You’re at the zoo and have a big red 3.20 L helium balloon. The barometric pressure is 790.0 mm Hg. Then you hear the roar of a lion. Startled, you accidently release the balloon. It flies away. By the time it reaches the clouds the atmospheric pressure is only 0.300 atm. What is the new volume of the balloon?

The new volume, when the pressure drops to 0.300 atm, is 11.09 L

Explanation:

Step 1: Data given

The volume of the helium balloon = 3.20 L

The pressure in the balloon is 790.0 mmHg = 790 / 760 = 1.03947 atm

The pressure decreases to 0.300 atm

Step 2: Calculate the new volume

P1*V1 =P2*V2

⇒with P1 = the initial pressure = 1.03947 atm

⇒with V1 = the initial volume = 3.20 L

⇒with P2 = the reduced pressure = 0.300 atm

⇒with V2 = the new volume = TO BE DETERMINED

1.03947 atm * 3.20 L = 0.300 atm * V2

V2 = (1.03947 atm * 3.20 L) / 0.300 atm

V2 = 11.09 L

The new volume, when the pressure drops to 0.300 atm, is 11.09 L

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%