- #1
ohms law
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My textbook (Burdge, Chemistry, 2nd edition) is pretty vague mathematically when it comes to calculating the examples, so I'm a bit confused about solving this example problem.
The equation is fairly straightforward:
[itex]ln{[A]_{t}/[A]_{0}}=-kt[/itex]
The 14.4 problem statement is (paraphrased):
2a -> b
[itex]k = 7.5 x 10^{-3}s^{-1}[/itex] (@ 110°c)
[A] = 2.25M,
What is [A] after 2.0 min?
So, 2.0 min = 120 sec
[itex]ln{[A]_{t}/2.25M}=-(7.5 x 10^{-3}s^{-1})(120s)[/itex]
The answer given is 0.91M. I could really use the proper steps to get there.
The thing that's really not clear is whether you take the natural log of each quotient value separately, or whether you divide the quotient and then take the natural log of that... and, then I'm getting my Properties of Natural Logs all confused while working the problem, so... I'm confused. Help, please?!
The equation is fairly straightforward:
[itex]ln{[A]_{t}/[A]_{0}}=-kt[/itex]
The 14.4 problem statement is (paraphrased):
2a -> b
[itex]k = 7.5 x 10^{-3}s^{-1}[/itex] (@ 110°c)
[A] = 2.25M,
What is [A] after 2.0 min?
So, 2.0 min = 120 sec
[itex]ln{[A]_{t}/2.25M}=-(7.5 x 10^{-3}s^{-1})(120s)[/itex]
The answer given is 0.91M. I could really use the proper steps to get there.
The thing that's really not clear is whether you take the natural log of each quotient value separately, or whether you divide the quotient and then take the natural log of that... and, then I'm getting my Properties of Natural Logs all confused while working the problem, so... I'm confused. Help, please?!