Class 12th

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New answer posted

10 months ago

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R
Raj Pandey

Contributor-Level 9

qE = mg ⇒ neE = (4/3)πr³ρg
⇒ n = (4πr³ρg) / (3eE) = (4 * 3.14 * (2 * 10? ³ )³ * 3 * 10³ * 9.81) / (3 * 1.6 * 10? ¹? * 3.55 * 10? ) = 1.73 * 10¹?

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

The structure of the given compound is as show below

 

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

ΔG° = –nFE°
∴ 17.37 x 10³ = –3 x 96500 x E°
∴ E° = –6 x 10? ² V

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

Ellingham diagram provides information on Gibb's free energy for formation of oxides as a function of temperature.

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10 months ago

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A
Aadit Singh Uppal

Contributor-Level 10

Industries manily select an optimal temperature which is high enough to increase the reaction rate (molecules colliding with each other at higher fequencies) and at the same time is balanced to avoid any side effect. This effectively yields high quality production without any wastage or leading to harmful results while ensuring safety of the consumers too.

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

Gd: [Xe]4f?5d¹6s²
∴ Gd³? is [Xe]4f?
Also μ = √n (n + 2) B.M
= √7x9 B.M = 7.9 B.M

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

Esters are susceptible to reaction in basic medium.

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