The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-1. If the initial pressure of N2O is 3.50 atm at 730°C, calculate the total gas pressure after one half-life. Assume that the volume remains constant. Insert your answer in decimal notation rounded to 3 significant figures.

 the total gas pressure after one half-life is 4.38 atm

Explanation:

The balanced equation for the decomposition of N2O to N2 and O2 is given as:

2N₂O(gas) ⇒ 2N₂(gas) + O₂(gas)

2 moles of N₂O produce 2 moles of N₂ and 1 mole of O₂

The change in pressure depends on the coefficient (number of moles) of the reactant and product.

                                                          N₂O                    N₂                     O₂

number of moles                                2                        2                        1

Initial pressure (atm)                           3.50                  0                        0

change in pressure                            -2x                    +2x                      x

Final pressure (atm)                            3.50 - 2x           2x                       x

The total final pressure is the sum of the individual total pressure. i.e.:

Total final pressure = final pressure of N₂O + final pressure of N₂ + final pressure of O₂

Total final pressure = (3.5 - 2x) + (2x) + x

Total final pressure = 3.5 + x

After one half life, the initial pressure of N₂O would be half its value.

Final pressure of N₂O = half of the initial pressure of N₂O

3.5 - 2x = 0.5(3.5)

3.5 - 2x = 1.75

2x = 1.75

x = 0.875 atm

Therefore, Total final pressure = 3.5 + x = 3.5 + 0.875

Total final pressure = 4.38 atm to 3 significant figures

i think it is true since you can test the answer and can use observations to complete it.

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.