Because Boron can absorb neutrons.

(i) Intermixing of particles 9ases or liquids spontaneously is called diffusion. Liquids diffuse at a slower rate as compared to gases.
(ii) Ammonia gas will reach first to a person in the centre of the laboratory because it is lighter than hydrochloric acid. 
(iii) The phenomenon of diffusion of gases is very useful in daily life. As a result of diffusion, the poisonous effect of the gases gets slowly diluted

If △G  is negative, process is spontaneous.
If △G is positive, the process is non-spontaneous.
If △G = 0, the process is in equilibrium.

\(\left[ P+\frac { { an }^{ 2 } }{ { V }^{ 2 } } \right] \)(V - nb) = nRT
Where 'a' and 'b' are van der waals constants.

Nitration of benzene is an electrophilic substitution reaction. Presence of electron releasing group such as - CH₃  activates the benzene nucleus towards electrophilic substitution while presence of electron withdrawing group such as -NO₂ deactivates the benzene nucleus towards electrophilic substitution.
Therefore, the ease of nitration decrease in the order
Toluene > benzene > m-dinitrobenzene
Thus, toluene will undergo nitration most easily.

sodium extract is boiled with nitric acid to decompose NaCN and Na₂S if present.
\(\mathrm{NaCN}+\mathrm{HNO}_{3} \longrightarrow \mathrm{NaNO}_{3}+\mathrm{HCN} \uparrow\)
\(\mathrm{Na}_{2} \mathrm{~S}+2 \mathrm{HNO}_{3} \longrightarrow 2 \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{~S} \uparrow\)
If cyanide and sulphide are not removed, they will react with AgNO₃ and hence, will interfere with the silver nitrate test for halogens
\(\mathrm{NaCN}+\mathrm{AgNO}_{3} \longrightarrow \underset{\text { White ppt }}{\mathrm{AgCN}}+\mathrm{NaNO}_{3}\)
\(\mathrm{Na}_{2} \mathrm{~S}+2 \mathrm{AgNO}_{3} \longrightarrow \underset{\text { Black } \mathrm{ppt}}{\mathrm{Ag}_{2} \mathrm{~S}}+2 \mathrm{NaNO}_{3}\)

CCI₄ will not give a white ppt of AgCI with AgNO₃ solution because CCI₄ is a covalent compound. It does not ionise to give CI^- ions required for the formation of AgCI precipate.

Hydron H⁺when used in relation to the isotopic mixture, is called hydron.

According to the given Reaction,
             S₈ (g)⇌4S₂ (g)
At start    1              0
At eqbm  1−0.29     4×0.29
=0.71 atm=1.16 atm
Kp\(=\frac{[pS_2]^4}{[pS_8]}\)
putting the values,
Kp\(=\frac{(1.16)^4}{(0.71)}\)=2.55

\({ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2NH_{ 3 }(g),\)
Given, at equilibrium, \({ P }N_{ 2 }=0.80\) atmosphere
\(P{ H }_{ 2 }=0.40\) atmosphere
\({ P }_{ { N }_{ 2 } }+{ P }_{ { H }_{ 2 } }+{ P }_{ { NH }_{ 3 } }=2.80\) atmosphere
\(\therefore { P }_{ { NH }_{ 3 } }=2.80-(0.80+0.40)=1.60\) atmosphere
From, K_p \(=\frac { { P }\overset { 2 }{ N{ H }_{ 3 } } }{ { P }_{ { N }_{ 2 } }\times P\overset { 3 }{ H_{ 2 } } } =\frac { (1.60)^{ 2 } }{ 0.80\times (0.40)^{ 3 } } =50.0\)

1×(+3)+2×(+2)+3x+7×(−2)=0
or  3+4+3x−14=0
3x=7; x=7/3

Formic acid exists as dimer because of hydrogen bonding

Because of hydrogen bonding, it pretends larger size as well as molecular mass.

SF₆ , S is the central atom with six valence electrons.
∴Number of hybrid orbitals=1/2[6+6−0+0]=6
Hence, the hybridisation involved in sp³d².

1 mole species = 6.022 x 10²³
Mass of 1 e^- = 9.11 x 10^-31kg
Charge on 1e^- = 1.602 x 10^.19C
\(\therefore \) Mass of 1 electron = 9.11 x 10^-31kg
Mass of 1 mole of electrons
= 9.11 x 10^-31 x 6.022 x 10^-23
= 54.86 x 10^-8 = 5.486 x 10^-7 kg
Charge on 1 electron = 1.602 x 10^-19
\(\therefore \) Charge on mole of electrons
= 1.602 x 10^-19 x 6.022 x 10²³ = 9.647 x 10⁴C

The energy of photon is sufficient to disturb a sub-automic particle so that there  is uncertainty in the measurement of position and momentum of the sub-atomic particle. However the energy is insufficient to disturb a macroscopic object.

1 molecular of oxygen ( O₂ ) = 2 atoms of oxygen
1 molecular of oxygen ( O₃ ) = 3 atoms of oxygen
In flask P, 1 mole of oxygen gas = 6.022 \(\times\) 10²³ molecules
\(\therefore\) 0.5 mole of oxygen gas  = 6.022 \(\times\)1023\(\times\)0.5 molecules
 =  6.022\(\times\)10²³\(\times\)0.5\(\times\)2 atoms
 =  6.022 \(\times\) 10²³ atoms
In flask Q, 1 mole of ozone gas = 6.022\(\times\)10²³ molecules
0.4 mole of ozone gas = 6.022\(\times\)10²³\(\times\) 0.4 molecules
=  6.022 \(\times\)10²³\(\times\)0.4\(\times\)3 atoms
= 7.23\(\times\)10²³ atoms
\(\therefore\) Flask Q has greater number of oxygen atoms as compared to the flask P.

No . it is useful our body

Temperature corresponding to 60 mm \(\simeq \) 313 K.

Let the total mass of solution = 100 g
Mass of water = 25g, Mass of ethanol = 25 g
Mass of acetic acid = 50 g
Moles of water 25/18 = 1388 \((\because \ molar \ mass \ of \ { H }_{ 2 }O=18)\)
Moles of ethanol = 25/46 = 0.543
\((\because \ molar \ mass \ of \ C_{ 2 }{ H }_{ 5 }OH=46)\)
Moles of acetic acid = 50/60 = 0.833
\((\because \ molar \ mass \ of \ CH_{ 3 }COOH=60)\)
Total number of moles = 1.388 + 0.543 + 0.833 = 2.764
Mole fraction of water = 1.388/2.764 = 0.196
Mole fraction of acetic acid = 0.833/2.764 = 0.302.

In aqueous solution, H₂SO₄ ionises to give H+(aq) and \({ SO }_{ 4 }^{ 2- }\)t(aq) ions.
H₂SO₄(aq) \(\rightarrow\) 7 2H⁺(aq) + \({ SO }_{ 4 }^{ 2- }\)(aq)
Thus, when electricity is passed, H+  (aq) ions move towards cathode while \({ SO }_{ 4 }^{ 2- }\) (aq) ions move towards anode. In other words, at cathode either H+(aq) ions or H₂O molecules are reduced. Their electrode potentials are:
2H⁺(aq) + 2e^- \(\rightarrow\)  H₂(g); E^o = 0.0 V
HzO(aq) + 2e- \(\rightarrow\) H₂(g) + 20H^-(aq); E^O = -0.83 V
Since the electron potential (i.e., reduction potential) of H+(aq) ions is higher than that of H₂O, therefore, at the cathode, it is W(aq) ions (rather than H₂O molecules) which are reduced to evolve H₂ gas.
Similarly at the anode, either \({ SO }_{ 4 }^{ 2- }\) (aq) ions or H₂O molecules are oxidised. Since the oxidation potential of \({ SO }_{ 4 }^{ 2- }\) is expected to be much lower (since it involved cleavage of many bonds as compared to those in H₂O than that of H₂O molecules, therefore, at the anode, it is H₂O molecules (rather than sot ions) which are oxidised to evolve O2 gas. From the above discussion, it follows that during electrolysis of an aqueous solution of H₂SO₄ only the electrolysis of H₂O occurs liberating H₂ at the cathode and O₂ at the anode.

Homolytic fission is breaking of a bond in such a manner that each atom takes one electron each to form free radicals.
\(A-B\longrightarrow A.+B\)
(b) Organic ions which contain a negatively charged carbon atom are called carbanions. e.g., \(\overset { \ominus }{ { CH }_{ 3 } } \) is carbanion.

Smaller the size, higher is the hydration enthalpy and size of alkali metal ions increases on moving down the group.
Sodium imparts a golden yellow colour to the flame.

For second dissociation constant,
\({ HS }^{ - }+{ H }_{ 2 }O\rightleftharpoons { H }_{ 3 }{ O }^{ + }+{ S }^{ 2- } \ (in \ absence \ of \ HCl)\)
\( \left[ { HS }^{ - } \right] \left[ { H }_{ 3 }{ O }^{ + } \right] =9.54\times { 10 }^{ -5 }M\)
\( Now,\ { K }_{ { a }_{ 2 } }=\frac { \left[ { { H }_{ 3 }{ O }^{ + } } \right] \left[ { S }^{ 2- } \right] }{ \left[ { HS }^{ - } \right] } \)
\(\left[ { H }_{ 3 }{ O }^{ + } \right] =\left[ { S }^{ 2- } \right] =\sqrt { { K }_{ { a }_{ 2 } }.C }\)
\( =\sqrt { 1.2\times { 10 }^{ -13 }\times 9.54\times { 10 }^{ -5 } } \)
\( =3.38\times { 10 }^{ -9 }M\)

For the given reaction,
2A(g)+B(g)→2D(g)
Δng=2−(3)=–1 mole
Substituting the value of ΔUθ the expression of Δ H:
Δ H^θ=Δ U^θ+ ΔngRT
=(−10.5kJ)+(−1)(8.314×10⁻³kJ K⁻¹mol⁻¹)(298 K)
=−10.5 kJ⁻².48 kJ
ΔH^θ=−12.98 kJ
Substituting the value of ΔH^θ and ΔS^θ in the expression of ΔG^θ:
ΔG^θ=ΔH^θ=TΔS^θ
=−12.98 kJ−(298K)(−44.1JK⁻¹)
=−12.98 kJ+13.14 kJΔ G^θ= +0.16kJ
Since Δ Gθ for the reaction is positive, the reaction will not occur spontaneously.

Given, m = 100g = 0.1kg
v = 100km/h = \(\frac { 100\times 1000 }{ 60\times 60 } =\frac { 1000 }{ 36 } { ms }^{ -1 }\)   
From de-Broglie equation, wavelength,
\(\lambda =\frac { h }{ mv } =\frac { 6.626\times { 10 }^{ -34 }{ kgm }^{ 2 }{ s }^{ -1 } }{ 0.1kg\times \frac { 100 }{ 36 } { ms }^{ -1 } } =238.5\times { 10 }^{ -36 }m\)
As the wavelength is very small so wave nature cannot be detected.

(i) Li₂O
(ii) 43.36 Kg