Chemistry

21. What does TNT stand for ?

What does TNT stand for ?

[amp_mcq option1=”Tri Nitro Toluene” option2=”Tri Nitro Tartum” option3=”Tetra Nitro Toluene” option4=”Thallium Nitrate Tetryl” correct=”option1″]

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TNT stands for Tri Nitro Toluene.
Tri Nitro Toluene (TNT) is a chemical compound with the formula C₆H₂ (NO₂)₃CH₃. It is a highly stable explosive material and one of the most widely used explosives, particularly in military and mining applications.
TNT is known for its relative insensitivity compared to other explosives, making it safer to handle. It is often used as a reference point for comparing the power of other explosives, measured by the “TNT equivalent”.

22. Which one of the following substances does not belong to class ‘B’ fir

Which one of the following substances does not belong to class ‘B’ fire ?

[amp_mcq option1=”Molten Sulphur” option2=”Alcohol” option3=”Solvent” option4=”Paint” correct=”option4″]

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Paint is the substance among the options that does not exclusively belong to Class ‘B’ fires, as it can also cause Class ‘A’ fires (e.g., dried paint) or be non-flammable (water-based paint).
– Class B fires involve flammable liquids and gases (e.g., petrol, oil, alcohol, propane).
– Alcohol and Solvents are standard examples of flammable liquids that cause Class B fires.
– Molten Sulphur, while solid at room temperature, burns as a flammable liquid when molten and is classified as a Class B fire hazard.
– Paint’s classification depends on its composition. Solvent-based paints cause Class B fires due to the flammable solvents. However, water-based paints contain minimal flammable solvents; their fire risk is primarily from the combustible solid components/residue, which would be a Class A fire. Dried paint is also a combustible solid (Class A).
Fire classification systems categorize fires based on the type of fuel involved, as different types of fires require different extinguishing agents. While many paints are Class B due to solvents, the general term ‘Paint’ is ambiguous and can include substances that result in Class A fires, making it the one least definitively belonging *only* to Class B among the options.

23. Which one of the following aqueous solutions will be neutral ?

Which one of the following aqueous solutions will be neutral ?

[amp_mcq option1=”NH₄Cl” option2=”NaCl” option3=”KCN” option4=”NaHSO₄” correct=”option2″]

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An aqueous solution is neutral if the salt dissolved in water is formed from the reaction of a strong acid and a strong base.
A) NH₄Cl is formed from a weak base (NH₄OH) and a strong acid (HCl). The solution will be acidic due to the hydrolysis of NH₄⁺ ions.
B) NaCl is formed from a strong base (NaOH) and a strong acid (HCl). The solution will be neutral as neither Na⁺ nor Cl⁻ ions undergo significant hydrolysis.
C) KCN is formed from a strong base (KOH) and a weak acid (HCN). The solution will be basic due to the hydrolysis of CN⁻ ions.
D) NaHSO₄ is sodium bisulfate. It is formed from a strong base (NaOH) and sulfuric acid (H₂SO₄), which is a strong acid. However, HSO₄⁻ is the conjugate base of the strong acid H₂SO₄, but it is also an acid itself that can donate a proton (HSO₄⁻ ⇌ H⁺ + SO₄²⁻). This makes the solution acidic.
Therefore, only NaCl forms a neutral aqueous solution.
The neutrality of an aqueous salt solution depends on the strength of the parent acid and base from which the salt is derived. Salts of strong acid and strong base yield neutral solutions. Salts of strong acid and weak base yield acidic solutions. Salts of weak acid and strong base yield basic solutions. Salts of weak acid and weak base yield solutions whose pH depends on the relative strengths of the weak acid and weak base.
Hydrolysis is the reaction of an ion with water. Cations of weak bases undergo acidic hydrolysis, releasing H⁺ ions. Anions of weak acids undergo basic hydrolysis, releasing OH⁻ ions. Cations of strong bases and anions of strong acids do not undergo significant hydrolysis.

24. Manganese is extracted from Manganese dioxide by reaction with Alumini

Manganese is extracted from Manganese dioxide by reaction with Aluminium as described by the following unbalanced chemical equation :
MnO₂(s) + Al (s) → Mn (l) + Al₂O₃ (s)
The number of moles of Al (s) required to form one mole of Mn from its oxide is

[amp_mcq option1=”1″ option2=”0.75″ option3=”1.33″ option4=”2″ correct=”option3″]

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The unbalanced chemical equation is MnO₂(s) + Al (s) → Mn (l) + Al₂O₃ (s). To determine the moles of Al required to form one mole of Mn, we first need to balance the equation.
Balancing the oxygen atoms (2 on the left, 3 on the right), we find the least common multiple is 6. Multiply MnO₂ by 3 and Al₂O₃ by 2:
3 MnO₂(s) + Al (s) → Mn (l) + 2 Al₂O₃ (s)
Now, balance the aluminium atoms (1 on the left, 4 on the right). Multiply Al by 4:
3 MnO₂(s) + 4 Al (s) → Mn (l) + 2 Al₂O₃ (s)
Finally, balance the manganese atoms (3 on the left, 1 on the right). Multiply Mn by 3:
3 MnO₂(s) + 4 Al (s) → 3 Mn (l) + 2 Al₂O₃ (s)
The balanced equation is 3 MnO₂(s) + 4 Al (s) → 3 Mn (l) + 2 Al₂O₃ (s).
According to the balanced equation, 4 moles of Al react to produce 3 moles of Mn. To produce 1 mole of Mn, the number of moles of Al required is (4 moles Al / 3 moles Mn) * 1 mole Mn = 4/3 moles Al.
The question requires balancing the given chemical equation and then using stoichiometry to find the mole ratio between reactants and products. The ratio of Al to Mn in the balanced equation is 4:3.
This reaction is a type of redox reaction, specifically a thermite reaction where a metal oxide is reduced by a more reactive metal (Aluminium). Aluminium is a strong reducing agent. The value 4/3 is approximately 1.333…

25. How many moles of water would be produced by the complete combustion o

How many moles of water would be produced by the complete combustion of one mole of natural gas, CH₄, in excess of oxygen ?

[amp_mcq option1=”1″ option2=”2″ option3=”3″ option4=”4″ correct=”option2″]

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The complete combustion of natural gas, which is primarily methane (CH₄), in excess oxygen is represented by the balanced chemical equation:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g or l)
The coefficients in the balanced equation represent the relative number of moles of reactants and products. According to the equation, one mole of methane (CH₄) reacts with two moles of oxygen (O₂) to produce one mole of carbon dioxide (CO₂) and two moles of water (H₂O).
The stoichiometry of a balanced chemical equation gives the mole ratios of reactants and products. For the combustion of methane, 1 mole of CH₄ produces 2 moles of H₂O.
Combustion is a rapid reaction between a substance with an oxidant, usually oxygen, to produce heat and light. Complete combustion occurs when there is sufficient oxygen, yielding carbon dioxide and water as products for hydrocarbons like methane. Incomplete combustion occurs with insufficient oxygen, producing carbon monoxide and/or carbon. The question specifies “complete combustion in excess of oxygen”, ensuring the reaction proceeds as shown in the balanced equation.

26. An element has an atomic number of 16. What is the principal quantum n

An element has an atomic number of 16. What is the principal quantum number (n) of its outermost electrons ?

[amp_mcq option1=”1″ option2=”2″ option3=”3″ option4=”4″ correct=”option3″]

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An element with atomic number 16 is Sulfur (S). To find the principal quantum number (n) of its outermost electrons, we need to write its electron configuration. The atomic number represents the number of protons and, in a neutral atom, the number of electrons. So, Sulfur has 16 electrons. The electron configuration is filled according to the Aufbau principle, Hund’s rule, and the Pauli exclusion principle.
Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, …
Electrons:
1s² (2 electrons)
2s² (2 electrons)
2p⁶ (6 electrons) – total 2+2+6 = 10 electrons
3s² (2 electrons) – total 10+2 = 12 electrons
3p⁴ (4 electrons) – total 12+4 = 16 electrons
The electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴. The outermost electrons are those in the highest principal energy level, which is n=3 in this case (3s and 3p orbitals).
The principal quantum number (n) of the outermost electrons corresponds to the highest energy level occupied by electrons in the atom’s electron configuration. For Sulfur (atomic number 16), the configuration is 1s² 2s² 2p⁶ 3s² 3p⁴, so the outermost electrons are in the n=3 level.
The outer electron shell is often referred to as the valence shell. The principal quantum number ‘n’ defines the energy level and size of the electron shell. n=1 is the first shell, n=2 is the second shell, n=3 is the third shell, and so on.

27. What is PETN?

What is PETN?

[amp_mcq option1=”Pentane epinephirine tantalum nitro” option2=”Pentaerythritol tetranitrate” option3=”Phenol erythritol tetranitrate” option4=”None of the above” correct=”option2″]

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PETN stands for Pentaerythritol tetranitrate.
PETN is a powerful explosive, known for its high brisance and sensitivity to initiation, making it commonly used in detonators, booster charges, and plastic explosives.
It is one of the most powerful explosive compounds known and has been used in various terrorist plots, including attempted airline bombings, often in relatively small amounts due to its potency.

28. Catalytic converter transforms waste gases from the engines of many ca

Catalytic converter transforms waste gases from the engines of many cars into carbon dioxide, nitrogen and water. The catalyst is made of

[amp_mcq option1=”platinum and copper” option2=”molybdenum and copper” option3=”platinum and rhodium” option4=”rhodium and molybdenum” correct=”option3″]

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Catalytic converters use precious metals like platinum, palladium, and rhodium as catalysts to convert harmful exhaust gases into less harmful substances.
Platinum and rhodium are commonly used together in catalytic converters, particularly in three-way converters, which target carbon monoxide (CO), unburnt hydrocarbons (HC), and nitrogen oxides (NOx).
Platinum is effective for oxidizing CO and HCs. Rhodium is effective for reducing NOx. Palladium is also used, often in conjunction with platinum, and is effective for oxidizing CO and HCs. Copper and molybdenum are not typically used as the primary catalysts in automotive catalytic converters.

29. The number of elements in the lanthanoids of the periodic table is

The number of elements in the lanthanoids of the periodic table is

[amp_mcq option1=”8″ option2=”18″ option3=”14″ option4=”32″ correct=”option3″]

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The number of elements in the lanthanoids of the periodic table is 14.
The lanthanoids (or lanthanides) are a series of chemical elements comprising the 15 elements from lanthanum (atomic number 57) through lutetium (atomic number 71). However, the term “lanthanoid series” usually refers to the 14 elements immediately following Lanthanum, from Cerium (atomic number 58) to Lutetium (atomic number 71), as these are the elements where the 4f subshell is progressively filled. Lanthanum itself has the electron configuration [Xe] 5d¹ 6s², not filling the 4f subshell, but it is chemically similar to the other lanthanoids and is often included in the series or considered the first element.

In the context of the f-block of the periodic table, which is typically displayed as two rows below the main body, the lanthanoids constitute the upper row. This row consists of 14 elements, starting after Barium (Z=56) and before Hafnium (Z=72) in the main table. These 14 elements are Cerium (58), Praseodymium (59), Neodymium (60), Promethium (61), Samarium (62), Europium (63), Gadolinium (64), Terbium (65), Dysprosium (66), Holmium (67), Erbium (68), Thulium (69), Ytterbium (70), and Lutetium (71).

Given the options, 14 is present and is the standard number of elements represented in the f-block lanthanoid series, corresponding to the filling of the 4f orbitals. While the definition can sometimes include Lanthanum (making it 15), 14 is the number that fits the typical arrangement and f-orbital filling.

The actinoids are the second series in the f-block, starting from Actinium (Z=89) or Thorium (Z=90) up to Lawrencium (Z=103), also containing 14 elements where the 5f subshell is filled. The lanthanoids and actinoids are sometimes collectively called the inner transition metals.

30. Out of the elements phosphorus (P), sulphur (S), chlorine (Cl) and flu

Out of the elements phosphorus (P), sulphur (S), chlorine (Cl) and fluorine (F), the elements having the most negative and least negative electron gain enthalpy, respectively are

[amp_mcq option1=”Cl and P” option2=”F and S” option3=”F and Cl” option4=”P and S” correct=”option1″]

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The elements having the most negative and least negative electron gain enthalpy among P, S, Cl, and F, respectively, are Cl and P.
Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom to form a negative ion. A negative value means energy is released, indicating the atom has an affinity for electrons. A more negative value signifies a stronger affinity.
We are comparing P (Group 15), S (Group 16), Cl (Group 17), and F (Group 17).
General trends:
– Across a period (left to right), electron gain enthalpy generally becomes more negative (atoms become smaller and nuclear charge increases, attracting the added electron more strongly). Group 18 (noble gases) have positive electron gain enthalpies as they have a stable electron configuration. Group 15 elements have relatively less negative (or slightly positive) values due to the stability of the half-filled p-subshell.
– Down a group, electron gain enthalpy generally becomes less negative (the added electron is further from the nucleus in a larger shell and experiences more shielding).

Let’s consider the given elements:
– F (Period 2, Group 17)
– Cl (Period 3, Group 17)
– S (Period 3, Group 16)
– P (Period 3, Group 15)

Comparing F and Cl (Group 17): Although the general trend is less negative down a group, Cl has a more negative electron gain enthalpy (-349 kJ/mol) than F (-328 kJ/mol). This is an anomaly due to the very small size of F, where the added electron experiences significant repulsion from existing electrons in the compact 2p subshell.
Comparing Cl and S (same period, different groups): Electron gain enthalpy becomes more negative across the period. Cl is in Group 17, S is in Group 16. So, Cl should have a more negative value than S. (Cl: -349 kJ/mol, S: -200 kJ/mol). This is consistent.
Comparing S and P (same period, different groups): S is in Group 16, P is in Group 15. Group 15 elements have unusually low (less negative) electron gain enthalpies due to the stable half-filled configuration. So, P should have a less negative value than S. (S: -200 kJ/mol, P: -74 kJ/mol). This is consistent.

Ranking the elements by electron gain enthalpy from most negative to least negative:
Cl (-349) > F (-328) > S (-200) > P (-74)

The element with the most negative electron gain enthalpy is Cl.
The element with the least negative electron gain enthalpy is P.

The pair is (Cl, P).

Electron gain enthalpy can be positive for some elements, meaning energy must be supplied to add an electron. Noble gases typically have positive electron gain enthalpies. Elements in Group 2 (alkaline earth metals) and Group 12 also tend to have values close to zero or positive for the first electron gain enthalpy.
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