Chemistry Electrochemistry

Kindly go through the solution 

[V (CO)₆]

EAN = 23 – 0 + 2 * 6

= 23 + 12

= 35 ≠ 36

⇒ so it does not obey EAN rule.

I₂ + conc. HNO₃ → HIO₃ + NO₂

Charge passed = It

= 0.015 * 15 * 60 C

Moles of electrons passed $=\frac{0.015*15*60}{96500}$  

Moles of Zn deposited = $\frac{1}{2}*\frac{0.015*15*60}{96500}$  

= 0.00007

Mass of Zn deposited = 0.00007 * 65.4 g = 4.58mg

–C º N $?\underset{\begin{array}{l}\left|\text{?}\right|\\ O\end{array}}{C}$ , – CH₃ are –R group which is deactivating

– $\ddot{N}H$ – $\stackrel{\begin{array}{l}O\\ \left|\text{?}\right|\end{array}}{C}$ – CH₃ and – $\ddot{N}H$ – CH₃ due to presence of lone pair in nitrogen atom behaves as activating (+R) group.

E_cell = 0 – log2

= – 0.03 (0.3)

 = – 0.009

 = – 9 * 10^–3 V

x = 9

$?=\frac{h}{p}\left[??\frac{1}{p}\right]$

Þ        lp = h (constant)

            So, the plot is a rectangular hyperbola.