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**To**:**Aquatic-Plants at actwin_com****Subject**:**needed accuracy to calculate and dose NPK****From**:**Neil Frank <aquarian.subjects at mindspring_com>**- Date: Mon, 14 Jan 2002 07:32:32 -0500

I wrote and then Tom Barr replied: >> Looks like the CaNO3 also helps to supply NO3 to make final N:K into 1:1 > >I was correct. > >This _was_ included in their elemental ratios. This is based on the total >elemental of all the additives not just one salt for one Nutrient source. >Just look at N from NO3 then. A 1 M solution of N from KNO3 would give 14 >grams/mole in a liter of N or 14000 mg of N in a liter. 14000ppm(or mg/l) x >.006liters(in 6mls) = 84ppm for KNO3 alone. Adding the other 3 sources of N >up you'll find it hits 224ppm. So the ratio first stated was correct. > >3.6 :1: 3.8 Tom, Yes of course you were correct. When I said 1:0:1, I was merely rounding your numbers for the nitrogen and potassium parts only. When we measure with teaspoons, it is more than sufficient to report values to the nearest integer! I was trying to emphasize my point that KNO3 may not be sufficient to provide all the needed nitrogen without leaving excess potassium to _potentially_ accumulate. Neil

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