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Tambaya 1 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 2 Rahoto
On the basis of the electrochemical series, which of these ions will show the greater tendency to be discharged at the cathode in an electrolytic cell
Bayanin Amsa
The electrochemical series is a list of metals and ions arranged in order of their decreasing tendency to lose or gain electrons, and thus, their ability to act as reducing or oxidizing agents. The higher the position of a metal or ion in the electrochemical series, the greater its tendency to lose electrons and undergo oxidation, while the lower its position, the greater its tendency to gain electrons and undergo reduction. In an electrolytic cell, the cathode is the electrode where reduction occurs, meaning that cations (positively charged ions) are attracted and gain electrons to form neutral atoms or molecules. Based on the electrochemical series, the ion with the higher position in the series will have a greater tendency to gain electrons and be discharged at the cathode, while the ion with the lower position will have a lower tendency and may not be discharged at all. Among the given options, the electrochemical series order is: Cu2+ > Sn2+ > Fe2+ > Zn2+ Therefore, Cu2+ has the highest tendency to be discharged at the cathode and undergo reduction, while Zn2+ has the lowest tendency. So, in an electrolytic cell, Cu2+ will be discharged at the cathode, while Zn2+ may not be discharged at all, depending on the conditions of the cell.
Tambaya 3 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 4 Rahoto
2-methylprop-1-ene is a structural isomer of?
Tambaya 5 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 6 Rahoto
Which of the following roles does sodium chloride play in preparation? It
Bayanin Amsa
The role that sodium chloride (NaCl) plays in soap preparation is to separate soap from glycerol. When fats or oils are hydrolyzed with an alkali, such as sodium hydroxide (NaOH), the result is a mixture of soap and glycerol. Adding NaCl to this mixture helps to induce the precipitation of the soap, allowing it to be separated from the glycerol. This process is known as "salting out" and is used to purify the soap and remove impurities. Sodium chloride does not react with glycerol or accelerate the decomposition of fat and oil. Also, it does not convert the fatty acid to its sodium salt as this conversion is done by the alkali (such as NaOH) during the saponification process.
Tambaya 7 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 8 Rahoto
Bayanin Amsa
The addition of charcoal to the filter bed of sand during water treatment for township supply is to remove odors and improve the taste of the water. Charcoal is a porous material that can adsorb impurities and chemicals from the water, such as dissolved organic matter that can contribute to unpleasant tastes and odors. This process helps to produce a better-quality drinking water that is free from unpleasant tastes and odors. It should be noted that while the addition of charcoal can help remove impurities, it does not kill germs or prevent tooth decay or goiter. Other water treatment methods, such as disinfection with chlorine or ultraviolet light, are required to kill harmful microorganisms and ensure the safety of the drinking water.
Tambaya 9 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 10 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 11 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 12 Rahoto
H+ + OH− → H2 O
The equation above illustrates
Tambaya 13 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 14 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 15 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 16 Rahoto
Wrought iron is obtained by heating cast iron in a furnace with?
Bayanin Amsa
Wrought iron is a type of iron that is very malleable and ductile, meaning it can be easily shaped and formed into various objects. It is obtained by heating cast iron in a furnace with haematite, also known as iron(III) oxide. When cast iron is heated with haematite in a furnace, a chemical reaction takes place where the haematite reacts with the carbon in the cast iron to produce carbon dioxide gas. This reaction also produces molten iron, which is then further heated to remove any impurities like sulfur and phosphorus. This molten iron is then poured into molds to form ingots of wrought iron. Therefore, haematite is essential in the process of obtaining wrought iron from cast iron.
Tambaya 18 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 19 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 20 Rahoto
A metal which can be used as sacrificial anode for preventing corrosion of length of iron pipe is
Tambaya 21 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 22 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 23 Rahoto
Which of the following does NOT contain aluminum as a component?
Tambaya 24 Rahoto
When heat is absorbed during a chemical reaction, the reaction is said to be
Bayanin Amsa
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.
Tambaya 26 Rahoto
A chemical widely used as a fertilizer is?
Bayanin Amsa
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.
Tambaya 27 Rahoto
6g of Mg was to 100cm3 of 1 moldm3 H2 SO4 . What mass of Mg remained undissolved? (Mg = 24)
Bayanin Amsa
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.
Tambaya 28 Rahoto
A certain liquid has a high boiling point. It is viscous, non-toxic, and miscible with water to be hygroscopic; this liquid most likely to be
Bayanin Amsa
The liquid is most likely to be option number 4: CH3OHCHOH2OH, which is also known as glycerol or glycerin. Glycerol has a high boiling point of 290°C, which is much higher than the boiling points of the other options. It is also a viscous liquid, which means it is thick and sticky. Glycerol is non-toxic, and it is often used in food, pharmaceuticals, and cosmetics. Furthermore, glycerol is miscible with water, which means that it can be easily mixed with water to form a homogeneous solution. It is also hygroscopic, which means that it can absorb water from the air. These properties make glycerol a useful substance in many applications, such as as a moisturizer in skincare products or as a humectant in food processing.
Tambaya 29 Rahoto
An organic compound which liberate carbon(iv)oxide from trioxocarbonate(iv) solution is likely to be?
Bayanin Amsa
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.
Tambaya 30 Rahoto
Which of the following substances is not a homogeneous mixture?
Bayanin Amsa
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.
Tambaya 31 Rahoto
A quantity of air passed through a weighted amount of alkaline pyrogallol. An increase in the weight of the pyrogallol would result from the absorption of
Bayanin Amsa
When air is passed through alkaline pyrogallol, the oxygen in the air is absorbed by the pyrogallol, resulting in an increase in the weight of the pyrogallol. The other gases in air, namely nitrogen, neon, and argon, do not react with pyrogallol under these conditions. Therefore, the answer is oxygen.
Tambaya 32 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 34 Rahoto
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
Bayanin Amsa
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
Tambaya 35 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 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
| 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 38 Rahoto
Which of the following constitutes a mixture? I. Petroleum II. Rubber latex III. Vulcanizer’s solution IV. Carbon (iv) sulphide
Bayanin Amsa
Tambaya 39 Rahoto
Complete dehydrogenation of ethyne yields
Tambaya 40 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.
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