Question 1

The nuclear activity (rate of radioactive decay) of a given radioisotope sample changes with which of the following external conditions, if any?

Accepted Answer

Introduction / Context: This question is about radioactivity and nuclear decay. It asks whether the activity of a radioisotope, meaning the rate at which nuclei disintegrate per unit time, depends on external factors such as temperature, pressure or chemical environment. Understanding this independence is important in nuclear chemistry, dating methods and practical use of radioisotopes in medicine and industry. Given Data / Assumptions: We are considering a particular radioisotope sample. Possible external influences listed are temperature, pressure and chemical environment. Activity is defined as the number of disintegrations per unit time and is related to half life and decay constant. We assume normal ranges of temperature and pressure encountered in laboratory and practical conditions, not extreme astrophysical environments. Concept / Approach: Nuclear decay is a random process governed by the internal structure of the atomic nucleus. The decay constant and half life of a radioisotope are characteristic nuclear properties and do not depend significantly on ordinary physical or chemical conditions. While chemical reactions involve electrons in outer shells and are sensitive to temperature, pressure and chemical environment, nuclear reactions involve protons and neutrons deep in the nucleus and are largely unaffected by such external factors. Therefore, for practical purposes, the activity of a radioisotope does not change when you heat, cool, compress or chemically react the substance, provided the nuclei remain the same. Step-by-Step Solution: 1) Recall that activity A of a radioactive sample is given by A = λN, where λ is the decay constant and N is the number of undecayed nuclei.2) The decay constant λ is a fundamental nuclear property related to the structure and energy levels of the nucleus.3) Unlike chemical reaction rates, which vary strongly with temperature and sometimes with pressure, the nuclear decay constant does not change significantly with these external conditions.4) Changing the chemical form of a radioisotope, for example by forming different compounds, alters the electron cloud but leaves the nucleus unchanged, so λ and thus the activity remain essentially the same.5) Normal laboratory changes in temperature, pressure or chemical environment do not appreciably affect nuclear decay rates.6) Therefore, none of the listed external conditions cause a meaningful change in the activity of a given radioisotope sample. Verification / Alternative check: Practical evidence from radiometric dating, medical imaging and industrial tracers shows that half lives of radioisotopes are extremely stable over wide ranges of physical conditions. For example, carbon 14 dating relies on the assumption that the half life of carbon 14 has been constant over thousands of years, regardless of temperature or chemical environment. Experiments with radioactive samples heated, cooled or dissolved in different chemicals show negligible changes in activity. Only in very rare cases involving electron capture and extreme conditions is there a tiny effect, which is not relevant to standard general knowledge questions. Thus, the best general statement is that the activity does not change with these external conditions. Why Other Options Are Wrong: Temperature of the sample: Chemical reaction rates depend on temperature, but nuclear decay rates remain essentially constant over normal temperature ranges. Pressure applied to the sample: Reasonable pressure changes compress matter slightly but do not significantly alter nuclear structure or decay constants. Chemical environment of the atoms: Changing compounds and bonding affects electrons but not the nucleus, so activity is not meaningfully changed. Both temperature and pressure only: This option implies some dependence, which contradicts the nuclear nature of radioactivity. Common Pitfalls: A common mistake is to transfer intuition from chemical kinetics to nuclear decay and assume that increasing temperature will speed up decay. Another confusion arises from hearing about extreme conditions in stars or reactors and imagining that everyday temperature or pressure changes have similar effects. For exam purposes, remember that radioactive decay is a nuclear phenomenon governed by internal nuclear forces and is essentially independent of normal physical and chemical conditions. Therefore, when such options are presented, selecting that none of these conditions change the activity is usually correct. Final Answer: The activity of a radioisotope sample does not change with ordinary temperature, pressure or chemical environment, so the correct choice is none of these external conditions.

Question 2

In nuclear chemistry, alpha (α) particles emitted from radioactive nuclei are best described as which of the following?

Accepted Answer

Introduction / Context: This question is about the nature of alpha particles in radioactivity. Alpha decay is a common type of nuclear decay, especially for heavy nuclei such as uranium and radium. Knowing what an alpha particle actually is helps in understanding its charge, mass, penetrating power and biological effects. Given Data / Assumptions: We are dealing with alpha particles emitted during radioactive decay. Options include electrons, hydrogen ions, X ray radiation, helium nuclei and gamma photons. We assume basic nuclear physics knowledge about types of radiation: alpha, beta and gamma. We consider the standard definition used in textbooks and examinations. Concept / Approach: An alpha particle is essentially the nucleus of a helium atom. It contains two protons and two neutrons, giving it a mass number of 4 and a positive charge of +2 (because of two protons and no electrons). When a heavy nucleus emits an alpha particle, its atomic number decreases by 2 and its mass number decreases by 4. Because alpha particles are relatively heavy and positively charged, they have low penetrating power but high ionising power. They are very different from electrons, X rays or gamma rays. Step-by-Step Solution: 1) Recall that alpha decay involves emission of an alpha particle from the nucleus of a heavy atom.2) Experimental evidence, such as deflection in electric and magnetic fields and measurement of mass and charge, shows that alpha particles carry a charge of +2 and have mass about four times that of a proton.3) This combination of charge and mass matches the helium nucleus, which has two protons and two neutrons.4) Therefore, an alpha particle is equivalent to a helium nucleus (He2+), often written as 4He2+ or simply α.5) High energy electrons correspond to beta minus particles, not alpha particles.6) X rays and gamma rays are forms of electromagnetic radiation with no rest mass and no charge, so they cannot be alpha particles.7) Thus, the correct description of an alpha particle is a double positively charged helium nucleus. Verification / Alternative check: When alpha particles are slowed and captured by matter, they often pick up two electrons and become neutral helium atoms, which can be detected in sealed radioactive sources over time. This observation further supports that alpha particles are helium nuclei. Rutherford scattering experiments used alpha particles and their known mass and charge to draw conclusions about atomic structure. These historical and experimental facts confirm that alpha particles are not electrons, hydrogen ions or photons, but rather helium nuclei with a +2 charge. Why Other Options Are Wrong: High energy electrons: These correspond to beta minus particles (β−), not alpha particles. Positively charged hydrogen ions (protons): A proton is a hydrogen nucleus and has charge +1 and mass 1, while an alpha particle has charge +2 and mass 4. High energy X ray radiations: X rays are photons of electromagnetic radiation without rest mass or charge. Neutral gamma ray photons: Gamma rays are also photons, massless and uncharged, and are distinct from particle radiation like alpha emissions. Common Pitfalls: Students sometimes mix up different types of radiation because all are emitted from radioactive substances. A good way to remember is: alpha equals helium nucleus, beta equals high speed electron or positron, and gamma equals high energy photon. Remembering that alpha particles are relatively heavy and doubly charged helps explain why they can be stopped by a sheet of paper while beta and gamma rays are more penetrating. This mental picture helps in quickly selecting the correct description in multiple choice questions. Final Answer: Alpha particles are double positively charged helium nuclei emitted from the nucleus during alpha decay.

Question 3

In basic atomic theory, the formation of a negatively charged ion (anion) from a neutral atom usually occurs by which process?

Accepted Answer

Introduction / Context: This question deals with the basic idea of ion formation in chemistry. It asks how a negatively charged ion, called an anion, is formed from a neutral atom. Understanding this process is fundamental for explaining ionic bonding, acid base reactions and the behaviour of salts in solution. Given Data / Assumptions: We are considering a neutral atom in its ground state. The atom can gain or lose subatomic particles to form an ion. We focus specifically on the formation of an anion, which has an overall negative charge. The options mention changes in electrons, protons and neutrons. Concept / Approach: An ion is formed when an atom gains or loses electrons, not when it changes the number of protons in the nucleus. The number of protons determines the identity of the element and is not normally altered in chemical reactions. A neutral atom has an equal number of protons and electrons. If it gains extra electrons, the total negative charge exceeds the positive charge of the nucleus, and the atom becomes a negatively charged ion known as an anion. If it loses electrons, it becomes a positively charged cation. Step-by-Step Solution: 1) Start with a neutral atom where the number of protons equals the number of electrons, giving net charge zero.2) An anion must carry an overall negative charge, which means the total negative charge from electrons is greater than the positive charge from protons.3) The simplest way to increase negative charge is for the atom to gain extra electrons, while the number of protons in the nucleus remains unchanged.4) Gaining one electron gives a charge of −1, gaining two electrons gives −2, and so on, depending on the valency of the element.5) Losing neutrons would change the mass number and isotope, but not the net charge, so it does not create an anion.6) Gaining protons would increase positive charge and move the atom to another element, which is not what happens in normal chemical processes forming anions.7) Therefore, formation of anions occurs by the gain of one or more electrons. Verification / Alternative check: Common examples illustrate this clearly. A neutral chlorine atom has 17 protons and 17 electrons. When it gains one electron, it has 17 protons and 18 electrons, giving a net charge of −1 and forming a chloride ion, Cl−. Similarly, an oxygen atom with 8 protons and 8 electrons can gain two electrons to form an oxide ion, O2−, with 8 protons and 10 electrons. In each case, the change in charge is due solely to electron gain. The nucleus, containing protons and neutrons, remains the same in ordinary chemical reactions. Why Other Options Are Wrong: Loss of neutrons from the nucleus: This changes isotope but not net electric charge, so it does not create a negative ion. Gain of protons by the nucleus: Adding protons would increase positive charge and actually make it harder to form an anion; it also changes the element identity and is not a typical chemical process. Simultaneous loss of protons and electrons: While this might leave the net charge similar, it is not a recognised mechanism for forming anions in basic chemistry. No change in number of electrons or protons: This would leave the atom neutral, not converted into an ion. Common Pitfalls: Some learners mix up cations and anions and may incorrectly think that losing electrons leads to a negative charge. It helps to remember the phrase: cation is positive (think of the plus sign in the letter t), and anion is negative. Also, students sometimes forget that protons are fixed inside the nucleus in chemical reactions and that only electrons are exchanged or shared. Focusing on electrons as the mobile charge carriers in chemical processes makes it easier to recall that gaining electrons produces an anion. Final Answer: The formation of an anion occurs by the gain of one or more electrons by a neutral atom.

Question 4

Among the following halogens and oxygen, which element is the most electronegative according to the Pauling electronegativity scale?

Accepted Answer

Introduction / Context: This general chemistry question tests knowledge of electronegativity, a key periodic property. Electronegativity measures the tendency of an atom in a molecule to attract shared electrons towards itself. Knowing which element is most electronegative is important for predicting bond polarity, molecular structure and reactivity patterns. Given Data / Assumptions: The elements listed are fluorine, bromine, iodine and oxygen, with an additional typical option like chlorine often considered. We are asked to identify the most electronegative element among them. We use the Pauling electronegativity scale, which is the most commonly quoted scale in textbooks. We assume standard values without considering extreme conditions. Concept / Approach: On the Pauling scale, fluorine has the highest electronegativity of all elements, with a value around 4.0. Oxygen is also very electronegative, but its value is lower than that of fluorine. Among the halogens, electronegativity decreases as we go down the group from fluorine to chlorine to bromine to iodine. Thus, in any list that includes fluorine and oxygen along with heavier halogens, fluorine will be the most electronegative element. Step-by-Step Solution: 1) Recall the general trend: electronegativity increases across a period from left to right and decreases down a group in the periodic table.2) In the halogen group (group 17), the order of electronegativity from highest to lowest is: fluorine > chlorine > bromine > iodine.3) Oxygen is very electronegative and lies to the right in period 2, but it still has a lower electronegativity than fluorine.4) Compare fluorine and oxygen directly: fluorine is at the extreme top right (excluding noble gases), so it is the most electronegative element in the entire periodic table.5) Bromine and iodine are below fluorine and chlorine in the same group, so their electronegativities are lower than that of fluorine.6) Therefore, the most electronegative element among fluorine, bromine, iodine and oxygen is fluorine. Verification / Alternative check: Pauling electronegativity values are approximately: fluorine 4.0, oxygen 3.5, chlorine 3.0, bromine 2.8 and iodine 2.5. These numerical values confirm the qualitative trend described. Fluorine value is the highest listed, not only among the given options but among all elements. This is why bonds such as H–F are very polar and why fluorine is an extremely reactive non metal. Why Other Options Are Wrong: Bromine: Its electronegativity is lower than that of fluorine and chlorine because it is further down the group. Iodine: Being even lower in the halogen group, iodine has an electronegativity significantly less than fluorine. Oxygen: Very electronegative and second only to fluorine in many tables, but it is still less electronegative than fluorine. Chlorine: Not in the original list but often present as a distractor; it is less electronegative than fluorine. Common Pitfalls: Students sometimes confuse high reactivity with high electronegativity and may mistakenly think oxygen is the most electronegative because of its role in combustion and oxidation. A simple way to avoid this error is to remember that fluorine occupies the topmost right position among the elements that form stable compounds and is explicitly identified in textbooks as the most electronegative element. Whenever fluorine appears as an option in such questions, it is almost always the correct answer for highest electronegativity. Final Answer: The most electronegative element among the options is fluorine.

Question 5

In solution chemistry, the number of moles of solute present in exactly one litre (1 dm^3) of solution is called the ______ of the solution.

Accepted Answer

Introduction / Context: This question tests a fundamental concept in solution chemistry: the different ways of expressing concentration. It asks for the term used when we express concentration as the number of moles of solute per litre of solution. Getting these definitions correct is crucial for solving stoichiometry problems in reactions that take place in solution. Given Data / Assumptions: The definition involves moles of solute. The volume considered is one litre (1 dm^3) of the final solution, not of pure solvent. Possible terms include normality, molarity, molality and formality. We consider standard definitions as used in introductory chemistry. Concept / Approach: Molarity (symbol M) is defined as the number of moles of solute dissolved in one litre of solution. Molality (symbol m) is the number of moles of solute per kilogram of solvent. Normality is based on gram equivalents of solute per litre of solution. Formality is similar to molarity but used for ionic solutes that dissociate. Therefore, when the question refers specifically to moles per litre of solution, the correct term is molarity. Step-by-Step Solution: 1) Read the definition in the question: number of moles of solute in one litre of solution.2) Recall the definition of molarity: Molarity = moles of solute / volume of solution in litres.3) Compare with molality: Molality = moles of solute / mass of solvent in kilograms, which is different because it uses solvent mass, not solution volume.4) Compare with normality: Normality = gram equivalents of solute / litre of solution, which is based on equivalents rather than moles.5) Compare with formality: Formality is sometimes used for ionic solutes and is based on formula units per litre, but the standard school level answer for moles per litre is molarity.6) Therefore, the term that exactly matches the given definition is molarity. Verification / Alternative check: Textbooks and laboratory manuals consistently define molarity as the primary concentration unit for reactions in solution. For example, a 1 M solution of sodium chloride contains 1 mole of NaCl in 1 litre of solution. When performing titrations, chemists often refer to molar solutions such as 0.1 M HCl or 0.2 M NaOH, clearly using moles per litre as the basis. This reinforces that the correct term is molarity and not any of the other alternatives. Why Other Options Are Wrong: Normality: Uses equivalents per litre, not moles per litre, and depends on the reaction type (acid base, redox and so on). Molality: Uses moles of solute per kilogram of solvent, so the volume of solution does not appear in the definition. Formality: Related concept for ionic solutes but not the standard name for moles per litre; it is used less commonly than molarity. Equivalent concentration: A descriptive phrase that relates more closely to normality than to molarity. Common Pitfalls: Students sometimes confuse molarity and molality because the names sound similar. A helpful memory aid is that molarity starts with letter M and involves volume in litres (think of measuring in a measuring flask), whereas molality starts with letter m and involves mass of solvent. Another trap is mixing up molarity and normality; remembering that normality depends on equivalents and the specific reaction helps keep the two concepts separate. Whenever you see moles per litre of solution, think of molarity. Final Answer: The number of moles of solute present in one litre of solution is called the molarity of the solution.

Question 6

In terms of classification of matter, milk taken as a whole (as we drink it) is best described as which of the following?

Accepted Answer

Introduction / Context: This question relates to the basic classification of matter into elements, compounds and mixtures. It asks how we should classify milk, a common household substance. Understanding that many everyday materials are mixtures rather than pure substances helps connect classroom chemistry with real life observations. Given Data / Assumptions: The sample in question is ordinary milk, such as cow milk or buffalo milk, as typically consumed. Milk contains water, fats, proteins, sugars and minerals. We are asked whether milk is an element, compound, metal or mixture. We assume the entire liquid, not just one isolated component of milk. Concept / Approach: Elements are pure substances that contain only one type of atom. Compounds are pure substances consisting of two or more elements combined in a fixed ratio and represented by a definite chemical formula, such as H2O or NaCl. Mixtures contain two or more substances physically combined in varying proportions and can often be separated by physical methods. Milk consists of water, fats (lipids), proteins, lactose sugar and various minerals. These are simply mixed and dispersed, not chemically combined in a fixed ratio, so milk is a mixture. More specifically, it is an emulsion and a colloidal system where fat globules are dispersed in water. Step-by-Step Solution: 1) Observe that milk contains visible and microscopic fat globules, dissolved sugars and suspended proteins.2) These constituents can be separated by physical processes such as centrifugation (to separate cream), evaporation and filtration in industrial processing.3) The composition of milk can vary from animal to animal and even from day to day, which means there is no single fixed chemical formula for milk as a whole.4) Because it has variable composition and its components are physically combined, milk cannot be classified as a pure compound.5) Milk is certainly not an element, since it contains many different elements in different molecules.6) It is also not a metal, as it is a liquid mixture and does not show metallic properties such as high electrical conductivity or metallic lustre.7) Therefore, milk is best described as a mixture, more precisely a colloidal mixture or emulsion. Verification / Alternative check: Scientific descriptions of milk refer to it as an oil in water emulsion where tiny fat droplets are dispersed in an aqueous solution of lactose, proteins and salts. When milk is processed to produce butter, cream and skimmed milk, physical processes such as churning and centrifugation are used, confirming that the constituents are not chemically bound in fixed ratios. These facts match the definition of a mixture and particularly that of a colloidal system, not an element or a pure compound. Why Other Options Are Wrong: A chemical element: Elements contain only one type of atom, such as copper or oxygen, whereas milk contains many compounds. A pure metal: Milk does not show metallic properties and is not composed of metal atoms in a lattice. None of these: This option is incorrect because mixture describes milk accurately. A simple compound with fixed formula: Milk composition is variable and includes many different compounds, so it is not a single chemical compound. Common Pitfalls: Some learners may think of milk as a pure substance because it looks uniform to the naked eye. However, uniform appearance does not always mean chemical purity; solutions and colloids can appear homogeneous but still be mixtures. Another confusion is between the term compound and general substances. Remember that a compound must have a definite formula and fixed ratio of elements, which does not apply to complex natural materials like milk. Thinking about how easily cream can be separated from milk reinforces that milk is a mixture. Final Answer: Milk, as commonly consumed, is best classified as a mixture (specifically a colloidal dispersion) rather than a pure substance.

Question 7

Diamond, a crystalline form of carbon used in jewellery and cutting tools, is classified in chemistry as which of the following?

Accepted Answer

Introduction / Context: This question explores how diamond is classified in chemistry. Diamond is widely known as a precious gemstone and an extremely hard material used in cutting tools. Scientifically, it is one of the crystalline forms of carbon, and understanding this helps clarify the difference between elements, compounds and mixtures, as well as the concept of allotropy. Given Data / Assumptions: Diamond is the substance in question. It consists of carbon atoms arranged in a specific three dimensional lattice. The options include element, compound, mixture and other categories. We assume that we are dealing with pure diamond, not jewellery that may contain metals and other materials. Concept / Approach: An element is a pure substance made up of only one type of atom. Carbon is an element that can occur in several different structural forms called allotropes, such as diamond, graphite and fullerenes. In diamond, each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral network, giving diamond its hardness. Because diamond contains only carbon atoms and no other elements in its ideal pure form, it is classified as an element, specifically an allotrope of carbon, not a compound or mixture. Step-by-Step Solution: 1) Recognise that diamond is composed solely of carbon atoms arranged in a three dimensional crystal lattice.2) No other atoms or elements are part of the ideal diamond structure, apart from very small impurities that may affect colour.3) By definition, a chemical compound must contain two or more different elements chemically combined in fixed proportions, which is not the case here.4) A mixture would involve a physical combination of two or more substances, which again does not describe pure diamond.5) Therefore, diamond belongs to the category of elements, but it is a particular structural form of carbon, known as an allotrope.6) Other allotropes of carbon include graphite, where carbon atoms form layers, and amorphous carbon forms like charcoal.7) This reasoning leads to the conclusion that diamond is an element, not a compound or mixture. Verification / Alternative check: Chemical analyses of pure diamond show that it is nearly 100 percent carbon. When diamond burns completely in oxygen under high temperatures, the only product is carbon dioxide gas, indicating that there are no other elements present. Textbooks on solid state chemistry classify diamond and graphite as allotropes of carbon, both being elemental forms. These facts confirm that the correct classification is element (allotrope of carbon), not compound or mixture. Why Other Options Are Wrong: A compound of carbon and hydrogen: This would describe hydrocarbons like methane or benzene, not pure diamond. A mixture of several substances: Diamond is a single crystalline substance of carbon atoms, not a mixture. A liquid hydrocarbon: Diamond is a solid crystalline form of carbon and does not contain hydrogen, nor is it a liquid. A metal: Diamond does not exhibit metallic properties such as high electrical conductivity or metallic lustre and is not classified as a metal. Common Pitfalls: Because diamonds are used in jewellery and often set in metals, some learners may confuse the gemstone with the entire piece of jewellery. It is important to focus on the material diamond itself, which is pure carbon in crystalline form. Another misunderstanding is to think that any solid with a regular lattice is automatically a compound, but elements can also form crystals. Remember that pure diamond is one of the elemental forms of carbon and is therefore classified as an element, specifically an allotrope of carbon. Final Answer: Diamond is an element, more precisely an allotrope of carbon in which carbon atoms form a three dimensional network.

Question 8

Stainless steel is an alloy of iron with small amounts of carbon and which major alloying element that primarily gives it corrosion resistance (stainless property)?

Accepted Answer

Introduction / Context: This question concerns alloys and materials science. Stainless steel is widely used in kitchenware, construction, medical instruments and many other applications. The question asks which major alloying element, added to iron and carbon, gives stainless steel its characteristic corrosion resistant or stainless property. Given Data / Assumptions: Stainless steel is based on iron, with small amounts of carbon. We are asked to identify the key added element responsible for corrosion resistance. Options include manganese, vanadium, chromium and other metals. We assume common grades of stainless steel used in industry. Concept / Approach: Stainless steel is defined as steel containing a significant proportion of chromium, usually at least about 10.5 percent by mass. Chromium forms a very thin, adherent oxide film on the surface that protects the underlying metal from further rusting. This phenomenon is called passivation. Without chromium, ordinary carbon steel rusts readily in moist air. Other elements like nickel, molybdenum or manganese may be added to tailor properties, but chromium is the essential alloying element that gives the stainless character. Step-by-Step Solution: 1) Recognise that stainless steel is a special type of steel designed to resist corrosion.2) In standard definitions, stainless steels contain iron, carbon and a significant percentage of chromium.3) Chromium in the alloy reacts with oxygen at the surface to form a very thin, invisible chromium oxide layer.4) This oxide layer is stable, tightly adherent and self healing if scratched, preventing further penetration of oxygen and moisture.5) Manganese and vanadium can also be present in some steel grades but do not by themselves provide the main stainless property.6) Therefore, the major alloying element responsible for corrosion resistance in stainless steel is chromium. Verification / Alternative check: Metallurgical descriptions state that a steel must contain at least a certain minimum percentage of chromium to be called stainless. Common grades such as 304 stainless steel contain around 18 percent chromium and 8 percent nickel, while 430 stainless steel contains about 16 to 18 percent chromium without nickel. The presence of chromium is the common feature in these grades, confirming its central role. In contrast, steels without chromium but with other alloying elements do not show the same high level of corrosion resistance. Why Other Options Are Wrong: Manganese: It is used to improve strength and hardenability in some steels, but it is not the primary cause of stainless behaviour. Vanadium: Often added to steel to increase strength and resistance to wear, but does not by itself create stainless steel. Tin: Used in some alloys and coatings, but not the defining element of stainless steel. Lead: Sometimes added in small amounts to free cutting steels to improve machinability, but not used to make stainless steels. Common Pitfalls: Students sometimes choose nickel because many stainless steels are described as chromium nickel alloys. While nickel improves toughness and resistance to certain types of corrosion, chromium is still the essential element that defines stainless steel. If you remember that the term stainless is closely associated with chromium oxide passivation, it becomes easier to identify chromium as the correct answer whenever the key corrosion resisting element in stainless steel is asked. Final Answer: Stainless steel contains iron and carbon along with chromium as the major alloying element that provides corrosion resistance.

Question 9

In air pollution control for large industrial cities, dust and smoke particles suspended in air can be effectively removed in electrostatic precipitators by using which physical principle?

Accepted Answer

Introduction / Context: This question comes from colloid chemistry and environmental engineering. It asks how dust particles in the air of large industrial cities are removed. Many power plants and factories use electrostatic precipitators to clean flue gases before they are released, and these devices rely on the principle of electrophoresis to separate charged particles from the gas stream. Given Data / Assumptions: Dust and smoke particles are dispersed in air, forming an aerosol or colloidal system. We want a method used on a large scale in industrial settings. Options include sedimentation, electrophoresis, Tyndall effect, Brownian movement and simple filtration concepts. We assume modern practices such as using electrostatic precipitators in chimneys and exhaust systems. Concept / Approach: Electrophoresis is the movement of charged particles under the influence of an electric field. In electrostatic precipitators, dust and smoke particles in flue gas are given an electric charge by passing the gas between charged electrodes. The charged particles then migrate towards oppositely charged plates and stick to them. This removes particles from the gas stream, which can then be released much cleaner. Sedimentation and Brownian motion are natural processes but are too slow or random for efficient industrial dust removal, and the Tyndall effect only explains light scattering, not removal. Step-by-Step Solution: 1) Recognise that the problem involves separating fine solid particles (dust, smoke) from a gas, which is often treated as a colloidal dispersion.2) In electrostatic precipitators, gas passes between wires and plates at high voltage, imparting charge to the dust particles.3) Once charged, these particles experience a force in the electric field and move (undergo electrophoresis) towards the oppositely charged collecting plates.4) The particles adhere to the plates, forming a layer that can later be removed mechanically.5) This process does not depend primarily on sedimentation or Brownian motion, but on directed movement of charged particles in an electric field, which is electrophoresis.6) Therefore, among the choices, electrophoresis best describes the principle used to remove dust particles from industrial exhaust air. Verification / Alternative check: Descriptions of large power station chimneys and industrial flue gas cleaning systems commonly mention electrostatic precipitators. These devices are specifically designed to utilise high voltage fields to charge and collect particles. The operating principle is explicitly described as the motion of charged particles in an electric field, that is, electrophoresis. In contrast, sedimentation and Brownian movement are not controllable enough for efficient large scale cleaning, and the Tyndall effect is just an optical phenomenon showing the presence of colloidal particles, not a removal method. Why Other Options Are Wrong: Sedimentation under gravity alone: Fine dust and smoke particles are too small and light to settle quickly by gravity in tall chimneys; the process would be inefficient. Tyndall effect (light scattering): This effect helps detect colloidal particles by scattering light but does not help remove them. Brownian movement of gas molecules: Brownian motion is random movement due to collisions with gas molecules and does not lead to systematic removal. Simple filtration through cloth bags only: While bag filters are used in some systems, the classic answer for large industrial dust removal in the context of colloid chemistry is electrophoresis in electrostatic precipitators. Common Pitfalls: Students may choose Tyndall effect because it is a well known colloid phenomenon, but it is only an optical test. Others may think of sedimentation because it is used in water treatment, but air dust particles are too small. A good approach is to link the word industrial chimneys with electrostatic precipitators and remember that these devices rely on applying high voltage and using charged plates, which is clearly related to electrophoresis. Whenever dust removal in industrial gases is mentioned in colloid chemistry, electrophoresis is the key principle. Final Answer: Dust particles in the air of large industrial cities are removed in electrostatic precipitators by electrophoresis in an electric field.

Question 10

A laboratory mixture of sand (non volatile solid) and naphthalene (volatile solid that sublimes) can be conveniently separated into its components by which method?

Accepted Answer

Introduction / Context: This question is about separation techniques in basic chemistry. It considers a mixture of sand and naphthalene and asks which laboratory method is best suited for separating them. Knowing the physical properties of the components, such as volatility and sublimation behaviour, is key to choosing the correct method. Given Data / Assumptions: The mixture contains sand, which is a non volatile solid, and naphthalene, which is a volatile organic solid. Naphthalene is known to sublime, changing directly from solid to vapour upon heating. We want to separate the two substances without decomposing them. Options include fractional distillation, sublimation, chromatography, separating funnel and simple filtration. Concept / Approach: Sublimation is a process in which certain solids directly convert to vapour on heating and can later re deposit as solid when cooled, without passing through a liquid state. Naphthalene is a classical example of a substance that sublimes easily, whereas sand (mainly silicon dioxide) does not sublime under ordinary laboratory conditions. If we gently heat the mixture, naphthalene vapours can be driven off, collected and re solidified on a cool surface, leaving sand behind. Therefore, sublimation is the ideal method to separate this mixture. Step-by-Step Solution: 1) Identify the key property of naphthalene: it sublimes on gentle heating, turning into vapour and then re forming solid on cooling.2) Recognise that sand is a non volatile, high melting solid that remains in the solid state when the mixture is gently heated.3) Set up a sublimation apparatus, where the mixture is placed in a dish and covered by an inverted funnel or cold surface.4) Upon gentle heating, naphthalene sublimes, and the vapour condenses as crystals on the cooler surface.5) After sufficient time, the apparatus is allowed to cool, and solid naphthalene is collected from the cold surface, leaving sand in the original container.6) This selective sublimation and re deposition separates naphthalene from sand effectively. Verification / Alternative check: Textbooks commonly use mixtures such as ammonium chloride and sand, or naphthalene and salt, to illustrate sublimation as a separation technique. In each case, one component sublimes while the other does not, making sublimation highly effective. Fractional distillation is used for miscible liquids, a separating funnel is used for immiscible liquids and chromatography is mainly used for separating soluble components in a liquid phase. None of those alternatives is as straightforward as sublimation for a simple solid solid mixture where one solid sublimes and the other does not. Why Other Options Are Wrong: Fractional distillation of the mixture: This method is meant for separating liquids with different boiling points, not a solid solid mixture with one subliming solid and sand. Paper chromatography: Useful for separating dissolved components in a liquid mixture, not for dry sand and naphthalene. Using a separating funnel: Separating funnels are used for immiscible liquids such as oil and water, not for solid solid mixtures. Simple filtration only: Filtration cannot distinguish between two solid substances when both are insoluble in the chosen solvent; it will not separate sand and naphthalene directly without exploiting sublimation. Common Pitfalls: Students sometimes pick filtration because sand is insoluble, forgetting that naphthalene is also a solid and may not be dissolved either. Another mistake is choosing fractional distillation simply because heating and vapour are mentioned, but that process is specific to liquids. A good rule is: when a mixture contains a sublimable solid and a non sublimable solid, sublimation is the preferred separation method. Remembering that naphthalene moth balls slowly disappear by sublimation can help you associate naphthalene with this technique. Final Answer: A mixture of sand and naphthalene is best separated by sublimation of naphthalene, leaving sand behind.

Question 11

“If a system A is in thermal equilibrium with system B and system B is in thermal equilibrium with system C, then system A and system C are also in thermal equilibrium with each other.” This transitive statement is an expression of which law of ther…

Accepted Answer

Introduction / Context: This question is about the basic laws of thermodynamics. The statement describes a transitive property of thermal equilibrium between systems A, B and C. Recognising which law formalises this idea helps in understanding how thermometers work and why temperature is a meaningful, measurable quantity. Given Data / Assumptions: Systems A, B and C are considered. A is in thermal equilibrium with B. B is in thermal equilibrium with C. We infer that A must be in thermal equilibrium with C. We must identify which thermodynamic law this statement represents. Concept / Approach: The zeroth law of thermodynamics states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law formalises the transitive nature of thermal equilibrium and provides the conceptual basis for defining temperature as a measurable property. The first law deals with conservation of energy, and the second law concerns the direction of spontaneous processes and entropy, not this transitive relation. Step-by-Step Solution: 1) Read the statement carefully: A is in thermal equilibrium with B, and B is in thermal equilibrium with C.2) From this, we conclude that A and C are also in thermal equilibrium.3) This transitive relation allows us to assign a common property, temperature, to all three systems when they are in equilibrium.4) The law that specifically states this transitive property is known as the zeroth law of thermodynamics.5) The first law deals with energy conservation and the equivalence of heat and work.6) The second law deals with entropy and the direction of heat flow, such as heat flowing spontaneously from hot to cold.7) Therefore, the correct choice for the given statement is the zeroth law of thermodynamics. Verification / Alternative check: Thermodynamics textbooks often introduce the zeroth law before the first law, even though it was formulated later historically. They usually phrase it similarly: if two systems are in thermal equilibrium with a third system, they are in equilibrium with each other. This forms the theoretical foundation for using thermometers; a thermometer in equilibrium with a body has the same temperature as that body. If two bodies are in equilibrium with the same thermometer, they are at the same temperature. This is exactly the content of the given statement. Why Other Options Are Wrong: Cyclic rule of thermodynamics: This term does not represent a fundamental law describing thermal equilibrium; it sometimes refers to relationships among partial derivatives in thermodynamics, not this statement. First law of thermodynamics: Expresses conservation of energy and the relation between heat, work and internal energy, not the transitive property of thermal equilibrium. Second law of thermodynamics: Concerns entropy and the direction of spontaneous processes, such as heat flowing from hot to cold, but does not state the above equilibrium relation. Third law of thermodynamics: Deals with behaviour of entropy as temperature approaches absolute zero, not with transitive equilibrium of three systems. Common Pitfalls: Some students confuse the numbering of the laws and may mistakenly think the first law must come first conceptually. The term zeroth law can sound strange, but it was named later to reflect its more fundamental character. Another confusion is to associate any statement about equilibrium with the second law, since it often mentions equilibrium states. To avoid these mistakes, remember that the zeroth law is specifically about the transitivity of thermal equilibrium and underlies the definition of temperature and the use of thermometers. Final Answer: The given statement about thermal equilibrium of systems A, B and C is an expression of the zeroth law of thermodynamics.

Question 12

A liquid mixture of ethyl alcohol (ethanol) and water forms a pair of completely miscible liquids with different boiling points. Which separation method is best suited to separate this mixture into its components?

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Introduction / Context: This question deals with separation techniques for liquid mixtures. Ethyl alcohol (ethanol) and water are completely miscible liquids. The question asks which method is suitable for separating such a mixture, highlighting the importance of differences in boiling points and the use of distillation techniques in chemistry and industry. Given Data / Assumptions: The mixture consists of ethyl alcohol and water. These two liquids are miscible in all proportions at ordinary temperatures. They have different boiling points (ethanol lower, water higher). We must choose among methods such as evaporation, sublimation, separating funnel, fractional distillation and chromatography. Concept / Approach: When two liquids are completely miscible but have different boiling points, the standard method of separation is fractional distillation. This technique uses a fractionating column, which allows repeated condensation and vaporisation cycles, enriching the more volatile component (ethanol) in the distillate. Simple evaporation would remove both components eventually, and a separating funnel cannot separate miscible liquids. Sublimation is only suitable when one component is a sublimable solid. Therefore, fractional distillation is the correct method. Step-by-Step Solution: 1) Recognise that ethanol and water form a homogeneous liquid mixture because they are mutually soluble in all proportions.2) Check the boiling points: ethanol boils at a lower temperature than water under standard pressure.3) Separation of miscible liquids with different boiling points is best achieved by fractional distillation, not by separating funnels or simple evaporation.4) In fractional distillation, the mixture is heated and vapour rises through a fractionating column, where repeated condensation and vaporisation enriches the vapour in the more volatile component.5) The vapour reaches the condenser, cools and is collected as distillate that is richer in ethanol, while water remains longer in the boiling flask.6) This process eventually allows collection of fractions with differing compositions and can approach pure ethanol and pure water with proper control. Verification / Alternative check: Laboratory manuals describe the fractional distillation of ethanol water mixtures as a classic experiment. Industrially, large distillation columns are used to concentrate ethanol from fermentation broths. In these examples, a fractionating column is essential for efficient separation due to repeated equilibration between liquid and vapour phases. Methods like using a separating funnel are explicitly stated to work only for immiscible liquid mixtures, such as oil and water, not for miscible pairs like ethanol and water. Why Other Options Are Wrong: Simple evaporation to dryness: This would remove ethanol first but would also eventually remove water, and it does not allow separate collection of both components as pure liquids. Sublimation of one component: Neither ethanol nor water is a subliming solid under ordinary conditions; both are liquids at room temperature. Using a separating funnel: A separating funnel relies on immiscible liquids forming layers; ethanol and water are completely miscible and form only one layer. Simple paper chromatography: Chromatography is useful for separating components in solution on a stationary phase, not for bulk separation of two miscible solvents. Common Pitfalls: Students sometimes think of separating funnels for any liquid mixture, forgetting the important condition of immiscibility. Another confusion is with simple distillation, which can separate some mixtures but is less efficient when boiling points are close. Fractional distillation with a column is the standard answer in exams for separating mixtures like ethanol and water. Remember that when you see miscible liquids with different boiling points, fractional distillation is usually the correct technique. Final Answer: A mixture of ethyl alcohol and water is best separated by fractional distillation of the liquid mixture.

Question 13

In a typical biology laboratory, which of the following chemicals is commonly used (usually as an aqueous solution called formalin) to preserve biological specimens such as animal organs and small organisms?

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Introduction / Context: This question is about preservation of biological specimens in laboratories and museums. Many students have seen jars containing preserved frogs, organs or other specimens in biology labs. The liquid in these jars usually contains a solution of a particular chemical that fixes and preserves tissues. Identifying this chemical connects practical laboratory experience with basic organic chemistry. Given Data / Assumptions: The setting is a biology laboratory where specimens are preserved long term in jars. We are asked which chemical is used to preserve such specimens. Options include phenol, benzaldehyde, formaldehyde and acetic acid. We assume standard school or college laboratory practice where formalin is the common preservative. Concept / Approach: Formaldehyde is the simplest aldehyde, with formula HCHO. In laboratories, it is commonly used as an aqueous solution of about 37 to 40 percent, known as formalin. Formaldehyde reacts with amino groups in proteins and cross links them, effectively fixing tissues and preventing decay. This makes formalin a widely used chemical for preserving anatomical specimens. Phenol and acetic acid have antiseptic or preservative properties but are not the standard specimen preservative in jars, and benzaldehyde is mainly a flavour or fragrance compound. Step-by-Step Solution: 1) Recall that preserved specimens in biology labs are often said to be stored in formalin.2) Formalin is an aqueous solution of formaldehyde, stabilised with small amounts of methanol.3) Formaldehyde reacts with proteins and nucleic acids in tissues, forming cross links that stabilise the structure and prevent decomposition by microorganisms.4) This property makes formaldehyde an effective tissue fixative and long term preservative.5) Phenol is a disinfectant and antiseptic, but it is not the usual choice for storing whole specimens in jars.6) Benzaldehyde is used mainly as an aromatic compound (almond like fragrance) and is not a standard specimen preservative.7) Acetic acid is used in some fixatives and buffering solutions but is not, by itself, the classic answer for preserving specimens in jars. Verification / Alternative check: Descriptions of histological and anatomical methods mention that organs and whole animals are fixed in formalin for preservation and later study. Formalin penetrates tissues and prevents autolysis and bacterial decay. School textbooks also refer to formalin jars in biology labs. Phenol based solutions may be used for disinfection, and acetic acid may be present in special fixative mixtures, but the widely known and tested preservative name is formaldehyde in the form of formalin. Why Other Options Are Wrong: Phenol: Used as an antiseptic and disinfectant but not as the main fluid in specimen jars for long term anatomical preservation. Benzaldehyde: Mainly a fragrance and flavouring agent with an almond like smell, not commonly used for preserving specimens. Acetic acid: Found in some specialised fixatives but is not the primary single chemical used in standard specimen jars. Ethyl alcohol: Can preserve small samples and is used in some collections, but in typical school biology contexts the textbook answer is formalin (formaldehyde solution). Common Pitfalls: Students sometimes choose phenol or acetic acid because they are aware these chemicals can kill microbes or are used in fixative formulations. However, exam questions at this level usually expect the specific name formaldehyde (or formalin) when asking about preserved specimens in jars. A useful memory aid is the phrase “formalin jars” that many textbooks and teachers use. Remember that formalin is simply aqueous formaldehyde, and that connection will guide you to the correct answer. Final Answer: Biological specimens in many laboratories are preserved in an aqueous solution of formaldehyde (formalin).

Question 14

In atomic physics, which scientist discovered the nucleus of the atom through the famous gold foil scattering experiment?

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Introduction / Context: The discovery of the atomic nucleus is a landmark moment in the history of modern physics and chemistry. Before this discovery, the prevailing plum pudding model proposed that positive charge was spread uniformly throughout the atom. The correct model, based on experimental evidence, showed that almost all the mass and positive charge is concentrated in a tiny central region called the nucleus. This question tests knowledge of the scientist associated with the famous gold foil scattering experiment that led to this breakthrough. Given Data / Assumptions: We are dealing with early atomic models and their experimental verification. The experiment involved alpha particles directed at very thin metal foil, typically gold. We assume standard school level descriptions of atomic theory. We assume that the name of the scientist is to be selected from the options. Concept / Approach: To answer the question, we recall which scientist designed and interpreted the gold foil scattering experiment. In this classic experiment, a beam of alpha particles was fired at a thin gold foil and the pattern of scattering was observed on a fluorescent screen. Most particles passed straight through, but a small fraction were deflected at large angles. This result could not be explained by the plum pudding model and led to the nuclear model of the atom. The scientist who led this work was Ernest Rutherford. Step-by-Step Solution: Step 1: Recall key early atomic models, such as the plum pudding model and the nuclear model. Step 2: Remember that the gold foil scattering experiment used alpha particles and thin gold foil to probe atomic structure. Step 3: Recognise that the unexpected large angle deflections of a few alpha particles suggested a small, dense, positively charged centre in the atom. Step 4: Identify the scientist who designed and interpreted this experiment as Ernest Rutherford. Step 5: Match the name Ernest Rutherford with the correct option provided. Verification / Alternative check: Many textbooks refer to this work as the Rutherford scattering experiment or Rutherford gold foil experiment. This naming itself confirms that Rutherford is the key figure connected with the discovery of the nucleus. Niels Bohr later refined this model to introduce quantised orbits but did not discover the nucleus. Thomson proposed the earlier plum pudding model and discovered the electron, not the nucleus. These cross checks confirm that Rutherford is the correct option. Why Other Options Are Wrong: Niels Bohr contributed the planetary model and quantised energy levels but did not perform the gold foil experiment. J J Thomson discovered the electron and proposed the plum pudding model, which was disproved by the gold foil results. Henry Moseley worked on atomic number and X ray spectra rather than discovering the nucleus. Max Planck introduced quantum theory related to energy quanta, not atomic structure via scattering experiments. Common Pitfalls: A frequent mistake is to confuse Bohr or Thomson with Rutherford, since all three are associated with atomic models. Another error is to assume that the scientist who discovered the electron must also have discovered the nucleus. Students sometimes remember the name of the experiment but forget the scientist and guess incorrectly based on familiarity with other names. Reading the question slowly and linking the gold foil scattering experiment specifically with Rutherford avoids this confusion. Final Answer: The nucleus of the atom was discovered through the gold foil scattering experiment carried out by Ernest Rutherford.

Question 15

In basic particle physics, what are the charge and mass values associated with a neutrino according to the traditional school level description?

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Introduction / Context: Neutrinos are extremely light, neutral particles that play important roles in nuclear reactions, beta decay and astrophysical processes. At the basic general knowledge and school level, students are usually introduced to neutrinos as neutral particles with almost negligible mass. This question checks understanding of the standard textbook description of neutrino charge and mass, which is often simplified to charge zero and very small or nearly zero mass. Given Data / Assumptions: A neutrino is a fundamental particle that interacts very weakly with matter. School level chemistry and physics typically treat neutrino mass as nearly zero. The answer choices compare different combinations of electric charge and mass. We assume that the required answer matches the conventional teaching approach. Concept / Approach: The key concept involves knowing that neutrinos carry no electric charge and are therefore electrically neutral. Historically, neutrinos were considered massless in many introductory texts. Modern experiments show that neutrinos have a very tiny mass, but it is still extremely small compared to protons and neutrons. For the purpose of most general knowledge examinations, a neutrino is described as having charge 0 and mass approximately 0. Therefore, among the options, we should look for the combination that reflects zero charge and negligible mass. Step-by-Step Solution: Step 1: Recall that neutrinos do not interact electromagnetically, which means they have no electric charge. Step 2: Eliminate any options where the charge is plus 1 or minus 1, since these would represent positively or negatively charged particles. Step 3: Recognise that their mass is extremely small compared with protons and neutrons and is often taken as approximately zero at basic level. Step 4: Identify the option that states charge 0 and mass approximately 0 as the best match. Step 5: Confirm that no other option provides both zero charge and negligible mass. Verification / Alternative check: A good cross check is to compare neutrinos with other fundamental particles. Protons have charge plus 1 and relatively large mass. Electrons have charge minus 1 but a much smaller mass than protons, yet they are still significantly heavier than the effective mass of a neutrino. Neutrons are neutral but have mass similar to protons, so they do not match the required description. Only neutrinos combine neutrality with very small mass, which is represented in school level options as charge 0 and mass approximately 0. Why Other Options Are Wrong: Options with charge plus 1 or minus 1 describe charged particles, not neutrinos. The pair charge 0 and mass 1 would describe a neutral particle with relatively large mass, closer to a neutron than to a neutrino. The combination charge plus 2 and mass 0 does not correspond to any standard fundamental particle model in basic physics. Common Pitfalls: Students sometimes confuse neutrinos with neutrons because both are neutral. However, neutrons have substantial mass and are part of atomic nuclei, while neutrinos are extremely light and rarely interact. Another pitfall is to focus only on the mass and ignore the charge. Always check both properties together. Remember that in school level examinations, the simplified description of the neutrino is usually enough to answer the question correctly. Final Answer: In the standard school level description, a neutrino has charge 0 and mass approximately 0.

Question 16

In nuclear chemistry, what name is given to atoms of different elements that have the same mass number but different atomic numbers?

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Introduction / Context: In chemistry and nuclear physics, it is important to distinguish between different classifications of atoms based on their numbers of protons, neutrons and mass numbers. Terms such as isotopes, isobars, isotones and nuclear isomers all refer to specific relationships between these quantities. This question focuses on the situation where the mass number is the same but the atomic number differs, and tests whether you can recall the correct term used for such atoms. Given Data / Assumptions: Mass number is the total number of protons and neutrons in the nucleus. Atomic number is the number of protons in the nucleus. Different elements have different atomic numbers. The question describes atoms that share the same mass number but have different atomic numbers. Concept / Approach: We begin by recalling the definitions. Isotopes are atoms of the same element with the same atomic number but different mass numbers. Isobars are atoms of different elements with the same mass number but different atomic numbers. Isotones are atoms that have the same number of neutrons but different atomic numbers and mass numbers. Nuclear isomers are atoms with the same atomic number and mass number but different energy states. The description in the question clearly matches the definition of isobars, so we should choose that term from the options. Step-by-Step Solution: Step 1: Note that the question specifically says same mass number but different atomic numbers. Step 2: Recall that mass number equals protons plus neutrons. Step 3: Write down the definition of isotopes and confirm they have same atomic number, which does not match the question. Step 4: Write down the definition of isobars, which states same mass number but different atomic numbers. Step 5: Match this definition to the wording in the question and select isobars as the correct term. Verification / Alternative check: A good way to verify is to use examples. Calcium 40 and argon 40 both have mass number 40 but different atomic numbers, so they are isobars. On the other hand, carbon 12 and carbon 14 are isotopes since they have the same atomic number but different mass numbers. Checking these known examples against the definitions confirms that the correct classification for atoms with equal mass number and different atomic number is isobars. Why Other Options Are Wrong: Isotopes have identical atomic numbers, which conflicts with the condition that atomic numbers are different. Isotones have equal numbers of neutrons rather than equal mass numbers, so that definition does not fit here. Isomers refer to nuclei with the same atomic and mass numbers but different energy states, which again does not match the question. Allotropes relate to different physical forms of the same element in the same physical state, for example different crystalline forms of carbon, and are not defined by mass and atomic numbers of individual atoms. Common Pitfalls: Many learners confuse the terms isotopes and isobars because both include the syllable iso. A useful memory aid is that isobars share the same "bar" or total mass number, while isotopes share the same atomic number. Another common error is to ignore that the question explicitly mentions different atomic numbers and to select isotopes by habit. Reading the condition carefully will prevent this mistake. Final Answer: Atoms of different elements that have the same mass number but different atomic numbers are called isobars.

Question 17

Which statement best describes the location and behaviour of electrons in a simple atomic model?

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Introduction / Context: Understanding how electrons are arranged in an atom is a foundation for explaining chemical bonding, periodic trends and many physical properties of matter. At the basic level, we use simplified models to describe electrons as moving around the nucleus in certain regions or shells. This question checks whether you can identify the best simplified statement about the position and behaviour of electrons when compared with several incorrect or misleading descriptions. Given Data / Assumptions: We are dealing with a simple school level atomic model, not a fully advanced quantum mechanical treatment. The nucleus contains protons and neutrons, with electrons outside the nucleus. Electrons have negative charge and are attracted to the positively charged nucleus. The options provide different statements about where electrons are and what they are made of. Concept / Approach: In the simplified Bohr style picture, electrons are described as moving around the nucleus in definite orbits or shells, each with a specific energy level. In more advanced quantum mechanics, electrons are described by orbitals and probability clouds instead of fixed circular paths, but they are still outside the nucleus. We can safely say that electrons orbit or move around the nucleus in specific energy regions, and they are not located inside the nucleus or made of protons and neutrons. Therefore, among the given options, the statement describing electrons as orbiting around the nucleus in specific energy regions is the most accurate for this level. Step-by-Step Solution: Step 1: Recall that the nucleus at the centre of the atom contains protons and neutrons. Step 2: Remember that electrons are located outside the nucleus in energy levels or shells. Step 3: Note that electrons are fundamental particles and are not composed of smaller particles in basic school level models. Step 4: Compare each option with this understanding to see which matches the standard description. Step 5: Select the option that refers to electrons orbiting around the nucleus in specific energy regions as the best description. Verification / Alternative check: Most introductory chemistry or physics textbooks include diagrams showing a central nucleus and electrons arranged in shells around it. These diagrams reinforce the idea that electrons are outside the nucleus and occupy energy levels. None of the standard diagrams or explanations claim that electrons sit inside the nucleus or that they are made from protons and neutrons. Cross checking with these visual representations confirms that the correct answer is the statement about electrons orbiting around the nucleus. Why Other Options Are Wrong: The statement that electrons are made of protons and neutrons is incorrect because electrons are separate fundamental particles, not composites of nucleons. The claim that electrons reside in the nucleus contradicts both experimental evidence and basic models of atomic structure. Describing electrons as positively charged is wrong because electrons have negative charge, which balances the positive charge of protons. The suggestion that electrons always move in perfect circles with no energy levels oversimplifies and misrepresents modern understanding and does not match standard textbook explanations. Common Pitfalls: Students sometimes confuse terms like orbit and orbital or think that electrons can be found inside the nucleus because of diagrams drawn out of scale. Another common pitfall is to assume that all subatomic particles share similar internal structure. Focusing on the key idea that electrons are negatively charged particles located outside the nucleus in distinct energy regions helps avoid these errors. Final Answer: The best simplified description is that electrons orbit around the nucleus in specific energy regions.

Question 18

In the group of noble gases, which element is considered the heaviest under normal conditions?

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Introduction / Context: Noble gases are a group of elements in the periodic table known for their very low chemical reactivity. They include helium, neon, argon, krypton, xenon and radon. One simple general knowledge question about these elements concerns their relative atomic masses and which noble gas is the heaviest. This is useful for understanding periodic trends and for answering basic chemistry questions in competitive examinations. Given Data / Assumptions: The noble gases in the conventional list are helium, neon, argon, krypton, xenon and radon. Heavier elements in a group have higher atomic numbers and larger atomic masses. The question asks specifically for the heaviest noble gas among the commonly discussed ones. We ignore super heavy synthetic elements not covered in standard school syllabi. Concept / Approach: In the periodic table, noble gases are placed in group 18. As we go down a group, atomic number and atomic mass increase. Helium is at the top with the smallest mass, followed by neon, argon, krypton, xenon and finally radon at the bottom of the group. Therefore radon is the noble gas with the highest atomic mass among the commonly encountered elements. Although advanced research has created even heavier synthetic elements, general knowledge questions usually consider radon as the heaviest noble gas. Step-by-Step Solution: Step 1: List the noble gases in order: helium, neon, argon, krypton, xenon and radon. Step 2: Recall that moving down the group increases both atomic number and atomic mass. Step 3: Identify the noble gas at the bottom of the group as the heaviest. Step 4: Recognise that radon lies below xenon and krypton on the periodic table. Step 5: Choose radon from the given options as the correct answer. Verification / Alternative check: A quick check using approximate atomic masses helps. For example, helium has an approximate mass of 4, neon around 20, argon about 40, krypton near 84, xenon around 131 and radon higher still, with a mass value above xenon. Comparing these approximate values confirms that radon is heavier than krypton and xenon, so radon is the heaviest noble gas in the conventional list. Why Other Options Are Wrong: Helium, neon and krypton are located above xenon and radon in the periodic table, and each has a smaller atomic mass. Xenon is heavier than krypton but is still above radon and therefore not the heaviest in the group. Any noble gas that is not at the bottom of the group cannot be the heaviest among the commonly studied noble gases. Common Pitfalls: Students sometimes get confused between xenon and radon because both are heavier noble gases and are used in special applications. Another mistake is assuming that xenon is the heaviest simply because it is more familiar. Remember that in a group the elements become heavier as you move downwards. Keeping the vertical order of the noble gases in mind makes it easy to recall that radon is the heaviest among them. Final Answer: Among the commonly known noble gases, the heaviest element is radon.

Question 19

In the composition of natural gas used as a fuel, which component is present as the primary or major constituent?

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Introduction / Context: Natural gas is an important fossil fuel widely used for cooking, heating and power generation. It is considered cleaner than many other fossil fuels because it produces less carbon dioxide per unit of energy. To understand its properties and uses, it is useful to know the major chemical component of natural gas. This question checks whether you know which hydrocarbon forms the largest fraction of natural gas in typical commercial supplies. Given Data / Assumptions: Natural gas is a mixture of light hydrocarbons and small amounts of other gases. The most important components are methane, ethane, propane and small amounts of higher hydrocarbons. There may also be trace amounts of nitrogen, carbon dioxide and other impurities. The question asks specifically for the primary or major constituent. Concept / Approach: Chemistry textbooks consistently describe methane as the main component of natural gas, usually accounting for more than 80 percent of the mixture. Other hydrocarbons such as ethane and propane are present in smaller amounts and may be separated for use as liquefied petroleum gas. Therefore, when asked for the primary component of natural gas, the correct answer is methane. Recognising this is important when studying combustion reactions, calorific values and environmental impacts of fuel use. Step-by-Step Solution: Step 1: Recall that natural gas is a fossil fuel mainly made up of light hydrocarbons. Step 2: Remember that methane is the simplest hydrocarbon with one carbon atom and four hydrogen atoms. Step 3: Note that typical natural gas compositions list methane as the majority component by volume. Step 4: Compare this with ethane and propane, which are present only in smaller percentages. Step 5: Select methane from the options as the primary component of natural gas. Verification / Alternative check: You can verify this by considering how natural gas is processed. In gas processing plants, methane is often left as pipeline gas, while other hydrocarbons such as propane and butane are extracted and sold as liquefied petroleum gas. This industrial practice reflects the fact that methane dominates the mixture. Also, combustion equations for natural gas are commonly written using methane as the representative fuel, which confirms its primary role in the mixture. Why Other Options Are Wrong: Ethane and propane are important minor components but they do not form the bulk of natural gas under normal conditions. Nitrogen and carbon dioxide are usually present only as impurities or in very small amounts and do not provide fuel value. Any gas that is not the main hydrocarbon in the mixture cannot be considered the primary constituent of natural gas. Common Pitfalls: Some students confuse natural gas with liquefied petroleum gas and may think propane or butane is the major component. Another mistake is to assume that nitrogen is primary because air contains a large amount of nitrogen, but natural gas is a separate fuel mixture. Always remember that natural gas is mainly methane, while liquefied petroleum gas contains a higher proportion of propane and butane. Final Answer: The major constituent and primary component of natural gas is methane.

Question 20

In molecular bonding theory, which of the following statements about bond order, bond length and stability is false?

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Introduction / Context: Bond order is a useful concept in chemical bonding that gives a rough idea of the number of bonding electron pairs between two atoms. It is related to bond length, bond strength and molecular stability. Understanding these relationships is important in predicting molecular properties and comparing different chemical bonds. This question asks you to identify the false statement among several that describe how bond order connects with bond strength, bond length and stability. Given Data / Assumptions: Bond order is defined as the number of chemical bonds between a pair of atoms, for example single, double or triple bonds. Bonds with higher bond order are generally shorter and stronger. Negative bond order values obtained from some calculations indicate that a molecule is not stable. We assume standard textbook relationships between bond order, bond length and bond energy. Concept / Approach: In simple terms, when bond order increases from a single bond to a double bond to a triple bond, the bond becomes stronger and the bond length decreases. This means bond order is directly related to bond strength but inversely related to bond length. If molecular orbital calculations give a negative bond order, it usually means that antibonding interactions dominate and the molecule is not stable. Therefore, any statement that claims bond order is directly proportional to bond length contradicts the usual understanding and is false. Step-by-Step Solution: Step 1: Recall that a single bond has bond order 1, a double bond has bond order 2 and a triple bond has bond order 3. Step 2: Note that triple bonds are generally stronger and shorter than double bonds, which in turn are stronger and shorter than single bonds. Step 3: Conclude that as bond order increases, bond strength increases and bond length decreases. Step 4: Examine each statement and compare it with this relationship between bond order, bond length and bond strength. Step 5: Identify the statement that claims bond order is proportional to bond length as false, since the relationship is actually inverse. Verification / Alternative check: You can verify by checking examples such as carbon carbon single, double and triple bonds. Typical bond lengths decrease from the single bond to the triple bond, while bond energies increase. This pattern matches the idea that bond order is inversely related to bond length. None of the standard examples or tables support a direct proportional relationship between bond order and bond length. Therefore, the statement that bond order is proportional to bond length is clearly incorrect. Why Other Options Are Wrong: The statement that bond strength increases with bond order is true and supported by many examples of double and triple bonds. The claim that molecules with negative bond order are highly unstable or do not exist reflects the usual interpretation of molecular orbital calculations. The statement that bond order is inversely proportional to bond length correctly summarises the trend that higher bond order corresponds to shorter bonds. Higher bond order generally means higher bond energy and therefore stronger bonds, so that statement is also correct. Common Pitfalls: A common mistake is to mix up direct and inverse proportionality when comparing bond order and bond length. Students may also forget that trends are general and may have small exceptions, but for basic questions it is safe to rely on the standard rule that higher bond order produces shorter, stronger bonds. Carefully reading each statement and matching it with known examples helps avoid confusion. Final Answer: The false statement is that bond order is proportional to the bond length.

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