An open manometer connected to a tank of argon has a mercury level of 83 mm higher in the atmospheric arm. if the atmospheric pressure is 76.9 kpa, what is the pressure of the argon in kpa?

- 88 kPa.

Solution:- The level of mercury in the atmospheric arm that is the open arm is 83 mm. It indicates the gas pressure is 83 mm higher than the atmospheric pressure.

gas pressure = height of mercury level in open tube in mm + atmospheric pressure

Let's convert this 83 mm pressure to kpa since the atmospheric pressure is given in kPa and the answer is also asked to report in kPa.

mmHg is converted to atm first and then atm is converted to kpa.

760 mmHg = 1 atm

and 1 atm = 101.325 kPa

= 11 kPa

Atmospheric pressure is 76.9 kPa.

gas pressure = 11 kPa + 76.9 kPa

gas pressure = 87.9 kPa

If we think about sig figs rule then 87.9 should be round to 88 as in addition we go with least number of decimal places. 11 kPa does not have any decimal places and so the gas pressure is 88 kPa.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.