@Randy Holmes-Farley Randy would you be able to help me out with a dosing calculation for Manganese and Molybdenum?
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@Rickyrooz my brother is gonna diy the alpha chemicals sodium molybdate dihydrate tonight, stay tuned ;Cat
reporting back, we diy'd molly with this, been dosing for 3 weeks now, and everything's looking fine
Amazon.com : 1 Pound - Sodium Molybdate - Na2MoO42H2O, 99% Pure : Garden & Outdoor
Amazon.com : 1 Pound - Sodium Molybdate - Na2MoO42H2O, 99% Pure : Garden & Outdoorwww.amazon.com
Improved, compared to not dosing, or was there not a “no dosing” period?
Do you have the recipe? How much do you use for what concentration ?re: manganese
omg...
with everyone's help and mental assistance (@ReefTeacher, @Randy Holmes-Farley), i just made my first DIY additive
using the alphachemicals.com "manganese sulfate" MnSO4 • H2O (i think the bag was 17 bucks), i made 1.5 liters, with a concentration that's 3.2 times the strength of triton's additive, with just this itsy-bitsy-tiny bit of powder...
it's such a physically small amount of dust... ridiculous!
p.s. it gained weight by the second, sucking in moisture from the air... so measure fast lol
Here's a recipe I gave for anhydrous sodium molybdate.
Vanadium pentoxide- need help from the big brains
Ok so ask them if it's anhydrous or its dihydrate yes? If yes I ask and let you know Yes. :)www.reef2reef.com
Yours is a dihydrate, so you need to use about 15% more if you want to be super accurate (so if it calls for 1 mL, dose 1.15 mL).
Sodium molybdate has a molecular weight of 206 grams per mole.
Molybdenum has a molecular weight of 96 g/mole.
Thus, sodium molybdate is 96/206 x 100 = 46.6% molybdenum by weight
If you dissolve 1 gram (1,000 mg) in 1 L (1000 mL), that has a concentration of 1 mg/mL sodium molybdate or 0.5 mg/mL molybdenum
Natural levels of molybdenum are close to 10 ug/L.
If you want to boost 100 liters of aquarium water by 10 ug/L, you need 100 L x 10 ug/L = 1,000 ug or 1 mg.
Thus, 2 mL of your stock solution (that is 0.5 mg/mL) will add that 1 mg to the 100 liters of aquarium water.
Holy wow that is an incredibly small amount. I'm waiting on the molybdate to come in still, i've got everything I need though. I hope you won't mind me potentially picking ur brain a bit, i'm going to send in icp to ensure i stay ontop of this.
Hence the description as a "trace" element. There's only a trace present.
Ok i've made the solution for the Molydbate, that was easy enough, thank you for the help.
I've been trying to apply the math you gave me for the Molybdate to the Manganese Sulfate Monohydrate i've gotten, which was referenced here:
"1 gram should contain 320mg of manganese. So 0.1 grams should have 32mg. If I dissolve 0.1 grams into 1000ml of rodi, I should have a solution with a concentration of 32ug/ml. My system volume is 100 gallons, or 379 liters. If I want a concentration of 1ug/l in my system, I need to dose 11.8ml of my solution."
But 379 liters x 1ug/l = 379 still so am I to assume 11.8ml x 32ug = 377 which is close enough to 379? Sorry if this seem like a stupid question, i'm still trying to learn the maths so I can do it on my own, referencing and understanding prior stuff is the best way for me to learn... Thank you for your help so far