Question 1

In electrochemistry, which acid is used as the electrolyte in lead storage cells (lead–acid batteries)?

Accepted Answer

Introduction / Context: This question deals with the chemistry of lead storage cells, commonly known as lead–acid batteries, which are used in automobiles and backup power systems. The electrolyte in these batteries is a dilute acid that allows ions to move between the electrodes during charging and discharging. Knowing which acid is used is important in both basic electrochemistry and practical safety, because it explains why battery electrolyte is corrosive and why proper handling is necessary. The standard acid used in lead storage cells is dilute sulphuric acid. Given Data / Assumptions: - The device mentioned is a lead storage cell or lead–acid battery. - The question asks which acid serves as the electrolyte in this battery. - Options include sulphuric, nitric, hydrochloric, and phosphoric acids. - It is assumed that you are familiar with the basic construction of a lead–acid battery: lead and lead dioxide plates immersed in an acidic electrolyte. Concept / Approach: In a lead–acid battery, the electrolyte is dilute sulphuric acid (H2SO4). The overall reactions during discharge involve the conversion of lead (Pb) and lead dioxide (PbO2) at the electrodes into lead sulfate (PbSO4), with sulphuric acid participating in the reaction and water being formed. The concentration of sulphuric acid decreases as the battery discharges and increases again during charging. Nitric acid, hydrochloric acid, and phosphoric acid are not used as electrolytes in standard lead storage cells; they would lead to different and often undesirable reactions and corrosion patterns. Therefore, the correct answer is sulphuric acid. Step-by-Step Solution: Step 1: Recall that the common car battery is a lead–acid battery containing lead and lead dioxide plates. Step 2: Remember that the liquid inside such a battery is corrosive and is described as dilute sulphuric acid. Step 3: Understand that during discharge, Pb and PbO2 react with sulphate ions from H2SO4 to form PbSO4 on both plates, and water is produced. Step 4: Recognise that the reversible electrochemical reactions rely specifically on sulphuric acid because it supplies sulphate ions and hydrogen ions in the correct stoichiometry. Step 5: Compare this with other acids: nitric acid and hydrochloric acid would lead to different side reactions, and phosphoric acid is not used in the classic lead–acid system. Step 6: Conclude that sulphuric acid is the acid used in lead storage cells. Verification / Alternative check: Automotive manuals and battery specifications describe the electrolyte as diluted sulphuric acid, often around 30–40 percent by mass. Electrochemistry textbooks that discuss the lead–acid cell explicitly write the cell notation with H2SO4 as the electrolyte. They explain that the state of charge of the battery can be monitored by measuring the specific gravity of the sulphuric acid solution. No standard reference suggests that nitric, hydrochloric, or phosphoric acid is used in conventional lead storage batteries, confirming that sulphuric acid is the correct answer. Why Other Options Are Wrong: Nitric acid is a strong oxidising acid and would rapidly corrode the lead plates in an uncontrolled way, making it unsuitable as a stable battery electrolyte. Hydrochloric acid forms soluble chlorides and presents corrosion issues with metals, again not matching the well known chemistry of lead–acid batteries. Phosphoric acid is used in some industrial processes and rust treatments but is not the electrolyte in typical automotive batteries. Therefore, options B, C, and D do not accurately describe the acid used in lead storage cells. Common Pitfalls: Students sometimes confuse sulphuric acid with other strong acids simply because all are corrosive. Others may only remember that the electrolyte is "acid" without recalling which specific acid is involved. To avoid such mistakes, it helps to associate the name lead–acid battery directly with sulphuric acid, and to remember that the formation of lead sulfate (PbSO4) during discharge is directly linked to the presence of sulphate ions from H2SO4 in the electrolyte. Final Answer: The acid used as the electrolyte in lead storage cells is sulphuric acid (H2SO4).

Question 2

In basic atomic structure, how many electron energy levels (shells) are occupied in a neutral magnesium atom?

Accepted Answer

Introduction / Context: This question checks your understanding of electron configurations and energy levels (shells) in atoms. Magnesium is a common element in school chemistry, and its atomic number and electron configuration are used frequently as examples. Knowing how many principal energy levels, or shells, an atom occupies helps you place it correctly in the periodic table and predict its chemical behaviour, including valence electrons and reactivity. The question asks specifically how many electron levels are filled or partially filled in a neutral magnesium atom. Given Data / Assumptions: - The element is magnesium, with atomic number 12. - A neutral magnesium atom has 12 electrons. - The question is about the number of electron levels (shells), not subshells or orbitals. - Standard electron configuration rules for main group elements apply (1st shell holds up to 2 electrons, 2nd and 3rd shells up to 8, and so on in the simple model). Concept / Approach: The atomic number of magnesium is 12, meaning a neutral atom has 12 protons and 12 electrons. In the Bohr model and basic electron configuration schemes, electrons fill shells in order of increasing energy: the first shell (n = 1) holds up to 2 electrons, the second (n = 2) holds up to 8, and the third (n = 3) begins to fill after the first two shells are filled. For magnesium, the electrons are arranged as 2 in the first shell, 8 in the second, and the remaining 2 in the third. Therefore, three distinct electron shells contain electrons in magnesium, so it has 3 occupied energy levels. Step-by-Step Solution: Step 1: Note that the atomic number of magnesium (Mg) is 12, so a neutral Mg atom has 12 electrons. Step 2: Fill the first shell (n = 1) with its maximum of 2 electrons, leaving 10 electrons. Step 3: Fill the second shell (n = 2) with its maximum of 8 electrons, leaving 2 electrons remaining. Step 4: Place the remaining 2 electrons in the third shell (n = 3). Step 5: Conclude that electrons occupy three shells in total: K (1st), L (2nd), and M (3rd), so the number of electron levels is 3. Verification / Alternative check: You can verify this using standard electron configuration notation. Magnesium has the configuration 1s2 2s2 2p6 3s2. Grouping by principal quantum number n, you have: n = 1 with 2 electrons (1s2), n = 2 with 8 electrons (2s2 2p6), and n = 3 with 2 electrons (3s2). This pattern confirms that shells 1, 2, and 3 are occupied. Additionally, in the simple distribution often taught in schools, magnesium is written as having 2, 8, 2 electrons in its shells, again showing three energy levels. No electrons occupy higher shells for the ground state of neutral magnesium, so the count remains three. Why Other Options Are Wrong: Option B (2) would correspond to an element whose electrons occupy only the first two shells, such as neon (2, 8). Option C (1) would suggest only the first shell is filled, as in helium (2). Option D (0) is impossible for any atom, as every atom must have at least one shell containing electrons. Therefore, options B, C, and D do not match the known electron configuration of magnesium. Only three occupied shells, as given by option A, are consistent with magnesium's atomic number and electron distribution. Common Pitfalls: Some students may confuse the number of valence electrons with the number of shells and mistakenly answer 2 because magnesium has two valence electrons in the third shell. Others may not recall its exact atomic number and try to guess based on its position in the periodic table. To avoid these mistakes, remember the simple shell distribution for magnesium as 2, 8, 2 and use it to count shells: each nonzero number in that sequence represents one occupied shell, giving a total of three shells for magnesium. Final Answer: A neutral magnesium atom has electrons in three energy levels (shells).

Question 3

In aromatic organic chemistry, arrange o-cresol, phenol, and o-nitrophenol in order of increasing acidity (lowest to highest) in aqueous solution.

Accepted Answer

Introduction / Context: This question is about comparing the acidity of substituted phenols in organic chemistry. Phenol and its derivatives show different acid strengths depending on the nature of substituents attached to the benzene ring. Electron-donating groups such as alkyl groups decrease acidity, while electron-withdrawing groups such as nitro groups increase acidity. Additionally, the position (ortho, meta, para) of the substituent influences resonance and inductive effects. Understanding how these substituents affect acidity helps students predict reaction outcomes, choose appropriate reagents, and rationalise trends in organic reactivity. Given Data / Assumptions: - The three compounds to be ordered by increasing acidity are o-cresol (ortho-methyl phenol), phenol, and o-nitrophenol (ortho-nitrophenol). - The order must be from least acidic to most acidic in aqueous solution. - It is assumed that you know methyl groups are electron donating and nitro groups are strongly electron withdrawing. - Basic principles of resonance, inductive effects, and stabilisation of the phenoxide ion apply. Concept / Approach: Acidity in phenols is largely determined by the stability of the phenoxide ion formed after deprotonation. Electron-donating groups like the methyl group in o-cresol push electron density towards the ring and the oxygen, destabilising the negative charge and decreasing acidity compared to unsubstituted phenol. Electron-withdrawing groups such as the nitro group in o-nitrophenol pull electron density away via both -I (inductive) and -R (resonance) effects, stabilising the phenoxide ion and increasing acidity relative to phenol. Therefore, in terms of acidity: o-cresol (with an electron-donating methyl group) is least acidic, phenol is intermediate, and o-nitrophenol (with a strongly electron-withdrawing nitro group) is most acidic. Step-by-Step Solution: Step 1: Consider o-cresol, which is ortho-methyl phenol. The methyl group is electron donating through its +I effect, increasing electron density on the ring and the oxygen, destabilising the negative charge that would form on oxygen in the phenoxide ion. Step 2: Compare this to phenol, which has no substituents on the ring. Its phenoxide ion is stabilised only by the normal resonance delocalisation of the negative charge over the aromatic ring. Step 3: Now consider o-nitrophenol, where the nitro group is a powerful electron-withdrawing group. It stabilises the phenoxide ion by pulling electron density away via both resonance and inductive effects. Step 4: The greater stabilisation of the phenoxide ion in o-nitrophenol means it more readily loses a proton, making it more acidic than phenol. Step 5: Conversely, the electron-donating methyl group in o-cresol makes its conjugate base less stabilised and therefore less acidic than phenol. Step 6: Putting these effects together, the acidity increases in the order: o-cresol (least acidic) < phenol < o-nitrophenol (most acidic). Verification / Alternative check: Experimental pKa values support this trend. Phenols with electron-donating alkyl groups generally have higher pKa values (weaker acids) than phenol, reflecting decreased acidity. Phenols with nitro substituents, especially in the ortho and para positions, show significantly lower pKa values, indicating stronger acidity. In o-nitrophenol, intramolecular hydrogen bonding between the nitro group and the hydroxyl hydrogen further stabilises the molecular form and influences acidity, but the dominant effect is the strong electron withdrawal by the nitro group stabilising the conjugate base. Reference tables consistently place nitrophenols as more acidic than unsubstituted phenol and alkyl-substituted phenols as less acidic, confirming the order o-cresol < phenol < o-nitrophenol. Why Other Options Are Wrong: The order phenol < o-nitrophenol < o-cresol implies that o-cresol is the most acidic, which contradicts the electron-donating nature of the methyl group and experimental data. The order o-nitrophenol < o-cresol < phenol suggests that nitro substitution makes the compound less acidic, which is opposite to the known strong electron-withdrawing effect of nitro groups. None of the above is incorrect because there is a correct and well established ordering based on substituent effects. Therefore, options B, C, and D do not match the real acidity trend. Common Pitfalls: Students sometimes assume that any substituted phenol must be more acidic than phenol, ignoring whether the substituent is electron donating or withdrawing. Another mistake is to focus only on the number of substituents rather than their nature, or to misremember that methyl is an electron donating group and nitro is electron withdrawing. Confusion can also arise from not distinguishing inductive and resonance effects. To avoid these pitfalls, always analyse whether a substituent donates or withdraws electron density and how that affects the stability of the conjugate base. In this case, methyl decreases acidity, nitro increases acidity, and phenol lies in between. Final Answer: The correct order of increasing acidity (lowest to highest) is o-cresol < phenol < o-nitrophenol.

Question 4

In basic chemical bonding, which of the following molecules contains a purely non polar covalent bond in which the bonding electrons are shared equally between identical atoms?

Accepted Answer

Introduction / Context: This question tests the idea of non polar covalent bonding in basic chemistry. In a non polar covalent bond, the bonding electrons are shared equally between the atoms. This usually happens when the two atoms have the same or very similar electronegativities. Understanding this concept helps learners distinguish between polar bonds, non polar bonds, and non polar molecules, which is important for predicting physical properties such as solubility and boiling point. Given Data / Assumptions: We are given several molecules: Br2, CO, CO2, CCl4, and H2O. We are asked which molecule contains a non polar covalent bond. Assume standard electronegativity values from the periodic table. Assume bonds are single or multiple covalent bonds between the atoms shown in the formula. Concept / Approach: The key concept is electronegativity difference. When two atoms have the same electronegativity, the electrons are shared equally and the bond is non polar covalent. When the electronegativity difference is significant, the bond becomes polar covalent. In diatomic molecules of the same element, such as Br2, the bond is always non polar because both atoms attract electrons equally. In heteronuclear molecules like CO, CO2, CCl4, and H2O, the bonds are generally polar because the atoms have different electronegativities. Step-by-Step Solution: Step 1: Consider Br2. It consists of two identical bromine atoms. The Br–Br bond has zero electronegativity difference, so this covalent bond is non polar. Step 2: Consider CO. Carbon and oxygen have different electronegativities, so the C–O bond is polar covalent. Step 3: Consider CO2. Each C–O bond is polar covalent for the same reason, even though the molecule as a whole is linear and can be non polar overall. Step 4: Consider CCl4. Each C–Cl bond is polar covalent, although the molecule is tetrahedral and overall non polar in shape. Step 5: Consider H2O. The O–H bonds are polar covalent because oxygen is more electronegative than hydrogen. Verification / Alternative check: An alternative way to check is to remember the simple rule that any molecule formed from two atoms of the same element, such as H2, N2, O2, F2, Cl2, Br2, and I2, will always contain a non polar covalent bond. All the other molecules listed are formed from different elements, so their bonds are polar. This confirms that Br2 is the only correct choice. Why Other Options Are Wrong: CO: The C–O bond is polar covalent due to a large difference in electronegativity between carbon and oxygen. CO2: Each C–O bond is polar covalent, even though the entire molecule can be non polar in shape. CCl4: Each C–Cl bond is polar covalent; the molecule is symmetric but the individual bonds are not non polar. H2O: The O–H bonds are polar covalent, which is why water is a highly polar molecule. Common Pitfalls: Students sometimes confuse a non polar molecule with a non polar bond. A molecule like CO2 or CCl4 can be overall non polar because of molecular geometry, even though the individual bonds are polar. In this question the focus is specifically on the non polar covalent bond itself, not the net polarity of the whole molecule. Forgetting to make this distinction is a frequent source of mistakes in introductory chemistry questions. Final Answer: The molecule that contains a purely non polar covalent bond is Br2.

Question 5

In basic chemistry, which one type of energy is stored within a chemical bond and released or absorbed when that bond is broken or formed?

Accepted Answer

Introduction / Context: This question checks understanding of the nature of energy stored in chemical bonds. In chemistry and thermodynamics, bonds are considered to store energy that can be released during reactions. Knowing whether that energy is potential or kinetic helps students understand how reactions power biological processes, engines, and many everyday phenomena. Given Data / Assumptions: We are asked what type of energy is stored in a chemical bond. Options mention potential energy, kinetic energy, thermal energy, light energy, and a combination. Assume normal school level chemistry definitions of energy types. Concept / Approach: Energy stored in the positions or arrangements of particles is called potential energy. In chemical bonds, electrons and nuclei are arranged in specific ways, and the energy in these arrangements is chemical potential energy. Kinetic energy, in contrast, is the energy of motion, such as molecules moving or vibrating. Thermal and light energies are forms in which potential energy can be converted and released, but they are not the primary form stored in a stable bond. Step-by-Step Solution: Step 1: Recall that chemical bonds represent stable arrangements of electrons and nuclei with a certain energy level. Step 2: Recognize that this stored energy depends on how particles are arranged relative to each other, which matches the definition of potential energy. Step 3: Understand that during a chemical reaction, these arrangements change and the difference in potential energy is released or absorbed as heat, light, or other forms. Step 4: Kinetic energy is present because particles move, but it is not the main stored energy of a bond itself. Step 5: Therefore, the correct classification of energy stored in the bond is chemical potential energy. Verification / Alternative check: Standard chemistry textbooks define chemical energy as a form of potential energy. For example, energy stored in food, batteries, or fuels is chemical potential energy. When a fuel burns, chemical potential energy is converted into thermal and sometimes light energy, confirming that the original reservoir is potential energy rather than kinetic energy. This supports selecting potential energy as the correct answer. Why Other Options Are Wrong: Kinetic energy: This is the energy of motion of particles, not the stored energy in the bond structure itself. Only thermal energy: Heat is usually a product or transfer of energy during reaction, not the inner stored energy in an unreacted bond. Only light energy: Light may be emitted in some reactions, but again it is a form of released energy, not the inherent storage within the bond. Both potential and kinetic energy in equal amounts: This is not how textbook chemistry defines bond energy; the stored component is classified as potential energy. Common Pitfalls: Students often think that since reactions give off heat, the energy in bonds must be thermal. Others believe kinetic energy is stored because molecules are always moving. The important idea is that the bond itself represents a particular arrangement of particles. That arrangement has associated potential energy. The kinetic and thermal energies mainly describe motion and temperature, not the inner storage. Keeping this distinction clear helps avoid confusion in more advanced topics like thermochemistry and Gibbs energy. Final Answer: The type of energy stored in a chemical bond is potential energy.

Question 6

The luster or shiny appearance of a metal surface is mainly due to the presence of free mobile electrons in the metallic structure. Is this statement correct or incorrect?

Accepted Answer

Introduction / Context: This question explores why metals have a characteristic shiny appearance known as metallic luster. In solid state chemistry and materials science, metallic bonding and free electrons play a crucial role in both electrical conductivity and optical properties. Understanding the reason for metallic luster helps connect abstract ideas about electron behavior with visible properties of everyday materials like copper, aluminium, and silver. Given Data / Assumptions: The statement claims that metal luster is due to free electrons. We need to decide if this statement is correct or incorrect. We assume a basic model of metallic bonding with a sea of delocalized electrons. Concept / Approach: In the metallic bond model, metal atoms release some of their valence electrons, which become delocalized and move freely through the crystal. These free electrons are able to absorb, reemit, and reflect light over a wide range of wavelengths. This interaction with light gives metals their shiny, reflective surfaces. The same free electrons also explain why metals are good conductors of electricity and heat. Therefore, the presence of free electrons is the key reason behind metallic luster. Step-by-Step Solution: Step 1: Recall that in metals, valence electrons are not bound to any single atom but are delocalized and form an electron sea. Step 2: When light strikes a metal surface, these free electrons can oscillate and reemit light, causing strong reflection. Step 3: This strong reflection of incident light makes the surface appear shiny or lustrous. Step 4: Because this explanation directly involves free electrons, the given statement aligns with the accepted model of metallic luster. Step 5: Therefore, the statement that metallic luster is due to the presence of free electrons is correct. Verification / Alternative check: Textbook discussions of metallic bonding usually mention three connected properties: luster, electrical conductivity, and thermal conductivity. All three are traced back to the same feature, namely the presence of delocalized electrons in the metallic lattice. Experiments show that metals with more free electrons generally have better conductivity and strong reflectivity. This supports the idea that free electrons are responsible for metallic luster and confirms that the statement is correct. Why Other Options Are Wrong: Option B: Tightly bound electrons in directional covalent bonds are more typical of non metals and do not produce the same shiny metallic reflection. Option C: Metals are conductors, not insulators; this option contradicts fundamental electrical behavior. Option D: Density can influence appearance but it is not the direct cause of luster; free electrons are the primary reason. Option E: The statement has a definite truth value; it is clearly correct based on metallic bonding theory. Common Pitfalls: Many learners mix up the roles of bound and free electrons. They sometimes think that any type of electron cloud can cause shine. However, metallic luster is strongly linked to the ability of electrons to move freely, which allows them to respond quickly to incoming light waves. Another mistake is to think that metallic luster and conductivity come from different causes, when in fact they share the same origin in the delocalized electron sea model. Keeping this connection in mind simplifies the understanding of metallic properties. Final Answer: The statement is correct because free mobile electrons in metals reflect incident light and produce metallic luster.

Question 7

In the periodic table, which properties are characteristic of group 1 alkali metals such as lithium, sodium, and potassium?

Accepted Answer

Introduction / Context: This question focuses on the typical properties of group 1 elements, also called alkali metals. Examples include lithium, sodium, potassium, rubidium, cesium, and francium. These metals share a number of distinctive physical and chemical properties that make them stand out in introductory inorganic chemistry. Recognizing these features helps in predicting their reactions and in understanding periodic trends. Given Data / Assumptions: We are considering group 1 metals in the periodic table. Listed properties include low melting points, production of hydrogen with water, and softness. The question asks which of the listed properties are characteristic of these metals. Concept / Approach: Alkali metals have a single valence electron, large atomic radii, and relatively weak metallic bonding compared with many transition metals. As a result, they tend to be soft, have relatively low melting points, and react vigorously with water to form metal hydroxides and hydrogen gas. To answer the question, we check each statement against standard descriptions of alkali metals from periodic table trends. Step-by-Step Solution: Step 1: Low melting points: Alkali metals have lower melting points than many other metals because their metallic bonds are relatively weak. Step 2: Soft nature: Lithium, sodium, and potassium are so soft that they can often be cut easily with a knife, which is a hallmark group 1 property. Step 3: Reaction with water: Alkali metals react vigorously with water to form alkaline hydroxide solutions and hydrogen gas, which is a well known chemical property. Step 4: Since each individual property is indeed characteristic, any option that gathers all of them together is likely correct. Step 5: Option D states that all of the listed properties are typical, which matches the analysis. Verification / Alternative check: In standard experiments, small pieces of sodium or potassium are dropped into water. The metals float, move rapidly, and fizz as hydrogen gas is released. Sometimes the hydrogen ignites, giving a flame. Teachers also demonstrate that sodium is easily cut, showing its softness. Reference tables for melting points show that these metals melt at temperatures significantly lower than many other common metals such as iron or copper. This collection of evidence supports option D as the correct answer. Why Other Options Are Wrong: Option A: This is true but does not capture all characteristic properties alone. Option B: This is also true but incomplete. Option C: This is true and very important but still only one part of the full description. Option E: This is incorrect because all three properties are actually known features of alkali metals. Common Pitfalls: A frequent mistake is to think that all metals are hard and have very high melting points. Group 1 metals are counterexamples, and this helps students appreciate periodic trends rather than overgeneralizing. Another misconception is that only transition metals show interesting reactions with water, but in fact alkali metals are among the most reactive. Remembering the specific behavior of group 1 metals helps in many competitive exams and school tests in chemistry. Final Answer: The characteristic properties of group 1 metals include all the listed features, so the correct answer is All of the above characteristics are typical for group 1 metals.

Question 8

In basic inorganic chemistry, which drying agent is suitable for drying ammonia gas without reacting with it?

Accepted Answer

Introduction / Context: This question tests knowledge of laboratory drying agents used in inorganic chemistry. When gases are prepared in the laboratory, they are often moist and need to be dried before further use or collection. The correct drying agent must remove water but should not react with the gas being dried. Ammonia is a basic gas, so its interaction with different drying agents is an important consideration in practical chemistry questions. Given Data / Assumptions: The gas to be dried is ammonia (NH3). We have several possible drying agents: sulphuric acid, calcium oxide, phosphorus pentoxide, calcium chloride, and concentrated nitric acid. We must choose an agent that absorbs moisture but does not react with ammonia. Concept / Approach: Ammonia is a basic gas and is very soluble in water. Strong acidic drying agents, or drying agents that chemically react with bases, are not suitable for drying ammonia. Calcium oxide (quicklime) is a basic oxide that can absorb water to form calcium hydroxide and does not react with ammonia in a problematic way. Therefore, it is commonly used in the laboratory for drying ammonia gas. Other strong dehydrating agents like concentrated sulphuric acid or phosphorus pentoxide can react with ammonia and are not appropriate. Step-by-Step Solution: Step 1: Recognize that the drying agent must remove moisture while remaining chemically inert toward ammonia. Step 2: Sulphuric acid is a strong acid and powerful dehydrating agent that reacts with ammonia to form ammonium sulfate, so it is unsuitable. Step 3: Phosphorus pentoxide is an extremely strong dehydrating agent that can also react with ammonia, so it is not used. Step 4: Calcium chloride is a common drying agent but it can form complexes with ammonia and is therefore not ideal. Step 5: Calcium oxide is a basic oxide that absorbs water to form calcium hydroxide without reacting significantly with ammonia, so it is the correct choice. Verification / Alternative check: Standard laboratory manuals and inorganic chemistry textbooks list quicklime (calcium oxide) as the recommended drying agent for ammonia. They also include cautionary notes that concentrated sulphuric acid, although good for many other gases, reacts with ammonia and should not be used. Practical laboratory setups often show a drying tower packed with quicklime placed after the apparatus that generates ammonia. This consistent recommendation across sources confirms that calcium oxide is the correct drying agent. Why Other Options Are Wrong: Sulphuric acid: Reacts with ammonia to form ammonium sulfate, so it is not inert. Phosphorus pentoxide: Very strong dehydrating agent that can chemically interact with ammonia. Calcium chloride: Can absorb ammonia and form complexes, which interferes with the purity and yield of the gas. Concentrated nitric acid: Strong oxidizing acid that would react vigorously with ammonia and is dangerous in this context. Common Pitfalls: Students sometimes think that any strong dehydrating agent is suitable as a universal drying agent. However, compatibility with the gas is crucial. Another common error is to choose calcium chloride simply because it is a familiar drying agent, forgetting its tendency to interact with ammonia. The key lesson is to select a drying agent that both removes water and remains chemically inert toward the gas being treated. Final Answer: The suitable drying agent for ammonia gas is Calcium oxide.

Question 9

In the classification of rocks, kaolin or china clay is chemically classified as which type of rock based on its clay rich composition?

Accepted Answer

Introduction / Context: This question links basic geology with applied chemistry by asking about the classification of kaolin, also known as china clay. Kaolin is widely used in ceramics, paper coating, and many industrial processes. Understanding its rock type helps students connect mineral composition with the broader rock classification system that includes argillaceous, silicious, calcareous, and other categories. Given Data / Assumptions: The material under discussion is kaolin or china clay. We are asked to choose its chemical rock classification. Options include argillaceous, silicious, metamorphic, calcareous, and igneous rock types. Concept / Approach: Argillaceous rocks are clay rich rocks that contain a high proportion of aluminosilicate minerals. Kaolin is essentially a clay mineral composed mainly of hydrated aluminium silicate. Silicious rocks are rich in silica, calcareous rocks are rich in calcium carbonate, metamorphic rocks are those altered by heat and pressure, and igneous rocks are formed from solidified magma. Since kaolin is a fine grained clay material, it is classified as argillaceous in nature. Step-by-Step Solution: Step 1: Recall that kaolin or china clay is composed mainly of the mineral kaolinite, a hydrated aluminium silicate. Step 2: Recognize that clay rich rocks and materials are grouped under argillaceous rocks. Step 3: Silicious rocks are dominated by silica, such as quartzites, which is not the primary identity of kaolin. Step 4: Calcareous rocks are rich in calcium carbonate, like limestone, which does not match the composition of kaolin. Step 5: Therefore, based on its clay content, kaolin is most appropriately classified as an argillaceous rock. Verification / Alternative check: Geology and ceramic engineering references describe kaolin as a residual clay that often forms from the weathering of feldspar rich rocks. It is consistently placed within the argillaceous category, because argilla refers to clay. Kaolin deposits are mapped and mined as clay resources rather than as silicious sandstones or calcareous limestones. This consistent classification supports the choice of argillaceous rock. Why Other Options Are Wrong: Silicious rock: These rocks are dominated by free silica, such as quartz, rather than clay minerals. Metamorphic rock: Kaolin is not primarily defined as a rock altered by heat and pressure; it is a clay deposit. Calcareous rock: These rocks are rich in calcium carbonate, which does not match the aluminosilicate composition of kaolin. Igneous rock: Igneous rocks solidify from magma or lava and do not describe a soft clay material like kaolin. Common Pitfalls: Students may confuse any rock involving silicate minerals with silicious rock, forgetting that silicious is reserved for rocks dominated by silica in the form of quartz. Others might pick calcareous simply because they associate white color with calcium carbonate. The key is to focus on the clay character of kaolin, which clearly places it in the argillaceous group. Paying attention to root words such as argilla for clay and calx for lime can help in remembering these classifications. Final Answer: Kaolin or china clay is chemically classified as an Argillaceous rock.

Question 10

For shock resisting steels used in tools and machine parts, which mechanical property must be particularly high so that the steel can absorb sudden impact loads without failure?

Accepted Answer

Introduction / Context: This question relates to material science and mechanical engineering, focusing on the design of shock resisting steels. Such steels are used in situations where components experience sudden impact loads, like hammers, chisels, springs, and certain machine parts. Knowing which property is most important helps in selecting and designing appropriate alloys for engineering applications. Given Data / Assumptions: The material is shock resisting steel. The application involves sudden impact or shock loads. We are asked which property this steel should possess. Options include toughness, low wear resistance, low tensile strength, low hardness, and very high brittleness. Concept / Approach: Toughness is the ability of a material to absorb energy before fracturing. It combines strength and ductility. For components that must withstand sudden shocks, toughness is critical because the material must absorb impact energy without cracking. Low wear resistance, low tensile strength, low hardness, or high brittleness are undesirable. Therefore, the correct answer is the property that allows energy absorption under shock loading, namely toughness. Step-by-Step Solution: Step 1: Recognize that shock loads are sudden and involve high energy transfer over a short time. Step 2: A material that is brittle will crack easily under such conditions, which is not acceptable. Step 3: Toughness measures the capacity to absorb energy and deform plastically before failure. Step 4: Shock resisting steels are designed with alloying and heat treatment to achieve high toughness. Step 5: Therefore, among the given options, toughness is the crucial property for shock resisting steels. Verification / Alternative check: Engineering handbooks and material data sheets for shock resisting tool steels frequently highlight their impact toughness and Charpy impact values. These steels are often tempered to achieve a balance of hardness and toughness. Testing methods such as impact tests explicitly measure toughness under shock loading. In contrast, properties like low wear resistance or low hardness are not features of high performance steels for shock applications. This confirms that toughness is the desired property. Why Other Options Are Wrong: Low wear resistance: Components exposed to repeated impact often also face wear; low wear resistance is not an advantage. Low tensile strength: Shock resisting steels must have adequate strength to carry loads, not low strength. Low hardness: While extreme hardness can reduce toughness, too low hardness will not provide durability or wear resistance. Very high brittleness: Brittle materials fail suddenly under shock, which is exactly what engineers want to avoid. Common Pitfalls: A common mistake is to confuse hardness with toughness. Hard materials resist scratching and indentation but are not always good at absorbing impact energy. Another error is to assume that any very strong material will also resist shock, even if it is brittle. In reality, a balance of strength and ductility is needed, and that combination is captured by the concept of toughness. Remembering this helps in many exam questions about materials used in tools, springs, and structural components. Final Answer: Shock resisting steels should have high Toughness so that they can absorb sudden impact loads without failure.

Question 11

Among the common non metals listed, which element is found in the liquid state at room temperature?

Accepted Answer

Introduction / Context: This question examines knowledge of the physical states of non metals at room temperature. Most elements are either solids or gases under standard conditions, and only a very small number are liquids. Knowing which non metal is liquid at room temperature is a classic general knowledge and chemistry question that often appears in school examinations and competitive tests. Given Data / Assumptions: The elements listed are chlorine, bromine, fluorine, iodine, and neon. Room temperature is assumed to be about 25 degrees Celsius. We must identify which of these non metals is liquid at this temperature. Concept / Approach: At room temperature, most halogens and noble gases have well known physical states. Fluorine and chlorine are gases, bromine is a heavy reddish brown liquid, and iodine is a solid that can sublime. Neon is a noble gas and remains gaseous at room temperature. Among the given options, bromine is the only non metal that occurs as a liquid under standard room conditions. Step-by-Step Solution: Step 1: Recall the halogens: fluorine, chlorine, bromine, and iodine form a group with gradually changing physical states. Step 2: Fluorine and chlorine are light, pale colored gases at room temperature. Step 3: Bromine is a dark reddish brown liquid that releases orange vapors and is the only liquid non metal at room temperature. Step 4: Iodine is a dark solid at room temperature which can sublime to give violet vapors. Step 5: Neon is a colorless noble gas, not a liquid at room temperature. Verification / Alternative check: Standard periodic table references and chemistry textbooks highlight bromine as one of the only two elements that are liquid at room temperature, the other being mercury, which is a metal. Charts of states of elements usually show bromine with a special color marking it as a liquid non metal. Laboratory bottles of bromine contain a heavy, fuming liquid, which is easily recognizable. All these confirmations support bromine as the correct answer. Why Other Options Are Wrong: Chlorine: A greenish yellow gas at room temperature. Fluorine: A very reactive pale yellow gas under standard conditions. Iodine: A solid at room temperature, forming violet vapors when heated. Neon: A noble gas that remains gaseous at room temperature and is used in discharge lamps. Common Pitfalls: Some learners misremember iodine as a liquid because it produces visible vapors, but it is actually a solid that sublimes. Others may think chlorine is a liquid due to images of pressurized cylinders, but in those cases it is compressed, not naturally liquid at room temperature. Remembering that the only common liquid non metal at room temperature is bromine helps answer many quick general knowledge questions accurately. Final Answer: The non metal found in liquid state at room temperature is Bromine.

Question 12

In water treatment and basic chemistry, permanent hardness of water can be removed by adding which chemical substance to precipitate calcium and magnesium ions?

Accepted Answer

Introduction / Context: This question focuses on the concept of water hardness in environmental and applied chemistry. Hard water contains dissolved calcium and magnesium salts. Permanent hardness is mainly due to chlorides and sulfates of calcium and magnesium, which do not disappear upon boiling. Understanding how to remove permanent hardness is important in water softening, boiler maintenance, and household applications such as laundry and soap use. Given Data / Assumptions: The type of hardness considered is permanent hardness. We are asked which chemical can remove this permanent hardness by addition. Options include alum, lime, sodium carbonate, potassium permanganate, and calcium sulfate. Concept / Approach: Permanent hardness due to calcium and magnesium chlorides and sulfates can be removed by adding washing soda, which is sodium carbonate. Sodium carbonate reacts with soluble calcium and magnesium salts to form insoluble carbonates that can be removed by filtration. Alum is used for coagulation, lime is more effective for temporary hardness, and potassium permanganate is an oxidizing agent. Therefore, sodium carbonate is the correct chemical for removing permanent hardness by precipitation. Step-by-Step Solution: Step 1: Recognize that permanent hardness is caused mainly by Ca2+ and Mg2+ in the form of sulfates and chlorides. Step 2: Recall that sodium carbonate (washing soda) reacts with these ions to form insoluble calcium carbonate and magnesium carbonate. Step 3: The reactions remove Ca2+ and Mg2+ from the water, thus softening it. Step 4: Alum is primarily a coagulant used to remove suspended particles and is not the main reagent for removing permanent hardness. Step 5: Lime is more commonly used in the lime soda process together with sodium carbonate and is often associated with temporary hardness removal. Verification / Alternative check: Standard water treatment processes include the lime soda process where sodium carbonate is specifically listed as the reagent used to remove the non carbonate (permanent) hardness. Textbooks and exam guides often state that washing soda can remove both temporary and permanent hardness by precipitation of metal carbonates. Experimental demonstrations of water softening frequently feature sodium carbonate as the main softening agent for permanently hard water. This evidence strongly supports sodium carbonate as the correct answer. Why Other Options Are Wrong: Alum: Works mainly as a coagulating agent to settle suspended impurities, not to precipitate calcium and magnesium from solution. Lime: It helps in softening but is especially used with sodium carbonate and is more closely associated with temporary hardness removal. Potassium permanganate: A strong oxidizing agent used for disinfection and oxidation of organic matter, not for hardness removal. Calcium sulfate: It is itself a salt that contributes to hardness rather than removing it. Common Pitfalls: Students sometimes confuse temporary and permanent hardness and may incorrectly think that boiling is always sufficient. Others may pick alum or lime because they are familiar water treatment chemicals. Remembering that washing soda, which is sodium carbonate, is the key agent for precipitating calcium and magnesium from permanently hard water helps answer many questions about water softening accurately. Final Answer: Permanent hardness of water may be removed by the addition of Sodium carbonate.

Question 13

Diamond is much harder than graphite mainly because diamond has a three dimensional covalent network structure, whereas graphite has weakly bonded layers. Which explanation best describes this difference in hardness?

Accepted Answer

Introduction / Context: This question compares two allotropes of carbon, diamond and graphite, and asks why their hardness is so different. Both substances are made only of carbon atoms, yet diamond is one of the hardest known materials, while graphite is soft and used as a lubricant and pencil lead. Understanding the structural reason for this difference is a classic topic in solid state chemistry and materials science. Given Data / Assumptions: Both diamond and graphite are allotropes of carbon. Diamond is experimentally found to be much harder than graphite. We are given several possible structural explanations and must choose the best one. Concept / Approach: The key concept is the arrangement and bonding of carbon atoms. In diamond, each carbon atom is sp3 hybridized and bonded covalently to four other carbon atoms in a tetrahedral three dimensional network. This creates a very rigid, strong structure throughout the crystal. In graphite, each carbon atom is sp2 hybridized and bonded to three other carbons in planar hexagonal layers. These layers are held together by weak forces, so they can slide over each other, making graphite soft. The correct explanation must mention the rigid three dimensional covalent network in diamond versus layered structure in graphite. Step-by-Step Solution: Step 1: Recall that diamond has a tetrahedral network where each carbon atom forms four strong covalent bonds. Step 2: This continuous three dimensional covalent network means that breaking the crystal requires breaking many strong covalent bonds, leading to very high hardness. Step 3: In graphite, carbon atoms form flat sheets with hexagonal patterns, with strong covalent bonds only within each sheet. Step 4: The sheets are stacked and held together by weak intermolecular forces, which allow the layers to slide easily. Step 5: The option that describes diamond as a rigid three dimensional network and graphite as weakly bonded layers is therefore the correct one. Verification / Alternative check: Solid state chemistry diagrams and models of diamond show a repeating tetrahedral framework, while graphite is represented as parallel sheets of hexagons. Measurements of mechanical properties show that diamond has extremely high hardness and high bulk modulus. Graphite, in contrast, is soft enough to leave marks on paper. These observations are consistent with strong three dimensional bonding in diamond and weak interlayer forces in graphite. Textbooks repeatedly emphasize this structural difference when explaining the contrasting properties of carbon allotropes. Why Other Options Are Wrong: Option B: Neither diamond nor graphite contains metal atoms; both are pure carbon. Option C: Density does not determine hardness by itself; many dense materials are not extremely hard and some less dense materials can be quite hard. Option D: Both diamond and graphite involve covalent bonding; diamond does not consist of purely ionic bonds. Option E: The number of atoms in the crystal is not the reason for hardness; it is the bonding pattern that matters. Common Pitfalls: Students sometimes believe that hardness is simply a result of density or the number of atoms, which is not correct. Others may think that diamond must contain some special element, but it is purely carbon, just like graphite. The key lesson is that the same element can form very different structures, called allotropes, and these structural differences lead to dramatically different physical properties. Focusing on bonding and geometry rather than just composition helps avoid these misunderstandings. Final Answer: Diamond is harder than graphite because Diamond has a rigid three dimensional covalent network structure, while graphite consists of weakly bonded planar layers.

Question 14

A sample of matter must be identified as copper if which one of the following conditions is definitely true for every atom in that sample?

Accepted Answer

Introduction / Context: This question tests understanding of how elements are defined in modern chemistry. While many properties such as color, reactivity, or melting point are associated with particular elements, the fundamental definition of an element is based on the number of protons in the nucleus of each atom. This number is called the atomic number. Copper has a specific atomic number, and that definition allows us to say with certainty when a sample must be copper. Given Data / Assumptions: We are asked when a sample must be copper. One option mentions that each atom has 29 protons. Other options mention properties such as reactivity with oxygen, conductivity, melting point, and color. We assume standard chemical definitions of atomic number and element identity. Concept / Approach: Every element is uniquely defined by its atomic number, which is the number of protons in the nucleus of its atoms. Copper has atomic number 29, meaning each copper atom has 29 protons. While many materials conduct electricity, react with oxygen, or have particular colors, these properties are not unique to copper. Only the proton count provides a certain identification. Therefore, the condition that every atom has 29 protons is the unambiguous test that the sample is copper. Step-by-Step Solution: Step 1: Recall that the atomic number of copper (Cu) is 29. Step 2: Understand that atomic number equals the number of protons in the nucleus of an atom. Step 3: Recognize that if every atom in a sample has 29 protons, by definition those atoms are copper atoms. Step 4: Check other properties like conductivity or melting point and note that they are not unique; many metals share them. Step 5: Therefore, the only condition that forces the sample to be copper is that each atom has 29 protons. Verification / Alternative check: Periodic tables show copper with symbol Cu and atomic number 29. Chemistry definitions state that if an atom has atomic number 29, it is copper irrespective of its isotopic mass or chemical state. Experimental methods such as mass spectrometry or nuclear measurements can determine the proton number. No other element shares the same atomic number, so there is a one to one relationship. This confirms that the proton count condition is the correct and precise criterion. Why Other Options Are Wrong: Atoms reacting with oxygen: Many elements form oxides, so this property is not unique to copper. Electrical conductivity: Many metals and some solutions conduct electricity; this does not uniquely identify copper. Melting at about 1768 K: While the melting point is characteristic, a single measured melting point could be subject to experimental error or alloy formation. Reddish brown color: Copper is famous for this color, but surface conditions, alloys, or coatings may change appearance, and color alone is not a fundamental definition. Common Pitfalls: Students often rely on familiar visual or practical properties to identify elements in questions. While those properties help in everyday identification, examination questions like this deliberately test conceptual knowledge. The key is to remember that atomic number is the only strict defining property of an element. By focusing on proton number rather than less precise features, you can answer similar questions about identity of elements with confidence. Final Answer: A sample must be copper if Each atom in the sample has 29 protons in its nucleus.

Question 15

In the context of drugs of abuse, PCP is more commonly known by which street name?

Accepted Answer

Introduction / Context: This general knowledge question relates to the street names of certain drugs. PCP stands for phencyclidine, which is a dissociative drug that has been abused in some regions. Law enforcement, medical staff, and public health educators often refer to drugs by both their chemical names and their street names. Knowing that PCP is commonly called angel dust can appear in competitive exams and public awareness materials. Given Data / Assumptions: The drug name given is PCP (phencyclidine). We are asked for its more common street name. Options include Acid, Angel dust, Ecstasy, None of the above, and Speed. Concept / Approach: Different illicit drugs have famous street names. For example, LSD is often called acid, MDMA is called ecstasy, some amphetamines are called speed, and PCP is most commonly known as angel dust. To answer the question, we match PCP with its widely known street name. General awareness articles and exam guides for health education usually list this mapping specifically. Step-by-Step Solution: Step 1: Recall that PCP stands for phencyclidine, a dissociative drug. Step 2: Note that LSD is commonly called acid, which is linked to hallucinogenic drugs, not PCP. Step 3: Recognize that MDMA is often called ecstasy, another distinct drug. Step 4: Speed is a common name for stimulant amphetamine type drugs. Step 5: The specific and widely documented street name for PCP is angel dust, so that is the correct option. Verification / Alternative check: Drug awareness literature from public health agencies, textbooks on pharmacology, and general encyclopedic sources list PCP under the street name angel dust. Lists of controlled substances and their street names repeat this association. Acid is listed for LSD, ecstasy for MDMA, and speed for some amphetamines. Since PCP is consistently identified as angel dust in these sources, the mapping is reliable. Why Other Options Are Wrong: Acid: This is a common street name for LSD, not for PCP. Ecstasy: This name is associated with MDMA, a different psychoactive drug. None of the above: Incorrect because a correct name is present among the options. Speed: Used for stimulant drugs like amphetamine and methamphetamine, not for PCP. Common Pitfalls: Students sometimes confuse different street names because they are all part of discussions about drugs of abuse. It is important to remember specific pairings for exam purposes: LSD with acid, MDMA with ecstasy, amphetamines with speed, and PCP with angel dust. Careful reading of health education material and practice through questions helps establish these connections clearly and avoid mix ups in multiple choice settings. Final Answer: PCP is more commonly known as Angel dust.

Question 16

German silver, also called nickel silver, is an alloy used in utensils and decorative items. How much real silver does German silver contain?

Accepted Answer

Introduction / Context: This question tests knowledge about a common alloy called German silver or nickel silver. Despite its name, it is not actually a silver alloy in the precious metal sense. Understanding the composition of such alloys is part of general knowledge in chemistry and helps avoid confusion created by traditional names that may not reflect true composition. Given Data / Assumptions: The material in question is German silver, also known as nickel silver. We are asked how much actual silver it contains. Options suggest various percentages of silver, including zero. Concept / Approach: German silver is an alloy typically made from copper, zinc, and nickel. It has a silvery appearance, which is why it gained this name, but it does not actually contain elemental silver. The name refers to its color rather than its composition. Therefore, the correct answer is that German silver contains no real silver, that is, 0 percent silver. Step-by-Step Solution: Step 1: Recall that German silver is also called nickel silver due to its nickel content. Step 2: The common composition is a mixture of copper, zinc, and nickel in various proportions. Step 3: There is no requirement for actual silver in this alloy; its silvery appearance comes from nickel and the metallic sheen. Step 4: Therefore, the percentage of true silver in German silver is zero. Step 5: Among the options, 0 percent silver matches this fact. Verification / Alternative check: Metallurgy and material science references list German silver compositions and clearly state that no silver is included. Jeweller guides warn that nickel silver items that look like silver are actually base metal alloys without precious metal content. This confirmation appears in multiple technical and consumer references, making the answer reliable. Why Other Options Are Wrong: 100 percent pure silver: That would simply be silver metal, not an alloy called German silver. About 10 percent silver: There is no standard composition that includes silver in German silver. About 1 percent silver: Even trace levels of silver are not part of the usual formulation. About 50 percent silver: This would be a high value precious metal alloy, not the base metal alloy sold as nickel silver. Common Pitfalls: The main confusion comes from the misleading name. People often assume that anything with the word silver in its name must contain silver. Similar confusion can occur with other trade names and alloys. It is important to rely on composition data rather than names alone when answering exam questions or making purchasing decisions about metals and jewellery. Final Answer: German silver contains 0 percent silver and is actually an alloy of copper, zinc, and nickel.

Question 17

Which of the following naming rules is correctly applied when writing names for binary covalent compounds formed between two non metals?

Accepted Answer

Introduction / Context: This question deals with the naming of binary covalent compounds in inorganic chemistry. These compounds consist of two different non metals, such as nitrogen and oxygen in nitrogen dioxide or carbon and oxygen in carbon monoxide. Correct application of naming rules, including suffixes and numerical prefixes, is important for clear communication and is frequently tested in examinations. Given Data / Assumptions: The focus is on covalent compounds formed between two non metals. Options describe different combinations of names, suffixes, and possible prefixes. We assume standard IUPAC style naming rules at basic school level. Concept / Approach: For binary covalent compounds between non metals, the name of the first element is written in full, and the second element is written with the suffix ide. Numerical prefixes such as mono, di, tri, tetra, and so on can be used to indicate the number of atoms of each element, especially when there is ambiguity. For example, CO is carbon monoxide and CO2 is carbon dioxide. The correct rule must reflect that both elements are non metals and that the second element ends in ide with possible use of prefixes. Step-by-Step Solution: Step 1: Recognize that the molecules under discussion are binary covalent compounds, not ionic compounds. Step 2: Recall that the first element is usually the less electronegative non metal and is named first without a suffix. Step 3: The second element is named with its root plus the suffix ide, and numerical prefixes may be used for both elements if needed. Step 4: Option A describes the rule as nonmetal name plus nonmetal name with ide and includes mention of prefixes, which matches the standard rule. Step 5: Other options either misuse the suffix ite or involve metal names, which is incorrect for purely covalent non metal compounds. Verification / Alternative check: Chemistry textbooks list simple examples such as nitrogen monoxide (NO), nitrogen dioxide (NO2), sulfur dioxide (SO2), sulfur trioxide (SO3), and dinitrogen pentoxide (N2O5). All follow the pattern where the first non metal name appears unchanged and the second adopts the ide ending along with prefixes. There is no use of the suffix ite in these standard binary covalent names. Therefore, the description given in option A is consistent with actual examples. Why Other Options Are Wrong: Option B: The suffix ite is generally used in naming oxyanions and related salts, not in simple binary covalent compounds. Option C: This pattern is typical for ionic compounds with a metal and non metal, not for two non metals. Option D: Again mixes metal plus non metal and uses the suffix ite, which does not match covalent naming rules. Option E: It is incorrect because both elements must be named; ignoring the first element would make the name incomplete. Common Pitfalls: Students sometimes transfer rules from ionic naming directly to covalent compounds or confuse suffixes such as ate and ite with ide. Another frequent error is forgetting to use numerical prefixes, especially for the second element, which can cause ambiguity. Practising examples and remembering that ide is used for the second non metal element in binary covalent compounds helps avoid these issues. Final Answer: The correct naming rule for binary covalent compounds is nonmetal name plus nonmetal name with suffix ide, often using numerical prefixes such as mono, di, or tri.

Question 18

Most commercial antifreeze solutions used in automobile radiators are dangerous if swallowed because of which hazardous component present in them?

Accepted Answer

Introduction / Context: This question concerns basic knowledge of antifreeze solutions used in vehicle cooling systems. These liquids prevent freezing and overheating but can be very dangerous if ingested by humans or animals. Understanding which chemical in antifreeze is hazardous is important in both general science and public safety awareness. Given Data / Assumptions: The question asks about the hazardous component in most commercial antifreeze. Options mention ethane alcohol, ethylene glycol, ethylene, ethene, and propane. We assume typical automotive antifreeze formulations. Concept / Approach: Ethylene glycol is a common base ingredient in antifreeze. It has suitable physical properties such as low freezing point and high boiling point. Unfortunately, it is highly toxic if swallowed and has a sweet taste, which can attract children and animals. Ethane alcohol, simple ethylene, ethene, and propane are not the primary components of standard antifreeze formulations. Therefore, ethylene glycol is the hazardous component referred to here. Step-by-Step Solution: Step 1: Recall that automobile antifreeze is commonly based on ethylene glycol mixed with water and additives. Step 2: Ethylene glycol has the formula HO–CH2–CH2–OH. Step 3: It is effective at lowering the freezing point, but ingestion can cause severe poisoning and kidney damage. Step 4: Other substances listed, such as propane or ethene, are gases and not typical liquid bases for antifreeze. Step 5: Therefore, the hazardous component in most antifreeze formulations is ethylene glycol. Verification / Alternative check: Product safety labels on antifreeze containers often highlight ethylene glycol as the active ingredient and warn about toxicity. Medical and toxicology references describe ethylene glycol poisoning as a serious emergency. Newer formulations sometimes use propylene glycol as a less toxic alternative, but the question refers to the most common traditional component, ethylene glycol. This confirms that option B is correct. Why Other Options Are Wrong: Ethane alcohol: This is not a standard term used in antifreeze formulations. Ethylene: A gaseous hydrocarbon, not used as the main liquid component in radiator antifreeze. Ethene: Another name for ethylene, again not the main liquid antifreeze component. Propane: A gas used as a fuel, not as a radiator coolant. Common Pitfalls: Some students may confuse similar sounding names like ethylene, ethanol, and ethylene glycol. It is important to remember that glycol indicates a diol structure, which gives the appropriate physical properties for antifreeze and also introduces toxicity. Being precise about chemical names helps in avoiding mistakes in both exams and real life situations involving chemical safety. Final Answer: The hazardous component in most antifreeze solutions is Ethylene glycol.

Question 19

In mineralogy, which mineral is commonly known by the name Horn silver and is an important ore of silver?

Accepted Answer

Introduction / Context: This question belongs to general knowledge in chemistry and geology. Many minerals have traditional names that differ from their chemical names. Horn silver is one such traditional name, and knowing its chemical identity helps connect common names with chemical formulas and uses in metallurgy, especially in silver extraction. Given Data / Assumptions: The common mineral name given is Horn silver. We must match it to the correct chemical compound. Options include silver chloride, zinc phosphate, silver iodide, silver sulfide, and lead chloride. Concept / Approach: Horn silver is the common name for the mineral consisting mainly of silver chloride, usually written as AgCl. It forms as a secondary mineral in the oxidation zones of silver deposits. Other silver compounds like silver iodide and silver sulfide have different common names and appearances. Therefore, identifying Horn silver as silver chloride is the correct choice. Step-by-Step Solution: Step 1: Recall that Horn silver is an old term used in mining and mineralogy for a silver rich mineral. Step 2: The mineral associated with this name is silver chloride, AgCl. Step 3: Zinc phosphate is a zinc compound without silver and is not called Horn silver. Step 4: Silver iodide and silver sulfide are different silver compounds with other uses and common names. Step 5: Therefore, Horn silver corresponds to silver chloride. Verification / Alternative check: Mineralogy references and mining dictionaries list Horn silver as the common name of the mineral cerargyrite, which is silver chloride. Photographs and descriptions show a gray to greenish mineral associated with oxidized silver ore deposits. Other silver minerals like argentite (silver sulfide) have their own specific names. This consistent identification confirms that silver chloride is the correct answer for Horn silver. Why Other Options Are Wrong: Zinc phosphate: Contains zinc and phosphorus, with no silver content. Silver iodide: A silver compound used in photography and cloud seeding, but not called Horn silver. Silver sulfide: Known as argentite or acanthite and has different properties and names. Lead chloride: A lead compound, not a silver ore mineral named Horn silver. Common Pitfalls: Students often confuse different silver minerals due to similar metallic associations. Remember that common names are historical and do not always follow systematic patterns. Linking each traditional name with one clear chemical identity through practice questions is a good way to avoid confusion in examinations and in general chemistry quizzes. Final Answer: The mineral known as Horn silver is Silver chloride.

Question 20

The working principle of an atomic bomb is mainly based on which nuclear process involving heavy atomic nuclei?

Accepted Answer

Introduction / Context: This general knowledge question relates to nuclear physics and the principle behind atomic weapons. Understanding whether an atomic bomb relies on nuclear fission or nuclear fusion is common in both physics education and general awareness. The term atomic bomb typically refers to devices using fission of heavy nuclei such as uranium or plutonium. Given Data / Assumptions: The device under discussion is an atomic bomb. We have options including nuclear fusion, nuclear fission, natural radioactivity, all options, and beta decay alone. We assume basic textbook descriptions of atomic and hydrogen bombs. Concept / Approach: In an atomic bomb, a supercritical mass of fissile material such as uranium 235 or plutonium 239 undergoes rapid uncontrolled chain reactions of nuclear fission. Each fission event splits a heavy nucleus into lighter nuclei, releasing a large amount of energy and more neutrons, which in turn cause more fission events. This chain reaction is the core principle of an atomic bomb. Fusion bombs, often called hydrogen bombs, use fusion of light nuclei but are a different category. Natural radioactivity and beta decay occur at much slower rates and do not by themselves produce the explosive energy of an atomic bomb. Step-by-Step Solution: Step 1: Recall that nuclear fission is the splitting of a heavy nucleus into smaller nuclei with energy release. Step 2: In an atomic bomb, a critical or supercritical mass of fissile material is assembled to allow an uncontrolled fission chain reaction. Step 3: Nuclear fusion involves combining light nuclei at extremely high temperatures and pressures and is used in hydrogen bombs, not basic atomic bombs. Step 4: Natural radioactivity and beta decay occur continuously at low rates and are not the main mechanisms in an atomic bomb explosion. Step 5: Therefore, the main principle for an atomic bomb is nuclear fission. Verification / Alternative check: Historical accounts of the first atomic bombs used in warfare describe devices based on uranium 235 and plutonium 239 fission. Physics textbooks clearly differentiate between fission bombs (atomic bombs) and fusion bombs (hydrogen bombs). Diagrams of the weapon designs show a fission core as the critical component. These sources consistently confirm that nuclear fission is the underlying principle for an atomic bomb. Why Other Options Are Wrong: Nuclear fusion: This is the main principle for hydrogen bombs or thermonuclear weapons, not the basic atomic bomb. Natural radioactivity only: Natural decay is too slow and low energy to create an atomic explosion. All options are correct: This is incorrect because fusion and simple natural radioactivity are not the primary principles in an atomic bomb. Beta decay alone: Beta decay is just one type of radioactive decay and cannot account for the massive energy release in an atomic bomb. Common Pitfalls: Students sometimes mix up the terms atomic bomb and hydrogen bomb and may think they are the same. It is important to remember that atomic bomb refers mainly to fission devices, while hydrogen bomb refers to fusion devices. Recognizing this difference helps answer questions correctly and improves understanding of nuclear physics concepts such as fission, fusion, and radioactivity. Final Answer: The principle of an atomic bomb is based on Nuclear fission.

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