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The hydrogen gas formed in a chemical reaction is collected over water at 30.0°C at a total pressure of 732 torr. What is the partial pressure of the hydrogen gas collected in this way? If the total volume of gas collected is 722 ml, what mass of hydrogen gas is collected? (Vapor pressure of water at 30.0°C is 31.82 torr).

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Final answer:

The partial pressure of the hydrogen gas collected is 700.18 torr, and the mass of the hydrogen gas collected is 17.589 g.

Step-by-step explanation:

When gas is collected over water, the total pressure is equal to the sum of the partial pressure of the gas and the vapor pressure of water. In this case, the total pressure is 732 torr, and the vapor pressure of water at 30.0°C is 31.82 torr. So, the partial pressure of the hydrogen gas collected can be calculated by subtracting the vapor pressure of water from the total pressure:

Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water = 732 torr - 31.82 torr = 700.18 torr

Now, to determine the mass of the hydrogen gas collected, we need to convert the volume to liters and then use the ideal gas law equation:

Volume of gas collected = 722 ml = 0.722 L

Using the ideal gas law equation:

Mass of gas = (Partial pressure × Volume) / (R × Temperature)

R is the ideal gas constant = 0.0821 L·atm/mol·K

Temperature = 30.0°C + 273.15 = 303.15 K

Now substituting the values:

Mass of hydrogen gas = (700.18 torr × 0.722 L) / (0.0821 L·atm/mol·K × 303.15 K)

Mass of hydrogen gas = 17.589 g

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