Question

@JFraser @thomaster

Please help!

Answer 1

rate is change in concentration over time, right?
\[rate = \frac{\Delta[H_2O_2]}{\Delta time}\]

Answer 2

Right. So I'm pretty sure the answer is A. Just need you to check.

Answer 3

so find the change in concentration from the first time interval, and compare it to the change in concentration over the second time interval

Answer 4

Well the first one is 6.7 x 10^-3 right?

Answer 5

it's not

Answer 6

the \(change\) in concentration is (2.0M - 1.6M), and the \(change\) in time is (300s - 0s)

Answer 7

Ohhh.

Answer 8

so the initial rate is\[\frac{1.6M - 2.0M}{300s - 0s}\]

Answer 9

and the second rate is the change in concentration over the second time interval

Answer 10

the rates do change

Answer 11

you're jumping without thinking

Answer 12

-0.4/300s ?

Answer 13

that's the first rate, now find the second rate

Answer 14

Can you tell me the formula for the second rate?

Answer 15

finding rate is always the same formula, change in concentration over change in time

Answer 16

-0.3/300s ?

Answer 17

the "second" rate isn't the actual second time window, it's the second set they ask you about, the time from 1800s to 3000s

Answer 18

0.72-0.15 ?

Answer 19

that's the change in concentration, what's the change in time?

Answer 20

1200

Answer 21

properly it's (3000s - 1800s) but yes

Answer 22

so\[rate_1 = \frac{1.6M - 2.0M}{300s - 0s}\] and \[rate_2 = \frac{0.15M - 0.72M}{3000s - 1800s}\]
are the two rates equal?

Answer 23

-0.4/300s and -0.57/1200s, no they're not equal.

Answer 24

so simplify the fractions and pick the correct option

Answer 25

Okay, so I got D. Is that correct?

Answer 26

that's what I get

Answer 27

Yay! Thank you so much! :)

Answer 28

YW

Answer 29

Can you please help me with this one?

http://openstudy.com/study#/updates/555f3747e4b023a7e9bab817