`i am holding a balloon containing 439 ml of gas over my fireplace. the temperature and pressure of the gas inside the balloon is 317.15 k and 0.959 atm, respectively. suppose i don't want the pressure to change, but i want to the volume to go down to 0.378 l. what is the temperature that i need to reach when i cool down the balloon? to what temperature (in celsius) must the balloon be cooled to reduce its volume to 0.378 l if the pressure doesn't change (remained constant)?

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law  

Given ,  

V₁ = 439 mL  = 0.439 L ( 1 L = 0.001 mL )

V₂ = 0.378 L

T₁ = 317.15 K

T₂ = ?

Using above equation as:

The conversion of T(K) to T( °C) is shown below:

T( °C) = T(K) - 273.15  

So, T = 273.08 - 273.15 °C = - 0.07 °C

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.

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.