Class 12th

${\Delta }_t=\frac{hc}{\lambda }=\frac{6.63*10^{-34}*3.08*10^8}{600*10^{-9}}$

$=\frac{6.63*3.08*10^{-17}}{600}$ J

$=765.765*10^{-21}J$

$?766*10^{-21}$ J

  $f\left(x\right)=\frac{4x^3-3x^2}{6}-2sinx+\left(2x-1\right)cosx$

$f\text{'}\left(x\right)=2x^2-x-2cosx+2cosx-\left(2x-1\right)sinx$

= $\left(2x-1\right)\left(x-sinx\right)\ge 0forx\ge 0$  

$f\text{'}\left(x\right)\ge 0\forall x\ge 1/2$           

$\therefore f\left(x\right)$ is increasing in $\left[\frac{1}{2},\infty \right)$  

The given value of µ (spin only)
2.84 = √n (n + 2) BM, So, n = 2
Among the given configurations, d? system in strong field ligand will have 2-unpaired e? in t? g set of orbitals as shown below.

$\frac{d\left[C\right]}{dt}=\left(\frac{20-10}{10}\right)=1mmold{m}^{-3}$

$+\frac{d\left[D\right]}{dt}=\left\{-\frac{d\left(B\right)}{dt}\right\}*1.5=\frac{3}{2}\left\{-\frac{d\left[B\right]}{dt}\right\}$

$\left\{-\frac{d\left[B\right]}{dt}\right\}=2*\left\{\frac{-d\left[A\right]}{dt}\right\}$

$\therefore \frac{1}{6}\left\{\frac{-d\left[B\right]}{dt}\right\}=\frac{1}{3}\left\{\frac{-d\left[A\right]}{dt}\right\}=\frac{1}{9}\left\{\frac{+d\left[D\right]}{dt}\right\}=\frac{d\left[C\right]}{dt}$

$\therefore$ rate of reaction = $\frac{+d\left[C\right]}{dt}$  = 1m.m dm^-3 S^-1

4-ethoxycarbonylpent-3-enoic acid

$F{e}^{+3}+l^{-}\to I_2+F{e}^{+2}$

Fe⁺² undergoes reduction & I₂ undergoes oxidation.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

= (0.77 – 0.54) V = 0.23 V

 = 23 * 10^-2 V

$\therefore$  x = 23

 x = 23

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

[as probability of getting odd number = probability of getting even number = ]

Probability of getting an odd number for odd number of times =

 

In S?2 reaction, inversion takes place, so configuration get reversed.

When Ferric chloride reacts with potassium thiocyanate a blood red colour of Ferric thiocyanate is formed
FeCl? + 3KSCN → Fe (SCN)? + 3KCl

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

                             = 0.00002 * 10³ mm