Class 12th

All monosaccharides which differ in configuration at C? and C? gives the same osazone. Since, glucose and fructose differ from each other only in configuration at C? and C? therefore, they give the same osazone. All other options given in the questions do not satisfy this condition and hence, do not from the same osazone.

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$           

$\Rightarrow \frac{dp}{P-900}=0.5dt$         

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$           

$\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$

$\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$           

T = 2 ln 18

Nitrogen, sulphur and halogens are tested in an organic compound by lassaigne's test.
The organic compound is fused with sodium metal as to convert these elements into ionisable inorganic substances.
Na + C + N → NaCN
2Na + S → Na? S
2Na + X? → 2NaX
The cyanide, sulphide or halide ions can be confirmed in aqueous solution by usual test.

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

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