Class 11th

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.

Kindly go though the solution 

$v=\sqrt[]{\frac{\gamma RT}{M}}=330m/s$

$I=\left(\frac{2}{5}M{R}^2+M{R}^2\right)*2$

$=\frac{14}{5}*2*\frac{9}{16}$

$=\frac{63}{20}$          

= 3.15 kg m²

[A] = L²

$B=\frac{x^2}{tE}\equiv \frac{L^2}{TM{L}^2{T}^{-2}}=\frac{1}{M{T}^{-1}}$            

[B] = M^–1T

[AB] = [M^–1L²T]

$P=\overrightarrow{F}\cdot \overrightarrow{v}$

= 18t⁴ – 18t²

->P (t = 2) =18 [16 − 4] = 216 W

f_K = μmg cosq

= 0.1 * $\frac{50*\sqrt[]{3}}{2}$

$=2.5\sqrt[]{3}N$

 F₁    = mg sinq + f_K

= 25 + 2.5 $\sqrt[]{3}$  

F₂    = mg sinq – f_K

= 25 – 2.5 

F₁ – F₂ = 5 $\sqrt[]{3}$  N