Determine the oxidation number of
Sulphur in Sodium thiosulphate

The oxidation number of  Sulphur = +2

General formula for determining Oxidation Numbers :

1. The number of oxidation of single element atoms = 0. Examples of Ar, Mg, Cu, Fe, N₂, O₂, etc. = 0

Group I : +1

Group II : +2

H in compound = +1, except metal hydride compounds (Hydrogen which binds I or II groups) oxidation number H = -1, for example, LiH, MgH₂, etc.

2. Oxidation number O in compound = -2, except OF2 = + 2 and in peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.

3 The oxidation number in a non-filled compound = 0,

The total amount of oxidation in ion = ion charge, Example NO₃⁻ = -1

Sodium thiosulphate - Na₂S₂O₃

Oxidation number Na = +1 ⇒x2 = +2

Oxidation number O = -2 ⇒x3 = -6

Oxidation number Na₂S₂O₃ = 0

2.Na+2.S+3.O=0

2.+1+2.S+3.(-2)=0

2+2S-6=0

-4+2S=0

2S=4

S=+2

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it