A sample of O2 with an initial temperature of 50°C and a volume of 105 L is called 225.0°C the new pressure is 105 point4K PA and the new volume is 55.0 L what was the initial pressure of the sample

35.81 KPa

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Initial temperature (T1) = 50°C

Initial volume (V1) = 105 L

Initial pressure (P1) =?

Final temperature (T2) = 225.0°C

Final pressure (P2) = 105.4 KPa

Final volume (V2) = 55.0 L

Step 2:

Conversion of celsius temperature to Kelvin. This is illustrated below:

K = °C + 273

T1 = 50°C + 237 = 323K

T2 = 225.0°C + 273 = 498K

Step 3:

Determination of the initial pressure of the O2 sample.

The initial pressure of the O2 sample can be obtained by using the general gas equation as follow:

P1V1/T1 = P2V2/T2

P1 x 105/ 323 = 105.4 x 55/ 498

Cross multiply to express in linear form

P1 x 105 x 498 = 323x 105.4 x 55

Divide both side by 105 x 498

P1 = (323x 105.4 x 55) /(105 x 498)

P1 = 35.81 KPa

Therefore the initial pressure of the O2 sample is 35.81 KPa

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.