What is the total pressure on the gas after the 760 mm hg has been added?
first phase atmospheric pressure= 760 torr gas volume= 60 ml (760 mm hg added to 2nd phase) second phase atmospheric pressure= 760 torr gas volume= 30 ml **image looks like a j. in the 2nd phase, there is 760 mm that is half of the j** if that makes any sense.
i am confused on how you solve the problem considering the torr and ml and mm and mmhg

1520 torr = 1520mmHg = 2 atm.

Explanation:

Please note that it is just basic conversion.

1 torr = 1mmHg

1 atm = 760mmHg = 760 torr

As per the question,

FIRST PHASE:

760 torr pressure is applied which gave = 60mL volume to gas.

SECOND PHASE:

760mmHg pressure is applied which gave = 30mL volume to gas

Since we know that 1mmHg = 1 torr.

Therefore, total gas pressure = 760 + 760 = 1520 torr OR 1520mmHg

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%

c. 75.7

i think that is the right