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PbS + 4H₂O_2->PbSO₄ + 4H₂O.

Work depends on the path followed by the system so work is path function.

[Ni (H₂O)₆]²⁺ Þ Ni²⁺ is octahedral and [Ni (CN)₄]^2– is square planar.

In [Ni (H₂O)₆]²⁺ Þ Ni²⁺ has two unpaired electrons and in [Ni (CN)₄]^2– Þ Ni²⁺ has no unpaired electrons. [Ni (H₂O)₆]²⁺ is coloured as it absorbs red light due to suitable d-d transition and complementary light emitted is green.

[Ni (CN)₄]^2– has strong field ligand so the electrons of Ni²⁺ pair up and it is colourless as it cannot absorb light from visible region.

Kindly go through the solution

Since V_Real > V_ideal

So, Z = V_Real > 1. V_ideal

KCN   + CN^–

            R – X + KCN  R – CN + KX

            R – X + AgCN  R – NC + AgX

            KCN is ionic therefore ionised and attack occurs through carbon.

            AgCN is covalent therefore attack starts with Nitrogen.

meq. of K₂Cr₂O₇ = meq. of NO^–₂

n₁ * 6 = 1 * 2

n₁=1/3

$C_{vm}=\frac{2\left(\frac{3R}{2}\right)+6\left(\frac{5R}{2}\right)}{2+6}$

$=\frac{3R+15R}{8}=\frac{18R}{8}$

Radial nodes = n –? – 1 = 3 – 1 – 1 = 1

Angular nodes =? = 1.

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$r=R_0{\left(192\right)}^{\frac{1}{3}}$

$\frac{r}{2}=R_0{\left(m\right)}^{\frac{1}{3}}$

$m=\frac{192}{8}=24$

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-> $V_{esc}=\sqrt[]{\frac{2GM}{R}}$

$Ratio=\sqrt[]{\frac{81}{9}}=3$