Question 1

Among the following types of electromagnetic radiation, which one has the maximum wavelength (that is, the lowest frequency and energy)?

Accepted Answer

Introduction / Context: The electromagnetic spectrum covers a wide range of radiation types, from very high energy gamma rays to very low energy radio waves. Each region of the spectrum can be described by its typical range of wavelength, frequency and photon energy. Many basic physics and chemistry questions ask you to compare these forms of radiation. Knowing which part of the spectrum has the longest wavelength helps build a clear mental picture of how different electromagnetic waves relate to each other. Given Data / Assumptions: The options given are ultraviolet, radio waves, X rays and infrared radiation. We are asked which of these has the maximum wavelength. We assume the standard ordering of the electromagnetic spectrum as taught in school level physics. There are no special conditions that would alter the usual wavelength ranges. Concept / Approach: In the electromagnetic spectrum, wavelength, frequency and energy are related. As wavelength increases, frequency and photon energy decrease. Starting from the shortest wavelengths, the order is gamma rays, X rays, ultraviolet, visible light, infrared, microwaves and radio waves. Therefore, among the four choices given, radio waves lie at the long wavelength end of the spectrum and must have the largest wavelength. Ultraviolet and X rays have much shorter wavelengths, while infrared lies between visible light and microwaves, still shorter than radio waves. Step-by-Step Solution: Step 1: Recall the standard order of electromagnetic radiation when arranged by increasing wavelength: gamma rays, X rays, ultraviolet, visible, infrared, microwaves, radio waves.Step 2: Identify the positions of the given options in this sequence.Step 3: X rays are near the short wavelength, high energy end of the spectrum, just after gamma rays.Step 4: Ultraviolet radiation has shorter wavelengths than visible light but longer than X rays.Step 5: Infrared radiation has longer wavelengths than visible light but shorter than microwaves and radio waves.Step 6: Radio waves occupy the region with the greatest wavelengths and lowest frequencies among the options listed.Step 7: Therefore, radio waves have the maximum wavelength among the given types of radiation. Verification / Alternative check: You can also check using frequency. Radio stations broadcast using waves whose frequencies are measured in kilohertz or megahertz, which are much lower than the frequencies of visible or ultraviolet light. Because the speed of light in a vacuum is essentially constant, lower frequency implies longer wavelength. In contrast, X rays and ultraviolet radiation are associated with very high frequencies and therefore very short wavelengths. This confirms that radio waves have the longest wavelength among the options given. Why Other Options Are Wrong: Option a, ultraviolet, lies between X rays and visible light and has shorter wavelengths than both visible and infrared radiation. Option c, X rays, are even more energetic than ultraviolet and have very short wavelengths. Option d, infrared, has longer wavelengths than visible light but is still shorter in wavelength than radio waves. Hence none of these can match the extremely long wavelengths characteristic of radio waves, which can span from meters to many kilometres. Common Pitfalls: Many students remember the order of colours in visible light but struggle with the full spectrum. A common error is to mix up infrared and ultraviolet, or to assume that infrared has the longest wavelength simply because it is associated with heat. Another pitfall is focusing on energy instead of wavelength and forgetting the inverse relationship. To avoid these mistakes, memorize a simple mnemonic for the entire spectrum and repeatedly visualise the progression from gamma rays to radio waves. Remember that radio waves always sit at the long wavelength end of this sequence. Final Answer: Among the listed types of electromagnetic radiation, radio waves have the maximum wavelength.

Question 2

Boron forms several different complex anions. Which one of the following boron containing anions cannot normally be formed or isolated under standard chemical conditions?

Accepted Answer

Introduction / Context: Boron chemistry includes a variety of oxoanions and complex anions with ligands such as hydroxide, fluoride and hydride. Some of these species are common and well characterised, while others are not chemically reasonable under normal conditions. Exam questions like this one test whether you can distinguish realistic boron anions from formulas that are unlikely or incorrect, based on your knowledge of common boron compounds and coordination tendencies. Given Data / Assumptions: The anions listed are BO2-, BF63-, BH4- and B(OH)4-. We assume typical aqueous or standard laboratory conditions. We rely on standard inorganic chemistry knowledge of borate, borohydride and borate related anions. The task is to identify which formula does not correspond to a stable or commonly known boron anion. Concept / Approach: Boron commonly forms oxoanions such as metaborate (BO2-) and tetrahydroxoborate (B(OH)4-), as well as hydride complexes like borohydride (BH4-). These anions are well known in inorganic and coordination chemistry. Fluoride complexes of boron do exist, but a very common and stable example is BF4-, not BF63-. The formula BF63- would involve six fluoride ligands around boron with a 3 negative charge, which is not a standard or stable coordination environment for boron. By recognising which anions are well established, you can spot the odd one out that is not normally formed. Step-by-Step Solution: Step 1: Consider BO2-. In borate chemistry, BO2- is a reasonable representation of metaborate, which can be found in salts such as sodium metaborate.Step 2: Consider B(OH)4-. This is the tetrahydroxoborate ion, commonly formed when boric acid reacts with hydroxide in aqueous solution.Step 3: Look at BH4-, known as the borohydride ion. It is widely encountered in sodium borohydride and other hydride reducing agents used in organic and inorganic reactions.Step 4: Examine BF63-. The most familiar boron fluoride complex in introductory inorganic chemistry is BF4-, the tetrafluoroborate ion, not BF63-.Step 5: Six coordination around boron with such a high negative charge is not consistent with the usual valence and bonding patterns of boron in simple compounds.Step 6: Therefore, BF63- is the anion among the options that boron does not normally form. Verification / Alternative check: You can cross check by recalling typical boron compounds that appear in textbooks or laboratory reagents. Borax, boric acid derivatives and borate salts involve BO3- or BO2- related units and occasionally B(OH)4-. Reducing agents such as sodium borohydride contain BH4-. Fluoroborate salts like sodium tetrafluoroborate contain BF4-. There is no standard example of a stable BF63- boron complex introduced at this level. This pattern confirms that BF63- is the unusual and chemically unreasonable formula in the list. Why Other Options Are Wrong: Option a, BO2-, corresponds to the metaborate ion, which is a recognised boron oxoanion. Option c, BH4-, is the well known borohydride ion used widely in reduction reactions. Option d, B(OH)4-, is a common anion formed when boric acid interacts with strong bases in water. Since all of these appear in standard boron chemistry, they cannot be the answer to a question asking which anion boron cannot form. Common Pitfalls: A common mistake is to focus only on the charge and assume that any unusual charge makes an ion impossible, without considering known examples like BH4-. Another pitfall is to confuse BF4- with BF63- and assume that any fluoroborate formula is acceptable. To avoid such errors, pay attention to the coordination number and recall at least a few concrete examples of boron anions you have encountered. When an unfamiliar formula appears that does not match known patterns, that is a strong clue that it is the incorrect one. Final Answer: The boron containing anion that is not normally formed is BF63-.

Question 3

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?

Accepted Answer

Introduction / Context: Entropy change during isothermal reversible processes is a key topic in thermodynamics. For an ideal gas undergoing an isothermal reversible expansion, the entropy change depends on the number of moles, the gas constant and the volume ratio. This question tests whether you can recall and correctly apply the formula for entropy change in such a process. It also checks your ability to handle natural logarithms and interpret units in J/mol/K for a multi mole system. Given Data / Assumptions: Number of moles, n = 2 mol of ideal gas. Initial volume, V1 = 10 dm3. Final volume, V2 = 100 dm3. Temperature is isothermal at 27°C, which is 300 K, but the formula for entropy change in volume based isothermal expansion does not require T directly. The process is reversible and the gas behaves ideally. Concept / Approach: For an ideal gas undergoing an isothermal reversible change, the entropy change of the gas can be written in terms of volume as ΔS = n * R * ln(V2 / V1). Here, n is the number of moles, R is the gas constant in J/mol/K, and ln denotes the natural logarithm. The temperature is constant, so this formula is appropriate. Our task is to compute the volume ratio V2 / V1, take the natural logarithm, multiply by n and R, and then match the result with the closest option given in the question. Step-by-Step Solution: Step 1: Write the formula for entropy change for isothermal reversible expansion of an ideal gas: ΔS = n * R * ln(V2 / V1).Step 2: Substitute the known values: n = 2 mol, V1 = 10 dm3 and V2 = 100 dm3.Step 3: Compute the volume ratio: V2 / V1 = 100 / 10 = 10.Step 4: Use R ≈ 8.314 J/mol/K.Step 5: Calculate ln(10). The natural logarithm of 10 is approximately 2.3026.Step 6: Multiply the terms: ΔS = 2 * 8.314 * 2.3026.Step 7: First compute 8.314 * 2.3026 ≈ 19.148.Step 8: Then multiply by 2: 2 * 19.148 ≈ 38.296 J/mol/K.Step 9: Round the result to one decimal place, giving ΔS ≈ 38.3 J/mol/K. Verification / Alternative check: As a quick check, you can estimate ln(10) as about 2.3 and R as about 8.3. Then ΔS ≈ 2 * 8.3 * 2.3. Multiplying 8.3 by 2.3 gives approximately 19.1, and multiplying this by 2 yields about 38.2. This rough estimate is very close to 38.3 J/mol/K, supporting the more precise calculation. None of the other options are as close to this value, confirming that 38.3 J/mol/K is the appropriate choice from the list. Why Other Options Are Wrong: Option a, 42.3 J/mol/K, would require a larger logarithm or different values of n or R, which are not consistent with the data. Options c, 35.8 J/mol/K, and d, 32.3 J/mol/K, are significantly lower than the calculated result and would correspond to smaller volume ratios or fewer moles of gas. Since we know the volume ratio is exactly 10 and the number of moles is 2, these values cannot be correct. Only 38.3 J/mol/K matches the formula based calculation for the given conditions. Common Pitfalls: One common mistake is to forget to use the natural logarithm and instead use base 10 logarithms, which leads to an incorrect value of ln(V2 / V1). Another error is to misinterpret the volume ratio, using V1 / V2 instead of V2 / V1, which would give a negative entropy change, contradicting the fact that entropy increases during expansion. Students may also forget to multiply by the number of moles, using n = 1 instead of n = 2. Carefully writing the formula and substituting each value step by step helps avoid these issues. Final Answer: The entropy change for the isothermal reversible expansion described is 38.3 J/mol/K.

Question 4

Bleeding from a small wound can be stopped by applying ferric chloride solution. Chemically, why does ferric chloride help to stop the bleeding?

Accepted Answer

Introduction / Context: Ferric chloride solution is sometimes used as a styptic agent to stop minor bleeding in laboratory and clinical settings. Its action is based on basic principles of protein chemistry and coagulation. Understanding why ferric salts help to stop bleeding links simple inorganic chemistry with human physiology. This kind of question is common in general science and pre medical examinations, where the focus is on the mechanism rather than on memorising a brand name product. Given Data / Assumptions: Ferric chloride (FeCl3) solution is applied to a small bleeding wound. The question asks for the chemical reason why bleeding stops. We assume minor capillary bleeding rather than deep arterial bleeding. We consider the interaction of ferric ions with blood components, especially proteins. Concept / Approach: Ferric chloride is a salt of trivalent iron (Fe3+) and functions as an astringent. When applied to a wound, the ferric ions react with proteins in the blood, particularly albumin and fibrinogen, causing them to denature and precipitate. This protein coagulation forms a clot like mass on the surface of the wound, which mechanically blocks further blood flow and helps seal small vessels. Therefore, the correct explanation must involve coagulation of blood and sealing of blood vessels, not simply reversal of flow or sealing without reaction. Step-by-Step Solution: Step 1: Recognise that blood contains many proteins that can be denatured or coagulated by certain chemicals.Step 2: Understand that ferric chloride provides Fe3+ ions, which interact strongly with protein molecules at the wound surface.Step 3: This interaction causes the proteins to denature and aggregate, forming a semi solid mass or clot.Step 4: The coagulated mass partially plugs the small blood vessels and physically seals the wound area.Step 5: As a result, blood loss slows down or stops because the pathway for flow has been blocked.Step 6: Therefore, the most accurate statement is that blood is coagulated and the blood vessels are effectively sealed. Verification / Alternative check: You can verify this understanding by comparing ferric chloride with other astringent agents that also work by protein coagulation, such as alum or tannic acid. In all such cases, the visible effect is formation of a dark or clotted layer on the surface, which reflects denatured proteins. Medical literature and first aid practices also describe the mechanism of many styptic pencils as causing local coagulation of blood and tissue proteins. This supports the idea that the key effect is coagulation leading to sealing of small vessels. Why Other Options Are Wrong: Option a suggests that ferric chloride simply seals vessels without any reaction, which ignores the crucial role of protein coagulation. Option b, claiming that blood flows in the opposite direction, has no physiological basis and does not help stop bleeding. Option c mentions that a solid is formed, which is partly correct, but it does not explicitly connect this to coagulation and sealing of vessels as clearly as option d does. Option d directly states that blood is coagulated and vessels are sealed, capturing both the chemical and physiological aspects of the process. Common Pitfalls: Students may confuse the action of ferric chloride with that of vasoconstrictor drugs, which narrow blood vessels by acting on smooth muscle, or with physical methods like bandaging. Another pitfall is to use very vague language, such as saying that the solution "dries" the blood, which does not explain the chemical interaction. To avoid these problems, remember that ferric salts are strong protein precipitants and that the word "coagulation" specifically refers to the process of converting liquid blood components into a semi solid mass that can block bleeding. Final Answer: Ferric chloride stops bleeding because it causes coagulation of blood, forming a clot that seals the small blood vessels.

Question 5

In discussions about clean energy, which gas is often referred to as the "fuel of the future" because of its high energy content and potential for low emission use?

Accepted Answer

Introduction / Context: As the world looks for cleaner alternatives to fossil fuels, several energy carriers are proposed as future solutions. Among them, hydrogen is frequently described as the "fuel of the future". This description appears in textbooks, competitive exams and media discussions about sustainable energy systems. The phrase highlights hydrogen's potential to provide high energy output with minimal direct emissions when used in fuel cells or combustion systems designed to manage pollutants. Given Data / Assumptions: The options include methane, oxygen, neon and hydrogen. The term "fuel of the future" refers to a gas that can serve as an energy carrier or fuel. We assume a general context of energy policy and clean technology, not a very narrow industrial use. The question focuses on common usage of the phrase in science and general knowledge sources. Concept / Approach: Hydrogen gas has a very high energy content per unit mass and, when used in a fuel cell, produces water as its main product, with no direct carbon dioxide emissions. This makes it attractive as a potential replacement for fossil fuels in transport, power generation and industrial processes. Methane is currently a widely used fossil fuel but still produces carbon dioxide when burned. Oxygen and neon are not fuels; oxygen supports combustion, and neon is an inert noble gas used mainly in lighting. Therefore, among the given options, hydrogen is the gas most commonly called the "fuel of the future". Step-by-Step Solution: Step 1: Eliminate gases that are not fuels. Oxygen is an oxidising agent rather than a fuel, and neon is chemically inert and not used as a fuel.Step 2: Consider methane. Although methane is used widely as natural gas, it is a hydrocarbon that releases carbon dioxide when burned, so it is not typically described as a future clean fuel.Step 3: Examine hydrogen. It can be produced from water and other sources and, in ideal systems, generates only water when used in fuel cells.Step 4: Note that both textbooks and discussions about the hydrogen economy often use the phrase "hydrogen, the fuel of the future".Step 5: Conclude that hydrogen is the gas being referenced by the phrase in the question. Verification / Alternative check: An additional check is to think about what governments and research programs are investing in for long term, low carbon energy systems. Many countries have national hydrogen strategies and pilot projects for hydrogen fuel cell vehicles, hydrogen powered trains and hydrogen blended natural gas networks. This focus is much stronger than for neon or oxygen as fuels. Natural gas (methane) remains important today but is not described as a fundamentally new fuel for a low carbon future in the same way hydrogen is. This confirms that hydrogen best fits the description. Why Other Options Are Wrong: Option a, methane, is a conventional fossil fuel. While it is cleaner than coal and oil in some respects, it still produces carbon dioxide and is not the main focus of "fuel of the future" discussions. Option b, oxygen, supports combustion but does not itself release energy through oxidation; it is therefore not classed as a fuel. Option c, neon, is an inert noble gas with limited specialised uses, not an energy carrier. None of these match the widespread association of hydrogen with future energy systems. Common Pitfalls: Students may confuse current prevalence with future potential and choose methane simply because natural gas is widely used today. Another mistake is to see oxygen as important in combustion and wrongly think of it as a fuel. To avoid these errors, remember that a fuel is a substance that is oxidised to release energy, and that the phrase "fuel of the future" in environmental and energy contexts almost always refers to hydrogen, especially when discussing fuel cells and hydrogen powered vehicles. Final Answer: The gas commonly referred to as the "fuel of the future" is hydrogen.

Question 6

In solid state chemistry, which of the following substances is an amorphous solid rather than a well ordered crystalline solid?

Accepted Answer

Introduction / Context: Solids can be broadly classified as crystalline or amorphous. Crystalline solids have a long range, regularly repeating arrangement of particles, while amorphous solids lack such long range order and behave more like very viscous liquids frozen in place. Distinguishing between these types of solids is important in materials science, glass technology and basic chemistry. This question asks you to identify which listed substance is amorphous. Given Data / Assumptions: The substances listed are quartz glass (SiO2), chrome alum, silicon carbide (SiC) and graphite. We assume standard forms of these materials as commonly encountered in laboratory and industrial contexts. We use the standard definitions of crystalline and amorphous solids from solid state chemistry. The focus is on structural order, not on conductivity or other properties. Concept / Approach: Quartz in its crystalline form is a well ordered crystal of silicon dioxide, but when SiO2 is rapidly cooled from the melt without time to crystallise, it forms glass. This glassy form of SiO2 is an amorphous solid, known as fused silica or quartz glass. Chrome alum is a crystalline double salt of chromium and potassium, silicon carbide is a covalent network crystal with a well defined lattice, and graphite is a crystalline allotrope of carbon with layered hexagonal sheets. Therefore, among the options, quartz glass is the amorphous solid. Step-by-Step Solution: Step 1: Recall that amorphous solids lack long range order, and glass is a classic example of an amorphous material.Step 2: Quartz glass (SiO2) refers specifically to the glassy, non crystalline form of silicon dioxide obtained by cooling the melt quickly.Step 3: Chrome alum is a double salt that crystallises from aqueous solution into well defined crystals, indicating a crystalline structure.Step 4: Silicon carbide (SiC) forms a highly ordered covalent network, used as an abrasive and in high temperature applications, and is crystalline.Step 5: Graphite, though soft, consists of carbon atoms arranged in well ordered hexagonal layers and is considered crystalline.Step 6: Thus, among the choices, only quartz glass (SiO2) is an amorphous solid. Verification / Alternative check: Another way to confirm is to think of how these materials behave with respect to melting and cleavage. Glass does not have a sharp melting point; instead, it softens over a range of temperatures and breaks with irregular fracture surfaces, which are typical properties of amorphous solids. Crystalline substances like chrome alum, silicon carbide and graphite have more definite melting points and cleave or fracture along specific planes related to their crystal lattices. These observations support the classification of quartz glass as amorphous and the others as crystalline. Why Other Options Are Wrong: Option b, chrome alum, is known to crystallise, often forming violet crystals used in laboratories. Option c, silicon carbide, occurs in well defined crystalline forms such as moissanite and is used as a hard, crystalline abrasive. Option d, graphite, has a layered crystal structure responsible for its lubricating properties and anisotropic electrical conductivity. None of these have the disordered, glassy structure characteristic of amorphous solids. Common Pitfalls: A common misunderstanding is to confuse softness or brittleness with being amorphous. For example, graphite is soft and easily marks paper, but it is still crystalline. Another pitfall is to assume that any material containing the word "quartz" must be crystalline, ignoring the fact that quartz glass refers specifically to a glassy, amorphous form of SiO2. To avoid such errors, focus on whether the solid has a regular long range ordering of its particle positions, not just on its mechanical strength or name. Final Answer: The amorphous solid among the given substances is quartz glass (SiO2).

Question 7

Natural rubber, found in latex and used in tyres and elastic materials, is a polymer formed by the repeated addition of which monomer?

Accepted Answer

Introduction / Context: Natural rubber is an important polymer obtained from the latex of rubber trees. It has been used for many years to manufacture tyres, elastic bands and numerous flexible products. In polymer chemistry, natural rubber is described in terms of the small unsaturated hydrocarbon molecules (monomers) that join together to form the long chains. Knowing the correct monomer helps you understand addition polymerisation in organic chemistry and distinguishes natural rubber from synthetic rubbers based on other monomers. Given Data / Assumptions: The question is about natural rubber, not synthetic variants. The options include butadiene, 2-methyl-1,3-butadiene, ethylene and isoprene. We assume standard IUPAC naming: isoprene is another name for 2-methyl-1,3-butadiene. The task is to match natural rubber with its correct monomer unit. Concept / Approach: Natural rubber is cis-1,4-polyisoprene, meaning it is made of many isoprene units joined together in a particular orientation. Isoprene is the common name for 2-methyl-1,3-butadiene. While butadiene itself can form synthetic rubbers such as polybutadiene, and ethylene forms polyethylene, natural rubber specifically arises from the polymerisation of isoprene units in biological systems. Therefore, the correct answer must either be isoprene or its IUPAC description, which both refer to 2-methyl-1,3-butadiene. Step-by-Step Solution: Step 1: Recall the structural formula of isoprene: CH2=C(CH3)-CH=CH2, which corresponds to 2-methyl-1,3-butadiene.Step 2: Recognise that natural rubber consists mainly of cis-1,4-polyisoprene chains.Step 3: Compare the options. Butadiene is CH2=CH-CH=CH2 and is used in synthetic rubbers but is not the monomer of natural rubber.Step 4: Ethylene, CH2=CH2, polymerises to form polyethylene, not rubber.Step 5: Both 2-methyl-1,3-butadiene and isoprene refer to the same molecule, but the standard textbook statement is that natural rubber is a polymer of isoprene.Step 6: Select isoprene as the correct monomer for natural rubber. Verification / Alternative check: Chemistry textbooks and exam guides typically summarise natural rubber in a short statement such as "Natural rubber is cis-1,4-polyisoprene". They often show the repeating unit as a segment derived from isoprene. Synthetic rubbers like styrene butadiene rubber (SBR) and polybutadiene are clearly distinguished from this natural material. Cross checking with these standard descriptions confirms that the monomer for natural rubber is isoprene, not plain butadiene or ethylene. Why Other Options Are Wrong: Option a, butadiene, is an important monomer for synthetic rubbers but does not describe natural rubber. Option b, 2-methyl-1,3-butadiene, is chemically the same as isoprene but the question and standard phrasing specifically highlight isoprene by name; using the common name emphasises the link that students are expected to remember. Option c, ethylene, is the monomer of polyethylene and is unrelated to natural rubber. Therefore, these options either point to different polymers or use less common naming in this context. Common Pitfalls: A frequent confusion arises because the names butadiene and isoprene sound similar and both are conjugated dienes. Students sometimes wrongly assume that plain butadiene must be the monomer of natural rubber. Another pitfall is to memorise that rubber comes from "a diene" without remembering which specific diene. To avoid this, associate the word "isoprene" directly with natural rubber and remember the phrase "cis-1,4-polyisoprene" as a key descriptor. Final Answer: Natural rubber is a polymer of the monomer isoprene (2-methyl-1,3-butadiene).

Question 8

In biochemistry, ATP is an important energy carrier molecule in cells. What does the abbreviation ATP stand for?

Accepted Answer

Introduction / Context: ATP is often called the energy currency of the cell because many biochemical reactions depend on it for energy transfer. It is a nucleotide with a specific structure, and knowing its full chemical name is basic knowledge in biology, biochemistry and general science. Exam questions frequently test the expansion of common abbreviations such as ATP, DNA and RNA to ensure that students connect the short forms with their detailed meanings. Given Data / Assumptions: The abbreviation in question is ATP. The options include adenosine triphosphate, alanine triphosphate, adrenalin triphosphate and asparagine triphosphate. We assume standard biochemical nomenclature for nucleotides and amino acids. The task is to match ATP with its correct full form used in biology and chemistry. Concept / Approach: ATP is a nucleotide composed of the nucleoside adenosine (adenine plus ribose) and three phosphate groups attached in a chain. The name reflects these components: "adenosine" for the base sugar part and "triphosphate" for the three phosphate units. Amino acids such as alanine and asparagine do not form standard triphosphate molecules called ATP in basic biochemistry. Adrenaline is a hormone and neurotransmitter, not part of the ATP molecule. Thus, the only scientifically correct expansion of ATP in cell biology is adenosine triphosphate. Step-by-Step Solution: Step 1: Recall that ATP is structurally related to nucleotides in nucleic acids, consisting of a base, sugar and phosphate groups.Step 2: Identify that the base sugar combination in ATP is adenosine, made from adenine attached to ribose.Step 3: ATP contains three phosphate groups linked in series, hence the term triphosphate.Step 4: Combine these two parts into the full name: adenosine triphosphate.Step 5: Compare with the options and recognise that alanine, adrenaline and asparagine are unrelated molecules (two amino acids and one hormone).Step 6: Select "Adenosine triphosphate" as the correct expansion of ATP. Verification / Alternative check: You can verify by thinking of related nucleotides such as ADP (adenosine diphosphate) and AMP (adenosine monophosphate). These names follow the same pattern: adenosine plus the count of phosphate groups. Textbooks consistently refer to ATP, ADP and AMP in discussions of energy metabolism, muscle contraction and cellular respiration. This consistent naming pattern strongly supports the conclusion that ATP stands for adenosine triphosphate. Why Other Options Are Wrong: Option b, alanine triphosphate, would imply a triphosphate form of the amino acid alanine, which is not the meaning of ATP in standard biology. Option c, adrenalin triphosphate, incorrectly mixes the hormone adrenaline with phosphate groups and is not used in cellular energy discussions. Option d, asparagine triphosphate, similarly misuses the amino acid asparagine. None of these alternatives match the structure and function of the molecule that actually carries the ATP abbreviation in living cells. Common Pitfalls: Students sometimes confuse adenosine (a nucleoside) with adenine (a nitrogenous base) or with adrenaline (a hormone) because the names sound similar. Another pitfall is to assume that any familiar biological molecule name combined with "triphosphate" might be correct. To avoid these errors, remember that ATP belongs to the nucleotide family, not to amino acids or hormones, and that the standard form is adenosine triphosphate. Associating ATP directly with terms like "energy currency of the cell" and "adenosine" helps reinforce the correct name. Final Answer: In biochemistry, ATP stands for adenosine triphosphate.

Question 9

Ethanol can be converted into ethoxyethane (diethyl ether) in the laboratory. Under which conditions is ethanol heated with concentrated H2SO4 to produce ethoxyethane?

Accepted Answer

Introduction / Context: The conversion of ethanol to ethoxyethane (diethyl ether) is a classic reaction in organic chemistry that illustrates how reaction conditions can control the product. Concentrated sulfuric acid acts as a catalyst and dehydrating agent, and the temperature at which the reaction is carried out determines whether ether or ethene is formed. This question checks whether you remember the specific temperature and reactant conditions required for ether formation. Given Data / Assumptions: The starting alcohol is ethanol. The desired product is ethoxyethane (diethyl ether). Concentrated H2SO4 is used in the reaction. Different options give different temperatures and relative amounts of ethanol and acid. We assume standard textbook conditions for laboratory preparation. Concept / Approach: When ethanol is heated with concentrated sulfuric acid at about 413 K (around 140°C) and excess ethanol is present, the major product is diethyl ether formed by intermolecular dehydration of ethanol. At higher temperatures, around 443 K (about 170°C), the same mixture instead undergoes intramolecular dehydration to form ethene. At lower temperatures or room temperature, the reaction does not proceed effectively to ether. Therefore, the correct choice must mention heating excess ethanol with concentrated H2SO4 at approximately 140°C. Step-by-Step Solution: Step 1: Recall that ethanol + concentrated H2SO4 at around 140°C leads to the formation of diethyl ether.Step 2: Recognise that "excess ethanol" is important to favour bimolecular ether formation rather than elimination to form ethene.Step 3: Note that at around 170°C (443 K), the major product shifts to ethene, not ethoxyethane.Step 4: Options that mention only room temperature or 273 K correspond to conditions that are too mild to drive the dehydration to ether.Step 5: Compare the options and find that only option a describes heating excess ethanol with concentrated H2SO4 at 140°C.Step 6: Conclude that option a gives the correct conditions for preparing ethoxyethane from ethanol. Verification / Alternative check: Textbook mechanisms show that sulfuric acid protonates ethanol to form an ethyl hydrogen sulfate intermediate. A second ethanol molecule can then attack this intermediate to give diethyl ether and regenerate the acid, but this bimolecular step is favoured at moderate temperatures and high ethanol concentration. At higher temperatures, elimination to ethene dominates. Many exam preparation books summarise this by stating: "At 413 K, ethanol plus concentrated H2SO4 gives ether; at 443 K, it gives ethene." This widely repeated rule helps confirm the choice of 140°C with excess ethanol for ether formation. Why Other Options Are Wrong: Option b and option c describe treatment with concentrated H2SO4 at room temperature or 273 K, which are not the standard conditions for significant ether formation; the reaction rate would be too low. Option d, heating ethanol with excess concentrated H2SO4 at 443 K, corresponds to a higher temperature where elimination to ethene is favoured over ether formation. Therefore, these alternatives do not match the known conditions for preparing ethoxyethane from ethanol. Common Pitfalls: A typical mistake is to remember that "concentrated H2SO4 plus heat" is used but to forget the difference between the two temperature ranges and their corresponding products. Another pitfall is to assume that more heat is always better for completing the reaction, which leads students to choose the higher temperature option that actually produces ethene. To avoid this, memorise the approximate values and link them with the product: around 140°C for ether and around 170°C for ethene. Final Answer: Ethanol is converted into ethoxyethane by heating excess ethanol with concentrated H2SO4 at about 140°C.

Question 10

In water chemistry, temporary hardness of water is mainly due to the presence of which dissolved salt?

Accepted Answer

Introduction / Context: Hardness of water is an important topic in environmental chemistry and everyday life because it affects soap consumption, scale formation and industrial processes. Hardness can be classified as temporary or permanent based on whether it can be removed by simple boiling. Temporary hardness is usually associated with bicarbonate salts of calcium and magnesium, while permanent hardness involves chlorides and sulphates. This question focuses on identifying the specific salt responsible for temporary hardness. Given Data / Assumptions: The question asks specifically about temporary hardness. The options include calcium carbonate, calcium bicarbonate, magnesium sulphate and calcium sulphate. We assume typical natural water containing dissolved salts from rocks and soil. We use the standard definitions: temporary hardness is removable by boiling, permanent hardness is not. Concept / Approach: Temporary hardness is mainly caused by the presence of soluble bicarbonates of calcium and magnesium, such as Ca(HCO3)2 and Mg(HCO3)2. When such water is boiled, these bicarbonates decompose to form insoluble carbonates, carbon dioxide and water, thereby removing hardness. In contrast, sulphates and chlorides of calcium and magnesium cause permanent hardness because they do not decompose upon boiling. Calcium carbonate itself is largely insoluble and does not generally remain dissolved in water to cause hardness; instead, it often precipitates as scale. Step-by-Step Solution: Step 1: Recall that temporary hardness is removed by boiling water, which suggests the presence of bicarbonate salts.Step 2: Calcium bicarbonate, Ca(HCO3)2, is soluble in water and is a classic cause of temporary hardness.Step 3: On boiling, calcium bicarbonate breaks down to form calcium carbonate, carbon dioxide and water, and the insoluble calcium carbonate can be removed.Step 4: Calcium carbonate, CaCO3, is not very soluble in water and, when it forms, usually precipitates rather than remaining in solution to cause hardness.Step 5: Magnesium sulphate and calcium sulphate contribute to permanent hardness, not temporary hardness, because they do not decompose on boiling.Step 6: Therefore, the correct answer for temporary hardness is calcium bicarbonate. Verification / Alternative check: You can verify by recalling the standard chemical equation for removal of temporary hardness: Ca(HCO3)2 (aq) on boiling gives CaCO3 (s) + CO2 (g) + H2O (l). This reaction is widely shown in textbooks to explain why boiling can soften water containing bicarbonate hardness. There is no similar simple decomposition for sulphates or chlorides at normal boiling temperatures. Thus, calcium bicarbonate fits perfectly with the concept of temporary hardness, whereas the other salts do not. Why Other Options Are Wrong: Option a, calcium carbonate, is mostly insoluble and is more commonly associated with scale deposits rather than dissolved hardness. Option c, magnesium sulphate, and option d, calcium sulphate, both cause permanent hardness and cannot be removed just by boiling. Since the question specifically refers to temporary hardness, any salt associated with permanent hardness must be ruled out. Only calcium bicarbonate matches the definition and behaviour of temporary hardness in water. Common Pitfalls: Students sometimes confuse the terms carbonate and bicarbonate, or they assume that any calcium salt must cause the same type of hardness. Another mistake is to think that calcium carbonate itself is responsible for temporary hardness because it appears in the boiling reaction. In fact, calcium carbonate is the precipitated product of boiling, not the original cause of hardness. To avoid confusion, always link temporary hardness with bicarbonates and permanent hardness with sulphates and chlorides of calcium and magnesium. Final Answer: Temporary hardness of water is mainly due to the presence of calcium bicarbonate in solution.

Question 11

The hard calcium carbonate shell of an egg is dissolved in hydrochloric acid. If the de shelled egg is then placed in a concentrated NaCl solution, what observable change will occur in the egg and why?

Accepted Answer

Introduction / Context: This question illustrates osmosis, a key concept in biology and chemistry. When an egg shell is removed with acid, the remaining membrane behaves like a semipermeable membrane, allowing water to move in or out depending on the surrounding solution. Placing the de shelled egg in a concentrated salt solution creates an osmotic gradient that drives water movement. Recognising whether the egg shrinks or swells tests your understanding of hypertonic and hypotonic solutions and the direction of osmotic flow. Given Data / Assumptions: The hard shell, composed mainly of calcium carbonate, is dissolved by hydrochloric acid, leaving behind the soft membrane and interior of the egg. The egg is then placed in a concentrated NaCl solution, which is hypertonic compared with the egg interior. The membrane around the egg acts as a semipermeable membrane. We assume no mechanical damage to the membrane, so osmosis can occur normally. Concept / Approach: Osmosis is the movement of water across a semipermeable membrane from a region of lower solute concentration (higher water potential) to a region of higher solute concentration (lower water potential). When the egg, whose interior fluid has a lower salt concentration, is placed in a concentrated NaCl solution, the outside solution has a higher solute concentration. The solution is hypertonic to the egg contents. Water will move out of the egg and into the salt solution. As water leaves, the egg loses volume and appears shrunken or shrivelled. Step-by-Step Solution: Step 1: Recognise that dissolving the eggshell in HCl exposes the semipermeable membrane surrounding the egg white and yolk.Step 2: Understand that concentrated NaCl solution has a higher solute concentration than the fluids inside the egg.Step 3: Identify the concentrated NaCl solution as hypertonic relative to the egg interior.Step 4: Recall that in a hypertonic environment, water moves out of the cell or structure through osmosis.Step 5: Apply this principle to the egg: water will move from inside the egg through the membrane into the surrounding NaCl solution.Step 6: As water leaves, the egg loses volume and becomes smaller and wrinkled, meaning it shrinks. Verification / Alternative check: You can verify your reasoning by comparing with a related experiment where a de shelled egg is placed in pure water. In that case, water moves into the egg because the external solution is hypotonic, and the egg swells. The opposite behaviour is seen in a concentrated salt or sugar solution, where the egg shrinks due to water loss. These classic demonstrations are used in schools to teach osmosis and support the conclusion that a concentrated NaCl solution makes the egg shrink rather than swell. Why Other Options Are Wrong: Option b, "The egg will swell", would be correct for a hypotonic external solution such as pure water, not for a concentrated salt solution. Option c, "The egg will become harder", confuses mechanical hardness with osmotic effects; the membrane remains soft, only the volume changes. Option d, "There will be hardly any visible change", is incorrect because concentrated NaCl typically produces a noticeable shrinking effect over time as water leaves the egg. Common Pitfalls: A common mistake is to mix up the terms hypertonic and hypotonic and their effects on cells or tissues. Students sometimes assume that the presence of any solution around the egg will cause swelling, forgetting that direction of water movement depends on relative solute concentrations. Another pitfall is to overlook the role of the membrane and think only the dissolved shell matters. To avoid these mistakes, always compare the solute concentration inside and outside and then apply the rule that water moves from lower solute concentration to higher solute concentration across a semipermeable membrane. Final Answer: When the de shelled egg is placed in a concentrated NaCl solution, the egg will shrink because water moves out of it into the hypertonic salt solution by osmosis.

Question 12

In a nuclear reactor, the rate of the fission chain reaction is controlled by inserting rods made of which material that absorbs neutrons effectively?

Accepted Answer

Introduction / Context: Nuclear reactors generate energy by sustaining a controlled chain reaction of nuclear fission, typically in fuel rods containing uranium or plutonium. To keep this chain reaction stable and prevent it from running out of control, reactors use control rods that absorb neutrons and moderate the number of neutrons available for further fission events. This question asks which material is used for such control rods in many reactor designs. Given Data / Assumptions: The context is a nuclear reactor where a chain reaction must be controlled. Options include rods made of cadmium, iron, platinum and graphite. We assume a typical thermal reactor design used in basic physics and chemistry discussions. The key property we seek is strong neutron absorption. Concept / Approach: Control rods in nuclear reactors are made of materials with high neutron absorption cross sections, so they can remove neutrons from the system when inserted into the reactor core. Common materials include cadmium, boron and hafnium. Cadmium, in particular, is well known for its ability to absorb thermal neutrons. Graphite, by contrast, is used as a moderator in some reactor types to slow down neutrons, not primarily as a control rod material. Iron and platinum do not have the same combination of properties and are not standard choices for control rods in basic textbook descriptions. Step-by-Step Solution: Step 1: Recall that controlling a chain reaction requires adjusting the number of free neutrons available for fission.Step 2: Control rods must therefore be made of materials that absorb neutrons efficiently.Step 3: Cadmium is a classic neutron absorbing material used in control rods because of its high capture cross section for thermal neutrons.Step 4: Graphite is usually used as a moderator to slow down neutrons, not as the main control rod material in basic discussions.Step 5: Iron and platinum do not feature in standard lists of control rod materials for controlling chain reactions.Step 6: Therefore, the correct choice for controlling the chain reaction is the cadmium rod. Verification / Alternative check: Textbooks on nuclear physics typically list cadmium and boron as materials used in control rods. Illustrations of reactor cores often show thin rods of these materials being raised or lowered to adjust neutron flux. Graphite is mentioned separately in descriptions of graphite moderated reactors, where it surrounds the fuel and slows down neutrons to thermal energies. This clear separation of functions between control rods and moderators reinforces that cadmium rods are associated with direct control of the chain reaction. Why Other Options Are Wrong: Option b, iron rod, refers to a structural metal that does not have particularly high neutron absorption and is not used primarily as a control rod material. Option c, platinum rod, is expensive and not standard in reactor control applications. Option d, graphite rod, might be used as a moderator or structural component, but in basic nuclear reactor questions, graphite is associated with slowing neutrons rather than absorbing them to shut down the reaction. Therefore, these alternatives do not match the typical control rod material described in introductory courses. Common Pitfalls: Students may confuse the roles of moderator and control rod or assume that any material present in the reactor can control the chain reaction. Another pitfall is focusing on the mechanical strength or cost of the material rather than its neutron absorption properties. To avoid these mistakes, remember that control rods must "soak up" neutrons, and cadmium is one of the classic examples of a strong neutron absorber used for this purpose. Final Answer: In a nuclear reactor, the chain reaction is controlled by inserting cadmium rods, which absorb neutrons effectively.

Question 13

During metal extraction, which of the following metallurgical processes is specifically used to remove reducible metal oxides from a molten metal by using a wooden pole or similar reducing agent?

Accepted Answer

Introduction / Context: In metallurgy, various refining processes are used to remove impurities from metals after their initial extraction from ore. Some impurities appear as oxides dissolved in the molten metal and can be removed by specific treatments. The process of poling is one such method, particularly used for refining metals like copper and tin. This question checks whether you can match the description of removing reducible oxides using a reducing agent with the correct metallurgical term. Given Data / Assumptions: The goal is to remove reducible metal oxides from a molten metal. A wooden pole or similar material is introduced into the melt to act as a reducing agent. The options include liquation, cupellation, poling and electro refining. We assume standard descriptions of these processes from metallurgy texts. Concept / Approach: Poling involves stirring molten metal with green wood poles or other carbonaceous materials. The gases and carbon released reduce metal oxides present in the melt back to the metal, improving purity. This method is especially common in the refining of blister copper, where cuprous oxide impurities are reduced to copper metal. Liquation separates metals with different melting points, cupellation oxidises impurities like lead and removes them as oxides, and electro refining uses electrolytic cells to purify metals. The description in the question clearly aligns with poling, not with the other processes. Step-by-Step Solution: Step 1: Interpret the description: reducible oxides are being removed from molten metal by reduction, not by oxidation or electrolysis.Step 2: Recall that in poling, a wooden pole is inserted into the molten metal, and the released gases act as reducing agents.Step 3: Recognise that this reduces metal oxides, for example Cu2O, back to the metal, such as copper.Step 4: Compare with liquation, which relies on differential melting points and does not involve such reduction by wooden poles.Step 5: Cupellation involves strong oxidation in a furnace to remove base metals, especially in silver refining, and does not match the reducing description.Step 6: Electro refining uses electrical current and electrolyte solutions, which again is completely different from the described process.Step 7: Therefore, the process described is poling. Verification / Alternative check: As an alternative check, think specifically about copper refining. Blister copper contains copper oxide impurities that make the metal brittle. In the poling step, a wooden pole is thrust into the molten copper, releasing reducing gases that convert Cu2O into Cu and CO2. This is a classic textbook example and matches the question stem almost word for word. No such description appears when explaining liquation, cupellation or electro refining, reinforcing that poling is the correct answer. Why Other Options Are Wrong: Option a, liquation, is used to separate metals based on differences in melting point and is not focused on reducing oxides. Option b, cupellation, is an oxidative process used mainly in refining silver or lead by oxidising and removing impurities, not by reducing oxides. Option d, electro refining, uses an electrolytic cell to transfer pure metal from an impure anode to a cathode and does not rely on wooden poles or gaseous reduction. Therefore, these options do not match the reduction based process described. Common Pitfalls: Students may confuse poling with cupellation because both can appear in the context of refining, or they may think any purification process could fit the description. Another pitfall is to focus on terms like "refining" without paying attention to whether the process is oxidative or reductive. To avoid such mistakes, link specific keywords with each process: "wooden pole and reduction" with poling, "oxidation of lead" with cupellation and "electrolytic cell" with electro refining. Final Answer: The metallurgical process used to remove reducible oxides from molten metal by using a wooden pole as a reducing agent is called poling.

Question 14

Among the following elements, which one is well known as a naturally radioactive element discovered by Marie and Pierre Curie?

Accepted Answer

Introduction / Context: Radioactivity is a fundamental concept in nuclear chemistry and physics. Some elements are naturally radioactive, emitting radiation due to unstable nuclei. Famous discoveries in this field were made by Marie and Pierre Curie, who identified several radioactive elements. Many general knowledge and chemistry questions ask you to recall which elements are particularly associated with natural radioactivity and with the work of the Curies. Polonium is one of these historically significant elements. Given Data / Assumptions: The elements listed are sulphur, polonium, tellurium and selenium. We are asked which is a radioactive element, not merely an element with radioactive isotopes. We assume a general knowledge context where polonium is highlighted as a radioactive element. The historical discovery by the Curies is relevant to the question. Concept / Approach: Polonium is a highly radioactive element discovered by Marie and Pierre Curie during their investigations of uranium ores. It has no stable isotopes and is known for intense alpha emission. Sulphur, tellurium and selenium, although they may have isotopes with different stabilities, are not typically cited as radioactive elements in introductory courses. When exam questions list polonium alongside ordinary non radioactive elements, the intention is usually to select polonium as the radioactive one. Step-by-Step Solution: Step 1: Recall that polonium, along with radium, was discovered by the Curies in their work on radioactivity.Step 2: Recognise that polonium has no stable isotopes and is strongly radioactive, primarily emitting alpha particles.Step 3: Consider sulphur, tellurium and selenium. These elements are generally treated as non radioactive in basic chemistry courses.Step 4: Match this information with the options and identify polonium as the element specifically known for being radioactive.Step 5: Therefore, select polonium as the correct answer. Verification / Alternative check: Historical accounts of the discovery of radioactivity often list uranium, thorium, polonium and radium as key radioactive elements. Polonium is named after Poland, the homeland of Marie Curie. It features in many examples concerning radioactivity in general science textbooks. Sulphur, selenium and tellurium are discussed mainly for their chemical properties and uses, not as examples of radioactivity. This historical and educational emphasis confirms that polonium is the intended answer. Why Other Options Are Wrong: Option a, sulphur, is a common non metal used in industry and has stable isotopes; it is not described as a radioactive element at the high school level. Option c, tellurium, and option d, selenium, are chalcogen elements with interesting semiconductor properties, but again they are not typically highlighted as radioactive in basic courses. Therefore, these options do not match the description of a well known radioactive element discovered by the Curies. Common Pitfalls: Students sometimes overthink the question and recall that many elements have some radioactive isotopes, leading them to doubt the obvious choice. However, in exam settings, "radioactive element" usually refers to elements like uranium, thorium, radium and polonium where radioactivity is a dominant property. Another pitfall is mixing up polonium with other p block elements because of their similar sounding names. To avoid this, remember the strong historical link: polonium was named by Marie Curie and is a classic example of a naturally radioactive element. Final Answer: The radioactive element among the options is polonium.

Question 15

In the history of nuclear chemistry, who is generally given credit for the discovery and systematic study of transuranic elements (elements beyond uranium in the periodic table)?

Accepted Answer

Introduction / Context: Transuranic elements are those with atomic numbers greater than that of uranium (greater than 92). They do not occur in significant amounts in nature and are typically synthesized in nuclear reactors or particle accelerators. The systematic discovery and organisation of many of these elements was a major achievement in twentieth century nuclear chemistry. Questions like this one ask you to connect the name of a key scientist with pioneering work on transuranic elements. Given Data / Assumptions: The scientists listed are Hahn, Rutherford, Seaborg and Curie. The focus is specifically on transuranic elements, not on radioactivity in general or nuclear fission alone. We assume standard historical accounts used in chemistry and physics education. The credit sought is for discovery and systematic study of several transuranic elements. Concept / Approach: Glenn T. Seaborg is widely credited with the discovery or co discovery of several transuranic elements, including plutonium, americium, curium, berkelium and californium. He also proposed the actinide concept, reorganising the periodic table to place actinides in a separate series. Other scientists in the options are associated with different landmark achievements: Otto Hahn with nuclear fission, Ernest Rutherford with the nuclear model of the atom, and the Curies with radioactivity and elements like polonium and radium. Therefore, when the question is specifically about transuranic elements, Seaborg is the correct choice. Step-by-Step Solution: Step 1: Recall that transuranic elements are artificially produced elements with atomic numbers beyond 92.Step 2: Identify Glenn T. Seaborg as the chemist who played a central role in discovering many of these elements at the University of California, Berkeley.Step 3: Recognise that Seaborg introduced the actinide concept, which reorganised the periodic table to include the actinide series of heavy elements.Step 4: Compare with Otto Hahn, whose major contribution was the discovery of nuclear fission in uranium, not the broader set of transuranic elements.Step 5: Note that Rutherford is famous for the nuclear model and scattering experiments, while the Curies discovered polonium and radium and studied radioactivity.Step 6: Conclude that Seaborg is the scientist specifically linked to the discovery and systematisation of transuranic elements. Verification / Alternative check: Historical summaries of the periodic table often highlight Seaborg as a Nobel Prize winning chemist who added multiple elements to the table and changed its layout. His name is sometimes seen in connection with element 106 (seaborgium), which further commemorates his work in heavy element research. In contrast, although Hahn, Rutherford and the Curies made crucial contributions to nuclear science, they are not described as the primary discoverers of transuranic elements. This comparison confirms Seaborg as the correct answer. Why Other Options Are Wrong: Option a, Hahn, is associated with the discovery of nuclear fission, which enabled later work but did not by itself catalogue transuranic elements. Option b, Rutherford, laid the theoretical foundations of nuclear physics but did not focus on discovering a wide range of transuranic elements. Option d, Curie, refers to Marie or Pierre Curie, whose main contributions were the discovery of radium and polonium and the study of radioactivity. None of these scientists match Seaborg's specific role in transuranic element discovery. Common Pitfalls: Students sometimes pick Hahn or Curie because they strongly associate those names with nuclear science and radioactivity. Another pitfall is to overlook Seaborg because his work is more specialised compared to the more famous stories of early radioactivity. To avoid these errors, remember that the phrase "transuranic elements" should immediately bring Glenn T. Seaborg to mind, along with his work on the actinide series and the many elements he helped discover. Final Answer: The credit for the discovery and systematic study of transuranic elements goes to Glenn T. Seaborg.

Question 16

Some reactive elements must be stored under water or other liquids to prevent dangerous reactions with air. Which one of the following elements is commonly preserved under water for safety?

Accepted Answer

Introduction / Context: Certain chemical elements are so reactive with oxygen in the air that they must be stored under liquids to prevent spontaneous combustion or rapid oxidation. This is a practical safety issue in laboratories and is frequently tested in basic chemistry questions. Knowing which elements require special storage conditions helps you understand their reactivity and the precautions needed when handling them. Given Data / Assumptions: The elements listed are sodium, sulphur, red phosphorus and white phosphorus. The question asks which element is preserved under water. We assume common laboratory practice for storing these substances. We distinguish between storage under water and storage under oil or in ordinary containers. Concept / Approach: White phosphorus is highly reactive and can ignite spontaneously in air at relatively low temperatures. To prevent contact with oxygen, it is stored under water, which acts as a barrier between the solid and the air. Sodium is also reactive with air and especially with water, but it is typically stored under kerosene or mineral oil, not under water, because water would react violently with sodium. Sulphur and red phosphorus are much less reactive and can be stored in ordinary containers under normal conditions. Step-by-Step Solution: Step 1: Consider sodium. Although it is reactive, especially with water, safe practice is to store it under kerosene, not under water.Step 2: Consider sulphur. It is relatively stable in air and is stored in closed containers without special liquid protection.Step 3: Consider red phosphorus. It is less reactive than white phosphorus and can be stored safely in closed bottles without needing water cover.Step 4: Consider white phosphorus. It is extremely reactive with oxygen and likely to ignite spontaneously when exposed to air.Step 5: To prevent this, white phosphorus is stored under water so that it does not contact air until deliberately removed.Step 6: Therefore, the element commonly preserved under water is white phosphorus. Verification / Alternative check: Laboratory manuals often specifically instruct that white phosphorus must always remain covered with water and should be handled with forceps under water for cutting or transferring. In contrast, sodium is shown stored in jars filled with kerosene. General chemistry texts also explain the difference in reactivity between white and red phosphorus, emphasising the need for underwater storage of the white allotrope. These consistent safety recommendations confirm that white phosphorus is the element preserved under water. Why Other Options Are Wrong: Option a, sodium, is not stored under water because the reaction between sodium and water produces hydrogen gas and sodium hydroxide and can be explosive. Option b, sulphur, does not require such precautions and is stable in air under normal conditions. Option c, red phosphorus, is more stable and safer than white phosphorus and does not need to be stored under water. Therefore, none of these options match the specific storage requirement described in the question. Common Pitfalls: Students sometimes remember that sodium is stored under a liquid and incorrectly assume that the liquid is water rather than kerosene. Another pitfall is to mix up red and white phosphorus and forget which allotrope is highly reactive. To avoid these errors, associate white phosphorus with high reactivity, low ignition temperature and underwater storage, while associating sodium with storage under kerosene and red phosphorus with relatively safe storage in bottles. Final Answer: The element that is commonly preserved under water is white phosphorus.

Question 17

Heated sawdust catches fire when a drop of concentrated nitric acid is added to it. Chemically, which type of process is mainly responsible for this combustion?

Accepted Answer

Introduction / Context: Sawdust is largely made of organic material such as cellulose, which contains carbon and hydrogen. Concentrated nitric acid is a strong oxidising agent. When the two come into contact, especially with some heating, a vigorous chemical reaction can occur that releases heat and may cause ignition. Understanding this reaction involves recognising the role of oxidation and the ability of nitric acid to act as an oxidiser in organic reactions. Given Data / Assumptions: Sawdust is heated and a drop of concentrated nitric acid is added. The sawdust catches fire after the acid is applied. Concentrated nitric acid is known to be a strong oxidising agent. The question asks which type of process (dehydration, oxidation, reduction or dehydrogenation) is mainly responsible for ignition. Concept / Approach: Oxidation, in a broad sense, involves the gain of oxygen, loss of hydrogen or loss of electrons by a substance. Organic materials such as sawdust can be oxidised vigorously by strong oxidising agents, releasing heat and sometimes leading to combustion. Concentrated nitric acid can oxidise many organic compounds rapidly, sometimes explosively. While dehydration and dehydrogenation can occur in some reactions, the key process that leads to catching fire is rapid oxidation of the organic matter, which releases enough heat to ignite it. Step-by-Step Solution: Step 1: Recognise that sawdust consists mainly of cellulose, an organic compound rich in carbon and hydrogen.Step 2: Recall that concentrated nitric acid is a strong oxidising agent capable of oxidising organic materials.Step 3: When concentrated nitric acid comes into contact with heated sawdust, it promotes rapid oxidation of the cellulose.Step 4: This oxidation reaction is highly exothermic, meaning it releases a large amount of heat.Step 5: The heat generated can raise the temperature of the sawdust to its ignition point, causing it to catch fire.Step 6: Therefore, the dominant process responsible for the combustion is oxidation. Verification / Alternative check: Strong oxidising acids such as nitric acid and concentrated sulphuric acid are known hazards when mixed with organic materials like paper, sugar or sawdust. Safety manuals warn that such combinations can cause charring, ignition or even explosions due to rapid oxidation. If dehydration alone were the main effect, you might see drying or charring without open flames. The occurrence of actual fire indicates that substantial oxidation and combustion are taking place, not just loss of water or hydrogen in a controlled way. Why Other Options Are Wrong: Option a, dehydration, refers to the removal of water from a substance. Although concentrated acids can dehydrate some organics, dehydration by itself usually leads to charring rather than vigorous ignition in this context. Option c, reduction, would require the sawdust to gain electrons, which is the opposite of what happens when it is oxidised. Option d, dehydrogenation, is a specific type of oxidation involving loss of hydrogen, but the broader process here is overall oxidation and combustion, not just selective dehydrogenation. Hence these options do not capture the main cause of the fire. Common Pitfalls: Students sometimes focus on the strong acidic nature of nitric acid and conclude that dehydration must be the key process, confusing it with concentrated sulphuric acid, which is more famously a dehydrating agent. Another pitfall is to think of any reaction with acid as "acid base" rather than redox. To avoid these mistakes, remember that nitric acid is a powerful oxidising agent and that burning is a form of rapid oxidation. When organic matter catches fire after treatment with concentrated nitric acid, oxidation is the central process. Final Answer: Heated sawdust catches fire when concentrated nitric acid is added mainly due to rapid oxidation of the organic material by the strong oxidising agent.

Question 18

In basic chemistry, which one of the following one mole samples is heaviest in terms of mass: water molecules, ammonia molecules, sodium atoms, or hydrogen molecules?

Accepted Answer

Introduction / Context: This chemistry question tests your understanding of the concept of the mole and molar mass. Although one mole always represents the same number of particles, namely Avogadros number, the mass of one mole depends on the molar mass of the substance. Being able to compare molar masses quickly is a fundamental skill in stoichiometry, chemical calculations and exam problem solving. Given Data / Assumptions: - Four different one mole samples are given: water molecules, ammonia molecules, sodium atoms and hydrogen molecules. - Each sample contains exactly one mole, that is the same number of particles. - We use standard approximate atomic masses: H ≈ 1 u, O ≈ 16 u, N ≈ 14 u, Na ≈ 23 u. - We are asked to identify which one mole sample has the greatest mass in grams. Concept / Approach: The key ideas are the mole concept and molar mass. One mole of any substance contains the same number of particles, but the gram mass of that mole equals the molar mass in grams per mole. Therefore, we must calculate or recall the molar masses of water, ammonia, sodium and hydrogen gas and compare them. The substance with the highest molar mass will give the heaviest one mole sample. Step-by-Step Solution: Step 1: Calculate molar mass of water (H2O): 2 * 1 (for two H atoms) + 16 (for one O atom) = 18 g mol^-1. Step 2: Calculate molar mass of ammonia (NH3): 14 (for N) + 3 * 1 (for three H atoms) = 17 g mol^-1. Step 3: Determine molar mass of sodium atoms (Na): approximately 23 g mol^-1 for metallic sodium. Step 4: Calculate molar mass of hydrogen molecules (H2): 2 * 1 = 2 g mol^-1. Step 5: Compare results: 18 g, 17 g, 23 g and 2 g. The largest value is 23 g mol^-1, which corresponds to one mole of sodium atoms. Verification / Alternative check: You can quickly verify by remembering periodic table trends. Sodium is a metal in the third period with atomic mass around 23, while the other species involve mostly light non metals (H, N, O). The molar mass of Na is clearly larger than that of typical small covalent molecules like H2O, NH3 and H2. Therefore, one mole of sodium atoms has the highest mass in grams among the given options. Why Other Options Are Wrong: 1 mole of water (18 g) is lighter than 1 mole of sodium (23 g). 1 mole of ammonia (17 g) is also lighter than 1 mole of sodium. 1 mole of hydrogen molecules (2 g) is the lightest of all the given samples. Common Pitfalls: A common mistake is to assume that one mole of any substance always has the same mass because the particle count is the same. In reality, the mass depends on the atomic or molecular composition. Another error is to miscalculate molar masses by forgetting to multiply by the number of atoms in the formula. To avoid such issues, always write out the formula, multiply each atomic mass by the number of atoms and sum accurately. Final Answer: The heaviest one mole sample among the options is 1 mole of sodium (Na) atoms.

Question 19

In everyday chemistry, which one of the following substances is an example of a natural colloid rather than a true solution: cane sugar solution, blood, sodium chloride solution or urea solution?

Accepted Answer

Introduction / Context: This question examines your understanding of colloidal systems versus true solutions in chemistry. Colloids are mixtures where the dispersed particles are intermediate in size between those in true solutions and suspensions. Many important biological and industrial materials are colloids. Knowing which common substances are colloidal helps you connect classroom theory with real life examples. Given Data / Assumptions: - Four common mixtures are listed: cane sugar solution, blood, sodium chloride solution and urea solution. - We assume all solutions mentioned are prepared in water. - You are asked to identify which one behaves as a natural colloid rather than as a true solution of small particles. Concept / Approach: In a true solution, solute particles are extremely small (typically less than 1 nanometre), completely dissolved and cannot be seen even under a powerful microscope. They do not scatter a beam of light significantly. In a colloid, dispersed particles are larger (about 1 to 1000 nanometres), do not settle under gravity and can scatter light (Tyndall effect). Blood is a classic natural colloid, consisting of plasma as the dispersion medium and cells and proteins as dispersed particles. In contrast, cane sugar, sodium chloride and urea dissolved in water form true solutions with molecular or ionic size particles. Step-by-Step Solution: Step 1: Consider cane sugar solution. Sugar molecules dissolve completely in water and form a homogeneous true solution with very small particles. Step 2: Consider sodium chloride solution. Sodium and chloride ions are fully dissolved and form an ionic true solution where particles are at the ionic scale. Step 3: Consider urea solution. Urea molecules also dissolve completely, giving another example of a true molecular solution. Step 4: Consider blood. It contains red blood cells, white blood cells, platelets and large proteins suspended in plasma. These dispersed particles are much larger than ordinary solute molecules and give blood colloidal properties. Step 5: Conclude that blood is the natural colloid among the four substances. Verification / Alternative check: You can think about practical observations. When a beam of light passes through a strong sugar solution, you do not see the beam path clearly, indicating a true solution. However, when light passes through milk or blood, the path of the beam becomes visible due to scattering by colloidal particles. Blood also shows settling behaviour only under centrifugation rather than under normal gravity. These are characteristic features of colloidal systems, confirming that blood is a colloid. Why Other Options Are Wrong: Cane sugar solution: sugar molecules are small and fully dissolved, so this is a true solution, not a colloid. Sodium chloride solution: the dissolved ions form a typical ionic solution, again not colloidal. Urea solution: urea forms a homogeneous molecular solution with no colloidal particle size range. Common Pitfalls: Learners sometimes classify any cloudy liquid as a colloid and any clear liquid as a true solution without considering particle size and behaviour. Although blood appears opaque and red, the reason lies in suspended cells rather than simple dissolution. Another pitfall is to think only of milk as a natural colloid and to forget that blood is an equally important biological colloidal system. Remember that colloids can be gas liquid, liquid liquid or solid liquid systems and that many body fluids fall into this category. Final Answer: The natural colloid among the options is blood.

Question 20

In organic chemistry, the classical Silver Mirror (Tollens) test is given by which of the following carbonyl compounds: 1. Benzophenone, 2. Acetaldehyde, 3. Acetone, 4. Formaldehyde?

Accepted Answer

Introduction / Context: This question tests your understanding of the Tollens test, also called the Silver Mirror test, in organic chemistry. The test is used to distinguish aldehydes from ketones based on their ability to be oxidised by Tollens reagent. Recognising which functional groups give a positive Silver Mirror test is essential when identifying unknown carbonyl compounds in qualitative organic analysis. Given Data / Assumptions: - Four carbonyl compounds are listed: benzophenone, acetaldehyde, acetone and formaldehyde. - Tollens reagent (ammoniacal silver nitrate) is used in the Silver Mirror test. - A positive test is indicated by the deposition of a shining silver mirror on the inner surface of the reaction vessel. - We assume standard laboratory conditions and definitions where typical aldehydes are easily oxidised, whereas simple ketones are not. Concept / Approach: Tollens reagent is a mild oxidising agent that oxidises aldehydes to corresponding carboxylic acids while itself being reduced to metallic silver. Ordinary ketones are generally not oxidised by such a mild reagent and therefore do not give the Silver Mirror. Among the given compounds, acetaldehyde and formaldehyde are aldehydes, while benzophenone and acetone are ketones. Therefore, acetaldehyde and formaldehyde should give a positive Tollens test and produce the silver mirror. Step-by-Step Solution: Step 1: Identify the functional group in benzophenone. It contains a carbonyl group attached to two phenyl rings (a diaryl ketone), so it is a ketone. Step 2: Identify the functional group in acetaldehyde (CH3CHO). It has a terminal aldehyde group, so it is an aldehyde. Step 3: Identify the functional group in acetone (CH3COCH3). It is a simple dialkyl ketone, not an aldehyde. Step 4: Identify the functional group in formaldehyde (HCHO). It is the simplest aldehyde with a formyl group. Step 5: Apply Tollens test rule: aldehydes (acetaldehyde and formaldehyde) are oxidised and give the Silver Mirror, while ordinary ketones (benzophenone and acetone) typically do not. Step 6: Conclude that compounds 2 and 4 give a positive Silver Mirror test, so the correct option is 2 and 4. Verification / Alternative check: In the presence of Tollens reagent, acetaldehyde is oxidised to acetic acid, and formaldehyde is oxidised to formic acid. During these redox reactions, silver ions are reduced to metallic silver, producing the characteristic mirror. On the other hand, acetone and benzophenone resist oxidation by this mild reagent because the carbonyl carbon in ketones is less easily oxidised without breaking carbon–carbon bonds. This difference in reactivity between aldehydes and ketones is a standard result in qualitative organic chemistry and confirms that compounds 2 and 4 are correct. Why Other Options Are Wrong: Only 1: Benzophenone is a ketone and does not normally give a Silver Mirror, so this option is incorrect. Only 2: Acetaldehyde does give the Silver Mirror, but formaldehyde also does, so 2 alone is incomplete. 1 and 3: Both benzophenone and acetone are ketones and typically do not respond positively to Tollens reagent. Any option that excludes formaldehyde or acetaldehyde is inconsistent with the known behaviour of aldehydes. Common Pitfalls: Students sometimes think that all carbonyl compounds respond similarly to Tollens reagent. Another common mistake is to confuse Fehling test and Tollens test or to forget that many ketones require stronger oxidising agents for further reaction. To avoid confusion, remember that Tollens reagent is specifically used for aldehydes and that formaldehyde, being very easily oxidised, gives one of the strongest Silver Mirror responses. Final Answer: The Silver Mirror (Tollens) test is given by acetaldehyde and formaldehyde, that is compounds 2 and 4.

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