The S-block Elements

An alkali metal oxide of formula MO₂ is super oxide which is formed by K and Rb. KO₂ has pale yellow colour and paramagnetic nature, while alkaline earth metal oxides formula MO₂ are peroxide and diamagnetic nature.

Lithium salts are extensively hydrated due to high hydration enthalpy of Li⁺

$L{i}^{+}>N{a}^{+}>K^{+}>R{b}^{+}>C{s}^{+}$ (order of polarizing power)

Moles of sodium = $\frac{8}{23}mol$  

Number of atoms = $\frac{8}{23}*6.02*10^{23}$  

$\approx 2*10^{23}$                                            

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

$N{a}_{2}O+H_{2}O\to 2NaOH$

w = 20 g

$V_{H_{2}O}=500mL$        

Mole of Na₂O=   $\frac{20}{62}$

$?$ 1 mole of Na₂O gives 2 mole of NaOH

$\therefore \left(\frac{20}{62}\right)$ mole of Na₂O gives $\left(2*\frac{20}{62}\right)$ moles of NaOH

$=\frac{20}{31}mole$    

Molarity of NaOH solution

$=\frac{\frac{20}{31}*100}{500}M=\frac{20}{31}*2M=\frac{40}{31}M$

= 1.29 M

=  $12.9*10^{-1}M\approx 13*10^{-1}M$

Ans. = 13

Portland cement contains

Dicalcium silicate = 26%

Tricalcium silicate = 51%

Tricalcium aluminate = 11%

Hence major percentage is of tricalcium silicate

(1) Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.

(2) Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

(3) In case of Csl, lattice energy is less but Cs⁺ having less hydration energy due to which it is less soluble in water.

(4) For alkali metal fluorides, the solubility in water increases from Li to Cs. LiF is least soluble in water.

Number of water molecule

Gypsum :- CaSO₄.2H₂O  =>          2

Plaster of paris :- CaSO₄.    =>          0.5

Dead burnt plaster : CaSO₄          =>          0

$\Delta H_f^0=\left(+\Delta H_{sub}^0\right)+\left(+\Delta H_{IE}^0\right)+\left(+\frac{1}{2}\Delta H_{BDE}^0\right)+\left(\Delta H_{EA}^0\right)+\left(-\Delta H_{LE}^0\right)$

$\Delta H_{LE}^0=89.2+419+121.5-348.6+436.7kJmo{l}^{-1}$

$\therefore$ Lattice energy = 717.8