Class 11th

The second ionization energy of K is maximum, because the second electron is removed from fully filled 3p? subshell. The second ionization energies of group 2 elements decrease down the group.
Hence, second I.E. of Ca > second I.E. of Ba.

The process of moving sodium and potassium ions across the cell membrane is an active transport process involving the hydrolysis of ATP to provide the necessary energy.

$h=\frac{a\left(1+t^2\right)}{2}.......(i)$

k = at        ……… (ii)

From (i) & (ii) $\frac{2h}{a}=1+\frac{k^2}{a^2}$  

$\therefore$ required locus of Q is y² = 2a (x – a/2)

Equation of directrix x – a/2 = -a/2Þ x = 0

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2*0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}*1\right)}^2}{\left(\frac{0.5}{1.5}*1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

As Zn (OH)? , BeO, Al? O? are amphoteric, so they can react with both HCl and NaOH.

I (6)       F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =   

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$

i. Zn + 2NaOH → Na? ZnO? + H?
ii. 4Au + 8NaCN + O? + 2H? O → 4Na [Au (CN)? ]+ 4NaOH
iii. Cu + 4HNO? → Cu (NO? )? + 2NO? + 2H? O
(conc.)

$E_{1_H}=-2.2*10^{-18}J$

$Li\to L{i}^{+2}+2e^{-},n=2$

$E_{L{i}^{+2}}=E_{1H}*\frac{Z^2}{n^2}=-2.2*10^{-18}*\frac{3^2}{2^2}$

$\lambda =4*10^{-8}$ m

Molar mass of C₇H₅N₃O₆ = 12 * 7 + 1 * 5 + 14 * 3 + 16 * 6 = 84 + 42 + 96 = 227 g/mol

Number of moles =  $\frac{681}{227}$  = 3 mol

$\therefore$  number of N-atoms

$=3*6.02*10^{23}*3=9*6.02*10^{23}$  

$=5418*10^{21}$  

$\therefore x=5418$