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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. |
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There is no answer at this time. |
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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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