You will be collecting a gas over water in this experiment. If the class baromet

Question

You will be collecting a gas over water in this experiment. If the class barometer reports an atmospheric pressure of 29.86 in Hg, what is the pressure of the gas mixture collected above the water in units of mm Hg? Remembering that the pressure of the gas in the head space includes the vapor pressure of water, how would you determine the pressure of the collected dioxygen gas alone? Please describe function of a catalyst in a chemical reaction. What is the catalyst in this experiment? In this lab, only the solutions that contain lead are collected as hazardous waste. Please list at least two reasons lead nitrate is treated with caution in the lab. You might be interested in the safety data sheet (SDS) for lead nitrate.

Explanation / Answer

Q1.

Pbar = 29.86 in Hg

gas of mixture in mm Hg

note that

1 in = 2.54 cm = 25.4 mm

so

29.86 in = 29.86 *25.4 = 758.444 mm Hg

This is atmospheric

We need to consider the apor pressu of water.. then... assume

water is at 25°C so Pvap = 23.7695 torr = 23.7695 mm Hg

Total P = Pgas + Pvap + Patm

Total P (should be kown)

Pgas = Total P - Pvap - Patm = X - 23.7695 + 758.444

Pgas = X - 782.2135 mm Hg

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