g A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was mm Hg. Selected Incorrect 773 Answers: 882 793 661 773 Correct 96.0

882mmHg

Explanation:

In a closed-end manometer, the gas added is under a pressure that is measured against atmospheric pressure. The difference in pressure between both gases is equal to the difference in mercury levels. Thus, the pressure of the argon can be:

783 mmHg - 96.0mmHg = 687mmHg

783 mmHg - 96.0mmHg = 879mmHg

Based on the options, the possible answer (The nearest to the second value) is:

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%