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Q: Synthetic Feed preparation ( No Answer,   1 Comment )
Question  
Subject: Synthetic Feed preparation
Category: Science > Chemistry
Asked by: godfather999-ga
List Price: $20.00
Posted: 08 May 2004 16:56 PDT
Expires: 12 May 2004 05:15 PDT
Question ID: 343336
I found in an article the following synthetic media formulation:Carbon
source 6.4 g/l, NH4HCO3 1.28 g/l, NaHCO3 2.48 g/l, KHCO3 2.98
g/l,(NH4)2SO4 0.32 g/l, K2HPO4 0.17 g/l, KH2PO4 0.13 g/l. The ratio
C:N:P mentioned is 100:5:1. My question is how they got those
values?...I have tried to get them but no succes so far...please
explain.
Answer  
There is no answer at this time.

Comments  
Subject: Re: Synthetic Feed preparation
From: stuyshick-ga on 09 May 2004 17:32 PDT
 
Here is how I think the calculation was made although I am a bit busy
to do it myself:
[6.4 / (mass of carbon)] + [1.28 / (molar mass NH4HCO3)] + [2.48 /
(molar mass of NaHCO3)] + [2.98 / (molar mass of KHCO3)] =A

[1.28 / (molar mass NH4HCO3)] + 2*[0.32 / molar mass of (NH4)2SO4)]=B

[0.17 / (molar mass of K2HPO4)] + [0.13 / (molar mass of KH2PO4)]=C

A:B:C should be 100:5:1 if all the other information that you gave me
was correct. Good luck and hope I helped.

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