Class 11th

BF₃ : sp2 hybridised hence shape will be trigonal planar.

Kindly go through the answers

$A_1=\sqrt[]{{\left(5\right)}^2+{\left(5\sqrt[]{3}\right)}^2}=10$

$A_2=5$

Then, $\frac{A_1}{A_2}=\frac{10}{5}=\frac{2}{1}$

Z=109 : Unnilennium = Meitnerium

Sn = [Kr]4d105s25p2

Sn (Z = 50) belongs to p-block element.

H2SO5

: x + 2 (+1) + 3 (–2) + 2 (–1) = 0

x = + 6.

$\underset{acid}{H_{3}B{O}_3}+2H_{2}O\underset{conjugatebase}{B{\left(OH\right)}_4^{-}+}{H}_3{O}^{+}$

CH₃COONa is salt of weak acid with strong base

$\therefore pH=7+\frac{1}{2}p{K}_a+\frac{1}{2}logC$

$pH=7+\frac{1}{2}*4.74+\frac{1}{2}*log1$

$pH=9.37\approx 9.4$

On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.