The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):
0.1 mole of compound S will weigh ...... g, (given the molar mass in g mol\(^{-1}\) C = 12, H = 1, O = 16)
If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
Given below are some nitrogen containing compounds: Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ...... mg of HCl.(Given Molar mass in g mol\(^{-1}\): C = 12, H = 1, O = 16, Cl = 35.5.)
1.24 g of \(AX_2\) (molar mass 124 g mol\(^{-1}\)) is dissolved in 1 kg of water to form a solution with boiling point of 100.105°C, while 2.54 g of AY_2 (molar mass 250 g mol\(^{-1}\)) in 2 kg of water constitutes a solution with a boiling point of 100.026°C. \(Kb(H)_2\)\(\text(O)\) = 0.52 K kg mol\(^{-1}\). Which of the following is correct?
Match List - I with List - II:
Choose the correct answer from the given below options
Match List - I with List - II: List - I: (A) Amylase (B) Cellulose (C) Glycogen (D) Amylopectin List - II: (I) β-C1-C4 plant (II) α-C1-C4 animal (III) α-C1-C4 α-C1-C6 plant (IV) α-C1-C4 plant
Match List - I with List - II:List - I: (A) \([ \text{MnBr}_4]^{2-}\) (B) \([ \text{FeF}_6]^{3-}\) (C) \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\) (D) \([ \text{Ni(CO)}_4]\) List - II: (I) d²sp³ diamagnetic (II) sp²d² paramagnetic (III) sp³ diamagnetic (IV) sp³ paramagnetic
The product (P) formed in the following reaction is:
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)
In the following substitution reaction:
The correct increasing order of stability of the complexes based on \( \Delta \) value is:
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm.