Chemistry

In sulphide ore, depressants selectively prevent impurity from coming to the fourth.

Except Te, all are metals.

Due to higher extent of polarization by Li⁺ and Mg²⁺, LiCl and Mgcl₂ have covalent character. Therefore they are soluble in ethanol.

Due to very high value of lattice energy, LiF is having very less solubility in water.

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having manufacturing of N-based fertizers)

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having  manufacturing of N-based fertizers)

% of C in organic compound

$=\frac{12}{14}*\frac{WC{O}_2}{W.O.C.}*100$  

$=\frac{952.56}{21.648}=44\mathrm{\%}$  

$\mathrm{\%}ofH=\frac{2}{18}*\frac{W{H}_{2}O}{W.O.C.}*100$  

= $\frac{20}{18}*\frac{0.4428}{0.492}*100$  

=  $\frac{88.56}{8.856}=10\mathrm{\%}$  

= 100 – 54 = 46%

$\left[Cu{\left(en\right)}_2{\left(SCN\right)}_2\right]$

More stable isomers = 3 (trans isomers)

Invertase → Cane sugar to glucose and fructose

Zymase → Glucose to ethanol

Diastase → Starch to Maltose

Maltase → Maltose to glucose

In 4d orbital, n = 4 and $\mathcal{l}=2$

Radial nodes = $n-\mathcal{l}-1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  $\mathcal{l}=2$

$Mn{O}_4^{-2}\to A+B$  

Oxidation state of Mn in B < A 

$Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$  

B is MnO₂

$\therefore$  Oxidation state of Mn = +4

${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$  

  $\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$  

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$  

$=3.87\approx 4$