Aballoon with a pressure of 1 atm has a volume of 2 liters. if you were to immerse the balloon into water where it experienced a pressure of 4 atm. what would the new volume of the balloon be?

1/2 liters
8 liters
2 liters
7 liters

½ L  

Step-by-step explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p₁V₁ = p₂V₂    Divide each side by p₂

 V₂ = V₁ × p₁/p₂

Data:

p₁ =  1 atm; V₁ = 2 L

p₂ = 4 atm; V₂ = ?

Calculations:

V₂= 2 L × (1 atm/4 atm)

V₂= 2 L × ¼

V₂ = ½ 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.

i'm pretty sure its b i might be wrong though!