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Ajụjụ 1 Ripọtì
The electronic configuration of Mg 2+ is
Akọwa Nkọwa
The electronic configuration of Mg2+ is: Is2 2s2 2p6. This configuration can be determined by first looking at the electron configuration of a neutral magnesium atom, which is 1s2 2s2 2p6 3s2. When magnesium loses two electrons to become Mg2+, the two electrons that are removed come from the 3s orbital, leaving behind the completely filled 2p orbital. Thus, the electronic configuration of Mg2+ is 1s2 2s2 2p6, which can be abbreviated as Is2 2s2 2p6. To understand this configuration, we can break it down into the following parts: - "Is2" represents the completely filled 1s orbital, which contains two electrons. - "2s2" represents the completely filled 2s orbital, which also contains two electrons. - "2p6" represents the completely filled 2p orbital, which contains six electrons. Overall, the electronic configuration of Mg2+ has a total of 12 electrons, with all the lower energy orbitals completely filled and the 2p orbital partially filled with six electrons.
Ajụjụ 2 Ripọtì
The highest rate of production of carbon (lV) oxide can be achieved using
Akọwa Nkọwa
The highest rate of production of carbon dioxide (CO2) can be achieved using which is 0.10 mol/L HCl and 5g powdered CaCO3. This is because the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3) produces carbon dioxide gas (CO2), water (H2O), and calcium chloride (CaCl2). The rate of the reaction is dependent on the concentration of the acid, and the powdered form of CaCO3 has a larger surface area compared to the lump form, which allows for more efficient reaction and faster production of CO2.
Ajụjụ 3 Ripọtì
2 - methyl propan -2- ol is an example of a
Ajụjụ 4 Ripọtì
A compound contains 40.0% C, 6.7% H 53.3% O. If the molecular mass of the compound is 180, its molecular formula is [C =12, H =1,O]
Akọwa Nkọwa
The first step to solving this problem is to determine the empirical formula of the compound. We assume that we have 100 grams of the compound, so we have 40.0 g of C, 6.7 g of H, and 53.3 g of O. To find the empirical formula, we need to convert these masses to moles. To do this, we divide each mass by the respective atomic weight: - C: 40.0 g / 12.01 g/mol = 3.33 mol - H: 6.7 g / 1.01 g/mol = 6.63 mol - O: 53.3 g / 16.00 g/mol = 3.33 mol The mole ratio of these elements can then be expressed as 1:2:1 (C:H:O). Therefore, the empirical formula is CH2O. To find the molecular formula, we need to know the molecular mass of the compound. In this case, it is given as 180 g/mol. We can calculate the molecular formula by dividing the molecular mass by the empirical formula mass and then multiplying each subscript by that factor. - Empirical formula mass: 12.01 + 2(1.01) + 16.00 = 30.03 g/mol - Factor: 180 g/mol ÷ 30.03 g/mol = 5.994 Multiplying each subscript by 5.994 gives us the molecular formula of the compound: C6H12O6, which is option (iv).
Ajụjụ 5 Ripọtì
The number of hydroxonium ions produced by one molecule of an acid in aqueous solution is its
Akọwa Nkọwa
The number of hydroxonium ions produced by one molecule of an acid in aqueous solution is its acid strength. Acids are substances that release hydrogen ions (H+) when dissolved in water. When an acid dissolves in water, it forms hydronium ions (H3O+) by combining with water molecules. The acid strength is defined as the ability of an acid to donate H+ ions in aqueous solution. Strong acids are those that completely dissociate into ions when dissolved in water, while weak acids only partially dissociate into ions. The acid strength is directly proportional to the concentration of H+ ions in solution, which determines the pH of the solution. The pH is a measure of the acidity or basicity of a solution and is defined as the negative logarithm of the concentration of H+ ions in solution. Basicity refers to the number of H+ ions that can be accepted by a base rather than donated by an acid. Concentration refers to the amount of an acid in a given volume of solution and does not directly relate to the number of hydroxonium ions produced by one molecule of an acid in aqueous solution.
Ajụjụ 6 Ripọtì
Which of the following is used to study the arrangement of particles in crystal lattices?
Akọwa Nkọwa
To study the arrangement of particles in crystal lattices, X-rays are used. X-rays are high-energy electromagnetic radiation that can penetrate solid materials and reveal the internal structure of an object. By shining X-rays through a crystal and measuring how they diffract, or bend, scientists can determine the arrangement of atoms in the crystal lattice. This information can help us understand the properties and behavior of materials and is used in a variety of fields, including materials science, chemistry, and biology.
Ajụjụ 8 Ripọtì
An increase in entropy can best be illustrated by
Akọwa Nkọwa
An increase in entropy refers to an increase in the amount of disorder or randomness in a system. The best illustration of an increase in entropy can be seen in the mixing of gases. When gases are mixed, the particles become more randomly distributed, and their movements become more disordered, resulting in an increase in entropy. On the other hand, freezing of water, the condensation of vapor, and solidifying of candle wax are all processes that involve a decrease in entropy. When water freezes or vapor condenses, the particles become more ordered, resulting in a decrease in entropy. Similarly, when candle wax solidifies, the molecules become more tightly packed, resulting in a decrease in entropy.
Ajụjụ 9 Ripọtì
Chlorination of water for town supply is carried out to
Akọwa Nkọwa
Chlorination of water for town supply is carried out to remove germs from the water. Chlorine is added to the water to kill harmful microorganisms such as bacteria and viruses, making the water safe to drink. Chlorine also helps to prevent the growth of algae and other organisms in the water supply system. This process is an important step in ensuring that the water supply is clean and safe for human consumption. It does not necessarily make the water taste better or remove its odor, but it does make it safe to drink.
Ajụjụ 10 Ripọtì
Which of the following methods can be used to obtain pure water from a mixture of sand, water and methanoic acid?
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Which of the following pollutants is associated with brain damage?
Akọwa Nkọwa
Carbon(II)oxide fuses with the haemoglobin of red blood cells. This can lead to brain damage.
Ajụjụ 13 Ripọtì
Commercial bleaching can be carried out using
Akọwa Nkọwa
Commercial bleaching can be carried out using chlorine and/or sulphur (IV) oxide. Chlorine is a strong oxidizing agent that reacts with color-causing molecules in stains, breaking them down into smaller, less visible molecules. It is often used to bleach paper, textiles, and other materials. Sulphur (IV) oxide, on the other hand, is a reducing agent that removes oxygen from color-causing molecules, thereby decolorizing them. It is commonly used to bleach food products such as flour and sugar. Both chlorine and sulphur (IV) oxide are highly reactive and must be handled with care to avoid safety hazards.
Ajụjụ 14 Ripọtì
Mineral acids are usually added to commercial hydrogen peroxide to
Akọwa Nkọwa
Mineral acids are usually added to commercial hydrogen peroxide to stabilize it and minimize its decomposition. Hydrogen peroxide is a highly reactive compound that can decompose easily, breaking down into water and oxygen. The addition of a mineral acid, such as phosphoric acid, helps to stabilize the hydrogen peroxide and slow down its decomposition, allowing it to be stored and transported more effectively. This is particularly important for commercial applications of hydrogen peroxide, such as in the production of household cleaning products, where it is important to ensure the stability and longevity of the product.
Ajụjụ 15 Ripọtì
50 cm3 of gas was collected over water at 10oC and 765 mm Hg. Calculate the volume of the gas at s.t.p. if the saturated vapour pressure of water at 10 oC is 5mm Hg
Akọwa Nkọwa
Ajụjụ 16 Ripọtì
Which of the following will produce a solution with pH less than 7 at equivalent point?
Akọwa Nkọwa
The substance that will produce a solution with pH less than 7 at the equivalent point is an acid. At the equivalent point, the moles of acid will equal the moles of base added, which means that the pH will be determined by the dissociation of the conjugate acid or base that remains in solution. Out of the given options, HNO3 is a strong acid and NaOH is a strong base, so their reaction will produce a salt (NaNO3) and water, and the solution will have a pH of 7 at the equivalent point. Similarly, H2SO4 is a strong acid and KOH is a strong base, so their reaction will produce a salt (K2SO4) and water, and the solution will have a pH of 7 at the equivalent point. HC is a weak acid, and Mg(OH)2 is a weak base, so their reaction will produce a salt (MgC2O4) and water. However, since HC is a weak acid, some of it will remain undissociated at the equivalent point, resulting in an acidic solution with a pH less than 7. HNO3 is a strong acid and KOH is a strong base, so their reaction will produce a salt (KNO3) and water, and the solution will have a pH of 7 at the equivalent point. Therefore, the substance that will produce a solution with pH less than 7 at equivalent point is HC + Mg(OH)2.
Ajụjụ 17 Ripọtì
An example of oxidation - reduction enzyme is
Akọwa Nkọwa
An example of an oxidation-reduction enzyme is dehydrogenase. Dehydrogenase is an enzyme that catalyzes the transfer of electrons from one molecule to another, often using the coenzyme NAD+/NADH. During this transfer, oxidation occurs as the molecule losing electrons becomes more positively charged, while the molecule gaining electrons becomes more negatively charged. This process is known as an oxidation-reduction reaction. In contrast, amylase, protease, and lipase are enzymes that catalyze the breakdown of carbohydrates, proteins, and lipids, respectively, but they do not participate in oxidation-reduction reactions.
Ajụjụ 18 Ripọtì
Alkanals can be distinguished from alkanones by the reaction with
Akọwa Nkọwa
Alkanals and alkanones are both types of organic compounds that contain carbonyl functional groups. However, they can be distinguished from each other by their reaction with Tollens' reagent (also known as silver mirror test) or Fehling's solution. Alkanals are oxidized by Fehling's solution to form a brick-red precipitate of copper(I) oxide, while alkanones do not react with Fehling's solution. On the other hand, Tollens' reagent is used to distinguish between alkanals and alkanones by producing a silver mirror when it reacts with alkanals, but not with alkanones. Therefore, the correct answer to the question is Fehling's solution.
Ajụjụ 19 Ripọtì
\(\mathrm{CO(g) + H_2O(g) \rightarrow CO_2(g) + H_2(g)}\)
Calculate the standard heat change of the reaction above, if the standard enthalpies of formation of \(\mathrm{CO_2(g)}\), \(\mathrm{H_2O(g)}\) and \(\mathrm{CO(g)}\) and \(\mathrm{CO(g)}\) in \(\mathrm{KJ\ mol^{-1}}\) are - 394 -242 and - 110 respectively.
Akọwa Nkọwa
Standard heat change = Product - Reactant
Product = -394
Reactant = CO + H2O = -242 + (-110) = -352
Heat change = -394 - (-352) = -42kj/mol
Therefore the answer is D
Ajụjụ 20 Ripọtì
I. It has a varied composition from one place to another.
II. It constituents can be separated by physical means.
III. It contains unreactive noble gases
Which of the above shows that air is a mixture?
Akọwa Nkọwa
The characteristics of a mixture are that it has a varied composition from one place to another and its constituents can be separated by physical means. Air meets both of these criteria, as it has a different composition depending on where it is sampled from and its components (oxygen, nitrogen, carbon dioxide, etc.) can be separated by physical means such as fractional distillation or filtration. The presence of unreactive noble gases such as helium and neon in air is not necessarily an indication that air is a mixture, as pure samples of these gases can also be obtained through other means. Therefore, the options that show that air is a mixture are I and II only.
Ajụjụ 21 Ripọtì
Group VII elements are
Akọwa Nkọwa
Group VII elements, also known as halogens, are a group of non-metallic elements found in the periodic table. These elements include fluorine, chlorine, bromine, iodine, and astatine. Firstly, these elements are not monoatomic, which means they exist as diatomic molecules such as F2, Cl2, Br2, I2, and At2. Secondly, Group VII elements are known to be good oxidizing agents because they have a high tendency to gain electrons to form negative ions. This makes them reactive and capable of removing electrons from other substances, making them oxidizing agents. Thirdly, these elements are highly electronegative, not electropositive. This means they have a high attraction for electrons and can easily pull electrons away from other atoms. Lastly, they are electron acceptors, not donors. They tend to gain electrons during chemical reactions to achieve a stable electronic configuration. In summary, Group VII elements are diatomic, good oxidizing agents, highly electronegative, and electron acceptors.
Ajụjụ 22 Ripọtì
A piece of filter paper moistened with lead (ll) ethanoate solution turns black when the paper is dropped into a gas likely to be
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
An increase in the pressure exerted on gas at a constant temperature results in
Akọwa Nkọwa
An increase in the pressure exerted on gas at a constant temperature results in a decrease in volume. This is due to the fact that gas molecules are constantly in motion and colliding with each other and with the walls of their container. When the pressure is increased, the gas molecules are pushed closer together, resulting in a decrease in the volume of the gas. Conversely, if the pressure is decreased, the gas molecules have more room to move around, resulting in an increase in volume. It's important to note that the temperature of the gas remains constant in this scenario, so the average kinetic energy of the gas molecules also remains constant.
Ajụjụ 24 Ripọtì
Potassium trioxochlorate (V)has a solubility of 1.5 mol dm -3 at 45 oC.On cooling this solution to a temperature of 20 oC, the solubility was found to be 0.5 mol dm -3. What mass of KCIO 3 was crystalized out?
[K = 39, Cl = 35.5 O =16].
Akọwa Nkọwa
To calculate the mass of KCIO3 crystallized out, we can use the following formula: Mass = (initial solubility - final solubility) x volume x molar mass of KCIO3 First, let's calculate the difference in solubility: 1.5 mol dm^-3 - 0.5 mol dm^-3 = 1.0 mol dm^-3 Next, we need to convert the volume from dm^3 to liters: 1 dm^3 = 1 liter Now we can substitute the values into the formula: Mass = 1.0 mol dm^-3 x 1 liter x 122.5 g mol^-1 Mass = 122.5 g Therefore, the mass of KCIO3 crystallized out is 122.5 g. Note that the molar mass of KCIO3 can be calculated as follows: Molar mass = (1 x atomic mass of K) + (1 x atomic mass of Cl) + (3 x atomic mass of O) Molar mass = (1 x 39.1 g mol^-1) + (1 x 35.5 g mol^-1) + (3 x 16.0 g mol^-1) Molar mass = 122.5 g mol^-1 The correct answer is: 122.50g.
Ajụjụ 25 Ripọtì
A major factor coonsidered in selecting a suitable method for preparing a simple salt is its
Akọwa Nkọwa
When preparing a simple salt, various methods can be used. One of the main factors that should be considered when selecting the most suitable method is the salt's solubility in water. Solubility refers to the ability of a substance to dissolve in water. If a salt is highly soluble in water, it means that it can easily dissolve in water to form a homogeneous solution. On the other hand, if a salt is not very soluble in water, it may be difficult to dissolve it and the resulting solution may be heterogeneous or even unstable. Therefore, the solubility of a salt in water is an important factor that should be considered when selecting a suitable method for preparing a simple salt. The other options, such as crystalline form, melting point, and reactivity with dilute acids, are also important factors to consider but are not as critical as solubility in water.
Ajụjụ 26 Ripọtì
Which of the following metals is the least reactive?
Akọwa Nkọwa
The least reactive metal among the options provided is Gold (Au). Gold is a noble metal and is generally unreactive, meaning it does not readily react with other substances. This is due to its electron configuration, which gives it a stable outer shell of electrons that is difficult to disturb. In contrast, other metals like lead (Pb), tin (Sn), and mercury (Hg) are more reactive than gold and can readily react with other substances, especially acids and oxygen in the air. Therefore, gold is often used in applications where a stable and unreactive metal is required, such as in jewelry and electronic components.
Ajụjụ 27 Ripọtì
What number of moles of Cu 2+ will be deposited by 360 coulombs of electricity?
[f = 96500 C mol -1]
Akọwa Nkọwa
The question is asking us to calculate the number of moles of Cu2+ that will be deposited by 360 coulombs of electricity. We are given a constant "f" which is 96500 C mol-1. To solve this problem, we need to use Faraday's laws of electrolysis which states that the mass of substance deposited during electrolysis is proportional to the quantity of electricity passed. The formula is: Mass of substance = (Current × Time × Atomic weight) / (No. of electrons × Faraday's constant) In this question, we are not given the current, time, or atomic weight, but we are given the quantity of electricity (360 C) and Faraday's constant (96500 C mol-1). The number of electrons involved in the reaction is 2 (from Cu2+), so we can rearrange the formula to solve for the number of moles of Cu2+: Number of moles of Cu2+ = (Quantity of electricity / (No. of electrons × Faraday's constant)) Substituting the values, we get: Number of moles of Cu2+ = (360 C / (2 × 96500 C mol-1)) Number of moles of Cu2+ = 1.87 × 10-3 mol Therefore, the answer is the second option: 1.87 x 10-3 mole.
Ajụjụ 28 Ripọtì
The chemicals used to soften hard water involves the addition of
Ajụjụ 29 Ripọtì
2H2(g) + O2(g) → 2H2 O(g)
In the reaction above, what volume of hydrogen would be left over when 300 cm3 of oxygen and 1000 cm3 of hydrogen are exploded in a sealed tube?
Ajụjụ 30 Ripọtì
A hydrogen atom which has lost an electron contains
Akọwa Nkọwa
A hydrogen atom consists of one proton and one electron. When the hydrogen atom loses its electron, it becomes a positively charged ion called a proton. Therefore, the correct option is "one proton only." The proton has a positive charge and is located in the nucleus of the atom. The loss of the negatively charged electron leaves the proton with an overall positive charge, making it an ion.
Ajụjụ 31 Ripọtì
The isomers of C3H8O ARE
Ajụjụ 32 Ripọtì
A metal M displaces zinc from ZnCl, solution, solution
This shows that
Akọwa Nkọwa
The observation that metal M displaces zinc from ZnCl2 solution shows that M is more electropositive than zinc. Electropositivity is a measure of an element's tendency to lose electrons and form positive ions. In this case, M is able to displace zinc from the solution, which means that M has a greater tendency to lose electrons and form a positive ion compared to zinc. Therefore, option B, "M is more electropositive than zinc," is the correct answer.
Ajụjụ 33 Ripọtì
The saponification of an alkanoate to produce soap and alkanol involves
Akọwa Nkọwa
The saponification of an alkanoate to produce soap and alkanol involves hydrolysis. Hydrolysis is a chemical reaction in which water is used to break apart a molecule into two or more parts. In the case of saponification, an alkanoate molecule is broken down by the addition of water, resulting in the formation of soap and an alkanol. Soap is made up of the fatty acid portion of the alkanoate molecule and the alkanol is produced from the alcohol portion. The reaction is an example of a hydrolysis reaction, which is a type of chemical reaction in which water is used as a reactant to break apart a molecule into smaller parts.
Ajụjụ 34 Ripọtì
\(2\mathrm{Na_2O_{2(s)}} + 2\mathrm{H_2O_{2(1)}} \rightarrow 4\mathrm{NaOH_{(s)}} + \mathrm{O_{2(s)}}\)
The substance that is oxidized in the reaction above is
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
The solubilites of different solutes in a given solvent can be compared by
Akọwa Nkọwa
The solubility of a solute in a solvent refers to the maximum amount of solute that can dissolve in the solvent at a given temperature and pressure. To compare the solubilities of different solutes in a given solvent, it is best to plot their solubility curves on the same axes. This allows for a direct comparison of the solubility of each solute in the solvent and can reveal trends or patterns in the solubility of different solutes. Plotting the solubility curves on separate axes or on different parts of the same graph can make it more difficult to compare the solubilities of different solutes.
Ajụjụ 36 Ripọtì
Which of the following compounds will burn with a brick-red colour in a non-luminous Bunsen flame?
Akọwa Nkọwa
The question is asking which of the given compounds will burn with a brick-red color in a non-luminous Bunsen flame. The correct answer is CaCl2. When a compound is burned in a Bunsen flame, the heat causes the electrons in the compound to become excited and emit light. The color of the flame depends on the types of electrons present in the compound and the energy required to excite them. Calcium chloride (CaCl2) is known to burn with a brick-red color in a non-luminous Bunsen flame. This is because the calcium ions (Ca2+) emit light at a specific wavelength, which appears as a brick-red color. The other compounds listed (LiCl, NaCl, MgCl2) do not emit light in the same way and will not produce a brick-red color in a Bunsen flame.
Ajụjụ 37 Ripọtì
Ethanol reacts with concentrated tetraoxosulphate (V) acid at a temperature above 170 oC to form
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
How many atoms are present in 6.0g of magnesium?
[Mg = 24, NA = 6.02 x 1023 mol-1]
Akọwa Nkọwa
6.0 grams of magnesium contains 1.51 x 10^23 atoms of magnesium. This can be calculated using the following steps: 1. Determine the number of moles of magnesium present in 6.0 grams using the formula: moles = mass/molar mass 2. The molar mass of magnesium is 24 g/mol 3. So, 6.0 g / 24 g/mol = 0.25 moles of magnesium 4. Using Avogadro's number (6.02 x 10^23 atoms/mol), we can determine the number of atoms of magnesium present: 5. 0.25 moles * 6.02 x 10^23 atoms/mol = 1.51 x 10^23 atoms. Therefore, 6.0 grams of magnesium contains 1.51 x 10^23 atoms of magnesium.
Ajụjụ 39 Ripọtì
The purest from of iron which contains only about 0.1% carbon is
Akọwa Nkọwa
The purest form of iron that contains only about 0.1% carbon is called "wrought iron." Wrought iron is produced by melting iron ore in a furnace and then purifying it with iron oxide. The impurities are removed through hammering, rolling, and squeezing the hot iron into shape. Wrought iron is relatively soft and malleable, making it easy to work with for blacksmiths and metalworkers. Pig iron is an intermediate product of iron processing that contains high amounts of carbon and other impurities, and it is not suitable for use in its raw form. Cast iron is another type of iron that is produced by melting pig iron with scrap iron and adding alloys to create a stronger, more brittle product. Iron pyrite is a mineral that contains iron and sulfur but is not used for industrial purposes because it is not a suitable source of iron. Therefore, the option that refers to the purest form of iron which contains only about 0.1% carbon is "wrought iron."
Ajụjụ 40 Ripọtì
L . Evaporation. ll. Sublimation. lll. Diffusion. IV. Brownian motion.
Which of the above can correctly be listed as evidences for the particulate nature of matter?
Akọwa Nkọwa
The evidence for the particulate nature of matter is based on the fact that matter is made up of particles. - Evaporation and Sublimation are evidence for the particulate nature of matter because they involve a change in state from liquid or solid to gas, without breaking the particles themselves. This shows that the particles in a liquid or solid are not continuous, but rather have spaces between them. - Diffusion is another evidence for the particulate nature of matter. It is the movement of particles from an area of high concentration to an area of low concentration. This shows that particles are constantly in motion, and that they have kinetic energy. - Brownian motion is also an evidence for the particulate nature of matter. It is the random movement of particles suspended in a liquid or gas. This motion is caused by the collision of the particles with each other and with the molecules of the medium. Brownian motion shows that particles are constantly moving, and that they interact with each other. Therefore, the correct option is: - l, ll, lll and lV, which means that all the given processes are evidences for the particulate nature of matter.
Ajụjụ 41 Ripọtì
Carbohydrates are large molecules with the molecular formula C x (H2O)y. In which of the following pairs is x not equal to y?
Akọwa Nkọwa
Carbohydrates have the molecular formula Cx(H2O)y, which means they have x number of carbon atoms and y number of water molecules. The number of carbon atoms determines the size of the carbohydrate molecule, while the number of water molecules determines the degree of hydration. In glucose and starch, x is equal to y, so they have the same number of carbon atoms and water molecules. Maltose and sucrose are also examples of carbohydrates where x is equal to y. However, in fructose and sucrose, x is not equal to y. Fructose has the same molecular formula as glucose (C6H12O6), but its atoms are arranged differently. Sucrose, on the other hand, is made up of glucose and fructose molecules bonded together. Thus, the molecular formula for sucrose is C12H22O11, which means x is 12 and y is 11. In summary, the pair where x is not equal to y is sucrose and fructose.
Ajụjụ 42 Ripọtì
Which of the following gases has a characteristic pungent smell, turns red litmus paper blue and forms dense white fumes with hydrogen chloride gas?
Ajụjụ 43 Ripọtì
During a titration experiment, 0.05 moles of carbon (lV) oxide is liberated. What is the volume of gas liberated?
Akọwa Nkọwa
In a titration experiment, a certain amount of a substance is reacted with a known volume of a solution of known concentration until the reaction is complete. The volume of the solution required to complete the reaction is measured, and this information is used to calculate the amount of substance present in the original sample. In this particular titration experiment, 0.05 moles of carbon (IV) oxide (CO2) is liberated. The question is asking for the volume of gas liberated, which means we need to find the volume of CO2 gas that corresponds to 0.05 moles. To do this, we can use the ideal gas law, which relates the volume of a gas to its pressure, temperature, and number of moles. The ideal gas law is expressed as: PV = nRT where P is the pressure of the gas, V is its volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature of the gas. Assuming that the experiment was carried out at standard temperature and pressure (STP), which is defined as 0°C and 1 atm pressure, we can simplify the ideal gas law to: V = nRT/P At STP, the pressure (P) is 1 atm and the temperature (T) is 0°C or 273.15 K. The universal gas constant (R) is 0.08206 L atm K^-1 mol^-1. Substituting these values into the ideal gas law, we get: V = (0.05 mol) x (0.08206 L atm K^-1 mol^-1) x (273.15 K) / (1 atm) Simplifying the expression gives us: V = 1.12 L Therefore, the volume of gas liberated in this titration experiment is 1.12 dm^3 or 1.12 liters. Option D is the correct answer, which states "1.12 dm^3".
Ajụjụ 44 Ripọtì
A common characteristic between zinc and the other transition elements is the ability to
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