Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

The rate law for a hypothetical reaction is rate = k [A]^2. When the concentration is 0.10 moles/liter, the rate is 2.7 × 10^-5 M*s^-1. What is the value of k?

Chemistry
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

rate = k [A]^2 just plug in 2.7*10^(-5) = k*.1^2 ^rate ^.1 molles/liter maybe and then solve for k :)
k = 0.27e-2
umhm so wait... would it be the same as 2.7*10^-3M^-1s^-1?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

well unites should be M*s^-1 so ( moles/liter)^2* k's units= M*s^-1
umhm alright. Thank you!
is moles/liter=M?
remember the unit M=mol/L \[ 2.7\times10^{-5}\,{\rm M/s}=k\times \left(0.1\,{\rm M}\right)^2\\ k=\frac{2.7\times10^{-5}}{0.01}\times \frac{{\rm M}}{\rm s\cdot M^2}\\ \boxed{k=2.7\times10^{-3}{\rm M^{-1}\cdot s^{-1}}} \] the reaction is thus, a second order reaction.
Soo, I was right? :D woho! Thanks alot electrokid!
yaya. good job.

Not the answer you are looking for?

Search for more explanations.

Ask your own question