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hydrogen gas was collected in a burette in a water bath with a water height difference of 15.0 cm. please calculate the pressure in the burette if the atmospheric pressure in the room at that time was 765.0 torr.

Sagot :

The pressure in the burette if the atmospheric pressure in the room at that time was 765.0 torr is 717.44 torr.

Calculation:-

We know that density of water = 1g/mL

Given : Density of mercury = 13.6 g / mL

Also, 1 torr = 1mmHg

So 1 torr = 13.6 mm of H2O

So 1mm of water = 1/13.6 torr

Height of water given = 14.1 cm = 141 mm

so 14.1 cm = 141 / 13.6 torr = 10.36 torr

Atmospheric pressure = 753 torr

Therefore pressure of H2 and pressure of water column = 753 torr

Hence pressure of hydrogen = 753 - 10.36 = 742.64 torr

A further correction needs to be made because the H2 gas also contains water vapor.

We know from standard values that vapour pressure exerted by water at 26 0C = 25.2 Torr

So pressure of dry hydrogen  = Pressure of hydrogen - pressure of vapours = 742.64 - 25.2 = 717.44 torr

Learn more about Atmospheric pressure here:-https://brainly.com/question/19587559

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