Question 1

For an isothermal reversible expansion of 2 moles of an ideal gas from 10 dm³ to 100 dm³ at 27°C (300 K), what is the entropy change ΔS of the gas (in J K⁻¹)?

Accepted Answer

Introduction: This chemistry and thermodynamics question asks you to calculate the entropy change for an isothermal (constant temperature) reversible expansion of an ideal gas. Entropy calculations are common in physical chemistry and chemical thermodynamics, especially for ideal gas processes. Given Data / Assumptions: Number of moles n = 2 mol. Initial volume V1 = 10 dm³. Final volume V2 = 100 dm³. Temperature T = 27°C = 300 K. Gas behaves ideally and the process is isothermal and reversible. Universal gas constant R ≈ 8.314 J mol⁻¹ K⁻¹. Concept / Approach: For an isothermal reversible expansion of an ideal gas, the entropy change of the gas is given by: ΔS = n * R * ln(V2 / V1)where n is the number of moles, R is the gas constant, and V2/V1 is the ratio of final to initial volume. Because the process is reversible and isothermal, only this simple expression is needed. We then compute ΔS numerically and match it with the closest option. Step-by-Step Solution: Step 1: Compute the volume ratio V2 / V1 = 100 / 10 = 10.Step 2: Use the formula ΔS = n * R * ln(V2 / V1).Step 3: Substitute values: ΔS = 2 * 8.314 * ln(10).Step 4: Use ln(10) ≈ 2.3026.Step 5: First multiply 2 * 8.314 = 16.628.Step 6: Now multiply by ln(10): 16.628 * 2.3026 ≈ 38.3 J K⁻¹.Step 7: Therefore, ΔS ≈ 38.3 J K⁻¹ for the gas. Verification / Alternative check: We can confirm the magnitude approximately: R is about 8.3, n is 2, so n*R ≈ 16.6. Since ln(10) is about 2.3, the result should be near 16.6 * 2.3 ≈ 38.2. This rough estimate is consistent with the more precise calculation of 38.3 J K⁻¹. Why Other Options Are Wrong: 42.3 J K⁻¹ and 35.8 J K⁻¹ are somewhat close but do not match the correct product of 16.628 and 2.3026. 32.3 J K⁻¹ and 30.0 J K⁻¹ are noticeably smaller and would correspond to a smaller volume ratio or fewer moles. Only 38.3 J K⁻¹ aligns with the correct calculation from the thermodynamic formula. Common Pitfalls: Typical mistakes include using log base 10 instead of natural log, forgetting to convert temperature to Kelvin (although T does not appear explicitly in this formula for ΔS of an ideal gas in an isothermal process), or using V1/V2 instead of V2/V1 in the logarithm. Always use ln(V2/V1) and ensure that volumes are in the same units so their ratio is dimensionless. Final Answer: The entropy change for the isothermal reversible expansion is approximately 38.3 J K⁻¹.

Question 2

In basic chemistry and metallurgy, which of the following is considered the purest commercial form of iron used in engineering applications?

Accepted Answer

Introduction / Context: Different forms of iron such as wrought iron, steel, and pig iron vary in their carbon content and impurity levels. This is a standard topic in general science and basic metallurgy. Many school-level and competitive exam questions ask which form of iron is the purest in the commercial sense, meaning it has the lowest proportion of carbon and other impurities while still being used in practical applications. This question checks whether you can identify wrought iron as that purest form. Given Data / Assumptions: - The question is about the purest commercial form of iron, not about laboratory purity. - Options list wrought iron, steel, pig iron, and nickel steel. - We assume standard definitions of these materials from chemistry and metallurgy textbooks. - Purest means the lowest amount of carbon and other impurities under practical industrial conditions. Concept / Approach: Pig iron contains a high percentage of carbon and various impurities, so it is considered the crude form of iron straight from the blast furnace. Steel has less carbon than pig iron but more than wrought iron and may contain several alloying elements. Wrought iron has minimal carbon and impurities and is thus known as the purest commercial form of iron. Nickel steel is an alloy where nickel is deliberately added, making it less pure in terms of iron content. The correct approach is to choose the option that represents iron with minimal impurities, which is wrought iron. Step-by-Step Solution: Step 1: Recall that pig iron directly obtained from the blast furnace has high carbon content (around 3 to 4 percent) and many impurities like silicon, phosphorus, and sulphur, so it is far from pure. Step 2: Remember that steel is an alloy of iron with carbon and sometimes other elements; while it is cleaner than pig iron, it still contains controlled amounts of carbon and alloys. Step 3: Wrought iron is produced by further refining, resulting in very low carbon content (typically less than 0.1 percent) and removal of most impurities. Step 4: Nickel steel deliberately adds nickel to steel to improve strength and toughness, which means it is even less pure as iron in a chemical sense. Step 5: Therefore, among the listed options, wrought iron is recognised as the purest form of commercial iron. Verification / Alternative check: Chemistry and engineering textbooks commonly state that wrought iron is the purest form of commercial iron, emphasising its very low carbon and impurity content. Its structure contains slag in fibrous form but chemically it is closer to pure iron than steel or pig iron. Wrought iron has been used historically for decorative work, chains, and certain structural uses where ductility and corrosion resistance are important, further reinforcing its identity as relatively pure iron. Why Other Options Are Wrong: Steel, although widely used and cleaner than pig iron, is still an alloy with significant carbon and sometimes other metals, so it is not the purest. Pig iron has the highest carbon and impurity content and is most definitely a crude form, not a pure one. Nickel steel includes nickel as a deliberate alloying element to improve properties, again reducing iron purity. These materials may have superior mechanical properties for specific uses, but they are all less pure than wrought iron in terms of iron content. Common Pitfalls: Students sometimes mistakenly pick steel because it is widely used and considered higher quality in many contexts. Others confuse the idea of strength with purity and think stronger alloys must be purer. It is important to separate mechanical quality from chemical purity. In exams, when asked about the purest commercial form of iron, you should remember the phrase wrought iron and associate it with minimal carbon and low impurities. Final Answer: The purest commercial form of iron is wrought iron, which contains very little carbon and impurities.

Question 3

In nuclear chemistry and physics, a hydrogen bomb (thermonuclear bomb) is based primarily on which fundamental principle?

Accepted Answer

Introduction / Context: Nuclear weapons and energy involve two key processes: nuclear fission and nuclear fusion. Understanding the difference between them is a common requirement in general science and competitive exams. This question asks specifically about the hydrogen bomb, also known as a thermonuclear bomb, and which fundamental nuclear process it uses. Recognising that a hydrogen bomb is based on nuclear fusion of light nuclei, rather than on fission alone, is crucial. Given Data / Assumptions: - The device in question is a hydrogen bomb, also called a thermonuclear bomb. - Several nuclear processes are listed: nuclear fusion, nuclear fission, natural radioactivity, and artificial radioactivity. - We assume standard textbook definitions and descriptions of nuclear weapons. - The task is to identify the principle primarily responsible for the explosive energy of a hydrogen bomb. Concept / Approach: An atomic bomb based on uranium or plutonium uses nuclear fission, where heavy nuclei split into lighter fragments, releasing energy. A hydrogen bomb, in contrast, uses extremely high temperatures and pressures to fuse light nuclei such as isotopes of hydrogen (for example, deuterium and tritium) into heavier nuclei, releasing enormous energy. While a fission device is often used as a trigger to create the conditions for fusion, the defining principle of a hydrogen bomb is nuclear fusion. Therefore, the correct approach is to look for the option that mentions fusion of light nuclei. Step-by-Step Solution: Step 1: Recall that nuclear fusion is the process where two light atomic nuclei combine to form a heavier nucleus, with a loss of mass and a large release of energy. Step 2: Remember that in a hydrogen bomb, isotopes of hydrogen are fused under extreme conditions, which is why it is called a thermonuclear device. Step 3: Check option a, which states nuclear fusion of light nuclei into heavier nuclei. This matches the definition and the known working principle of a hydrogen bomb. Step 4: Option b, nuclear fission, correctly describes the principle of an atomic (fission) bomb, not a thermonuclear hydrogen bomb, so it is not the best answer here. Step 5: Options c and d refer to radioactivity in general, whether natural or artificially induced, which are related but not the defining process responsible for the hydrogen bomb's massive energy output. Step 6: Therefore, select option a as the correct answer. Verification / Alternative check: You can verify this by recalling that the Sun and other stars generate their energy mainly through nuclear fusion of hydrogen into helium. Hydrogen bombs attempt to mimic this stellar process on a very small and destructive scale, using a fission bomb as an initial trigger. Physics and chemistry textbooks consistently describe hydrogen bombs as thermonuclear fusion devices, not as pure fission or simple radioactive sources, confirming that nuclear fusion is the governing principle. Why Other Options Are Wrong: Option b, nuclear fission, is the correct principle behind atomic bombs such as those using uranium-235 or plutonium-239. Hydrogen bombs may use a fission stage to achieve the required conditions, but the main destructive energy comes from the subsequent fusion stage. Option c, natural radioactivity, refers to spontaneous decay of unstable nuclei and is not sufficient to explain the explosive yield of a hydrogen bomb. Option d, artificial radioactivity, involves creating radioactive isotopes through reactions, which is again not the central principle behind thermonuclear explosions. Thus, none of these options accurately captures the defining mechanism of a hydrogen bomb. Common Pitfalls: Students often confuse atomic bombs and hydrogen bombs and may pick nuclear fission out of habit. Another pitfall is to treat any nuclear process as equivalent and not distinguish between fission, fusion, and radioactivity. To avoid such mistakes, remember that atomic bombs are fission-based, while hydrogen bombs are fusion-based. The word hydrogen in the name should remind you that isotopes of hydrogen are being fused in the device. Final Answer: A hydrogen bomb is based on the principle of nuclear fusion of light nuclei into heavier nuclei.

Question 4

In inorganic chemistry, what is the correct molecular geometry (shape) of iodine heptafluoride, IF7, according to the VSEPR model?

Accepted Answer

Introduction / Context: Molecular geometry is an important topic in inorganic chemistry and is often analysed using the Valence Shell Electron Pair Repulsion (VSEPR) model. Iodine heptafluoride, IF7, is a classic example of a hypervalent molecule in which the central atom has more than eight electrons in its valence shell. This question asks you to recall or deduce the correct geometry of IF7, which is a commonly tested example in exams dealing with molecular shapes. Given Data / Assumptions: - The central atom in the molecule is iodine (I). - There are seven fluorine (F) atoms bonded to iodine, giving the formula IF7. - Options include trigonal bipyramidal, pentagonal bipyramidal, square pyramidal, and tetrahedral geometries. - We assume the use of the VSEPR model for predicting molecular shape. Concept / Approach: According to VSEPR theory, the shape of a molecule is determined by the repulsion between electron pairs around the central atom. In IF7, iodine uses an expanded valence shell to accommodate seven bonding pairs with no lone pairs. The arrangement that minimises repulsion among seven bonding pairs is a pentagonal bipyramidal geometry, where five fluorine atoms occupy the vertices of a pentagon in one plane and two fluorine atoms occupy axial positions above and below this plane. The correct approach is to count the electron domains and match them to the known geometries for seven domains. Step-by-Step Solution: Step 1: Determine the total number of electron pairs around the central iodine atom in IF7. There are seven I–F bonds and, in this structure, no lone pairs on iodine. Step 2: Recognise that seven bonding pairs with no lone pairs correspond to a coordination number of seven for the central atom. Step 3: Recall the standard VSEPR shapes: 5 electron pairs give trigonal bipyramidal, 6 give octahedral, and 7 give pentagonal bipyramidal geometry when there are no lone pairs. Step 4: Compare each option. Option a explicitly mentions pentagonal bipyramidal geometry with five fluorine atoms in a plane and two axial, which matches the VSEPR prediction. Step 5: Option b, trigonal bipyramidal, is appropriate for five bonding pairs, not seven, so it cannot describe IF7 correctly. Step 6: Option c, square pyramidal, is associated with five bonding pairs and one lone pair, such as in BrF5, not with seven bonds and no lone pairs. Step 7: Option d, tetrahedral, only accounts for four bonding pairs and is obviously not suitable for a heptafluoride. Verification / Alternative check: Textbooks in coordination chemistry and main group chemistry list IF7 as a classic pentagonal bipyramidal molecule. Structural studies support this geometry, showing five fluorine atoms arranged in a regular pentagon around iodine, with two more fluorine atoms aligned axially. This arrangement gives each bond approximately equal repulsion and stabilises the molecule. No alternative geometry among the options can accommodate seven equivalent bond pairs with minimal repulsion as effectively as the pentagonal bipyramidal shape. Why Other Options Are Wrong: Trigonal bipyramidal geometry is associated with five electron pairs, not seven, so option b is inconsistent with the formula IF7. Square pyramidal geometry normally corresponds to six electron domains (five bonds and one lone pair) and fits compounds like BrF5, not IF7. Tetrahedral geometry is limited to four bonding pairs and cannot describe a seven-coordinate molecule. Therefore, options b, c, and d do not match the coordination number and electron pair arrangement present in IF7. Common Pitfalls: Students sometimes incorrectly generalise that any molecule with more than four bonds must be trigonal bipyramidal or octahedral and forget that coordination numbers greater than six are possible for heavier elements. Another pitfall is mixing up examples: BrF5 is square pyramidal, SF6 is octahedral, and IF7 is pentagonal bipyramidal. Memorising a small set of standard examples with their shapes, and relating them back to the number of bonding pairs and lone pairs, helps avoid confusion in exam situations. Final Answer: The correct molecular geometry of IF7 is pentagonal bipyramidal, with five fluorine atoms in a plane and two in axial positions.

Question 5

In general chemistry, which of the following is a non-metal that exists as a liquid at room temperature under normal atmospheric pressure?

Accepted Answer

Introduction / Context: Most elements are solids at room temperature, with some being gases and very few being liquids. Among the non-metals, only a very small number are liquid at ordinary room temperature and normal atmospheric pressure. This question tests whether you can correctly identify bromine as the non-metal that is a liquid under these conditions, distinguishing it from other common non-metals like chlorine, phosphorus, and helium. Given Data / Assumptions: - We focus on non-metallic elements and their physical state at room temperature. - Room temperature is assumed to be about 25 degrees Celsius under normal atmospheric pressure. - Options include chlorine, phosphorus, bromine, and helium. - We assume standard textbook information on physical states of elements. Concept / Approach: At room temperature, chlorine (Cl2) is a greenish-yellow gas, phosphorus in its common forms (such as white and red phosphorus) is a solid, and helium (He) is a colourless noble gas. Bromine (Br2), however, is a reddish-brown liquid and is one of the very few elements that exist as a liquid at room temperature, along with mercury (a metal). The approach is to recall the physical states of these elements and select the one that is a liquid non-metal under normal conditions. Step-by-Step Solution: Step 1: Recall that chlorine is a halogen gas at room temperature and is stored in pressurised cylinders due to its gaseous nature. Step 2: Remember that phosphorus exists in forms like white, red, and black phosphorus, all of which are solids at room temperature. Step 3: Bromine is another halogen, but unlike chlorine and fluorine, it is a volatile liquid at room temperature, giving off reddish-brown fumes. Step 4: Helium is a noble gas that remains gaseous at room temperature and is used in balloons and as a cooling medium at very low temperatures. Step 5: Among all four options, only bromine matches the description of a non-metal that is liquid at room temperature. Verification / Alternative check: You can verify this fact from the periodic table and standard chemistry references, which note that bromine is one of the only elements that is liquid at room temperature, the other common example being mercury, which is a metal. Bromine's melting point is just below room temperature and its boiling point is above, so it remains liquid in the typical classroom or laboratory environment. Routine safety instructions also emphasise that bromine is a corrosive liquid, reinforcing its liquid state under normal conditions. Why Other Options Are Wrong: Chlorine is a gas at room temperature and standard pressure, so option a is incorrect. Phosphorus, in its various allotropes, is a solid and must be heated significantly to become liquid, so option b is also wrong. Helium is a noble gas and remains a gas at room temperature, becoming liquid only at extremely low temperatures near absolute zero, so option d does not satisfy the condition of being liquid at ordinary room temperature. Only bromine satisfies both conditions of being a non-metal and liquid at room temperature. Common Pitfalls: A frequent mistake is to confuse bromine with chlorine or to think that all halogens must be gases at room temperature. Another pitfall is to overlook that helium, although often discussed in the context of low-temperature liquids, is not liquid at normal room conditions. To avoid errors, remember the specific list: mercury (a metal) and bromine (a non-metal) are the standard textbook examples of elements that are liquid at room temperature. Final Answer: The non-metal that remains liquid at room temperature is bromine, a reddish-brown liquid.

Question 6

When formaldehyde (methanal) is heated with potassium hydroxide (KOH), which of the following organic products is formed as a result of the Cannizzaro reaction?

Accepted Answer

Introduction / Context: The reaction of aldehydes without alpha hydrogen, such as formaldehyde, with a strong base like potassium hydroxide is a classic example of the Cannizzaro reaction in organic chemistry. In this disproportionation reaction, one molecule of the aldehyde is oxidised to a carboxylate salt and another is reduced to an alcohol. This question asks you to identify the main organic product obtained when formaldehyde is heated with potassium hydroxide, which is a common textbook example. Given Data / Assumptions: - The aldehyde used is formaldehyde (also called methanal, HCHO). - The base used is potassium hydroxide (KOH). - Heating is involved, suggesting a base-catalysed reaction. - Options include methyl alcohol, acetylene, methane, and ethyl formate. - We assume the standard Cannizzaro reaction mechanism for aldehydes lacking alpha hydrogen. Concept / Approach: In the Cannizzaro reaction, two molecules of an aldehyde without alpha hydrogen react in the presence of a concentrated base. One aldehyde molecule is reduced to an alcohol, while the other is oxidised to a carboxylate salt. For formaldehyde, this means that one molecule is reduced to methanol (methyl alcohol), and another is oxidised to potassium formate. The correct approach is to recognise formaldehyde as a Cannizzaro-active aldehyde and identify methanol as one of the principal organic products. Step-by-Step Solution: Step 1: Write a simplified equation for the Cannizzaro reaction of formaldehyde: 2 HCHO + 2 KOH → CH3OH + HCOOK. Step 2: In this equation, CH3OH is methyl alcohol (methanol), and HCOOK is potassium formate, the salt of formic acid. Step 3: Among the options, only option a explicitly mentions methyl alcohol formed along with a salt of formic acid, which matches the known reaction. Step 4: Option b, acetylene, is an alkyne and is not produced in this base-induced disproportionation. Step 5: Option c, methane, would require a different type of reduction and is not the product of the Cannizzaro reaction of formaldehyde. Step 6: Option d, ethyl formate, is an ester and does not arise from simply heating formaldehyde with KOH. Step 7: Therefore, the correct answer is methyl alcohol along with a salt of formic acid (potassium formate). Verification / Alternative check: Organic chemistry textbooks consistently list formaldehyde as a key example of an aldehyde that undergoes the Cannizzaro reaction when treated with concentrated alkali. The reaction products are always described as methanol and a formate salt. This pattern also fits the general rule that in Cannizzaro reactions, the aldehyde is simultaneously oxidised and reduced: one molecule gives an acid salt, and another gives an alcohol, with no unsaturated hydrocarbons or esters directly produced under these simple conditions. Why Other Options Are Wrong: Acetylene (option b) is produced by quite different reactions, such as the hydrolysis of calcium carbide, and is not a product of base-catalysed disproportionation of aldehydes. Methane (option c) would require hydrogenation or decarboxylation mechanisms that are not relevant here. Ethyl formate (option d) is an ester of formic acid and ethanol and is formed by esterification, not by the Cannizzaro reaction of formaldehyde. Hence, these options do not match the known reaction pathway and products. Common Pitfalls: Students sometimes confuse the Cannizzaro reaction with other organic reactions and may guess any common organic molecule. Another typical error is to forget that aldehydes without alpha hydrogen, such as formaldehyde and benzaldehyde, undergo disproportionation rather than aldol condensation. Remembering the simple balanced equation and the pattern aldehyde → alcohol + salt of acid will help you quickly identify methyl alcohol as the main organic product here. Final Answer: When formaldehyde is heated with KOH, the main organic product is methyl alcohol (methanol), formed along with a salt of formic acid via the Cannizzaro reaction.

Question 7

In modern physics, which scientist clearly stated and formulated the principle that matter can be converted into energy and energy into matter?

Accepted Answer

Introduction / Context: One of the most famous ideas in modern physics is that matter and energy are interchangeable, captured in the equation E = m * c^2. This relation shows that a small amount of mass can be converted into a large amount of energy and vice versa. The question asks which scientist is associated with clearly stating and formulating this principle in the form of a precise theory, which is a well-known fact in general science and competitive exams. Given Data / Assumptions: - The principle in question is the equivalence of mass and energy: matter can be converted into energy. - Scientists listed include Boyle, Lavoisier, Avogadro, and Einstein. - We assume standard historical attributions from physics and chemistry. - We must connect the principle with the correct scientist. Concept / Approach: Albert Einstein, through his special theory of relativity, derived the mass-energy equivalence formula E = m * c^2, where E is energy, m is mass, and c is the speed of light in vacuum. This formula mathematically expresses the possibility of converting matter into energy and explains phenomena like nuclear reactions. Boyle is associated with gas laws, Lavoisier with the law of conservation of mass in chemical reactions, and Avogadro with the concept of the mole and volume relationships of gases. The correct approach is to link the concept of mass-energy equivalence to Einstein. Step-by-Step Solution: Step 1: Recall that Albert Einstein developed the special theory of relativity in 1905, which led to the equation E = m * c^2. Step 2: Understand that this equation means mass can be viewed as a form of energy; thus matter can be converted into energy. Step 3: Recognise that Robert Boyle is best known for Boyle's law relating pressure and volume of a gas, not for mass-energy equivalence. Step 4: Recognise that Antoine Lavoisier established the law of conservation of mass in chemical reactions, stating that mass is conserved, but did not introduce mass-energy equivalence. Step 5: Recognise that Amedeo Avogadro proposed Avogadro's hypothesis about equal volumes of gases containing equal numbers of molecules, which is unrelated to matter-energy conversion. Step 6: Therefore, only Albert Einstein matches the principle that matter can be converted into energy and vice versa. Verification / Alternative check: Physics textbooks and even general knowledge sources frequently mention Einstein's equation E = m * c^2 as one of the most famous scientific formulae in history. It is used to explain the enormous energy released in nuclear fission and fusion reactions and underlies the working of atomic and hydrogen bombs. These applications are direct examples of matter being converted into energy. No other scientist among the options is credited with this specific theoretical insight, which confirms Einstein as the correct answer. Why Other Options Are Wrong: Boyle's work focused on the behaviour of gases under pressure and temperature changes, not on the equivalence of mass and energy. Lavoisier emphasised that mass is conserved in chemical reactions, which is approximately true in chemical processes where mass changes are extremely small; he did not propose that mass could be transformed into energy. Avogadro dealt with gas volumes and particle numbers, providing the foundation for the mole concept. None of these contributions involve the mass-energy conversion principle described in the question. Common Pitfalls: Some students may confuse Lavoisier's conservation of mass with Einstein's mass-energy equivalence. While both deal with mass, they address different aspects: conservation of mass in chemical reactions versus convertibility of mass and energy in nuclear and relativistic contexts. Another pitfall is simply guessing based on familiarity with names. To avoid mistakes, remember that whenever you see E = m * c^2 or the idea of matter converting to energy, you should immediately think of Albert Einstein. Final Answer: The scientist who clearly stated that matter can be converted into energy is Albert Einstein.

Question 8

Among the major nutrients listed, milk is considered a relatively poor source of which vitamin or nutrient in a normal human diet?

Accepted Answer

Introduction / Context: Milk is often described as a nearly complete food because it contains proteins, fats, carbohydrates, vitamins, and minerals. However, it is not perfectly balanced for all nutrients, and some vitamins are present only in small amounts. This question examines your knowledge of which nutrient milk is a poor source of, an important point in basic nutrition and general science used in many competitive exams. Given Data / Assumptions: - Milk is known to supply various nutrients including calcium, protein, and carbohydrates. - The options list calcium, protein, vitamin C, and carbohydrates. - We assume normal cow's milk as commonly consumed by humans. - We must select the nutrient for which milk is relatively deficient compared with other food sources. Concept / Approach: Cow's milk is rich in calcium, which supports bone and tooth health. It also provides high-quality protein and a significant amount of carbohydrates in the form of lactose. However, milk has only very small amounts of vitamin C (ascorbic acid); it is not considered a good source of this vitamin. Humans typically need to obtain vitamin C from fruits and vegetables. The approach is therefore to choose the option that represents vitamin C as the nutrient in which milk is poorest. Step-by-Step Solution: Step 1: Consider calcium. Milk is famously recommended for calcium intake because it contains a high concentration of this mineral, so milk cannot be a poor source of calcium. Step 2: Consider protein. Milk contains complete proteins like casein and whey, making it a valuable source of high-quality protein. Step 3: Consider carbohydrates. Milk contains lactose, a disaccharide, which contributes significant carbohydrate calories to the diet. Step 4: Now consider vitamin C. While milk has trace amounts of vitamin C, the quantities are small compared with what is found in citrus fruits, berries, and many vegetables. Step 5: Nutritional science frequently notes that milk is a poor source of vitamin C; infants and adults alike need additional vitamin C-rich foods. Step 6: Therefore, among the options, vitamin C is the correct choice for the nutrient that milk poorly supplies. Verification / Alternative check: Standard nutrition tables show that a typical serving of milk provides considerable calcium (often 200 milligrams or more per glass), several grams of protein, and carbohydrates from lactose. However, the vitamin C content is very low, especially after processing and storage. Dietary guidelines often mention that while milk can help meet many nutrient requirements, vitamin C must be obtained primarily from fruits and vegetables. This confirms that vitamin C is the correct answer. Why Other Options Are Wrong: Calcium is one of the first nutrients associated with milk and dairy products; they are recommended specifically to meet daily calcium needs. Protein intake from milk and milk products plays a major role in growth, particularly in children. Carbohydrates in the form of lactose make milk an energy-dense food. Because milk is a good source of these three, none of them can be the correct answer to the question about a poor nutrient source. Only vitamin C fits the description of being relatively lacking in milk. Common Pitfalls: Students may sometimes think that milk must be poor in carbohydrates because it does not taste as sweet as sugary drinks, but lactose still contributes significantly to energy intake. Others might be confused by the idea that milk is a complete food and may assume it has all vitamins in abundance. Remembering that milk is low in vitamin C and that scurvy prevention relies on fruits and vegetables will help you quickly identify vitamin C as the nutrient for which milk is a poor source. Final Answer: Milk is a relatively poor source of vitamin C (ascorbic acid).

Question 9

In basic chemistry, which one of the following substances is not a mixture but instead consists of a single pure substance?

Accepted Answer

Introduction / Context: Chemistry distinguishes between pure substances and mixtures. Pure substances have uniform composition and fixed properties, whereas mixtures contain two or more substances physically combined and can be separated by physical methods. This question asks you to identify which among the listed examples is not a mixture and instead represents a pure substance. It tests your understanding of everyday examples like air, milk, cement, and metallic elements such as mercury. Given Data / Assumptions: - Four options are provided: air, mercury, milk, and cement. - We must decide which one is not a mixture under normal conditions. - Pure substance here means a single element or compound with uniform composition. - Mixtures can be homogeneous or heterogeneous combinations of different substances. Concept / Approach: Air is a mixture of gases, primarily nitrogen and oxygen, along with small amounts of other gases. Milk is an emulsion and colloidal mixture of water, fats, proteins, and minerals. Cement is a mixture of different powdered compounds like calcium silicates and aluminates. Mercury, on the other hand, is a chemical element, symbol Hg, and is a pure liquid metal under normal conditions. The logical approach is to identify the one option that refers to a single element rather than to a combination of substances. Step-by-Step Solution: Step 1: Consider air. It contains nitrogen, oxygen, carbon dioxide, argon, and traces of many other gases. This clearly makes it a mixture of gases. Step 2: Consider milk. It contains water, fat globules, proteins like casein, lactose, and minerals, forming a heterogeneous mixture or emulsion. Step 3: Consider cement. It is produced by grinding together various compounds such as clinkers of calcium silicates, calcium aluminates, and sometimes other additives, so it is also a mixture. Step 4: Consider mercury. It is an element with atomic number 80 and is present as a single type of atom in metallic form, not a blend of different substances under ordinary conditions. Step 5: Since mercury is a single pure substance, and the others are combinations, the correct choice is mercury. Verification / Alternative check: You can verify this by looking at the periodic table and standard descriptions in chemistry books. Mercury is listed as an element and is often used as the textbook example of a metal that is liquid at room temperature. Air, milk, and cement are routinely listed as examples of mixtures when explaining homogeneous and heterogeneous systems. This confirms that mercury is the only option that qualifies as a pure substance rather than a mixture. Why Other Options Are Wrong: Air (option a) is a homogeneous mixture of gases, even though it may appear uniform to our senses. Milk (option c) is a complex colloidal mixture and can be separated into cream, whey, and other components by physical methods such as centrifugation. Cement (option d) is a heterogeneous mixture of several compounds and does not have a single fixed chemical formula. Thus, all three are mixtures and cannot be the correct answer to a question asking which one is not a mixture. Common Pitfalls: Students sometimes think that because air and milk look uniform at first glance, they must be pure substances. This confusion arises from not distinguishing between homogeneous mixtures and true pure substances. Another pitfall is assuming that any industrial material like cement must be a compound; in reality, it is a mixture of several substances. Remembering that elements on the periodic table, such as mercury, represent pure substances will help you avoid these errors. Final Answer: The substance that is not a mixture but a pure substance is mercury, a pure elemental metal.

Question 10

Brass gets discoloured or tarnished in air mainly because of the presence of which gas, which reacts to form a surface coating?

Accepted Answer

Introduction / Context: Brass is an alloy of copper and zinc and is widely used in decorative items, utensils, and fittings. Over time, brass articles tend to lose their shine and develop a discoloured surface. This tarnishing is a common observation in everyday life and can be explained using basic chemistry. This question asks which gas in the air is primarily responsible for the discolouration of brass, testing your understanding of corrosion and surface reactions. Given Data / Assumptions: - The metal alloy in question is brass, containing copper and zinc. - The discolouration occurs when brass is exposed to air over time. - Options include hydrogen sulphide, oxygen, nitrogen, and carbon dioxide. - We assume typical atmospheric composition with trace gases present. Concept / Approach: Copper and its alloys, like brass, are sensitive to sulphur-containing gases. Hydrogen sulphide present in small amounts in the atmosphere can react with copper in brass to form copper sulphide, a dark compound that leads to tarnishing. Oxygen and carbon dioxide do play roles in general corrosion, but the characteristic dark discolouration is mainly due to sulphide formation. Nitrogen is relatively inert under normal conditions and does not significantly react with brass. The approach is to link the observed discolouration with the formation of sulphide layers from hydrogen sulphide in the air. Step-by-Step Solution: Step 1: Recall that brass contains copper, which readily reacts with sulphur compounds in the environment. Step 2: Hydrogen sulphide (H2S) is a trace gas in the atmosphere, often arising from decaying organic matter and industrial pollution. Step 3: When brass is exposed to air containing H2S, a chemical reaction occurs forming copper sulphide on the surface, which is dark and causes discolouration. Step 4: Oxygen alone tends to form oxides, which may also dull the surface, but the specific tarnishing often associated with brass and silver is linked to sulphides. Step 5: Nitrogen is largely inert at ordinary temperatures and does not react significantly with metals like copper under normal atmospheric conditions. Step 6: Carbon dioxide can react with metal oxides to form carbonates in some cases, but it is not the primary cause of the characteristic sulphide tarnish on brass. Step 7: Therefore, hydrogen sulphide is the gas most directly responsible for discolouring brass in air. Verification / Alternative check: You can verify this by recalling that silver jewellery and cutlery also tarnish in the presence of hydrogen sulphide, forming black silver sulphide. Similar chemistry applies to the copper component of brass. Real-world cleaning products for brass and silver often advertise their ability to remove sulphide tarnish, further confirming that sulphur-containing gases are the main culprits rather than oxygen or carbon dioxide alone. Why Other Options Are Wrong: Oxygen (option b) does promote oxidation, but the dull oxide film it forms is generally different in colour and does not fully explain the characteristic dark tarnish associated with brass. Nitrogen (option c) is very stable as N2 and typically does not react with metals in ordinary air. Carbon dioxide (option d) is involved in the formation of carbonates in some metals but is not the main factor causing the specific discolouration observed in brass due to sulphide formation. Therefore, these gases are less significant than hydrogen sulphide in tarnishing brass. Common Pitfalls: A common mistake is to assume that oxygen must be responsible for all forms of corrosion and discolouration because rusting of iron is widely known as an oxidation process. However, for alloys like brass and metals like silver, sulphur-containing gases such as hydrogen sulphide are the major causes of tarnish. Remembering this difference between oxide formation and sulphide tarnish will help you answer such questions correctly. Final Answer: Brass gets discoloured in air mainly because of the presence of hydrogen sulphide, which forms dark sulphide coatings on its surface.

Question 11

In the context of nuclear materials, the term Yellow cake, which is sometimes mentioned in reports of smuggling across borders, usually refers to which substance?

Accepted Answer

Introduction / Context: Yellow cake is a term that often appears in discussions about nuclear energy, nuclear proliferation, and smuggling of strategic materials. It is not related to narcotic drugs or precious metals but to nuclear fuel. This question tests whether you recognise Yellow cake as a uranium oxide concentrate used as an intermediate step in the production of nuclear fuel, making it a sensitive material in international security and non-proliferation efforts. Given Data / Assumptions: - The phrase Yellow cake is specifically highlighted. - It is described as an item involved in smuggling across borders. - Options include crude heroin, crude cocaine, uranium oxide, and unrefined gold. - We assume the term is used in its standard nuclear industry sense. Concept / Approach: In the nuclear industry, Yellow cake refers to a concentrated form of uranium ore, typically uranium oxide (often U3O8), produced by milling and chemical processing of uranium-bearing ores. Because it can be further processed to produce enriched uranium for reactors or potentially for weapons, it is a material of strategic importance and is strictly regulated. The approach is to identify the option that correctly associates Yellow cake with uranium oxide rather than with drugs or gold. Step-by-Step Solution: Step 1: Recall that nuclear fuel production begins with the mining of uranium ore, which is then processed into a concentrated powder called Yellow cake. Step 2: This Yellow cake is mainly composed of uranium oxides and is a key intermediate in the nuclear fuel cycle. Step 3: Consider option c, which explicitly states uranium oxide concentrate used in the nuclear fuel cycle. This matches the standard definition. Step 4: Options a and b, crude heroin and crude cocaine, refer to illegal drugs that have completely different terminology and are never called Yellow cake in professional usage. Step 5: Option d, unrefined gold, is sometimes nicknamed gold ore or bullion, but not Yellow cake in nuclear and security contexts. Step 6: Therefore, the correct answer is uranium oxide concentrate. Verification / Alternative check: News reports and international security discussions about nuclear proliferation frequently mention seizures or tracking of Yellow cake, always in connection with uranium supplies. Science and general knowledge references describe Yellow cake as uranium oxide obtained from processing uranium ore, which is then sent to conversion and enrichment facilities. No reputable source uses this term for drugs or gold, confirming its nuclear association. Why Other Options Are Wrong: Heroin (option a) and cocaine (option b) are illegal narcotic drugs derived from opium poppies and coca plants respectively, and while they may have street nicknames, Yellow cake is not one of the standard names for them in scientific or official contexts. Unrefined gold (option d) is a precious metal in ore form or as nuggets but is described with different terminology by miners and customs authorities. Thus, these options are distractors and do not match the technical meaning of Yellow cake. Common Pitfalls: Some candidates may read only the word smuggling and immediately think of narcotic drugs like heroin or cocaine. Others might associate the term cake with gold due to its value and the idea of something solid. To avoid these traps, focus on the context of nuclear fuel and recall that Yellow cake appears often in stories about nuclear materials, not drug shipments. Remembering the link between uranium mining and Yellow cake will help you answer correctly. Final Answer: Yellow cake is the name given to uranium oxide concentrate used as an intermediate in the nuclear fuel cycle.

Question 12

An electrochemical cell (galvanic cell) stops working after some time because which of the following conditions is eventually reached between its electrodes and solutions?

Accepted Answer

Introduction / Context: Electrochemical cells, such as galvanic or voltaic cells, convert chemical energy into electrical energy through redox reactions. Over time, however, such cells eventually stop delivering current, and the lamp or device they power goes out. This question asks you to identify the main reason why a cell stops working, focusing on the concept of electrode potentials and cell electromotive force (EMF). Understanding this helps link practical battery behaviour with theoretical electrochemistry. Given Data / Assumptions: - We are dealing with an electrochemical cell that initially produces a potential difference (EMF). - Over time, the cell stops working, meaning it can no longer drive a current through an external circuit. - Options mention electrode potentials becoming equal or zero, electrode disappearance, and reaction reversal. - We assume a typical galvanic cell with two half-cells and a salt bridge. Concept / Approach: The EMF of a galvanic cell arises from the difference in electrode potentials between the two half-cells. As the cell operates, reactants are consumed and products accumulate, gradually changing the concentrations in each half-cell. Eventually, the system approaches equilibrium, the forward and reverse reaction rates become equal, and the difference in electrode potentials falls to zero. At that point, there is no driving force for electron flow, and the cell can no longer deliver current. The correct approach is therefore to identify the option that states the electrode potentials equalise, eliminating net EMF. Step-by-Step Solution: Step 1: Recall that cell EMF is given by E_cell = E_cathode - E_anode. Step 2: As the cell discharges, ion concentrations in the two half-cells change and the Nernst equation shows that electrode potentials shift. Step 3: When the reaction reaches equilibrium, the cell EMF becomes zero because the driving force for further reaction disappears. Step 4: Mathematically, this corresponds to E_cathode = E_anode, so their difference is zero. Step 5: Option a states that the electrode potential of both electrodes becomes equal, so no net EMF remains. This matches the theoretical explanation. Step 6: Options b, c, and d either exaggerate or misrepresent the behaviour and do not correctly explain the underlying equilibrium condition. Verification / Alternative check: Electrochemistry textbooks explain that a galvanic cell functions as long as its overall redox reaction is thermodynamically favourable, which means it is not at equilibrium. Over time, as the reaction proceeds, the Gibbs free energy change approaches zero and so does the cell EMF. At equilibrium, the reaction quotient equals the equilibrium constant, and the Nernst equation yields zero EMF. This theoretical description aligns perfectly with the idea that the electrode potentials equalise, causing the cell to stop working. Why Other Options Are Wrong: Option b suggests that one of the electrodes completely vanishes, which may happen in some extreme cases over long periods, but a cell can stop working even before that happens, once equilibrium is reached. Option c claims that the reaction reverses its direction, but at equilibrium the forward and reverse reactions occur at equal rates, resulting in no net change, not continuous reversal with net current. Option d implies that both electrode potentials are zero from the start, which would mean the cell never worked; in reality, they only converge towards equality over time, not necessarily to zero. Hence, these explanations are either incomplete or incorrect. Common Pitfalls: A typical mistake is to imagine physical disappearance of an electrode as the main reason for cell failure, ignoring the crucial role of equilibrium. Another pitfall is thinking that the reaction somehow stops abruptly for no reason, without linking it to changes in concentration and electrode potential. Remembering that cell EMF is driven by a difference in electrode potentials, and that this difference goes to zero as equilibrium is approached, will help you understand and answer such questions correctly. Final Answer: An electrochemical cell stops working because the electrode potentials of both electrodes become equal and the net EMF falls to zero.

Question 13

A breath test used by traffic police to check drunken driving commonly uses which chemical system to detect the presence of alcohol in a driver's breath?

Accepted Answer

Introduction / Context: Drunken driving is a major safety concern, and traffic police often use chemical breath tests to estimate the presence of alcohol (ethanol) in a driver's breath. One traditional chemical method employs an oxidising agent that changes colour when it reacts with ethanol. This question asks which chemical system is commonly used in such breath analysers, testing your knowledge of applied chemistry and basic redox reactions. Given Data / Assumptions: - The test is a breath test for detecting alcohol in exhaled air. - Options include turmeric, silver nitrate on silica gel, potassium permanganate-sulphuric acid, and potassium dichromate-sulphuric acid. - We assume older or classical chemical breath analysers, not modern electronic fuel-cell devices. - The test relies on a noticeable colour change caused by the presence of ethanol. Concept / Approach: Traditional breath analysers often used acidified potassium dichromate solution as the oxidising agent. Ethanol in the breath is oxidised to acetic acid, and during this process, orange dichromate ions are reduced to green chromium(III) ions, causing a visible colour change. Potassium permanganate can also oxidise alcohols but is less commonly associated with standard breathalyser kits in general knowledge questions. Turmeric and silver nitrate have entirely different uses. The correct approach is to recall that potassium dichromate in sulphuric acid is the classic system used to detect ethanol through a colour change. Step-by-Step Solution: Step 1: Recall that potassium dichromate (K2Cr2O7) in acidic medium is a strong oxidising agent that can oxidise ethanol to acetic acid. Step 2: During this oxidation, the chromium in dichromate is reduced from the +6 oxidation state to the +3 state, producing a distinct colour change from orange to green. Step 3: In a breathalyser, this colour change is correlated with the amount of alcohol in the breath, giving an indirect measure of blood alcohol content. Step 4: Turmeric is an acid-base indicator that changes colour primarily with pH, not specifically with ethanol, so option a is not suitable. Step 5: Silver nitrate on silica gel can be used to detect halide ions, not alcohol, so option b is unrelated to breath tests. Step 6: Potassium permanganate in acidic medium is also an oxidising agent, but general knowledge questions about breath tests traditionally refer to acidified potassium dichromate, not permanganate. Step 7: Therefore, the correct choice is potassium dichromate-sulphuric acid, option d. Verification / Alternative check: Descriptions of classical breathalyser designs in chemistry texts and general science references explicitly mention orange potassium dichromate in sulphuric acid solution being used to oxidise ethanol, with the extent of colour change indicating how much alcohol was present. While modern devices may use fuel cells or infrared sensors, exam questions about basic chemistry usually focus on this traditional dichromate-based test. This consistency across sources supports selecting potassium dichromate-sulphuric acid as the answer. Why Other Options Are Wrong: Turmeric (option a) is used as a simple indicator in some acid-base demonstrations and in testing household substances, but it does not specifically detect ethanol. Silica gel coated with silver nitrate (option b) is used for detecting halides such as chloride and bromide, not alcohol. Potassium permanganate with sulphuric acid (option c) is indeed a strong oxidising mixture, but it is not the classic breathalyser reagent given in school-level chemistry and general knowledge texts, which emphasise potassium dichromate instead. As such, these options do not match the standard description of breath tests for drunken driving. Common Pitfalls: Some candidates may confuse potassium permanganate and potassium dichromate because both are strong oxidising agents and undergo colour changes in reactions. Another pitfall is to think of any familiar indicator, like turmeric, as a possible answer without considering whether it is sensitive to alcohol. To avoid errors, remember that the iconic school-level example of a breath test uses acidified potassium dichromate which turns from orange to green upon reacting with ethanol. Final Answer: The breath test commonly used by traffic police relies on potassium dichromate mixed with sulphuric acid to detect alcohol in the driver's breath.

Question 14

Cobalt-60 is commonly used in radiation therapy for cancer treatment because it predominantly emits which type of penetrating radiation?

Accepted Answer

Introduction / Context: Cobalt-60 is a radioactive isotope widely used in medical applications, especially in radiotherapy units for treating cancer. Different types of nuclear radiation (alpha, beta, and gamma) have different penetrating powers and biological effects. This question asks which type of radiation emitted by cobalt-60 makes it particularly suitable for targeting tumours deep inside the body, a key point in nuclear medicine and general science. Given Data / Assumptions: - The radioactive isotope mentioned is cobalt-60 (Co-60). - It is used in radiation therapy for cancer. - Options include alpha rays, beta rays, gamma rays, and X-rays. - We assume standard knowledge of radioactive emissions and their properties. Concept / Approach: Cobalt-60 decays by beta emission, producing an excited nickel nucleus that then emits high-energy gamma rays. These gamma rays have strong penetrating power and can pass through body tissues to reach tumours located below the surface. Alpha particles, by contrast, are stopped by a sheet of paper or skin, and beta particles have only moderate penetration. X-rays are electromagnetic radiation like gamma rays but are usually generated by electronic transitions or X-ray tubes. The correct approach is to identify gamma rays as the key therapeutic radiation from cobalt-60 sources. Step-by-Step Solution: Step 1: Recall that cobalt-60 undergoes beta decay to form nickel-60, and in the process, the daughter nucleus emits gamma rays. Step 2: These gamma rays are very penetrating and can be carefully collimated and directed toward a tumour from outside the body. Step 3: Alpha rays (helium nuclei) are heavy and carry a double positive charge; they have strong ionising power but very low penetrative ability and are stopped by skin. Step 4: Beta rays (electrons or positrons) have more penetration than alpha particles but still cannot penetrate deep tissues as effectively as gamma rays. Step 5: X-rays are also high-energy electromagnetic waves, but in the context of cobalt-60 therapy, the radiation actually used is gamma radiation from nuclear transitions. Step 6: Therefore, the correct choice is gamma rays, option c. Verification / Alternative check: Medical physics and radiology textbooks describe cobalt therapy units as gamma-ray sources with characteristic energies around 1 megaelectron volt. These gamma rays are used in external beam radiotherapy to deliver controlled doses to tumours. Although other modalities like linear accelerators now generate high-energy X-rays for similar purposes, the specific association with cobalt-60 in exam questions is gamma radiation. This confirms the selection of gamma rays as the correct answer. Why Other Options Are Wrong: Alpha rays (option a) have very low penetration and are not practical for external beam therapy; they might be considered in some internal radiotherapy approaches but not for cobalt-60 units. Beta rays (option b) emitted in cobalt-60 decay are mostly absorbed in the source and shielding and do not form the main therapeutic beam. X-rays (option d) are generated differently and, although used in radiotherapy from other machines, are not the primary radiation directly emitted from cobalt-60 nuclei. Hence, these options do not correctly describe the radiation from cobalt-60 used in therapy. Common Pitfalls: Students sometimes confuse X-rays and gamma rays because both are electromagnetic radiation and may be used for imaging or therapy. The key distinction is their origin: X-rays arise from electronic transitions or machine-generated beams, while gamma rays come from nuclear transitions. Another pitfall is thinking that the initial beta emission is what treats the tumour; in practice, it is the gamma radiation that has the right combination of penetration and controllability for external therapy. Keeping this distinction clear will help you answer questions accurately. Final Answer: Cobalt-60 is used in radiation therapy because it emits highly penetrating gamma rays suitable for treating deep-seated tumours.

Question 15

In the history of measurement systems, the M.K.S. system of units (metre–kilogram–second) was first introduced by which scientist or engineer?

Accepted Answer

Introduction / Context: Measurement systems evolved over time from various local and traditional units to more standardised systems like the M.K.S. (metre–kilogram–second) and ultimately the S.I. system. Understanding who contributed to these developments is part of general science and physics history. This question asks who first introduced or proposed the M.K.S. system, which later formed the basis for the modern S.I. units used worldwide in science and engineering. Given Data / Assumptions: - The specific system mentioned is the M.K.S. system of units: metre, kilogram, and second. - Options include Archimedes, Einstein, Newton, and Giorgi. - We assume standard historical accounts of the development of unit systems. - The task is to connect the M.K.S. system with its proposer. Concept / Approach: Giovanni Giorgi, an Italian engineer and physicist, proposed a rationalised system of units that combined the M.K.S. system with an electrical unit, leading to the foundation of the international system (S.I.). Archimedes, Newton, and Einstein made enormous contributions to science but were not responsible for this particular system of units. The correct approach is to recall that the M.K.S. system and its subsequent generalisation into the S.I. system are historically linked to Giorgi's proposal. Step-by-Step Solution: Step 1: Recognise that the M.K.S. system uses metre for length, kilogram for mass, and second for time, and became a precursor to the S.I. system. Step 2: Recall that Giovanni Giorgi proposed combining mechanical units with an electrical unit to form a coherent system, building on M.K.S. units. Step 3: Evaluate Archimedes. He worked in ancient Greece on hydrostatics and geometry, long before modern scientific unit systems were developed. Step 4: Evaluate Newton. He established the laws of motion and gravity and used existing units but did not create the M.K.S. system as a formal standard. Step 5: Evaluate Einstein. He revolutionised physics with relativity and quantum ideas but did not introduce the M.K.S. system. Step 6: Giorgi is historically credited with proposing the rationalised M.K.S. system, so option d is correct. Verification / Alternative check: Histories of unit systems and the S.I. system note that in the early 20th century, Giovanni Giorgi suggested that a coherent system based on the metre, kilogram, second, and one electrical unit (eventually the ampere) could unify mechanical and electrical measurements. This Giorgi system later evolved into the modern S.I. system. No such attribution is made to Archimedes, Newton, or Einstein, confirming that Giorgi's name is the correct one to associate with the introduction of the M.K.S. framework. Why Other Options Are Wrong: Archimedes lived in a period when measurement units were local and not standardised globally; he did not develop the M.K.S. system. Newton's work predated modern unit standardisation and, while highly influential in mechanics, did not include a formal proposal of the M.K.S. system. Einstein focused on theoretical physics and used existing units in his work but did not create this unit system. Therefore, associating the M.K.S. system with any of these three scientists would be historically inaccurate. Common Pitfalls: Students sometimes guess Newton or Einstein whenever a physics-related question arises, ignoring the specific context of measurement systems and unit standardisation. Another pitfall is being unfamiliar with the name Giorgi and assuming that a more famous scientist must be the right answer. To avoid these mistakes, it is helpful to remember a few key names like Kelvin for temperature scales and Giorgi for the rationalised M.K.S. system that paved the way for S.I. units. Final Answer: The M.K.S. system of units was first introduced and rationalised by Giovanni Giorgi.

Question 16

In biochemistry and fermentation, which enzyme complex is responsible for converting glucose into ethyl alcohol (ethanol) during alcoholic fermentation by yeast?

Accepted Answer

Introduction / Context: Alcoholic fermentation is a biological process in which sugars such as glucose are converted into ethyl alcohol (ethanol) and carbon dioxide by the action of enzymes in yeast. Several enzymes play roles in carbohydrate digestion and metabolism, but only one specific enzyme complex from yeast is traditionally associated with alcoholic fermentation. This question tests whether you can correctly identify zymase as the enzyme complex that converts glucose into ethanol during fermentation. Given Data / Assumptions: - The process discussed is the conversion of glucose into ethyl alcohol. - The context is alcoholic fermentation by yeast. - Options include the enzymes maltase, zymase, diastase, and invertase. - We assume standard textbook definitions and roles of these enzymes in carbohydrate metabolism. Concept / Approach: Zymase is the name given to the enzyme complex present in yeast that catalyses the series of reactions converting glucose to ethanol and carbon dioxide during fermentation. Maltase acts on maltose, diastase acts on starch, and invertase acts on sucrose. None of these other enzymes by themselves complete the fermentation pathway that produces ethanol. The correct approach is to match the specific function converting glucose to ethanol with zymase, rather than with enzymes that merely break down larger sugars into glucose or other monosaccharides. Step-by-Step Solution: Step 1: Recall the overall alcoholic fermentation reaction: glucose → ethanol + carbon dioxide, carried out anaerobically by yeast. Step 2: Recognise that zymase is the established name for the mixture of enzymes in yeast cells that catalyses this conversion. Step 3: Consider maltase. It hydrolyses maltose into two glucose molecules but does not by itself convert glucose into ethanol. Step 4: Consider diastase. It breaks down starch into maltose and smaller sugars but is not the fermenting enzyme complex producing ethanol. Step 5: Consider invertase. It splits sucrose into glucose and fructose; again, it prepares sugars for metabolism but does not complete the fermentation step to ethanol. Step 6: Therefore, among the options, the enzyme complex that actually performs alcoholic fermentation and produces ethyl alcohol is zymase. Verification / Alternative check: Biology and biochemistry textbooks describe the role of yeast in baking and brewing, emphasising that the zymase enzyme complex converts glucose into ethanol and carbon dioxide. Historically, zymase was one of the first enzyme complexes to be discovered and studied in the context of fermentation. While other enzymes like invertase and maltase help release glucose from larger carbohydrates, zymase is specifically linked with the final fermentation steps, confirming it as the correct choice. Why Other Options Are Wrong: Maltase (option a) only acts on maltose, not directly on glucose to produce ethanol. Diastase (option c) acts on starch, reducing it to smaller sugars such as maltose, but again does not complete the fermentation to alcohol. Invertase (option d) catalyses the hydrolysis of sucrose into glucose and fructose but stops there; the subsequent fermentation to ethanol requires zymase. Therefore, these enzymes play auxiliary roles in carbohydrate digestion or preparation but are not the main fermentation enzyme complex. Common Pitfalls: Students sometimes confuse the enzymes that prepare sugars (like invertase and diastase) with those that carry out fermentation itself. Another pitfall is to see the word malt in maltase and associate it with brewing, even though maltase only breaks maltose into glucose. To avoid such mistakes, remember that zymase is the classic name specifically linked to yeast fermentation of glucose into ethyl alcohol and carbon dioxide. Final Answer: The enzyme complex that converts glucose to ethyl alcohol during alcoholic fermentation is zymase.

Question 17

In basic general science, which one of the following substances is not a metal, while the others listed are typical metals?

Accepted Answer

Introduction / Context: Identifying metals and non-metals is one of the first skills taught in basic chemistry and general science. Metals have properties such as lustre, malleability, ductility, and good electrical and thermal conductivity, while non-metals often lack these traits and may be brittle or gaseous. This question asks which one of the listed substances is not a metal, testing your ability to recognise a common non-metal among everyday metallic elements. Given Data / Assumptions: - The options include iron, copper, sulphur, and aluminium. - Three of these are well-known metals used in industry and daily life. - One is a non-metal that does not exhibit typical metallic properties. - We assume standard periodic table classifications. Concept / Approach: Iron, copper, and aluminium are all typical metallic elements found on the left and central parts of the periodic table. They are solid at room temperature, shiny when polished, malleable, and good conductors of heat and electricity. Sulphur, however, is a non-metal located in the p-block and exists as a brittle yellow solid at room temperature; it is a poor conductor and lacks metallic lustre. The approach is to identify the odd one out based on these physical and chemical characteristics. Step-by-Step Solution: Step 1: Consider iron. It is widely used in construction and engineering and is a typical metal with high strength and good conductivity. Step 2: Consider copper. It is a reddish metal known for excellent electrical and thermal conductivity and is used in electrical wiring and plumbing. Step 3: Consider aluminium. It is a light, silvery metal used in aircraft, packaging, and many structural applications. Step 4: Consider sulphur. It is a yellow, brittle, non-metallic solid that is not malleable or ductile and is a poor conductor of heat and electricity. Step 5: Based on these properties, sulphur clearly stands out as the non-metal, while the other three are metals. Verification / Alternative check: The periodic table lists iron (Fe), copper (Cu), and aluminium (Al) as metallic elements. They are located in regions associated with metals and form positive ions (cations) in many of their compounds. Sulphur (S) is located among the non-metals and typically forms negative ions (anions) such as sulphide (S^2-) in ionic compounds. It also forms covalent compounds like sulphur dioxide and sulphuric acid. These periodic table positions and chemical behaviours confirm that sulphur is a non-metal while the others are metals. Why Other Options Are Wrong: Iron (option a) is unmistakably a metal, used extensively for making steel and other alloys. Copper (option b) is a classic metal used in conductors and heat exchangers. Aluminium (option d) is a lightweight metal widely employed where low density and corrosion resistance are important. None of these substances fits the description of a non-metal, so they cannot be the correct answer to a question asking which one is not a metal. Common Pitfalls: Some learners may be unfamiliar with sulphur in its elemental form and focus mainly on metal uses, leading them to overthink the question. Another pitfall is confusing metalloid and non-metal categories, but sulphur is clearly categorised as a non-metal, not a metalloid like silicon. Remembering basic periodic table groups and the common physical appearance of sulphur as a yellow, brittle substance will help you quickly identify it as the non-metal among the choices. Final Answer: The substance that is not a metal is sulphur, a non-metallic element.

Question 18

Among the following metal chlorides, which compound shows the maximum covalent character: MgCl2, FeCl2, SnCl2 or AlCl3?

Accepted Answer

Introduction / Context: This chemistry question tests your understanding of how to compare the covalent and ionic character of different metal chlorides using Fajans rules. Being able to predict whether a bond is more ionic or more covalent is important for explaining melting points, solubility, and volatility of inorganic compounds. Given Data / Assumptions: The compounds to compare are MgCl2, FeCl2, SnCl2, AlCl3 and CaCl2. All are metal chlorides with chloride ions Cl- as the common anion. We assume normal oxidation states: Mg2+, Fe2+, Sn2+, Al3+, Ca2+. Covalent character is judged qualitatively using Fajans rules. Concept / Approach: Fajans rules state that the covalent character of an ionic bond increases when the cation has high charge and small size, and when the anion is large and easily polarised. A small, highly charged cation strongly polarises the electron cloud of the anion, pulling it towards itself and creating sharing of electrons similar to covalent bonding. Since the anion is the same in all options (chloride), we mainly compare the size and charge of the metal cations to decide which compound is most covalent in nature. Step-by-Step Solution: 1) List cations: Mg2+, Fe2+, Sn2+, Al3+, Ca2+ all combined with Cl- ions.2) According to Fajans rules, greater positive charge and smaller ionic radius make a cation more polarising.3) Among the given cations, Al3+ has a +3 charge and is quite small in size compared with Mg2+, Fe2+, Sn2+ and Ca2+.4) Because Al3+ is highly polarising, AlCl3 shows strong polarisation of chloride ions and therefore has the maximum covalent character among the listed chlorides.5) The other cations with +2 charge are less polarising, so their chlorides are comparatively more ionic than AlCl3. Verification / Alternative check: We can cross check this conclusion with physical properties. AlCl3 has a relatively low melting point and exists as a covalent molecular solid or dimeric Al2Cl6 in the vapour phase. In contrast, typical ionic salts like MgCl2 and CaCl2 have high melting points and exist as extended ionic lattices. This supports the prediction from Fajans rules that AlCl3 has more covalent character than the other metal chlorides listed in the options. Why Other Options Are Wrong: MgCl2: Magnesium ion Mg2+ has lower charge and a larger radius than Al3+, so MgCl2 is more ionic and less covalent. FeCl2: Iron(II) ion Fe2+ has the same charge as Mg2+ but is not as small and highly charged as Al3+, so FeCl2 has less covalent character than AlCl3. SnCl2: Tin(II) ion Sn2+ is larger and has lower charge density than Al3+, so SnCl2 is less covalent compared to AlCl3. CaCl2: Calcium ion Ca2+ is a typical alkaline earth metal cation with large size and +2 charge, giving mainly ionic character and therefore not the most covalent among the options. Common Pitfalls: Many students look only at the metal being a transition element or a p block metal and ignore the charge and size together. Some think that FeCl2 or SnCl2 must be more covalent because transition metals often form coloured compounds. However, for covalent character prediction Fajans rules focus on charge density, not only on block of the periodic table. Another common error is to confuse covalent character with simple electronegativity difference, but here the more subtle factor is polarising power of the cation since all anions are chloride ions. Final Answer: The compound with the maximum covalent character among the given metal chlorides is AlCl3.

Question 19

A radioactive substance has a half life of 4 months. After how many months will three-fourths of the original sample have decayed?

Accepted Answer

Introduction / Context: This numerical chemistry question checks your understanding of the basic idea of half life in radioactive decay. Half life problems are very common in nuclear chemistry, physics and environmental science because they help estimate how long a radioactive material remains significantly active. Given Data / Assumptions: Half life of the radioactive substance, t(1/2) = 4 months. We want the time required for three-fourths (3/4) of the original sample to decay. This means only one-fourth (1/4) of the original sample will remain undecayed. Radioactive decay follows first order kinetics and the definition of half life applies repeatedly. Concept / Approach: Half life is defined as the time required for half of the radioactive nuclei in a sample to decay. After one half life, the activity or amount becomes 1/2 of the initial value. After two half lives, it becomes 1/2 * 1/2 = 1/4 of the original amount. By expressing the desired remaining fraction in terms of powers of 1/2, we can easily count how many half lives have passed and then multiply by the given half life time to get the total elapsed time. Step-by-Step Solution: 1) Let the initial amount of the radioactive substance be N0.2) After one half life (4 months), the amount left is N0 / 2.3) After two half lives, the remaining amount is (N0 / 2) / 2 = N0 / 4.4) N0 / 4 corresponds to one-fourth of the original amount, which means three-fourths have decayed.5) Two half lives correspond to 2 * 4 months = 8 months. Verification / Alternative check: We can verify with the general formula for first order decay: N = N0 * (1/2)^(t / t(1/2)). We want N = N0 / 4, so N0 / 4 = N0 * (1/2)^(t / 4). Cancelling N0 and writing 1/4 as (1/2)^2 gives (1/2)^2 = (1/2)^(t / 4). This implies t / 4 = 2, so t = 8 months, which matches the earlier reasoning from repeated half lives. Why Other Options Are Wrong: 3 months: This is less than one half life, so more than half of the substance would still remain, not just one fourth. 4 months: After one half life only 1/2 of the substance has decayed, not 3/4, so this is insufficient time. 12 months: This corresponds to three half lives (12 / 4 = 3), at which point the remaining amount would be N0 / 8, meaning 7/8 decayed, more than required. 16 months: This is four half lives, leaving N0 / 16 remaining, so far more than three-fourths would have decayed. Common Pitfalls: A frequent mistake is to think that three-fourths decayed means three half lives, because of the number three in the fraction, but this is incorrect. Students must focus on the remaining fraction, not the decayed fraction. Another error is trying to subtract half life linearly from 100 percent, which ignores the exponential nature of radioactive decay. Always express remaining fraction as (1/2)^n to find the correct number of half lives that have passed. Final Answer: Three-fourths of the radioactive substance will have decayed after 8 months.

Question 20

The two common isotopes of chlorine with mass numbers 35 and 37 exist in approximately what ratio in nature?

Accepted Answer

Introduction / Context: This general chemistry question focuses on isotopes of chlorine and their natural abundance. Understanding isotope ratios is important when calculating average atomic masses and when interpreting data from mass spectrometry and nuclear chemistry. Given Data / Assumptions: Chlorine has two main stable isotopes with mass numbers 35 and 37. The question asks for their approximate natural abundance ratio. We consider the common textbook approximation used for basic chemistry calculations. The options are different simple integer ratios such as 1 : 2, 2 : 3, 3 : 1, 3 : 4 and 1 : 1. Concept / Approach: The atomic mass of chlorine listed in the periodic table is about 35.5 atomic mass units. This value is a weighted average of the masses of its isotopes based on their relative abundances. Since the average is much closer to 35 than to 37, the isotope with mass 35 must be more abundant. By considering a simple ratio that gives an average near 35.5, we can deduce the approximate ratio Cl 35 : Cl 37 as about 3 : 1, which corresponds closely to the known natural abundance values (roughly 75 percent for Cl 35 and 25 percent for Cl 37). Step-by-Step Solution: 1) Let the ratio of Cl 35 to Cl 37 be a : b.2) The average atomic mass can be approximated as (35a + 37b) / (a + b).3) We want this to be around 35.5, the accepted average atomic mass of chlorine.4) If we try a simple ratio 3 : 1, the average becomes (35 * 3 + 37 * 1) / 4 = (105 + 37) / 4 = 142 / 4 = 35.5.5) Therefore, the approximate natural abundance ratio of the isotopes Cl 35 : Cl 37 is 3 : 1. Verification / Alternative check: From more detailed data, the abundance of Cl 35 is about 75.8 percent and that of Cl 37 is about 24.2 percent. If we divide 75.8 by 24.2, we get a value a little above 3, which again supports an approximate ratio of 3 : 1. This confirms that the commonly used textbook approximation is correct for basic multiple choice questions and average mass calculations. Why Other Options Are Wrong: 1 : 2: This would give an average mass much closer to 36.3, which does not match the periodic table value of about 35.5. 2 : 3: This ratio also favours the heavier isotope too much and would not yield an average close to 35.5. 3 : 4: Here the heavy isotope fraction is still too large, shifting the average atomic mass away from the accepted chlorine value. 1 : 1: Equal amounts of Cl 35 and Cl 37 would give an average of 36, which again disagrees with the actual atomic weight of chlorine. Common Pitfalls: Students sometimes guess that isotopes must be present in equal proportions, which is rarely true in nature. Another common error is to memorise a wrong ratio or to ignore the link between the ratio and the average atomic mass. Always check whether a proposed ratio gives an average mass consistent with the value printed in the periodic table, as this is an easy way to verify your choice in such questions. Final Answer: The isotopes chlorine 35 and chlorine 37 exist in nature in the approximate ratio 3 : 1.

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