Vanadium pentoxide- need help from the big brains

Randy Holmes-Farley

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Hey mate how are u can you help me with DIY dose for
Nickel sulfate cas number 10101-97-0 dihydrate
Very appricate for help

Be very careful to not overdose nickel. It can be quite toxic.

Do you know the hydration state? or can you give a link or picture of the product label?
 
AS

dani1970

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Be very careful to not overdose nickel. It can be quite toxic.

Do you know the hydration state? or can you give a link or picture of the product label?
The chemical store that sell me give this cas number:10101-97-0 and I attach now poc of chemical how is looks like

1629290062850783841336806225636.jpg
 

Randy Holmes-Farley

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OK, let's first make a dosing solution.

10101-97-0 is nickel sulfate hexahydrate (molecular weight 262.85 g/mole)

Nickel weighs 58.7 grams per mole.

Thus, the solid is 58.7/262.85 x 100% = 22.3% nickel by weight.

If we dissolve one gram of solid (= 223 mg nickel) in 1 L of RO/DI, it will contain 223 mg/L or 223 ug/mL

If you are detecting none by ICP, I think a reasonable dose is 0.5 ug/L

Thus to raise 100 L of seawater from 0 to 0.5 ug/L you need 50 ug of nickel, which is attained by dosing 0.22 mL (0.22 g) of dosing solution to the tank.

If that is hard for you to accurately measure, you can dilute the dosing solution by a factor of 10 by taking 100 mL and dilution with 900 mL of RO/DI. Then dose 10x the amount above, or 2.2 mL.
 
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dani1970

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OK, let's first make a dosing solution.

10101-97-0 is nickel sulfate hexahydrate (molecular weight 262.85 g/mole)

Nickel weighs 58.7 grams per mole.

Thus, the solid is 58.7/262.85 x 100% = 22.3% nickel by weight.

If we dissolve one gram of solid (= 223 mg nickel) in 1 L of RO/DI, it will contain 223 mg/L or 223 ug/mL

If you are detecting none by ICP, I think a reasonable dose is 0.5 ug/L

Thus to raise 100 L of seawater from 0 to 0.5 ug/L you need 50 ug of nickel, which is attained by dosing 0.22 mL (0.22 g) of dosing solution to the tank.

If that is hard for you to accurately measure, you can dilute the dosing solution by a factor of 10 by taking 100 mL and dilution with 900 mL of RO/DI. Then dose 10x the amount above, or 2.2 mL.
So need to dose 0.22ml per 100 liters right ?
 

Randy Holmes-Farley

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So need to dose 0.22ml per 100 liters right ?

If you have an ICP test that shows no nickel, that is what I would dose if I were to dose nickel. I'm not sure it will be useful, but it's a fine experiment.

Without an ICP test before dosing, I would not dose it.
 

dani1970

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If you have an ICP test that shows no nickel, that is what I would dose if I were to dose nickel. I'm not sure it will be useful, but it's a fine experiment.

Without an ICP test before dosing, I would not dose it.
I dose it only after icp you right I risky
The doses 0.22ml is per 100 liters ?
 
Aquarium Specialty - dry goods & marine livestock

dani1970

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As long as you buy the same type of product, it doesn't matter from who. I wanted to see the chemical form, degree of hydration, etc.

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

That material looks fine.

It is 46% molybdenum by weight.

If you dissolve one gram in one liter of water, that gives a concentration of 460 mg/L or 0.46 mg/mL.

If you want to add 10 ug/L to a 100 L aquarium, you need 10 ug/L x 100 L = 1,000 ug or 1 mg.

Thus, 2 mL added to a 100 L aquarium will boost molybdenum by about 10 ug/L, which is close to the NSW level.
Hey can you tell me please if I wanna add 1ug/L once a day , what dose I need with this solution for 100 liters ?
 

dani1970

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That material looks fine.

It is 46% molybdenum by weight.

If you dissolve one gram in one liter of water, that gives a concentration of 460 mg/L or 0.46 mg/mL.

If you want to add 10 ug/L to a 100 L aquarium, you need 10 ug/L x 100 L = 1,000 ug or 1 mg.

Thus, 2 mL added to a 100 L aquarium will boost molybdenum by about 10 ug/L, which is close to the NSW level.
Hey can you tell me please if I wanna add 1ug/L once a day , what dose I need with this solution for 100 liters ?
 

Randy Holmes-Farley

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I already did the calculations for 10 ug/L in the quote above.

Use 1/10th of the amount if you want to dose 1 ug/L.

"Thus, 2 mL added to a 100 L aquarium will boost molybdenum by about 10 ug/L, which is close to the NSW level."

To dose 1 ug/L, use 0.2 mL of the fluid above dosed to 100 L of aquarium water.
 

dani1970

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I already did the calculations for 10 ug/L in the quote above.

Use 1/10th of the amount if you want to dose 1 ug/L.

"Thus, 2 mL added to a 100 L aquarium will boost molybdenum by about 10 ug/L, which is close to the NSW level."

To dose 1 ug/L, use 0.2 mL of the fluid above dosed to 100 L of aquarium water.
Thank you
 
Zoanthids

Randy Holmes-Farley

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Hey mate could you please tell me how dissolve sodium bentonite I try dissolve him it not dissolve for 100% maybe I can add some chemical to help him dissolved ?
Like sodium hydroxide ?
Thanks for advanced

Bentonite is a clay.

Why are you trying to dissolve it?

Surely you do not intend to dose it to an aquarium?
 

Randy Holmes-Farley

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I'm at a loss to understand why you would want to dose it.

Sodium bentonite contains aluminum, sodium, and silicate.

I do not recommend dosing it. It is hard to dissolve and contains aluminum. There's never a reason to add aluminum to a reef tank (unless you are doing some sort of toxicity testing; in that case, use aluminum trichloride).

If you want to dose silicate, use sodium silicate solution.
 

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