Class 11th

$pH=\frac{p{K}_w+p{K}_a?p{K}_b}{2}$

pK_a = pK_b

$pH=\frac{p{K}_w}{2}=7$

Adenine base pairs with thymine with 2 hydrogen bonds and cytosine base pairs with guanine with 3 hydrogen bonds.

$T=2\pi \sqrt[]{\frac{I}{g}}$

$g=\frac{4{\pi }^{2}I}{T^2}$

$\frac{\Delta g}{g}=\frac{\Delta I}{I}+\frac{2\Delta T}{T}$

$=\frac{0.2}{20}+2\left(\frac{1}{40}\right)$

= 6%

The magnitude of resultant vector (R)

$=\sqrt[]{a^2+b^2+2abcos\theta }$            

here a = b = A

then R = $\sqrt[]{A^2+A^2+2A^{2}cos\theta }$  

$=A\sqrt[]{2}\sqrt[]{1+cos\theta }$

$=\sqrt[]{2}A\sqrt[]{2co{s}^2\frac{\theta }{2}}$

$=2Acos\frac{\theta }{2}$      

q = 0 as adiabatic process is given

            W = 0 as p_axt = 0

            q + W = DU

            q = 0

            W = 0

            Þ DU = 0

PF₅, PCl₅, PBr₅, Fe (CO)₅ Þ Trigonal bipyramidal

BrF₅ Þ Square pyramidal

[PtCl₄]^2– Þ Square planar

SF₆ Þ Octahedral

$a=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2\right)}=\frac{g}{8}$

8m₁ – 8m₂ = m₁ + m₂

7m₁ = 9m₂

$\frac{m_1}{m_2}=\frac{9}{7}$          

U = nC_vT

-> $U=n_1{C}_{v_1}T+n_2{C}_{v_2}T$  

-> $8*\frac{5R}{2}*T+4*\frac{5R}{2}*T$

= 30RT

Boiling point: $\stackrel{\left(239.7\right)}{N{H}_3}>\stackrel{\left(185.5\right)}{P{H}_3}$  due to hydrogen bonding in NH₃.

CuSO₄ acts as catalyst in Kjeldahl's method of estimation of nitrogen.