Ana loda....
Latsa & Riƙe don Ja Shi Gabaɗaya |
|||
Danna nan don rufewa |
Tambaya 1 Rahoto
Which of the following statement is TRUE of the complete hydrolysis of a glyceride by sodium hydroxide?
Bayanin Amsa
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.
Tambaya 2 Rahoto
Which of the following will act as both oxidizing agents and reducing agents?
Bayanin Amsa
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.
Tambaya 3 Rahoto
Addition of sodium chloride to water to form a solution would lead to?
Bayanin Amsa
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.
Tambaya 4 Rahoto
Bayanin Amsa
Carbon dioxide (CO2) has a linear molecular geometry, with two oxygen atoms bonded to the central carbon atom. Each bond between carbon and oxygen is a double bond, consisting of two pairs of electrons shared between the atoms. Therefore, there are two bonding pairs in each of the carbon-oxygen double bonds, giving a total of four bonding pairs in CO2. The answer is 4.
Tambaya 5 Rahoto
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?
Bayanin Amsa
Tambaya 6 Rahoto
An organic functional group which can likely decolorize ammoniacal silver nitrate is?
Bayanin Amsa
The organic functional group that can likely decolorize ammoniacal silver nitrate is an alkyne. When ammoniacal silver nitrate is added to a solution containing an alkyne functional group, a white or yellowish precipitate of silver acetylide is formed. Silver acetylide is a highly explosive compound and is sparingly soluble in water, causing it to appear as a white or yellowish solid precipitate. This reaction is used as a test to detect the presence of an alkyne functional group in an organic compound. In contrast, alkanes, alkenes, and alkanols do not react with ammoniacal silver nitrate, so they cannot decolorize it. Therefore, an organic functional group that can likely decolorize ammoniacal silver nitrate is an alkyne.
Tambaya 7 Rahoto
Calcium forms complexes with ammonia because
Bayanin Amsa
The reason why calcium forms complexes with ammonia is that it has empty d-orbitals.
Tambaya 8 Rahoto
The reactions below represent neutralization reaction, in which of them is the value of ΔH highest?
Bayanin Amsa
The reaction with the highest ΔH (change in enthalpy) would be the reaction between HCL and NaOH, which forms NaCL and H2O. This is because the formation of water releases energy in the form of heat, which is reflected in the positive ΔH value for this reaction. When an acid and a base react, they neutralize each other and form a salt and water, with the release of heat being a sign of an exothermic reaction.
Tambaya 9 Rahoto
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
Tambaya 10 Rahoto
Zn + 2HCL → ZnCl2 + H2
What happens to zinc in the above reaction?
Bayanin Amsa
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."
Tambaya 11 Rahoto
Using the metal activity series, the metal that can liberate hydrogen gas from steam is?
Bayanin Amsa
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.
Tambaya 12 Rahoto
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
Tambaya 13 Rahoto
How many neutrons are present in atom with mass number and atomic number 37 and 17 respectively?
Bayanin Amsa
The atomic number of an atom represents the number of protons in the nucleus of the atom. Since the atomic number given is 17, it means that there are 17 protons in the nucleus. The mass number of an atom represents the total number of protons and neutrons present in the nucleus. Therefore, if the mass number is given as 37, it means that the total number of protons and neutrons in the nucleus is 37. To determine the number of neutrons in the nucleus, we can subtract the atomic number (which represents the number of protons) from the mass number (which represents the total number of protons and neutrons). Thus, the number of neutrons in the atom with a mass number of 37 and an atomic number of 17 is: Number of neutrons = Mass number - Atomic number = 37 - 17 = 20 Therefore, the answer is 20.
Tambaya 15 Rahoto
Chlorine is a common bleaching agent. This is not true with
Bayanin Amsa
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.
Tambaya 16 Rahoto
In the preparation of salts, the method employed will depend on the?
Bayanin Amsa
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.
Tambaya 17 Rahoto
The IUPAC nomenclature of the structure is
Bayanin Amsa
The IUPAC nomenclature of the structure is "2-chloro-2-methylbutane". The name is derived by first identifying the longest carbon chain, which in this case contains four carbon atoms (butane). The carbon chain is numbered from one end to the other, giving the substituents the lowest possible numbers. Starting from either end, we can see that the first carbon atom has a chlorine atom attached to it, which is represented by the prefix "chloro-". Moving along the chain, the second carbon atom has a methyl group attached to it, which is represented by the prefix "methyl-". Since the substituents are in the second position from each other, we use the prefix "di-" to indicate two substituents in this position. Finally, we use the suffix "-ane" to indicate that the molecule is an alkane. Therefore, the correct name for this molecule is "2-chloro-2-methylbutane".
Tambaya 18 Rahoto
In the extraction of iron, hot air is introduced into the blast furnace through?
Bayanin Amsa
In the extraction of iron, hot air is introduced into the blast furnace through tuyeres. Tuyeres are nozzles that are located at the bottom of the blast furnace, and they are used to blow hot air into the furnace. The hot air helps to burn the coke (a fuel made from coal) which provides the heat needed to melt the iron ore. The air also helps to remove the waste gases that are produced during the reaction, allowing the iron to be extracted more efficiently.
Tambaya 20 Rahoto
A piece of radioactive element has initially 8.0×10^22 atoms. The half life of two days after 16 days the number of atom is
Tambaya 21 Rahoto
Which of the following will precipitate in dil. HCl
Bayanin Amsa
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.
Tambaya 22 Rahoto
The sulphide that is commonly used in coating electric fluorescent tubes is?
Bayanin Amsa
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.
Tambaya 24 Rahoto
Alkanes are used mainly?
Tambaya 25 Rahoto
The IUPAC name for CICH2-CH2-CH2-OH is
Bayanin Amsa
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.
Tambaya 26 Rahoto
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.
Tambaya 27 Rahoto
A certain volume of gas at 298k is heated such that its volume and pressure are now four times the original values. What is the new temperature?
Bayanin Amsa
We can use the ideal gas law to solve this problem, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in kelvin. If the volume and pressure are both increased by a factor of 4, then the new volume V' and new pressure P' are given by: V' = 4V P' = 4P Substituting these values into the ideal gas law, we get: (4P)(4V) = nR(T') Simplifying this equation, we get: 16PV = nRT' Dividing both sides by PV, we get: 16 = nRT' / PV Since n, R, and P are constant, we can simplify this to: 16 = T' / T Solving for T', we get: T' = 16T Therefore, the new temperature is 16 times the original temperature. Substituting T = 298 K, we get: T' = 16 x 298 K = 4768 K So the correct answer is 4768.0K.
Tambaya 28 Rahoto
Which of the following is stable to heat
Bayanin Amsa
Out of the given options, K2CO3 is stable to heat.
Tambaya 29 Rahoto
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.
Bayanin Amsa
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:
Tambaya 30 Rahoto
A metal which can be used as sacrificial anode for preventing corrosion of length of iron pipe is
Tambaya 31 Rahoto
There is a large temperature interval between the melting point and the boiling point of metal because:
Bayanin Amsa
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.
Tambaya 32 Rahoto
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?
Bayanin Amsa
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.
Tambaya 33 Rahoto
A colored gas that is known to be poisonous and can readily damage the mucous lining of the lungs is?
Bayanin Amsa
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.
Tambaya 34 Rahoto
SO2 + O2 → 2SO3
In the reaction above, the most suitable catalyst is?
Bayanin Amsa
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.
Tambaya 35 Rahoto
Which of the following is used as a moderator to control nuclear fission?
Bayanin Amsa
Heavy water (D2O) is used as a moderator to control nuclear fission. A moderator is a substance that is used to slow down the neutrons produced in a nuclear reaction, making them more likely to be captured by the fuel nuclei and causing further fission. Heavy water is a type of water that contains a larger amount of the isotope deuterium (D) than regular water. Deuterium has an extra neutron compared to the more common hydrogen isotope, and this makes heavy water more effective at slowing down neutrons than regular water. Lead, iron, and chromium are not typically used as moderators in nuclear reactors. Lead can be used as a shield to absorb radiation, while iron and chromium are used in the construction of the reactor vessel and other components.
Tambaya 36 Rahoto
One of the active components of baking powder is
Bayanin Amsa
The active component of baking powder is sodium bicarbonate (NaHCO3). It is responsible for the leavening or rising of baked goods by releasing carbon dioxide gas when it reacts with an acid. Other ingredients in baking powder, such as monocalcium phosphate and sodium aluminum sulfate, provide the acid component for the reaction to occur. Magnesium sulfate (MgSO4) and calcium sulfate (CaSO4) are not typically used in baking powder, and sodium chloride (NaCl) is simply table salt and not an active ingredient in leavening.
Tambaya 37 Rahoto
Which of the following increases as boiling water changes to steam?
Bayanin Amsa
The degree of disorder of the system increases as boiling water changes to steam. When water is boiled and changes to steam, the water molecules gain energy and become more disordered, which means that the molecules move more rapidly and the entropy of the system increases. The temperature of the system also increases during this process, but the degree of disorder is the factor that specifically increases as the water changes to steam. The number of molecules and activation energy remain constant during this phase transition.
Tambaya 38 Rahoto
The following non-metal form acidic oxides with oxygen except?
Bayanin Amsa
An acidic oxide is an oxide that reacts with water to form an acidic solution. Non-metals have a greater tendency to form acidic oxides than metals. Therefore, among the given options, the non-metal that does not form an acidic oxide with oxygen would be the one that does not react with water to form an acidic solution. Out of the given options, chlorine is the non-metal that does not form acidic oxides with oxygen. Chlorine reacts with oxygen to form a number of oxides such as chlorine monoxide (Cl2O), chlorine dioxide (ClO2), and chlorine trioxide (ClO3), but none of these oxides react with water to form an acidic solution. Instead, they react with water to form oxyacids or oxoacids such as hypochlorous acid (HClO), chlorous acid (HClO2), and chloric acid (HClO3), which are stronger acids than the oxides. Therefore, the correct answer is chlorine.
Tambaya 39 Rahoto
An organic compound with fishy smell is likely to have a general formula?
Bayanin Amsa
The organic compound with a fishy smell is most likely to have the general formula RNH2, which represents a primary amine. Amines are organic compounds that contain a nitrogen atom bonded to one or more carbon atoms. Primary amines have one alkyl or aryl group and two hydrogen atoms bonded to the nitrogen atom. Some primary amines have a fishy smell, which is caused by the presence of volatile amines. These amines are small molecules that can easily evaporate and have a strong odor, similar to that of fish. Examples of compounds that have a fishy smell include trimethylamine, which is found in fish, and butylamine, which is used in the production of rubber and pharmaceuticals. In summary, the organic compound with a fishy smell is likely to have the general formula RNH2, which represents a primary amine.
Tambaya 40 Rahoto
Ethene, when passed into concentrated H2SO4, is rapidly absorbed. The product is diluted with water and then warmed to produce
Bayanin Amsa
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.
Za ka so ka ci gaba da wannan aikin?