Cyclobutane (C4H8) can decompose according to the following reaction. The concentration of cyclobutane versus time for the reaction is plotted below in three ways, including linear trend lines for each plot. C4H8(g) → 2 C2H4(g) a. Write the rate law for the reaction (including the value and units of the rate constant). Explain how you got your answer.   b. Determine the concentration of C4H8 after 120 seconds have passed.
 Use 3 significant figures.   c. Determine the half-life of the reaction after 120 seconds have passed. 
Use 3 significant figures.

Using this chart : Time (seconds) [PHT] ln[PHT] 1/[PHT] 0 0.005000 -5.298 200.0 10.5 0.004500 -5.404 222.2 22.3 0.004000 -5.521 250.0 35.7 0.003500 -5.655 285.7 51.1 0.003000 -5.809 333.3 69.3 0.002500 -5.991 400.0 91.6 0.002000 -6.215 500.0 120.4 0.001500 -6.502 666.7 160.9 0.001000 -6.908 1.000E3 230.3 0.0005000 -7.601 2.000E3 *Your x-axis should be "time"zero order: time vs [PHT]first order: time vs ln[PHT]second order: time vs 1/[PHT]Enter the absolute value of both slope and correlation into the table below. Order slope correlation 0     1     2

Question 21

How did you get this? k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs   0.693 a 102.0 nanogram(ng) sample of sodium-24 was stored in a lead-lined cabinet for 2.5 days How much sodium-24 remained? The half-life os sodium -24 is 15 hours    _________ ng sodium -24 remain  check_circle Expert Answer thumb_up   thumb_down Step 1 Radioactive decay is a first order reaction. Conversion of days into hours: 1 day2.5 days===24 hrs2.5×2460 hrs1 day=24 hrs2.5 days=2.5×24=60 hrs Calculation of k: k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs-1 Step 2 Radioactive decay is a first order reaction, therefore: [A]=[Ao] e−kt[A]=[Ao] e-kt where [A] is the final amount remained at time t, [Ao] is the initial amount. Putting values: [A]====102.0 ng×e−0.0462 hrs−1×60 hrs102.0 ng×e−2.772102.0 ng×0.062546.38 ng[A]=102.0 ng×e-0.0462 hrs-1×60 hrs=102.0 ng×e-2.772=102.0 ng×0.06254=6.38 ng Thus 6.38 ng of sodium-24 remained.

Use the graph and the best fitting line to predict the concentration of NO2NO2 at 2000 ss . Express the concentration to two significant figures and include the appropriate units.

What is the average rate over the first 80 minutes, in M • min^-1 a. 0.81 b. 0.94 c. 1.0 d. 1.23

The second law of thermodynamics has been called the arrow of time. Explain why this is so.

Fill the rest out for me and please explain how you did it. (Including the 0)

Please answer fast it’s very important and urgent I say very urgent so please answer super super fast please For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466

A flight is delayed one day and 4 hours. What is that delay in seconds?