Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 atm, what volume will the gas occupy? (r = 0.08206 l×atm/k×mol)

Step 1: Change density from g/mL to g/L;

                                  0.807 g/mL  =  807 g/L

Step 2: Find Moles of N₂;
As,
             Density  =  Mass / Volume
Or,
             Mass  =  Density × Volume

Putting Values,

            Mass  =  807 g/L × 1 L

            Mass  =  807 g
Also,
            Moles  =  Mass / M.mass

Putting values,

            Moles  =  807 g / 28 g.mol⁻¹

            Moles  =  28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
                        P V  =  n R T

                            V  =  n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

                 V  =  (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

                 V  =  704.76 L

The correct name for n3-ion should be azide.

Yes she is correct. the answer is completely correct