Class 12th

In n-type semiconductor majority charge carriers are e- and P type semiconductor majority charge carriers are holes.

$\mathrm{I}={\mathrm{n}\mathrm{e}\mathrm{A}\mathrm{V}}_{\mathrm{d}}=$ neA $\left(\mu \mathrm{E}\right)$

${\mu }_{\mathrm{e}}>{\mu }_{\mathrm{h}}\Rightarrow {\mathrm{I}}_{\mathrm{e}}>{\mathrm{I}}_{\mathrm{h}}$

E_cell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
= 1.05 - (0.059 * 3) / 2 = 1.05 - 0.0885 = 0.9615 volt
There is a misprint in the question. The E? _cell is incorrectly given as 10.5 V. This should have been 1.05 volt.

Statement I is true but statement II is incorrect as 3°-alcohols are the most reactive and give immediate turbidity with Luca's reagent.

Hematite: Fe? O?
Magnetite: Fe? O?
Calamine: ZnCO?
Kaolinite: [Al? (OH)? Si? O? ]

kc = [O? ]² / [O? ]³
3 * 10? = [O? ]² / [4 * 10? ²]³
[O? ]² = 192 * 10? ⇒ [O? ] = 4.38 * 10? ³²

Gravitational constant $=\left. [{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]$

Gravitational potential energy $=\left. [{\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$

Gravitational potential $=\left. [{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$

Gravitational intensity $=\left. [{\mathrm{L}\mathrm{T}}^{-2}\right]$

k = (2.303 / 5) log (0.1 / 0.001) = (2.303 * 2) / 5 = 4.606 / 5 = 0.921 min? ¹

When acetone & 2-pentanone are added in a base for an aldol condensation reaction there are several possibilities that may arise for product formation.

But structure in choice (2) is not possible. It will be possible with 3-pentanone.

Polar molecules have centres of positive and negative charges separated by some distance, so they have permanent dipole moment

$\stackrel{?}{\mathrm{v}}?\stackrel{?}{\mathrm{E}}*\stackrel{?}{\mathrm{B}},\stackrel{\mathrm{ˆ}}{\mathrm{v}}=\stackrel{\mathrm{ˆ}}{\mathrm{i}}$

Option (1) $\stackrel{?}{\mathrm{E}}*\stackrel{?}{\mathrm{B}}=\stackrel{?}{0}\stackrel{?}{\mathrm{E}}?\stackrel{?}{\mathrm{B}}$

Option (2) $\stackrel{?}{\mathrm{E}}*\stackrel{?}{\mathrm{B}}=2\stackrel{\mathrm{ˆ}}{\mathrm{i}}$ parallel to $\stackrel{?}{\mathrm{v}}$

Option (2) $\stackrel{?}{\mathrm{E}}*\stackrel{?}{\mathrm{B}}=\stackrel{?}{0}\stackrel{?}{\mathrm{E}}\Uparrow \stackrel{?}{\mathrm{B}}$

Option (2) $\stackrel{?}{\mathrm{E}}*\stackrel{?}{\mathrm{B}}=\stackrel{?}{0}\stackrel{?}{\mathrm{E}}?\stackrel{?}{\mathrm{B}}$