A mixture of He , N 2 , and Ar has a pressure of 16.7 atm at 28.0 °C. If the partial pressure of He is 1737 torr and that of Ar is 2087 mm Hg, what is the partial pressure of N 2 ?

The answer to your question is   P N₂ = 11.67 atm  

Explanation:

Data

He  1737 torr

N₂  = ?

Ar 2087 mmHG

Total pressure = 16.7 atm

Process

1.- Convert torr to atm

             1 atm 760 torr

             x         1737 torr

             x = (1737 x 1)/760

             x = 2.29 atm

2.- Convert mmHg to atm

             1 atm 760 mmHg

             x         2087 mmHG

             x = (2087 x 1)/760

             x = 2.74

3.- Find the Partial pressure of N₂

       Total pressure = He pressure + N₂ pressure + Ar pressure

- Substitution

        16.7 = 2.29 + P N₂ + 2.74

- Solve for P N₂

        P N₂ = 16.7 - 2.29 - 2.74

        P N₂ = 11.67 atm      

asking the goog might u get an answer

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.