Class 12th

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New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

P = (μ – 1) (1/R – 1/ (-R)
P = 2 (μ – 1)/R
1.5P = (μ – 1)/R') * 1.5 = 1/2R'
R' = R/3

New answer posted

10 months ago

0 Follower 9 Views

R
Raj Pandey

Contributor-Level 9

Correct IUPAC name is 4-Bromo-2-methylcyclopentane carboxylic acid

New answer posted

10 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

i = V/r {1 - e? /? }
i? = V/r

U = ½Li²
½ L (V²/r²) {1 - e? /? } = (1/n) x (LV²/2r²)
1 - e? /? = 1/√n ; e? /? = 1 - 1/√n = (√n - 1)/√n
e? /? = √n/ (√n - 1) ; rt/L = ln (√n/ (√n - 1)
T = L/r ln (√n/ (√n - 1)

New answer posted

10 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

In Ohm's law experiment, ammeter is used in series because in series same current will flow through it. But voltmeter is used in parallel to resistor to measure the potential difference across it.

 

New answer posted

10 months ago

0 Follower 8 Views

R
Raj Pandey

Contributor-Level 9

Two linkage isomers

 

New answer posted

10 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

Total mass of reactant should be greater than that of product.
This condition is only fulfilled in case- 3

New answer posted

10 months ago

0 Follower 14 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

10 months ago

0 Follower 19 Views

V
Vishal Baghel

Contributor-Level 10

i? = 8/8 = 1 A    

New answer posted

10 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

E = E? (1 − ax²)

F = qE?
acceleration = F/m = (qE? /m) (1 - ax²) = v (dv/dx)
(qE? /m) ∫? (1 - ax²)dx = ∫? vdv; (qE? /M) (x - ax³/3) = 0
x (1 - ax²/3) = 0; x = 0 & x = √ (3/a)

New answer posted

10 months ago

0 Follower 1 View

A
Aadit Singh Uppal

Contributor-Level 10

SN2 is a bimolecular reaction. This is because as per the rate law,
Rate = k [substrate] [nucleophile]

Here, the rate determining step is dependent on both substrate and nucleophile, unlike SN1 who only involves substrate. Therefore, SN2 is proved to be bimolecular.

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