[paul544]

Thanks for any help.

[freedumb2003]

You signed up in 2000 just to get help with chemistry???

(Other than mixology, I haven’t done chemistry in over 30 years...)

[rdl6989]

Lol, I’m not even smart enough to understand the question.

[cripplecreek]

[Tolerance Sucks Rocks]

Which part of the equation is the problem for you?

[JRandomFreeper]

I can do homework professionally (gack, I'm back to this? The economy really, reaaly needs to improve), but I charge steeply, depending on the degree you are seeking. Specials on term papers.

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

[Hot Tabasco]

If I'm not mistaken, that formula is used to determine the best brewing temperature of beer.........

I might be wrong tho so here's a pingster for someone who might be more knowledgable in these things.........

[Drango]

42

[devere]

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.

[allmost]

What are you absorbing?

[CharlesWayneCT]

First: Beers law: Beers Law

The web site shows how you get from a standard transmittal equation T=(P/Po) to the equation you are using.

A = log10 P0 / P
A = log10 1 / T 
A = log10 100 / %T
A = 2 - log10 %T 

As to your question, it seems you should just plug in the percent given, and you'll get an "A" value.

For example, for 25% transmission, the answer is:

A= 2 - log10 25
A= 2 - 1.39794
A= 0.60206

You use the percent number directly (that's the point of changing the 1/T to 100/%T -- 1/.25 is the same as 100/25).

If you had a different question, please ask again.

[Scutter]

I don’t know the chemistry, but I think I can answer your question.

I did a search on this, and the formula for absorbency seems to be -log(T). However, a percentage is often expressed as 100 times the fractional amount. Log base 10 of 100 is 2. And log(A x B) = log(A) + log(B).

So 2 - log(T) = -log(1/100) - log(T) = - log(T/100)

So if you express T as a percentage from 0 to 100, you use the formula (2 - LOG(T)). If instead you express T as a fraction, from 0 to 1.0, you would use the formula (-LOG(T)).

[martin_fierro]

Also check out Wolfram Alpha

[DBrow]

It’s usually A= log(1/T)

[Hoodat]

It looks like you are trying to do a Beer-Lambert equation for absorbancy. The spectrophotometer you used gave you the transmission percentage of light passing through your solution medium. No absorbance of light results in T-value of 100%. Complete absorbance of light results in T-value of 0%. You need to convert this into a value for absorbance - A. By your equation, A = 2 - log(%T)

If T = 25%, then A = 0.60206. You will need this value for the next equation:

A = εlc

A = absorbance
ε = molar absorbtivity [L/(mol.cm)]
l = length of test light path through medium [cm], (usually 1 cm)
c = molarity of solution [mol/L]

In a typical Beer's Law experiment, you are searching for the concentration of an unknown sample. You do this by calibrating the molar absorptivity through a series of experiments with known concentrations. You can plot these values of concentration v. absorbance and they should be linear. The slope will give you εl.

ε = A/lc

Once the absorptivity constant is calculated, you repeat the experiment for the unknown sample to determine it's molarity. Capisce?

[lwoodham]

All this math is eye opening. I just realized that LOG[The Won] = ZERO

[a6intruder]

Thank goodness I chose the much easier profession of landing 20,000 pound jets aboard an aircraft carrier at night.

No math required.