What does TNT stand for ?
Chemistry
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 ?
– 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).
23. Which one of the following aqueous solutions will be neutral ?
Which one of the following aqueous solutions will be neutral ?
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.
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
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.
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 ?
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).
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 ?
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).
27. What is PETN?
What is PETN?
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
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
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.
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
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).