Chemistry

Kindly consider the following figure

${\left[Fe{\left(CN\right)}_{5}NOS\right]}^{4-}$  ${{}_{}}^{}$ Violet colour

${{}_{}}^{}$  ${\left[Fe{\left(SCN\right)}_6\right]}^{4-}$ Blood red colour

$$ $F{e}_4{\left[Fe{\left(CN\right)}_6\right]}_3{H}_{2}O\to$  Prussian blue colour

$$ ${\left[Fe{\left(CN\right)}_6\right]}^{4-}\to$  Yellow colour

${\left[PtC{l}_4\right]}^{2-}+H_{2}O?{\left[Pt\left(H_{2}O\right)C{l}_3\right]}^{-}+C{l}^{-}$  

At equilibrium  $\frac{dx}{dt}=0,$ hence

$0=\frac{-d\left[{\left[PtC{l}_4\right]}^{2-}\right]}{dt}=4.8*10^{-5}\left[{\left[PtC{l}_4\right]}^{2-}\right]-2.4*10^{-3}\left[{\left[Pt\left(H_{2}O\right)C{l}_3\right]}^{-}\right]\left[C{l}^{-}\right]$           

$K_c=\frac{K_f}{K_b}=\frac{4.8*10^{-5}}{2.4*10^{-3}}$              

= 2 * 10^-2

= 0.02

For equation A + B $?C+D$  

  $K_C=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=100at298K$             

Here;    A +  B  $?$ C    +    D

t = 0       1M        1M

eq.         (1-x)      (1-x)      (1-x)      (1-x)

$Kc=\frac{\left(1+x\right)*\left(1+x\right)}{\left(1-x\right)*\left(1-x\right)}$

$\therefore x=\frac{9}{11}$  

At equilibrium, concentration of D is = 1 + $\frac{9}{11}=\frac{11+9}{11}=\frac{20}{11}$  

= 1.818 = 181.8 * 10^-2 M

Ans. = 182 (the nearest integer)

$\Delta H=3{\left(\Delta H_{comb}\right)}_{C_2{H}_2\left(g\right)}-{\left(\Delta H_{comb}\right)}_{C_6{H}_6\left(l\right)}$

$=-1300*3-\left(-3268\right)kJ/mole$

$=\left(-3900+3268\right)kJ/mole$

$=-632kJ/mole$

Cu₂Cl₂ + Water -> Insoluble

AgCl + Water -> Insoluble

ZnCl₂ + Water -> Colouless solution

CuCl₂ + Water -> Coloured solution due to 3d⁹ electronic configuration.

In photo-electric effect;

  $\frac{hc}{\lambda }=h{v}_0+\frac{1}{2}m{v}^2$             

  $\frac{6.63*10^{-34}*3*10^8}{500*10^{-9}}=\left(6.63*10^{-34}*4.3*10^{14}\right)+\frac{1}{2}*9.0*10^{-31}*v^2$            

  $v=\sqrt[]{\frac{\left(3.978*10^{-19}-2.85*10^{-19}\right)}{4.5*10^{-31}}}=\sqrt[]{25*10^{10}}=5*10^{5}m/s$             

Ans. = 5

Molar conductivity measured at infinite dilution is called limiting molar conductivity. Hence at infinite dilution stage, both strong electrolyte (KCl) and weak electrolyte (CH₃COOH) have identical value of conductivity. Therefore statement – I is false.

Statement – II is also false because molar conductivity increases by the decreasing the concentration of electrolyte.

Molar weight of BaSO₄ = 233 gram

233 gm of BaSO₄ has 32 gm of S content

Hence 1.44 gm of BaSO₄ has $\left(\frac{32*1.44}{233}\right)$ g of S content = 0.1977 gm of 'S'

Weight of sulphur = 0.2 gram (the nearest)

Weight of organic sample = 0.471 gram

% of Sulphur = $\frac{0.2*100}{0.471}=42.46\mathrm{\%}$  

the nearest = 42%

Kindly consider the following figure