Chemistry

Using, µ = √n (n+2) B.M, n = number of unpaired electrons.
? Ti? ³ = 4s? 3d¹, unpaired electron = 1
µ = 1.73 B.M
? V? ² = 4s? 3d³, unpaired electron = 3
µ = 3.87 B.M
? Sc? ³ = 4s? 3d? , unpaired electron = 0
µ = 0 B.M

Anions have larger radii than atoms. Also, higher the e/p ratio higher the ionic radii. So, N? ³ > O? ² > F?

Heating of cis-but-2-enedioic acid gives maleic anhydride as shown.
[Chemical reaction showing conversion of cis-but-2-enedioic acid to maleic anhydride upon heating]

CFC breakdown by visible light to give Cl radical which react with stratospheric ozone.
CFC, CF? Cl? - (hv)-> Cl• + •CF? Cl
Cl• (g) + O? → ClO• + O?
ClO• + O → Cl• + O?
Atmospheric ozone reacts with NO to give NO? and O?
O? + NO → NO? + O?

Bohr's model solved the instability problem by proving about stationary states. In such stats, the electrons move in fixed orbits. They do not emit energy. This contradicted classical electromagnetic theory of Maxwell, which says accelerating charges should emit radiation and collapse into the nucleus. Bohr simply assumed Maxwell's laws don't apply to these special orbits.

Bohr's model is too simple for atoms beyond hydrogen. In multi-electron atoms like helium, it fails because it ignores a couple of aspects. First is the electron-to-electron repulsion, and second is the shielding effect, where inner electrons reduce the nuclear pull on outer ones. Due to both, orbitals with the same principal quantum number don't have the same energy. Bohr's model of atom assumes that it should have the same energy.

${\Delta }_r{H}^o=80kJmo{l}^{-1}$

${\Delta }_r{S}^o=2TkJmo{l}^{-1}$

For a reaction to be spontaneous;

${\Delta }_r{S}^o>\frac{{\Delta }_r{H}^o}{T}$

$T>\frac{{\Delta }_r{H}^o}{{\Delta }_r{S}^o}$

$T>\frac{80*10^3}{2T}$

$T^2>40000K$

$T>200K$

So, minimum T at which reaction will be spontaneous is 200 K.

50000.020 * 10^-3

The significant figure in the given number is 8.

Moles of SO₂ = $\frac{224*10^{-3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 * 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18}=2$

Using ; relative lowering in V.P

$\frac{P^o-P}{P^o}=i{x}_B$

Where; $\Delta P=P^0-P$ is lowering in V.P

$\Delta P=P^{0}i{x}_B$

i for NaHSO₃ = 2

here; $x_B=\frac{n_B}{n_A}$ since solution is dilute

$\Delta P=24*2*\frac{0.01}{2}=0.24$

$\Delta P=24*10^{-2}mmHg$

So; x = 24

PV = nRT

1 * V = $\frac{3.12}{32}*0.0821*300$

V = 2.4 litre

Vol of O₂ adsorbed per gm = 2.4 / 1.2 = 2 litre