5 L OF H2 GAS WAS COLLECTED OVER WATER AT 25 CELCIUS AND 750 MM PRESSURE CALCULATE THE VOLUME OF DRY H2 AT STP AQUEOUS TENSION AT 25 CELCIUS IS 23 MM

Note the correct spelling of celsius.

Please don't use all caps. It makes it harder to read.

Use (P1V1/T1) = (P2V2/T2)
You know V1 and T1, P1 = 750 mm-23 mm = ?
P2 and T2 are at STP. I assume you know those. Solve for V2.

To calculate the volume of dry H2 gas at STP (Standard Temperature and Pressure), we need to apply the ideal gas law.

The ideal gas law is represented by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Given:
Initial volume of H2 gas collected over water = 5 L
Temperature = 25 degrees Celsius = 298 Kelvin (25 + 273)
Initial pressure = 750 mm (millimeters of water)
Aqueous tension at 25 degrees Celsius = 23 mm

We need to convert the pressure from mm to atm, as the ideal gas constant 'R' is commonly used with atm.

To convert mm to atm, divide the pressure by 760 (since 760 mm of Hg = 1 atm):
Initial pressure in atm = 750 mm / 760 = 0.9868 atm

Next, to determine the pressure of the dry H2 gas, we subtract the aqueous tension from the initial pressure:
Dry H2 gas pressure = Initial pressure - Aqueous tension = 0.9868 atm - 0.0303 atm (23 mm H2O to atm conversion) = 0.9565 atm

Now, let's calculate the number of moles of H2 gas:
PV = nRT
Rearranging the equation, n = PV / RT

Substituting the values into the equation:
n = (0.9565 atm * 5 L) / (0.0821 atm L/mol K * 298 K) ≈ 0.1939 moles

Finally, we can use the number of moles to calculate the volume of dry H2 gas at STP (Standard Temperature and Pressure) using the ideal gas law at STP conditions:
P = 1 atm
T = 273 K
n = 0.1939 moles

Rearranging the equation PV = nRT to solve for V:
V = nRT / P

Substituting the values into the equation:
V = (0.1939 mol * 0.0821 atm L/mol K * 273 K) / (1 atm)
≈ 4.87 L

Therefore, the volume of dry H2 gas at STP is approximately 4.87 liters.