anonymous
  • anonymous
It is believed that the rate of at see cricket's chirping is related to temperature. Studies have shown that the activation energy for the cricket's chirping is 22kj/mol and that the cricket chirps 10 times per minute at 27 degrees Celcius. How often does the cricket chirp at 42 degrees celcius?
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Xishem
  • Xishem
Here we need to use an equation which (I believe) is derived from the Arrhenius equation... \[\ln(\frac{k_2}{k_1})=\frac{-E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})\]\[\ln(\frac{k_2}{10s^{-1}})=\frac{-22kJ*mol^{-1}}{8.314J*K^{-1}*mol^{-1}}(\frac{1}{315K}-\frac{1}{300K})\]\[\ln(\frac{k_2}{10s^{-1}})=-2.6461*K^{-1}(-1.5873*10^{-4})\]\[\ln(\frac{k_2}{10s^{-1}})=4.2002\]\[\frac{k_2}{10s^{-1}}=e^{4.2002}\]\[\frac{k_2}{10s^{-1}}=1.0004\]\[k_2=10.004s^{-1}\]Since, we're really only justified in keeping 2 SFs...\[k_2=10.004s^{-1}\approx 10s^{-1}\]
Xishem
  • Xishem
Oh! I see the mistake I made here. Let me fix that.
Xishem
  • Xishem
The final units just need to be in inverse minutes, not inverse seconds: \[k_2=10mi n^ {-1}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.