Class 12th

Hydroboration oxidation : The reaction proceeds in an anti Markovnikov manner

=> No rearrangement is possible

Fe₃O₄ is a ferromagnetic substance

Malachite : CuCO3.Cu (OH)2

Azurite : 2CuCO3.Cu (OH)2

Cuprite : Cu₂O

Copper pyrite : Cu₂S.Fe₂S₃.

For BCC unit cell

$\sqrt[]{3}a=4r$

$a=\frac{4r}{\sqrt[]{3}}$

Y_A = p_a /p_a + p_b = 400 x ½ / 400 x ½ + 500 x ½ = 4/9

Higher the value of (ixM), higher will be the boiling point of solution.

For FCC lattice,

Packing efficiency = π/3√2

CsCl has BCC structure in which Cl^- is present at corners of cube and Cs+ at body centre.

$As{V}^2=V_R^2+C_C^2$

$?V_R=\sqrt[]{V^2-V_C^2}=\sqrt[]{{10}^2-8^2}=6V$

$\tan ?\theta =\frac{X_C}{X_R}=\frac{V_C}{V_R}=\frac{8}{6}=\frac{4}{3}$

R =? *L/A

This is the formula of calculating a wire's resistance. If you look closely at the formula,

A= cross sectional area

More the surface area, more will be the area for passage of free electrons.

Hence, we can easily conclude that thickness of a wire plays an jmportant role in determining it's conductivity.