Air is made up mostly of O2, N2, and CO2. If the total pressure of air is 765mmHg and P(N2) = 0.79 atm and P(CO2) = 0.05 atm, then what is the partial pressure of the O2?

the partial pressure of the O2 is 0.167 atm

Explanation:

The computation of the partial pressure of the O2 is shown below:

As we know that

P = P_N2 + P_O2 + P_CO2

P_O2 = P - P_N2 -  P_CO2

= (1.007 - 0.79 - 0.05)

= 0.167 atm

Hence, the partial pressure of the O2 is 0.167 atm

we simply applied the above formula

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.

explanation: no but i do not usually comment