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Class 12th Chemistry

Given below are two statements:

Statement I : In CuSO₄.5H₂O, Cu-O bonds are present

Statement II : In CuSO₄.5H₂O, ligands coordinating with Cu(II) ion are O- and S- based ligands

In the light of the above statements, choose Correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

From the given structure of CuSO₄. 5H₂O Cu (II) ion and oxygen bonds are present but ligands coordinating with Cu (II) ion are not O and S both.

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Class 12th Chemistry

Among the following structure, which will show the most stable enamine formation?

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alok kumar singh

Contributor-Level 10

Enamines are inter conversible and have low stability with respect to imine. Among all C is most stable due to steric factor.

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Class 12th Physics Ray Optics and Optical Instruments

A light ray is incident, at an incident angle q₁, on the system of two plane mirrors M₁ and M₂ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M₁ it gets reflected back by the mirror M₂ with an angle of reflection 30°. The total deviation of the ray will be ___________ degree.

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Raj Pandey

Contributor-Level 9

$δ_{t o t a l} = 360 ° - 2 θ$

= 360 – 2 * 75° = 210°

Total deviation = $δ = 150 °$

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Class 12th Chemistry

Match List – I with List – II:

List – I (Ore) List – II (Composition)

(A) Siderite

...more

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Vishal Baghel

Contributor-Level 10

Ore Formula

Siderite FeCO₃

Malachite &nb

...more

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Class 12th Chemistry

Which will have the highest enol content?

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alok kumar singh

Contributor-Level 10

Aromaticity drives the highest enolic percentage of given structure:

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Class 12th Electromagnetic Induction

A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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Raj Pandey

Contributor-Level 9

Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i₀ $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

...more

Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i₀ $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

= $= - L i_{0} \frac{- 1}{2} e^{- t / τ}$

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

$e_{a v g} = \frac{\int_{0}^{200 μ s} e d t}{\int_{0}^{200 μ s} d t} = \frac{\int_{0}^{200 μ s} \frac{L i_{0}}{τ} e^{- t / τ} d t}{\int_{0}^{200 μ s} d t}$