Class 12th

$Xe{F}_2+P{F}_5\to {\left[XeF\right]}^{+}{\left[P{F}_6\right]}^{-}$

  ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+},N{H}_3$ becomes strong ligand due to +3 oxidation state of cobalt ion. Hence electronic configuration of CO³⁺ is $t_{2g}^6{e}_g^0$  

Haemoglobin is a positive colloid. Hence greater is the charge of anion, more effective will be the coagulation of haemoglobin.

$$ Correct order of coagulating power is $P{O}_4^{3-}>S{O}_4^{2-}>C{l}^{-}$

Electrophoresis : The migration of colloidal particles under the influence of an electric field is known as electrophoresis.

$r_1=k{\left[x\right]}^n$ … (i)

$$ $2r_1=k{\left[4x\right]}^n$  … (ii)

$\left(ii\right)÷\left(i\right)$

$\Rightarrow \frac{2r_1}{r_1}=\frac{k{\left[4x\right]}^n}{k{\left[x\right]}^n}$

$\Rightarrow 2={\left(2\right)}^{2n}$

$\Rightarrow n=\frac{1}{2}$

XeF2 : Hybrid orbital = sigma bond + L.P

= 2 + 3

= 5 = sp3d

Shape will be linear

For a first order reaction.

$k=\frac{0.693}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{0.693}{231}=3*10^{-3}{s}^{-1}$   

$5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 2M{n}^{+}+8H_{2}O+10C{O}_2$

5 moles of $C_2{O}_4^{2-}$ ${}_{}{}_{}^{}$ is needed to reduce 2 moles of $Mn{O}_4^{2-}$ ${}_{}^{}$

1 mole of $C_2{O}_4^{2-}$  is needed to reduce $\frac{2}{5}$  moles of $Mn{O}_4^{2-}$

Yes, a material's resistivity will increase with the presence of impurities. This is because these impurities will occupy more surface area and disrupt the flow of electrons which will impact their free movement. As a result, the current cannot pass freely through the material and it will affect the resistivity.

Resistance is referred to the ability of an object to oppose the flow of electric current through itself. On the other hand, resistivity is the property of that particular material of resisting the passage of electric current over a particular unit area. Resistance has SI unit ohm whereas resistivity has ohm-metre.