Class 12th

Get insights from 11.9k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question
11.9k

Questions

0

Discussions

48

Active Users

0

Followers

New answer posted

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

For the complex K? [Cr (oxalate)? ], the central metal ion is Cr³?

Electronic configuration of Cr (24) is [Ar] 4s¹3d?

Electronic configuration of Cr³? is [Ar] 4s?3d³.

The number of unpaired electrons in Cr³? is 3.

New answer posted

10 months ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

Given the function:
f(x) = { x(2 - sin(1/x)), if x ≠ 0
{ 0, if x = 0

For x < 0: f(x) = x(2 - sin(1/x))

For x > 0: f(x) = x(2 - sin(1/x))

The derivative f'(x) for x ≠ 0 is:
f'(x) = 1*(2 - sin(1/x)) + x*(-cos(1/x))*(-1/x²) = 2 - sin(1/x) + (1/x)cos(1/x)

The text calculates the derivative differently:
For x < 0: f'(x) = -2 + sin(1/x) - (1/x)cos(1/x)
For x > 0: f'(x) = 2 - sin(1/x) + (1/x)cos(1/x)

To check if f'(0) is defined, we would need to use the limit definition of the derivative at x=0. As x approaches 0, the term (1/x)cos(1/x) oscillates and does not approach a finite limit. Therefore, f'(0) is undefined.

New answer posted

10 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Given K_f = 1.85 K kg mol? ¹ for a solution with molality of 2 m.

ΔT_f = I * K_f * m

3.885 = I * 1.85 * 2

The van't Hoff factor, I = 1.05.

i = 1 + (n-1)α. For an electrolyte dissociating into 2 ions, n=2.

1.05 = 1 + (2-1)α.

The degree of dissociation, α = 0.05 or 50 * 10? ³.

New answer posted

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Please consider the following Image

 

New answer posted

10 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Please consider the following Image

 

New answer posted

10 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

For an acidic buffer solution, pH = pKa + log ( [Base]/ [Acid]).

Given pH = 5.74 and pKa = 4.74.

5.74 = 4.74 + log ( [Base]/1).

1 = log ( [Base]).

[Base] = 10M.

New answer posted

10 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

For the reaction C? H? → C? H? + H? , calculate the enthalpy change (ΔH).

ΔH = [Bond energy (C-C) + 6 * Bond energy (C-H)] - [Bond energy (C=C) + 4 * Bond energy (C-H) + Bond energy (H-H)]

ΔH = 347 + 2 (414) - 611 - 436 = 128 kJ/mol.

New answer posted

10 months ago

0 Follower 16 Views

R
Raj Pandey

Contributor-Level 9

Benzyl amine (0.1 mole) reacts with 3 equivalents of CH? Br to form Benzyl trimethyl ammonium bromide.

Given 23g of Benzyl trimethyl ammonium bromide (molar mass 230 g/mol ), which is 0.1 mol.

Therefore, moles of CH? Br = 0.3 = 3 x 10? ¹. The value of n is 3.

New answer posted

10 months ago

0 Follower 3 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

When white light passes through a cool gas, atoms absorb specific wavelengths. This produces a continuous background with dark lines at those absorbed wavelengths. This is atomic absorption spectra. But when atoms are excited, they release photons at specific wavelengths, producing a dark background with bright lines. That is emission spectra. 

New answer posted

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

CH? (methane) is produced / generated from paddy fields. And methane leads to both global warming and photochemical smog.

CO? is used in photosynthesis, acid rain etc. but methane is not consumed.

So methane is a stronger global warming gas than CO?

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 702k Reviews
  • 1850k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.