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[APD] Re: hair additive in the substrate



 Does hair ever breakdown? I know I've read about the problems in water
treatment where so much human hair needs to be dealt with, and bits of fur
are always the last to go on roadkill.

I can tell you that cat fur will tend to wind around the impeller shaft. My
cat leaves a trail of fur wafting in the air as he moves and some always
finds its way into the tanks. It is sort of a pain to unwind. If anyone ever
does try this, I hope they mince the hair first. Maybe electric razors would
be a better source than cats or barbershops!

Ann Viverette

> Message: 8
> Date: Sun, 4 Jan 2004 23:57:27 -0600
> From: Paul Krombholz <krombhol at teclink_net>
> Subject: [APD] Re:
> To: aquatic-plants at actwin_com
>
>
>      * From: Joey Choochootrain <hubbahubbahehe at yahoo_com>
> >
> >i hope you guys are joking,  what if some guy takes it seriously and
> >actually starts adding hair to the substrate?
>
> I am serious.  What if, indeed?   We will never find out if somebody
> doesn't try it.  What could go wrong?  I am quite sure that hair does
> not decompose rapidly enough to make the substrate unhealthy for
> plant roots.  The only problem may be if it decomposes so slowly that
> it provides too little nitrogen to make much difference.
>
> I am not  interested in trying to keep nitrogen  available in the
> substrate but absent in the water column, but somebody who is, ought
> to try it!
>
> --
> Paul Krombholz in rainy central Mississippi
>
> ------------------------------
>
> Message: 9
> Date: Sun, 4 Jan 2004 22:57:57 -0800 (PST)
> From: "A.Kumar" <aar111 at yahoo_com>
> Subject: [APD] Re: Triple PO4 and Apatite solubility
> To: aquatic-plants at actwin_com
>
> Hi,
>         I did some calculations on the fluorapatite
> (Ca5(PO4)3F) solubility using my old textbooks. The
> solubility product, Ksp=3.16E-60 for fluorapatite.
> The ions (Ca++, PO4--, F-) are in the ratio 5:3:1. Let
> S represent one ionic concentration. Then we would get
> the equation Ksp=(5S)^5(3S)^3(S). If you do the math,
> you will get 6.94E-8 M. Multiply by 3 (because we have
> 3 phosphates) and you get 2.1E-7 M. The "M" stands for
> molar which is moles/Liter. PO4 weighs 95g/mol.
>         After some more number crunching, I end up
> with a value of .02mg/L, or .02ppm phosphate.
> This seems to be kind of low. However, the 0.02ppm
> phosphate will be constantly maintained (by
> dissolution) even as the plants uptake phosphate.
> Also, there will be some pH effects and I think an
> acidic pH will increase dissolution. Hope this helps,
> AK
>
>
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> End of Aquatic-Plants Digest, Vol 5, Issue 15
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