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(a) Nitrates of both group 1 and group 2 elements are soluble in water because hydration energy is more than the lattice energy.
Nitrates of both group 1 and group 2 elements are thermally unstable but they decompose differently except LiCO₃ e.g.

2NaNO₃ →2NaNO₂ + O₂

2KNO₃ →2KNO₂ + O₂

4LiNO₃ →2MgO + 4 NO₂ + O₂

2Mg (NO₃)₂ →2MgO + 4NO₂ + O₂

(b) Carbonates of group 1 elements are soluble in water except Li₂CO₃ They are also thermally stable except Li₂CO₃.

Li₂CO₃ →Li₂O + CO₂
Group 2 carbonates are insoluble in water because their lattice energy is higher than hydration energy.
Thermal stability of carbonates of group 2 increases down the group because lattice energy goes on increasing due to increase in ionic character.

(c) Sulphates of group 1 are soluble in water except Li₂SO₄. They are thermally stable.
Solubility of sulphates of group 2 decreases down the group because Lattice energy dominates over hydration energy.
Sulphates of group 2 elements are thermally stable and increasing down the group due to increases in Lattice energy.

13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $*{10}^5$  MW = 2 $*{10}^4$ MW = 2 $*{10}^4*{10}^6$  J/s

= 2 $*{10}^4*{10}^6*60*60*24*365$  J/y = 6.3072 $*{10}^{17}$ J/y

Heat energy release per fission of a ${}_{}{}^{235}U$  nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $*{10}^6$ eV = 50 $*{10}^6*1.6*{10}^{-19}$ J = 8 $*{10}^{-12}$  J

Therefore, number of atoms required per year = $\frac{6.3072*{10}^{17}}{8*{10}^{-12}}$  = 7.884 $*{10}^{28}$

1 mole of ${}_{}{}^{235}U$  = 235 gm of ${}_{}{}^{235}U$  contains 6.023 $*{10}^{23}$ atoms

Hence the mass of 7.884 $*{10}^{28}atoms$  =$\frac{235*{10}^{-3}}{6.023*{10}^{23}}$ $*$ 7.884 $*{10}^{28}$ = 30.76 $*{10}^3$ kg

Hence, the Uranium needed per year is 30.76$*{10}^3$ kg

13.30 Amount of hydrogen, m = 1 kg = 1000 g

1 mole of hydrogen, i.e. 1 g of hydrogen ( ${}_1{}^{1}H)$ contains 6.023 $*{10}^{23}$  atoms

1 kg of hydrogen contains = 1000 $*$ 6.023 $*{10}^{23}$  atoms = 6.023 $*{10}^{26}$ atoms

Within Sun, four ( ${}_1{}^{1}H)$  nuclei combine and forms 1 ${}_2{}^{4}He$  nucleus. In this process 26 MeV of energy is released.

Hence the energy released from fusion of 1 kg of ${}_1{}^{1}H$  is:

$E_1$  = $\frac{6.023*{10}^{26}*26}{4}$  MeV = 3.91495 $*{10}^{27}$ MeV

Amount of ${}_{92}{}^{235}U$ , m = 1 kg = 1000 g

1 mole of ${}_{92}{}^{235}U$ , i.e. 235 g of ${}_{92}{}^{235}U$  contains 6.023 $*{10}^{23}$ atoms

1 kg of  ${}_{92}{}^{235}U$ contains = $\frac{1000}{235}$ $*$  6.023 $*{10}^{23}$ atoms = 2.563 $*{10}^{24}$  atoms

It is known that the amount of energy released in the fission of 1 atom of ${}_{92}{}^{235}U$  is 200 MeV.

Hence the energy released from fusion of 1 kg of  ${}_{92}{}^{235}U$  is:

 = 2.563 $*{10}^{24}*200$  MeV = 5.126 $*{10}^{26}$ MeV

$\frac{E_1}{E_2}$ = $\frac{3.91495*{10}^{27}}{5.126*{10}^{26}}$ = 7.64

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times of the energy release during the fission of 1 kg of ${}_{92}{}^{235}U$ .

13.29 It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_1$ decays from 1.088 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to  decay is given as:

 = 1.088 – 0 = 1.088 MeV = 1.088 $\times {10}^6$  eV = 1.088 $\times {10}^6\times 1.6\times {10}^{-19}$ J

= 1.7408 $\times {10}^{-13}$ J

We know, $E_1=h{\nu }_1$ , where

${\nu }_1$ = Frequency of radiation radiated by ${\gamma }_1$  decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $\times {10}^{-34}$ Js

Hence,  ${\nu }_1$ = $\frac{E_1}{h}$  = $\frac{1.7408\times {10}^{-13}}{6.6\times {10}^{-34}}$  = 2.637 $\times {10}^{20}$ Hz

It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_2$ decays from 0.412 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to ${\gamma }_2$  decay is given as:

$E_2$ = 0.412 – 0 = 0.412 MeV = 0.412 $\times {10}^6$  eV = 0.412 $\times {10}^6\times 1.6\times {10}^{-19}$  J

= 6.592 $\times {10}^{-14}$ J

We know, $E_2=h{\nu }_2$ , where

${\nu }_2$ = Frequency of radiation radiated by ${\gamma }_2$ decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $\times {10}^{-34}$ Js

Hence, ${\nu }_2$  = $\frac{E_2}{h}$ = $\frac{6.592\times {10}^{-14}}{6.6\times {10}^{-34}}$  = 9.987 $\times {10}^{19}$  Hz

It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_3$ decays from 1.008 MeV energy level to the 0.412 MeV energy level.

Hence the energy corresponding to  decay is given as:

 = 1.088 - 0.412 = 0.676 MeV = 0.676 $\times {10}^6$  eV = 0.676 $\times {10}^6\times 1.6\times {10}^{-19}$  J

= 1.0816 $\times {10}^{-13}$  J

We know, $E_3=h{\nu }_3$ , where

${\nu }_3$ = Frequency of radiation radiated by ${\gamma }_3$  decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $\times {10}^{-34}$  Js

 Hence, ${\nu }_3$  = $\frac{E_3}{h}$  = $\frac{1.0816\times {10}^{-13}}{6.6\times {10}^{-34}}$ = 1.639 $\times {10}^{20}$ Hz

Mass of ( ${}_{79}{}^{198}\mathrm{A}\mathrm{u}$ ) = 197.968233 u

Mass of ( ${}_{79}{}^{198}\mathrm{A}\mathrm{u}$ ) =197.966760 u

Energy of the highest level is given as:

E = $\left[\mathrm{m}\left({}_{79}{}^{198}\mathrm{A}\mathrm{u}\right)-\mathrm{m}\left({}_{80}{}^{198}Hg\right)\mathrm{}\right]c^2$

= $\left[197.968233-197.966760\right]c^2$ u

= 1.473 $\times {10}^{-3}{c}^2$ u

= 1.473 $\times {10}^{-3}\times 931.5$  MeV

= 1.3720995 MeV

${\beta }_1^{-}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{s}$  from 1.3720995 MeV level to 1.088 MeV level

Hence maximum kinetic energy of the ${\beta }_1^{-}$ particles = 1.3720995 - 1.088 = 0.2840995 MeV

${\beta }_2^{-}$ from 1.3720995 MeV level to 0.412 MeV level

Hence maximum kinetic energy of the ${\beta }_2^{-}$ particles = 1.3720995 – 0.412 = 0.9600995 MeV

Solvay process is also known as ammonia soda process, wherein carbon dioxide is passed through a brine solution (containing about 28 % NaCl) which is saturated with ammonia to form sodium carbonate.

2NH₃ + H₂O + CO₂ → (NH₄)₂CO₃

(NH₄)₂CO₃ + H₂O + CO₂ → 2NH₄HCO₃

NH₄HCO₃ + NaCl → NaHCO₃↓ + NH₄Cl

The precipitate of sodium bicarbonate is filtered, dried and ignited to form sodium carbonate.
2NaHCO₃? →? Na₂?CO₃? + CO₂ ?+ H₂?O
The carbon dioxide required for the reaction can be obtained by heating limestone (calcium carbonate) to 1300 K in a lime klin. Lime dissolves in water to form calcium hydroxide which is then transferred to the ammonia recovery tower.
CaCO₃?→? ?CaO+CO₂?
CaO+H_2?O→Ca(OH)_2?
Ammonia required for the process can be prepared by heating ammonium chloride with calcium hydroxide.
2NH_4?Cl+Ca(OH)₂→ ?2NH₃?+CaCl₂?+H₂?O
Hence, the only by product of the reaction is calcium chloride.