In a lab, 2.20 L of gas is collected over water at a temperature of 30°C and a total pressure of 735.43 mmHg. Find the volume that the dry nitrogen of gas would occupy at STP.

V₂= 1.9 L

Explanation:

Given data:

Initial volume of gas = 2.20 L

Temperature = 30°C  (30+273 = 303 K)

Initial pressure = 735.43 mmHg (735.43 /760 = 0.97 atm)

Final volume of gas = ?

Final temperature = standard = 273 K

Final pressure = standard = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 0.97 atm × 2.20 L × 273 K / 303 K × 1 atm

V₂= 582.58 atm .L. K / 303 k.atm

V₂= 1.9 L

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.

the answer to your question is c