A chemical engineer studying this reaction fills a flask at with of nitrogen dioxide gas. he then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains of nitrogen dioxide gas. the engineer then adds another of nitrogen dioxide, and allows the mixture to come to equilibrium again.

Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric chemistry. It can react with itself to form another form of NOx, dinitrogen tetroxide. A chemical engineer studying this reaction fills a 25.0 L tank at 19. °C with 10. mol of nitrogen dioxide gas. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 3.9 mol of nitrogen dioxide gas. The engineer then adds another 3.3 mol of nitrogen dioxide, and allows the mixture to come to equilibrium again. Calculate the moles of dinitrogen tetroxide after equilibrium is reached the second time. Round your answer to 2 significant digits.

Explanation:

                                                                         2NO₂ →   N₂O₄

Molarity of NO₂ = 10/25 = 0.4M                     Ф 0.4              0 → I

Equilibrium = 3.9 / 25 = 0.16M                       -2x                  +x→ c

 Adding NO₂ = 3.3 / 25 = 0.132M                  0.4 - 2x             x→ E

                                                                         0.4 - 2x = 0.16

                                                                                    x = 0.12

Kc =  [N₂O₄]   =   (0.12 )

        [NO₂]²         (0.16)²

kc = 4.69

N total = NO₂ = 0.4 + 0.132 = 0.532M

                      2NO₂ → N₂O₄

               0.532 - 2x              0.12 + x

                4.69 =  x + 0.12

                         ( 0.532 - 2x )²

  (4 x¬ + 0.283 - 2.128 x ) 4.69 = x + 0.12

   18.76 x¬ - 10.98 x  + 1.23 = 0

          x = 0.414 , 0.171 ,    0.147

            [ N₂O₄ ] =  0.12 + 0.147 = 0.267M

              Moles of N₂O₄ = 0.267 ₓ 25 = 6.68  ⇒ Moles of N₂O₄ = 6.7

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: