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. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.