ajụjụ 7 Ripọtì
Ethene, when passed into concentrated H2SO4, is rapidly absorbed. The product is diluted with water and then warmed to produce
Akọwa Nkọwa
When ethene is passed into concentrated H2SO4, it undergoes electrophilic addition reaction to form ethyl hydrogen sulfate as the product. The reaction mixture is then diluted with water and warmed to produce ethanol as the main product. Therefore, the answer is ethanol.
Ajụjụ 8 Ripọtì
N2 O4 ? 2NO2 (? = -ve)
From the reaction above, which of these conditions would produce the highest equilibrium yield for N2 O4 ?
Akọwa Nkọwa
The highest equilibrium yield of N2O4 would be produced at low temperature and low pressure. In a chemical reaction, the position of the equilibrium can be influenced by changing the temperature or pressure. A decrease in temperature or an increase in pressure favors the side of the reaction with the fewer moles of gas (in this case, N2O4). This means that, if the temperature is low and the pressure is low, there will be more N2O4 at equilibrium, as the reaction will shift to the right to counteract the reduction in the concentration of N2O4. So, low temperature and low pressure would produce the highest equilibrium yield of N2O4.
Ajụjụ 9 Ripọtì
The oxidation number of oxygen in BaO2 is
Akọwa Nkọwa
To determine the oxidation number of oxygen in BaO2, we can use the fact that the overall charge of a compound must be zero. Barium (Ba) is a Group 2 element and has an oxidation state of +2. The compound BaO2 has no net charge, so the sum of the oxidation states of all the atoms must be zero. Let x be the oxidation state of oxygen in BaO2. Therefore, we have: (+2) + 2(x) = 0 Solving for x, we get: x = -1 Therefore, the oxidation number of oxygen in BaO2 is -1.
Ajụjụ 10 Ripọtì
The IUPAC name for CICH2-CH2-CH2-OH is
Akọwa Nkọwa
The IUPAC name for CICH2-CH2-CH2-OH is 3-chloropropan-1-ol. To name the compound using the IUPAC nomenclature system, we start by identifying the longest continuous chain of carbon atoms that contains the functional group (-OH). In this case, the longest chain contains three carbon atoms, so the root name is propane. Next, we identify the position of the substituent (-Cl) on the chain. The substituent is attached to the third carbon atom in the chain, so the name of the compound becomes 3-chloropropane. Finally, we add the suffix -ol to indicate that the compound contains an alcohol functional group (-OH), so the complete name of the compound is 3-chloropropan-1-ol. Therefore, the correct answer is 3-chloropropan-1-ol.
Ajụjụ 11 Ripọtì
Which of the following compound is NOT the correct formed compound when the parent metal is heated in air?
Akọwa Nkọwa
The compound that is NOT correctly formed when the parent metal is heated in air is: tri-iron tetraoxide (Fe2O). This is because the correct compound formed from the heating of iron in air is iron (III) oxide or Fe2O3. The formula for tri-iron tetraoxide is incorrect, as it implies that there are only three iron atoms in the compound when there should be four.
Ajụjụ 12 Ripọtì
30 cm3 of oxygen at 10 atmosphere pressure is placed in a 20 dm3 container. Calculate the new pressure if the temperature is kept constant.
Akọwa Nkọwa
Given:
First, convert all volumes to the same units. Since 1 dm3dm3 is 1000 cm3cm3:
𝑉2=20 dm3=20×1000 cm3=20000 cm3V2=20dm3=20×1000cm3=20000cm3
Now, using Boyle's Law:
𝑃1𝑉1=𝑃2𝑉2P1V1=P2V2
Substitute the known values into the equation:
10×30=𝑃2×2000010×30=P2×20000
300=𝑃2×20000300=P2×20000
Solve for 𝑃2P2:
𝑃2=30020000P2=20000300
𝑃2=0.015 atmospheresP2=0.015atmospheres
Therefore, the new pressure if the temperature is kept constant is:
Ajụjụ 13 Ripọtì
Silver chloride turns gray when exposed to sunlight because
Ajụjụ 14 Ripọtì
Using the metal activity series, the metal that can liberate hydrogen gas from steam is?
Akọwa Nkọwa
The metal that can liberate hydrogen gas from steam is iron. The metal activity series is a list of metals in order of their reactivity, with the most reactive metals at the top and the least reactive metals at the bottom. When a metal is placed in a solution of steam (water vapor), the metal will react with the steam if it is more reactive than hydrogen. In this case, iron is more reactive than hydrogen, so it can displace hydrogen from the steam to form hydrogen gas. This reaction can be represented by the equation: Fe + H2O (steam) → FeO (iron oxide) + H2 (hydrogen gas) So, when steam is passed over iron, hydrogen gas is liberated and iron oxide is formed.
Ajụjụ 15 Ripọtì
A sample of gas exerts a pressure of 8.2 atm when confined in a 2.93 dm3 container at 20c. The number of moles of gas in the sample is
Ajụjụ 16 Ripọtì
If the volume of a given mass of a gas at 0ºc is 29.5cm3 . What will be the volume of the gas at 15ºc, given that the pressure remains constant.
Ajụjụ 17 Ripọtì
The boiling point of water, ethanol, toulene and butan-2-ol are 373.0k, 351.3k, 383.6k and 372.5k respectively, which liquid has the highest vapour pressure at 323.0k
Ajụjụ 18 Ripọtì
Chlorine is a common bleaching agent. This is not true with
Akọwa Nkọwa
Chlorine is not a common bleaching agent for wet litmus paper, wet pawpaw leaf, and most wet fabric dyes. It is commonly used as a bleaching agent for printer's ink.
Ajụjụ 19 Ripọtì
Zn + 2HCL → ZnCl2 + H2
What happens to zinc in the above reaction?
Akọwa Nkọwa
In the above reaction, zinc (Zn) reacts with hydrochloric acid (HCl) to form zinc chloride (ZnCl2) and hydrogen gas (H2). The chemical equation for the reaction is: Zn + 2HCl → ZnCl2 + H2 During the reaction, zinc atoms lose two electrons each and get oxidized to form positively charged zinc ions (Zn2+), as they react with the hydrogen ions (H+) from the hydrochloric acid to form zinc chloride. The hydrogen ions, on the other hand, gain an electron each and get reduced to form hydrogen gas molecules (H2). Therefore, in the given reaction, zinc is getting oxidized, as it loses electrons and forms a positively charged ion. Hence, the correct option is "oxidized."
Ajụjụ 20 Ripọtì
There is a large temperature interval between the melting point and the boiling point of metal because:
Akọwa Nkọwa
The correct answer is: "melting does not break the metallic bond but boiling does." The metallic bond is the force of attraction between metal atoms, which holds them together to form a solid. When a metal is heated, its temperature increases, and at a certain point, the energy provided by the heat is enough to overcome the metallic bond and cause the metal to melt. However, even in the liquid state, the metallic bond remains intact, which is why metals have a very high melting point. On the other hand, when the temperature is further increased, the energy provided by the heat becomes enough to break the metallic bond, and the metal atoms become completely detached from one another. This results in the metal boiling and turning into a gas. Because the metallic bond is much stronger than other types of intermolecular forces, such as van der Waals forces, it requires a lot of energy to break, resulting in a large temperature interval between the melting point and boiling point of metal.
Ajụjụ 21 Ripọtì
In the preparation of salts, the method employed will depend on the?
Akọwa Nkọwa
The method employed in the preparation of salts will depend on the composition of the salt. Different salts have different chemical properties, and the method used to prepare them will depend on these properties. For example, some salts can be easily dissolved in water, while others are not very soluble and may require the use of a different solvent or special conditions to dissolve. The dissociating ability, stability to heat, and precipitating ability of the salt may also play a role in determining the preparation method, but the most important factor is the composition of the salt.
Ajụjụ 22 Ripọtì
2H2 + O2 → 2H2 O
From the equation above, calculate the volume of unreacted oxygen gas if a mixture of 50cm3 of hydroden and 75cm3 of oxygen are involved
Akọwa Nkọwa
The balanced chemical equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor. Therefore, the stoichiometric ratio of hydrogen to oxygen is 2:1. In this problem, there are 50cm3 of hydrogen gas and 75cm3 of oxygen gas. Since the gases are at the same temperature and pressure, their volumes are directly proportional to the number of moles of gas present. Using the stoichiometric ratio, we can calculate that the amount of oxygen gas required to react completely with 50cm3 of hydrogen gas is (1/2) * 50cm3 = 25cm3. Since there are 75cm3 of oxygen gas present, there must be (75cm3 - 25cm3) = 50cm3 of unreacted oxygen gas remaining. Therefore, the volume of unreacted oxygen gas is 50cm3. Answer: 50cm3
Ajụjụ 23 Ripọtì
which of these compounds exhibits resonance
Akọwa Nkọwa
The compound that exhibits resonance is benzene.
Ajụjụ 24 Ripọtì
A chemical widely used as a fertilizer is?
Akọwa Nkọwa
The chemical widely used as a fertilizer is nitrochalk. Nitrochalk is a type of fertilizer that contains a mixture of ammonium nitrate and calcium carbonate. Ammonium nitrate provides the necessary nitrogen for plant growth, while calcium carbonate helps to balance the soil's pH level. This combination of nutrients helps to promote healthy plant growth and increase crop yields. Nitrochalk is commonly used in agriculture and gardening to fertilize crops such as corn, wheat, and soybeans, as well as fruits and vegetables.
Ajụjụ 25 Ripọtì
Which of the following will precipitate in dil. HCl
Akọwa Nkọwa
Among the given options, only CuS will precipitate in dilute HCl. CuS is insoluble in dilute HCl, and hence it will precipitate when added to dilute HCl. However, the other options will dissolve in dilute HCl, and hence they will not precipitate. ZnS will dissolve in dilute HCl to form ZnCl2 and H2S. Na2S will react with dilute HCl to produce H2S and NaCl. FeS will dissolve in dilute HCl to form FeCl2 and H2S. Therefore, the correct answer is (4) CuS.
Ajụjụ 26 Ripọtì
Complete dehydrogenation of ethyne yields
Ajụjụ 27 Ripọtì
Which of the following will act as both oxidizing agents and reducing agents?
Akọwa Nkọwa
The oxidizing and reducing properties of a substance depend on its ability to gain or lose electrons. A substance that can gain electrons acts as an oxidizing agent, while a substance that can lose electrons acts as a reducing agent. Among the given options, both Cl2 (chlorine gas) and SO2 (sulfur dioxide) can act as both oxidizing and reducing agents depending on the reaction conditions. - Cl2 can act as an oxidizing agent when it gains electrons to form Cl- ions, and it can act as a reducing agent when it loses electrons to form Cl+ ions. For example, in the reaction Cl2 + 2KBr → 2KCl + Br2, chlorine gas is acting as an oxidizing agent since it is gaining electrons from bromide ions to form bromine gas. However, in the reaction 2Cl- + Cl2 → 2Cl2-, chlorine gas is acting as a reducing agent since it is losing electrons to form chloride ions. - SO2 can act as an oxidizing agent when it gains electrons to form sulfite ions (SO32-), and it can act as a reducing agent when it loses electrons to form sulfur trioxide (SO3). For example, in the reaction SO2 + 2H2S → 3S + 2H2O, sulfur dioxide is acting as a reducing agent since it is losing electrons to form elemental sulfur. However, in the reaction 2SO32- + O2 → 2SO42-, sulfur dioxide is acting as an oxidizing agent since it is gaining electrons to form sulfate ions. H2S (hydrogen sulfide) and NH3 (ammonia) are not likely to act as both oxidizing and reducing agents under normal conditions. H2S tends to act as a reducing agent by donating electrons to oxidizing agents, while NH3 tends to act as a reducing agent by donating electrons to oxidizing agents or as a base by accepting protons.
Ajụjụ 28 Ripọtì
What is the shape of a molecule of CCl4?
Akọwa Nkọwa
The shape of a molecule of CCl4 is tetrahedral.
Ajụjụ 29 Ripọtì
| GAS | CO2 | N2 | O2 |
| % BY VOLUME | 4 | 72 | 24 |
The above table shows the compositions of the atmosphere of planet X. Which of these gases are present in higher percentages on earth?
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
Which of the following statement is TRUE of the complete hydrolysis of a glyceride by sodium hydroxide?
Akọwa Nkọwa
The statement that is TRUE of the complete hydrolysis of a glyceride by sodium hydroxide is: - 3 moles of NaOH are required for each mole of glyceride. During the hydrolysis of a glyceride (a triglyceride), the ester bonds between the fatty acid chains and glycerol are broken by the action of a strong base like sodium hydroxide. This results in the formation of glycerol and the corresponding salts of fatty acids, which are commonly known as "soaps." The reaction can be represented by the following equation: Triglyceride + 3 NaOH → 3 soap + glycerol As per the equation, 3 moles of NaOH are required to hydrolyze one mole of glyceride, and 3 moles of soap and one mole of glycerol are produced. The use of concentrated sulfuric acid (H2SO4) is not essential for the completion of the reaction, but it can be used as a catalyst to speed up the reaction.
Ajụjụ 31 Ripọtì
The sulphide that is commonly used in coating electric fluorescent tubes is?
Akọwa Nkọwa
The sulphide commonly used in coating electric fluorescent tubes is Zinc Sulphide. Zinc Sulphide is a type of material that glows when it is exposed to ultraviolet light. When ultraviolet light is generated inside a fluorescent tube, it excites the Zinc Sulphide particles, causing them to emit visible light. This visible light is what we see as the bright light coming from the tube. So, Zinc Sulphide acts as a phosphor and helps in producing the bright light in fluorescent tubes.
Ajụjụ 32 Ripọtì
Addition of sodium chloride to water to form a solution would lead to?
Akọwa Nkọwa
The addition of sodium chloride to water to form a solution would lead to a decrease in freezing point and an increase in boiling point. This effect is known as colligative properties, which depend on the concentration of solute particles in a solution. When sodium chloride dissolves in water, it breaks down into sodium ions and chloride ions. These ions occupy space between water molecules and interfere with the formation of ice crystals during freezing. As a result, the freezing point of the solution is lowered below that of pure water. This is why we use salt to de-ice roads and sidewalks during the winter season. Similarly, the presence of solute particles in a solution also raises the boiling point of the solution. The increased concentration of solute particles in the solution causes a decrease in the vapor pressure of the solvent (water), making it harder for the solvent molecules to escape into the gas phase. This means that more energy is required to bring the solution to its boiling point compared to pure water. In summary, the addition of sodium chloride to water forms a solution with lower freezing point and higher boiling point compared to pure water.
Ajụjụ 34 Ripọtì
Mixing aqueos solution of barium hydroxide and sodium tetraoxocarbonate (iv) yields a white precipitate of
Akọwa Nkọwa
Mixing aqueous solutions of barium hydroxide and sodium tetraoxocarbonate (IV) would result in a chemical reaction that produces a white precipitate of barium tetraoxocarbonate (IV). The balanced chemical equation for this reaction is: Ba(OH)2(aq) + Na2CO3(aq) → BaCO3(s) + 2NaOH(aq) In the above equation, the barium hydroxide (Ba(OH)2) reacts with sodium tetraoxocarbonate (IV) (Na2CO3) to form barium tetraoxocarbonate (IV) (BaCO3), which is a white precipitate, and sodium hydroxide (NaOH). Therefore, the correct option is 4) Barium tetraoxocarbonate.
Ajụjụ 35 Ripọtì
Which of the following substances is not a homogeneous mixture?
Akọwa Nkọwa
The substance that is not a homogeneous mixture is flood water. Flood water is typically a mixture of various substances, such as sediment, dirt, debris, and organic matter, that have been carried along by the water. As such, flood water is usually a heterogeneous mixture, meaning that it does not have a uniform composition throughout. In contrast, filtered sea water, soft drinks, and writing ink are all examples of homogeneous mixtures, where the components are evenly distributed and the mixture has a uniform composition throughout.
Ajụjụ 36 Ripọtì
SO2 + O2 → 2SO3
In the reaction above, the most suitable catalyst is?
Akọwa Nkọwa
The most suitable catalyst for the given reaction is vanadium(V)oxide (V2O5). Vanadium(V)oxide is a commonly used catalyst for the oxidation of sulfur dioxide (SO2) to sulfur trioxide (SO3). The reaction is an exothermic reaction, and it occurs at high temperatures (around 450-500°C) in the presence of a catalyst. V2O5 is an effective catalyst for this reaction because it has a high surface area and can provide active sites for the reaction to occur. The vanadium ions in the V2O5 catalyst undergo redox reactions with the sulfur dioxide and oxygen molecules, which promotes the formation of sulfur trioxide. Chromium(VI)oxide and iron(III)oxide are not suitable catalysts for this reaction because they are not effective at promoting the oxidation of sulfur dioxide to sulfur trioxide. Copper(I)oxide can be used as a catalyst for the reaction, but it is not as effective as vanadium(V)oxide.
Ajụjụ 37 Ripọtì
Calcium forms complexes with ammonia because
Akọwa Nkọwa
The reason why calcium forms complexes with ammonia is that it has empty d-orbitals.
Ajụjụ 39 Ripọtì
When heat is absorbed during a chemical reaction, the reaction is said to be
Akọwa Nkọwa
When heat is absorbed during a chemical reaction, the reaction is said to be endothermic. Endothermic reactions are characterized by the absorption of heat energy from the surroundings. In other words, the reactants absorb energy from the environment, usually in the form of heat, to form the products. As a result, the temperature of the surroundings decreases, and the reaction feels cold to the touch. Endothermic reactions can be found in many natural processes, such as photosynthesis, melting of ice, and the evaporation of liquids. These processes require energy to occur, and they absorb heat from the surroundings to power the reaction.
Ajụjụ 40 Ripọtì
6g of Mg was to 100cm3 of 1 moldm3 H2 SO4 . What mass of Mg remained undissolved? (Mg = 24)
Akọwa Nkọwa
The balanced chemical equation for the reaction between magnesium (Mg) and sulfuric acid (H2SO4) is: Mg + H2SO4 -> MgSO4 + H2 According to the equation, one mole of Mg reacts with one mole of H2SO4 to produce one mole of magnesium sulfate (MgSO4) and one mole of hydrogen gas (H2). Since the concentration of the sulfuric acid is 1 moldm3, this means that there is one mole of H2SO4 in every 1 liter (1000 cm3) of solution. To determine the amount of Mg that reacts with the H2SO4, we need to use stoichiometry. One mole of Mg reacts with one mole of H2SO4, so the amount of Mg that reacts with 1 moldm3 of H2SO4 is given by: 6g / 24g/mol = 0.25 mol Since the reaction is 1:1, this means that 0.25 mol of H2SO4 is consumed in the reaction. The volume of the solution is 100cm3 (0.1 dm3), so the amount of H2SO4 in the solution is: 1 mol/dm3 x 0.1 dm3 = 0.1 mol The amount of H2SO4 that remains after the reaction is: 0.1 mol - 0.25 mol = -0.15 mol This negative value means that all of the H2SO4 was consumed in the reaction, and there is excess Mg left over. The mass of Mg that remains undissolved is given by: 0.15 mol x 24g/mol = 3.6g Therefore, the correct answer is 3.6g.
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