Posted on 11/13/2010 6:27:32 PM PST by paul544
Was wondering if there are any chemistry whizzes here who can help me with a problem in a lab I am working on?
We were given a list of %T values and told to use A=2-log %T to get the absorbency (A). I'm struggling with this. Can someone take me step by step through one? Say 25%?
You signed up in 2000 just to get help with chemistry???
(Other than mixology, I haven’t done chemistry in over 30 years...)
Lol, I’m not even smart enough to understand the question.
why don’t you answer his question and not be a smart ass?
The maximum number in % is 100%. This is the big clue.
Then split -log (%T) in -Log % and -Log T. Replace % with 100.
Log (100) = 2. After that you have A = 2-2-log T. And this becomes A=- log T. Therefore A = -log T.
Do an example by filling T=5.
Then you have 2- log (100*5) = -0.69897 or log (5) is also -0.69897.
Hope this helps.
* 3 years ago
Which part of the equation is the problem for you?
Work it dude. You can do it.
Asking for help for homework in class is ok. Maybe not so much on a public forum.
/johnny
I might be wrong tho so here's a pingster for someone who might be more knowledgable in these things.........
Someone got up on the wrong side of the bed today...
I’m assuming you don’t leave the % as a whole number... So, up to this point I’ve been converting it to, say .25. If that is right, then I’m going with the log of .25 but that doesn’t seem right. I’ve seen where it might be 1/.25 then log...
42
/johnny
Thanks. I saw that out there as well. I’ve been trying to use it and think it helps. Appreciate it.
Are you having trouble with the math or the chemistry?
For the math you just need to clarify if %T should be 25 or .25, and then compute the answer.
Geeze edcoil:
It’s Saturday Night and paul544 has been on here long enough to take a razzing.
Probably knew he was going to get a few anyway.
BTW paul544 the answer is log - 42 = %. /s
Now that’s funny.
getting some popcorn...
I get the chemistry for the most part.
The issue is when I see 2-log, I’m not sure of the math to apply to the %. If I make the 25%, .25, is the next step just applying the log function or is there another step?
Chem tip of the day: Never mix ammonia with bleach...produces lung searing gasses that can off you.
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