What is the gauge pressure exerted by sea-wo bottom of a pool that is 2.5 m deep, 12.0 m wide and 30.0 m long

Explanation:

Relation between gauge pressure and density and height is as follows.

The given data is as follows.

                = 1025

                        g = 9.81

                        h = 2.5 m

Therefore, substitute the given values into the above formula as follows.

                            =

                            =

or,                        =

                           = 25.138125 kPa

Thus, we can conclude that the gauge pressure exerted by sea-water is 25.138125 kPa.

it should be all of them because it asks if you included all of those in your answer

explanation:

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :