Homework Question 13.37A- Significant Figures for e


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Sabrina Smelser 3D
Posts: 8
Joined: Fri Sep 26, 2014 2:02 pm

Homework Question 13.37A- Significant Figures for e

Postby Sabrina Smelser 3D » Sat Feb 07, 2015 6:56 pm

I went through the process indicated in the solutions manual. My work matched the manual up until the final answer was given. I calculated that the final value for Q was 3.27x10^6 while the solutions manual indicates that Q is 10^6. Is there a reason for leaving out the 3.27 and simply rounding to 1,000,000?

Thank you for your help!

Sarah H Brown 1L
Posts: 56
Joined: Fri Sep 26, 2014 2:02 pm

Re: Homework Question 13.37A

Postby Sarah H Brown 1L » Sat Feb 07, 2015 10:52 pm

So when you do out the math, you get Q = 3.27x106. However, some fancy significant figure rules are happening here.

The solutions manual lists the last calculation as:
lnQ = 15
Take e of both sides
elnQ = e15
Q = ?
When you raise each side as a power of e, you have special significant figure rules. If you had Q = e15.12, for example, the significant figures would be different.
It can be broken up: (e15)(e0.12) so technically only the 0.12 is significant, and the answer would have two significant figures and would be 1.1.
Because the solutions manual lists e15, I suppose there are no significant decimal places so the answer becomes 106.

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

Re: Homework Question 13.37A- Significant Figures for e

Postby Justin Le 2I » Mon Feb 09, 2015 9:56 pm

Also, I think the solutions got the 15 from the two significant digits after doing this: 1.52 V - 1.33 V


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