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The Indian Institute of Management Calcutta packages offered to the PGPEX batch 2024 are presented below:

B-Cl bond has dipole moment because of polarity. In BCl_3,  since the molecule is symmetrical (planar), the polarities cancel out and hence the dipole moment is zero

No, candidates looking for course admission at CII Institute of Hospitality- ITC Gardenia, Bangalore cannot enrol for direct admissions. The institute admits students based on CII Aptitude Test + Industry Panel Viva. It also considers the candidates' academic performance for final selection. The final selection of the candidate will be based on the overall performance in the CII Institute of Hospitality- ITC Gardenia, Bangalore admission selection process.

1.6. A mass percent of 69% means that 100 g of nitric acid solution contains 69 g of nitric acid by mass.
Molar mass of nitric acid HNO₃= 1 + 14 + (3x16) = 63 gmol^-1

The difference in bond length is due to the difference in the state of hybridisation. In BF₃ 'B' is sp² hybridised and in BF₄^– 'B' is sp³ hybridised.

Yes,  BA (H) is available at Magadh University. This undergrad programme is offered via many affiliated colleges. Students who are willing to take admission can check the below table to know the key highlights:

Note: The above information has been sourced from various sources. Hence it is indicative.

2.4 Radius of the spherical conductor, r = 12 cm = 0.12 m

Uniformly distributed charge, q = 1.6 $\times {10}^{-7}$ C

The electrical field inside the conductor is zero.

Electric field E just outside the conductor is given by

E = $\frac{1}{4\pi {\epsilon }_0}\times \frac{q}{r^2}$ where ${\epsilon }_0$ = permittivity of free space = 8.854 $\times {10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

E = $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{1.6\mathrm{}\times {10}^{-7}}{{0.12\mathrm{}}^2}$ = 99863.8 N $C^{-1}$ $\cong {10}^5$ N $C^{-1}$

At a point 18 cm from the centre of the sphere. Hence r = 18 cm = 0.18 m

From the above equation we get

E at 18 cm from the centre = $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{1.6\mathrm{}\times {10}^{-7}}{{0.18\mathrm{}}^2}$ = 4.438 $\times {10}^4$ N $C^{-1}$

The key highlights of the IIM Calcutta placements for its PGPEX Class of 2023 and 2024 are presented below:

• PbCl₂ + Cl₂→ PbCl₄.
  This is because Pb can show +2 oxidation state more easily than +4 due to inert pair effect.
• PbCl₄→ PbCl₂ + Cl₂
  Because Pb²⁺ is more stable than Pb⁴⁺ due to inert pair effect.
• PbI₄does not exist because I^- ion being a powerful reducing agent reduces Pb⁴⁺ ion to Pb²⁺ ion in solution.

Pb⁴⁺ + 2I^– → Pb²⁺ + I₂

1.5

Since the molar mass of sodium acetate is 82.0245 g mol^-1.

⇒ 1 mol of CH₃COONa has a mass = 82.0245 g

∴Mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution

= 0.375 x 0.5 mol x 82.0245 g mol^-1

= 15.3795 g = 15.380 g

Yes, Impact Paramedical and Health Institute offers four-year long BSc paramedical courses. The available programmes as per the official website are MLT, Radiology and Imaging Technology, and Operation Theatre Technology. Admission is done on a merit-basis. Moreover, students who are Class 12 or equivalent pass with at least 50% aggregate can apply for admission.

The branch change is not possible in Jadavpur University.

2.3 The arrangement is represented in the adjoining figure.

An equipotential surface is the plane on which total potential is zero everywhere. The plane is normal to line AB. The plane is located at the midpoint of the line AB because the magnitude of the charge is same.

The direction of the electric field at every point on that surface is normal to the plane in the direction of AB.

(a) CO₃^2-  (sp²) (b) Diamond (sp³) (c) Graphite (sp²)

2.2

The charge at each corner, $q_1$ = 5 $\mu C=5\times {10}^{-6}$ C

The sides of the hexagon, AB = BC = CD = DE = EF =FA = 10 cm = 0.1 m

Let O be the centre. The electric potential at O is given by

V = $\frac{1}{4\pi {\epsilon }_0}\times \frac{6q_1}{d}$ , where ${\epsilon }_0$ = permittivity of free space = 8.854 $\times {10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

Hence V = $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{6\times 5\times {10}^{-6}}{0.1}$ = 2.696 $\times {10}^6$ V

Therefore, the potential at the centre is 2.7 $\times {10}^6$ V

1.3 We are given that the percentage of iron by mass is 69.9% and the percentage of oxygen by mass is 30.1%.

Since we have relative moles of both elements, we can calculate the simpler molar ratio of iron to oxygen

= 1.25: 1.88

Divide by the smaller value to both

= 1.25/1.25: 1.88/1.25

= 1: 1.5

= 2: 3

So, now we can write the empirical formula of iron oxide as Fe₂O₃.