A 10.0 L balloon contains helium gas at a pressure of 655 mmHg. What is the final pressure, in millimeters of mercury, if the final volume is 13.8 L?

474.64mmHg

Explanation:

From the question given, we obtained the following:

V1 = 10L

P1 = 655mmHg

V2 = 13.8L

P2 =?

P1V1 = P2V2

10 x 655 = P2 x 13.8

Divide both side by the coefficient P2 i.e 13.8

P2 = (10 x 655) / 13.8

P2 = 474.64mmHg

Therefore, the new pressure will be 474.64mmHg

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%