Chemistry

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

(1) 4 mol of $=4{}_A$ atoms

(2) 4 u of $=\frac{4u}{4u}=1$ atom

(3) 4 g of Helium $=\frac{4}{4}$ mole $=1$ mole $=N_A$ He atom

(4) $2.2710982$ of He at STP $=\frac{2.271}{22.710982}$ mole

$=0.1$

$=0.1{}_A$

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

As $Q_c>K_c$ , so reaction has a tendency to move backward.

Nucleophilic addition of sodium hydrogen sulphite to aldehyde or ketone is a;

$NaHS{O}_3?N{a}^{+}+HS{O}_3^{?}$                

So; nucleopilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.

Addition of hydrogen cyanide;

Final product is cyanohydrin.

All given oxide have nitrogen – nitrogen bond except N₂O₅ as ;

BOD is biological oxygen demand which represents the amount of oxygen required to degrade organic matter in water.

Higher the BOD more polluted the water is

So; here water sample with BOD = 3 ppm is the cleanest

Here, C₃H₆ is propene

Second reaction is iodoform reaction.

Given are the oxide of alkali and alkaline earth metals which are ionic in nature.

Simple oxide are Li₂O, CaO, MgO and K₂O.

Peroxide is Na₂O₂ and superoxide is KO₂.

All simple oxides are diamagnetic as it has no unpaired electron.

Density of metal a very specific and depends upon many factors.

Li -> 0.53 gm/cm³

Na -> 0.97 gm/cm³

K -> 0.86 gm/cm³

Rb ->1.53 gm/cm³

Cs -> 1.90 gm/cm³

Kindly consider the below image