Chemistry

H? S? O? , peroxodisulphuric acid (Two S-OH, Four S = O, One S – O – O – S )

H? SO? , sulphurous acid (Two S-OH, One S = O bond)

H? S? O? , pyrosulphuric acid (Two S-OH, Four S=O, One S-O-S bond)

H? SO? , sulphuric acid (Two S – OH, Two S = O )

Fe? O? + CO → 2FeO + CO?

The above reaction takes place at 500 – 800 K in blast furnace.

Pumice stone is an example of solid sol. In this type of colloid, the dispersion medium is solid and the dispersion phase is gas.

One edge of an cube is common to four unit cells. Hence One edge centre OV contribute 1/4 to one unit cell.

Eutrophication occurs when the water body becomes overly enriched with nutrients. Eutrophication leads to decrease in the level of dissolved oxygen (DO) in water bodies.

Kolbe's electrolysis is a process where an aqueous solution of sodium and potassium salt undergoes electrolysis to lead to the formation of a symmetrical alkene. Principle of this technique is that at anode, the carboxylate ion undergoes oxidation and loses carbon dioxide (CO? ). Similarly, at cathode the water decomposes to form hydrogen gas. The alkyl radicals formed in these two cases combine to form alkenes.

Acc. to Huckel rule, A compond/molecule is said to be aromatic if it have cyclic conjugate and [4n + 2]π electron system

⇒ Molecule must be sp² hybridised

⇒ Molecule must be planar.

Reduction half reaction.

Cr? O? ²? + 14H? + 6e? → 2Cr? ³ + 7H? O

Oxidation half reaction

SO? ²? + H? O → SO? ²? + 2e? ] * 3

Oxygen is balanced by adding water and hydrogen is balanced by adding H? and the charge is balanced by electrons.

Add ( eq. (i) + (3 * eq. (ii) )

Cr? O? ²? + 3SO? ²? + 8H? → 2Cr? ³ + 3SO? ²? + 4H? O

a = 1 b = 3 c = 8

Keq. = ( [C] [D])/ ( [A] [B])

Keq = (10 * 6)/ (2 * 3) = 10

ΔG° = −2.303RTlog Keq. = −2.303 (2) (300) (log 10)

= −1381.8cal (? R = 2cal/molk)

Kindly go through the answer

ΔH = ΔU + Δn_gRT