Chemistry

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

Hydroxyapatite is the calcium minerals whose molecular formula is $3C{a}_3{\left(P{O}_4\right)}_2.Ca{\left(OH\right)}_2$ ${}_{}{{}_{}}_{}{}_{}$ or $C{a}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2.$ ${}_{}{{}_{}}_{}{}_{}$ During contact with water dissolved F ion, hydroxyapatite converted into ${}_{}{{}_{}}_{}{}_{}$ $3C{a}_3{\left(P{O}_4\right)}_2.Ca{F}_2$ .

Yellow precipitate is Agl (s) which only produce by structure IV

${?}_0=h{\upsilon }_0$

$=6.6*10^{-34}*1.3*10^{15}J$

^{$=8.58*10^{-19}J$}