0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
How do you prepare 125 parts per million Fe from the salt Fe(NH4)2(SO4)2.6H2O?
Molar mass Fe is 55.845 mg/mmol, so 125 mg represents 125 / 55.845 = 2.238 mmol Fe.Since the molar mass of (NH4)2Fe(SO4)2·6H2O is 392.13 g/mol, 2.238 mmol of this Mohr's salt weights 392.13(mg/mmol) * 2.238 mmol = 878 mg Mohr's salt.This has to be dissolved in 1.00 L water to give the desired 125 ppm (=mg/L) Fe2+ solution
How many mg in 0.2 mmol of Te?
The answer is 0,02552 mmol.
How does adding solute to a solution affect the mass of the solution?
It increases the mass.
What is the formula to convert gm to mmol per liter?
formula for to convert gm to mmol
2.55 mmol to moles?
1000 mmol = 1 mol. So, what you do is 2.55mmol*(1mol/1000mmol). The mmol's cancel and you are left with mol. The "m" is a metric prefix. So, 1000mN = 1N just like 1000mmol = 1mol.
How do you prepare 125 parts per million Fe from the salt Fe(NH4)2(SO4)2.6H2O?
Molar mass Fe is 55.845 mg/mmol, so 125 mg represents 125 / 55.845 = 2.238 mmol Fe.Since the molar mass of (NH4)2Fe(SO4)2·6H2O is 392.13 g/mol, 2.238 mmol of this Mohr's salt weights 392.13(mg/mmol) * 2.238 mmol = 878 mg Mohr's salt.This has to be dissolved in 1.00 L water to give the desired 125 ppm (=mg/L) Fe2+ solution
How much sodium bicarbonate would be required to prepare 500ml of a 100mM solution?
500 mL * 100(mMol/mL) = 50 mMol NaHCO3 , hence50 mMol NaHCO3 = 50(mMol) * 84(mg/mMol) = 4200 mg = 4.2 g NaHCO3 in 500 mL
How do you convert mmol to milliters?
1 mol = 103 mmol Conversely, 1 mmol = 10-3 mol For example: 25 mol x 103 mmol/1 mol = 25000 mmol and, 3.2 mmol x 10-3 mol/1 mmol = 0.0032 mol
How do you convert mgd to mmol?
This means that the conversion factor depends on the molecular weight of the substance in question. To convert mgd to mmol we multiply by 0.055.
What is meant by the mass mass percent concentration of a solution?
This is (mass of solute) divided by (mass of total solution) expressed as a percentage. The solute is what you are dissolving into the solution. Example: you have 90 grams of water, and you add 10 grams of salt (sodium chloride). The water is the solvent, sodium chloride is the solute, and the solution is salt water. 90 grams + 10 grams = 100 grams (mass of total solution). (10 grams) / (100 grams) = 0.1 --> 10% mass mass percent concentration.
How you prepare Sodium Tartrate Buffer pH-5?
10 mM tartaric acid (sodium) buffer solution (pH=4.2) Tartaric acid (M.W.=150.09)..........................2.5 mmol (0.375 g) Sodium tart rate dihydrate (M.W.=230.08)........7.5 mmol (1.726 g) Add water to make up to 1 L. 10 mM tartaric acid (sodium) buffer solution (pH=2.9) Tartaric acid (M.W.=150.09)..........................7.5 mmol (1.13 g) Sodium tartrate dihydrate (M.W.=230.08)........2.5 mmol (0.58 g) Add water to make up to 1 L.
Convert mmol to mol?
1 mEq=1 mmol/valence e.g.For sodium, 1 mEq=1mmol/1 (valence of sodium=1) means, 1 mmol sodium=1 mEq of sodium take for calcium,valence=2 1 1 mEq of calcium=1mmol/2=0.5 mmol of calcium
Is mmol EDTA the same as mmol Ca2?
A mmol is a mmol but the amount of matter is different for each chemical.
What is the salt NaCl of Blood?
The concentration of sodium in blood plasma is 136-145 mmol/L.
How much sugar is there in 100mL of a10 percent aqueous sugar solution and what is the solvent in the solution?
percent concentration = (mass of solute/volume of solution) X 100 To solve for mass of solute, mass of solute = (percent concentration X volume of solution)/100 So, mass of solute = (10% X 100mL)/100 = 10g
What is a concentration of a solution?
The concentration of a solution is basically how strong the solution is.
what is the mass and the identity of the precipitate that forms when 55.0 ml of 0.100 M BaCl2 reacts with 40.0 ml of 0.150 M Na2CO3?
BaCl2 + Na2CO3 --> BaCO3 + 2NaCl Barium carbonate is the precipitate. 55ml x 0.1 M = 5.5 mmol Ba 40ml x 0.15 M = 6.0 mmol CO3 Thus limiting reagent is Ba. Molar mass barium carbonate = 137 + 12 + 3 x 16 = 197 g/mole = 197 mg/mmol 5.5 mmol x 197 mg/mmol = 1083.5 mg Round for sig figs .... 1080 mg barium carbonate
