Chemistry

Following are the possible isomers of dimethyl cyclopentane.

Mol wt A₂B = Mol wt of AB₃

 Hence (2A + B) = (A + 3B)

  $\therefore A=2B$

Atomic wt of A is two times of atomic wt of B.

Following is the reaction for oxidation of oxalic acid by KMnO₄

$2KMn{O}_4+5H_2{C}_2{O}_4+3H_{2}S{O}_4?K_{2}S{O}_4+2MnS{O}_4+10C{O}_2+8H_{2}O$       

Here, Mn²⁺ ion is in product form

Having 5 unpaired electron

Magnetic moment =   BM = 5.916

Cu⁺  Paramagnetic due to unpaired e^-

Cu⁺  Colourless from due to 3d¹⁰

Cu⁺  acts as readucing agent

Cu⁺  it's a product form of Fehling's solution

According to Arrhenius equation ;

$Ink=lnA-\frac{Ea}{RT}$                

Here lnk = 33.24  $-\frac{2.0*10^4}{T}$  

$\therefore \frac{Ea}{R}=2.0*10^4$      

$and{E}_a=2.0*10^4*R=2.0*10^4*8.3=16.6*10^{4}J=166kJ$      

${\lambda }_m^{\infty }$  of Nal = 12.7 mSm² mol^-1

${\lambda }_m^{\infty }$ of NaNO₃ = 12.0 m Sm² mol^-1

${\lambda }_m^{\infty }$ of AgNO₃ = 13.3 mSm² mol^-1

$\therefore {\lambda }_{Agl}^{\infty }=\left[{\lambda }_m^{\infty }\left(AgN{O}_3\right)+{\lambda }_m^{\infty }\left(Nal\right)\right]-{\lambda }_m^{\infty }NaN{O}_3$                

= (13.3 + 12.7) - 12.0 = 14.0 m Sm² mol^-1

$2NOC{l}_{\left(g\right)}?2NO\left(g\right)+C{l}_2\left(g\right)$

t = 0       2 mol/Lit             0            0

t_eq          2 – x                    x            x/2

$K_C=\frac{{\left[NO\right]}^2*\left[C{l}_2\right]}{{\left[NOCl\right]}^2}=\frac{x^2*x}{{\left(2-x\right)}^2*2}=\frac{x^3}{{\left(2-x\right)}^2*2}$      

$=\frac{0.4*0.4*0.4}{1.6*1.6*2}$ = 0.0125 = 125 * 10^-4

  $\Delta T_b=i*K_b*m$

 or $\Delta T_b=K_b*m$   (i = 1 for non-electrolyte)

  $or\frac{{\left(\Delta T_b\right)}_A}{{\left(\Delta T_b\right)}_B}=\frac{{\left(K_b\right)}_A}{{\left(K_b\right)}_B}$

  $or\frac{x}{y}=\frac{1}{8}\left|\therefore y=8\right|$              

$\Delta V=2.4*10^{-26}m/sec$

And $\Delta x=10^{-7}m$  

According to Heisenberg's uncertainty

$\Delta v*\Delta x=\frac{h}{4\pi *m}$  

$\therefore m=\frac{h}{4\pi *\Delta v*\Delta x}=\frac{6.626*10^{-34}}{4*3.14*2.4*10^{-26}*10^{-7}}$

$\therefore$  m = 0.02198 Kg

Or 21.98 gm

Or 22 gm in the nearest integer.

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x*\left(+2\right)+\left(1-x\right)*\left(+3\right)=$ O.S. of Fe atom

$or2x+3-3x=\frac{+200}{93}$

$\therefore \mathrm{\%}ofF{e}^{2+}=85\mathrm{\%}and\mathrm{\%}ofF{e}^{3+}=15\mathrm{\%}$