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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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