Match the LIST-I with LIST-II.
LIST-I | LIST-II | ||
---|---|---|---|
A. | Pnicogen (group 15) | I. | Ts |
B. | Chalcogen (group 16) | II. | Og |
C. | Halogen (group 17) | III. | Lv |
D. | Noble gas (group 18) | IV. | Mc |
Choose the correct answer from the options given below :
Let's match the families of elements with their corresponding symbols from the given lists, focusing on the most recently named elements in these groups:
A. Pnicogen (group 15): The elements in Group 15 are Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi), and Moscovium (Mc). Therefore, Pnicogen matches with Mc. A - IV
B. Chalcogen (group 16): The elements in Group 16 are Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), Polonium (Po), and Livermorium (Lv). Therefore, Chalcogen matches with Lv. B - III
C. Halogen (group 17): The elements in Group 17 are Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At), and Tennessine (Ts). Therefore, Halogen matches with Ts. C - I
D. Noble gas (group 18): The elements in Group 18 are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn), and Oganesson (Og). Therefore, Noble gas matches with Og. D - II
The correct matching is A-IV, B-III, C-I, D-II, which corresponds to option (2).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).