This depends on the concentration of the primary solution; for a 25 % ammonium hydroxide solution you need 85,15 mL.
You need 1,25 moles ammonia.
Ammonium molybdate is (NH4)2MoO4. Its molar mass is 148g/mol. A 0.5M solution will have half a mole per liter of water. Half a mole of ammonium molybdate is 74g, so you would measure out 74g of it, and dissolve it in a liter of water.
The formula is C6H12O6 which is 180g/mole. Divide that in half for 90g in one liter of water for a 0.5 molar solution
30 grams
Dissolve 1.0 mole gas (17 gram) in 1.0 Liter water
To prepare 6 nM ammonium hydroxide a 30 percent solution you need to know the volume of the 30 percent solution that you have and the volume of 6nM solution you would like to make. Then use the following formula: C1V1 = C2V2 where C = concentration in moles/Liter and V = volume in liters.
6N ammonium hydroxide (NH4OH) is the same as 6 M NH4OH. The molar mass of NH4OH is 35 g/mole. Dissolve 6 x 35 g = 210 g NH4OH in enough H2O to make 1 liter of solution.
You need 1,25 moles ammonia.
Ammonium molybdate is (NH4)2MoO4. Its molar mass is 148g/mol. A 0.5M solution will have half a mole per liter of water. Half a mole of ammonium molybdate is 74g, so you would measure out 74g of it, and dissolve it in a liter of water.
Ammonium molybdate is (NH4)2MoO4. Its molar mass is 148g/mol. A 0.5M solution will have half a mole per liter of water. Half a mole of ammonium molybdate is 74g, so you would measure out 74g of it, and dissolve it in a liter of water.
744 g/L of ammonium sulphate, at 20 0C
pH is 8
Molarity = moles of solute/liters of solution ( 300 ml = 0.300 liter ) 0.250 molar KOH = moles KOH/0.300 liters = 0.075 moles KOH
Molarity = moles of solute/liters of solution M = 10 mol NaOH/1 liter = 10 M -log(10 M) = -1 14 - (-1) = 15 pH sodium hydroxide
mw : 134 /2= 67
The formula is C6H12O6 which is 180g/mole. Divide that in half for 90g in one liter of water for a 0.5 molar solution
25