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The highest package offered during Symbiosis Institute of Operations Management, Nashik placements 2025 stood at INR 21.60 LPA. The table below presents the same:

The placements at SIOM Nashik are good. The following table presents the placement statistics for SIOM Nashik:

Candidates aspiring to take admission to the postgraduate courses at the University of Rajasthan will have to appear for the URAT PG entrance exam conducted before the session starts. The examination cell then releases the merit list based on the scores of the candidates in exam. Candidates whose names appear in the merit list will then have to visit the alloted college for document verification and confirmation of seat after the fee payment.

Following are the $\Delta H°$  hydration of given ions

Cr²⁺ - 1926 KJ/mole

Mn²⁺ - 1860 KJ/mole

Fe²⁺ - 2000 KJ/mole

Co²⁺ - 2078 KJ/mole

Acid = M (OH)₂ ® Salt + H₂O

M.E of Acid = me of Base

$\therefore 10\times \left(0.1\times nf\right)=\left(0.05\times 2\right)\times 30$

$\therefore n_f=3$

Thus basicity of acid = 3

$x=\frac{1}{2}$

$y=2{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left(\frac{y^2+3}{4}-\frac{y^2+1}{2}\right)dy}$

$=\frac{1}{2}{\left(\frac{y^3}{3}+3y\right)}_0^1-{\left(\frac{y^3}{3}+y\right)}_0^1$

$=\frac{1}{2}\left(\frac{10}{3}\right)-\frac{4}{3}=\frac{1}{3}$

Cobalt (II) meta borate is a blue colour of Bead.

Biuret test is given by proteins and dipeptide & compounds having following groups

Hence positive test given by tripeptide & biuret.

Monomer of novolac is

Molecular mass of monomer unit

= 12 × 7 + 1 × 8 + 16 × 2

= 84 + 8 + 32

= 124 g

Molecular mass of polymer = 963 g

$\therefore$ n × mass of monomer = mass of polymer + (n – 1) × 18

(n – 1) unit of H₂O is removed during formation

$\therefore 124\times n=963+18\left(n-1\right)$        

106n = 945

$\therefore n=\frac{945}{106}=8.91\approx 9$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{si{n}^{2}x\left(si{n}^{2}x-3sinx+2\right)}{co{s}^{2}x.\left(\sqrt[]{2si{n}^{2}x+3sinx+4}+\sqrt[]{si{n}^{2}x+6sinx+2}\right)}$  

$\downarrow$                                                       $\downarrow$                            

3                                                       3

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{9}.\frac{\left(sinx-1\right)\left(sinx-2\right)}{\left(1-sinx\right)\left(1+sinx\right)}=\frac{1}{6}\left(-1\right).\frac{-1}{2}=\frac{1}{12}$

Here 'X' is the Gypsum (CaSO₄.2H₂O) which is used to enhance the setting of time.

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FeCl₃. 3H₂O i.e. [Fe(H₂O)₃], K₃[Fe(CN]₆ and [Co(NH₃)₆]Cl₃; the last two complex are inner-orbital complex due to presence of strong ligand.

$\therefore {\Delta }_{0}of{K}_3\left[Fe{\left(CN\right)}_6\right]>{\Delta }_{0}of\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$       

Because CN^- is strong ligand

$\therefore {\Delta }_0\propto \frac{1}{\lambda }$         

More ${\Delta }_0$  smaller value of absorbed $\lambda$ .

$K_3\left[Fe{\left(CN\right)}_6\right]\to F{e}^{+3}=4s^{0}3d^{5}4p^0$       

$\therefore u.e=1$      

$\therefore \mu =\sqrt[]{1\times \left(1+2\right)}B.M=\sqrt[]{3}B.M$        

= 1.732 B.M

$\approx$ 2

Number of replaceable H is 2.

$\Delta =\left|\begin{array}{ccc}\hfill -k\hfill & \hfill 3\hfill & \hfill -14\hfill \\ \hfill -15\hfill & \hfill 4\hfill & \hfill -k\hfill \\ \hfill -4\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|$

$=\left(-k\right)\left(12-k\right)+3\left(4k+45\right)-14\left(-15+16\right)$

$\Delta =0\Rightarrow k=\pm 11$             

 For k = -11,

->11x + 3y – 14z = 25

-4x + y + 3z = 4

$\left\{\begin{array}{l}-11x+3y-14z=25\\ -15x+4y-11z=3\\ -4x+y+3z=4\end{array}\right\}\to$ No solution for k = $\pm$ 11

$F{e}^{+3}+3e^{-}\to Fe$  

3 Faraday is required to deposit 1 mole Fe

$?56g$ deposited by  $3\times 96500$ C charge

$\therefore 0.3482g$ deposited by  $\frac{3\times 96500}{56}\times 0.3482C=\frac{100803.9}{56}C=1800.7C$

    $\therefore Q=It$            

1800.07 = 1.5 t

$\therefore t=1200sec=\frac{120\cancel{0}}{6\cancel{0}}$  min = 20 min

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t_1/2 = 340 sec      P₀ = 55.5 kPa

t_1/2 = 170 sec      P₀ = 27.8 kPa

$?t_{1/2}?{\left(P_0\right)}^{1?n}$

$?1?n=1n=0$

$?$ zero order reaction